profiles = new Array(
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Objectives ¤ From a UMIST lecture... We have seen how an innovation strategy supports a business strategy by defining a stream of new and improved products. At any point in time, this stream constitutes a portfolio that must be balanced in terms of timing, key resource utilisation, risk and reward. On-going choices must be made between innovations...| since the firm has finite resources. Construction and review of the balanced portfolio is the key management control activity between the innovation strategy and the operational level stage-gate process for individual projects. The term 'portfolio' is central to this lecture but is also quite widely used in everyday language, particularly in the idea of a portfolio of investments. The analogy of having a range of investments to get a good return, but not risking everything on one project carries over directly to the way the term portfolio is used in this lecture. This lecture will cover: why you need to manage the portfolio of innovation projects what methods are used to measure or score the value of innovation projects what methods can be used to evaluate and maximise the value of a portfolio of innovation projects what techniques you can use to achieve balance within the portfolio of innovation projects what approaches you can take to achieve alignment between the portfolio of innovation projects and the innovation and business strategy.  |../MathematicsSession/Maths00005.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Objectives ¤ This session is intended to provide some groundwork in essential mathematical ideas that will be required throughout the course. The idea of a limit is introduced first as this concept is necessary for the definition of the derivative. The derivative of a function is the function that gives us the gradient rather than the value of the...| function itself. Once this has been covered, the process of finding areas under curves (integration) is presented. By the end of this session you will know: what limits, derivative and integrals are how to calculate limit how to calculate derivative how to calculate integral how to solve maximum/minimum problems.  |../MathematicsSession/Maths00010.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Observing Limits ¤ Consider the function: where x is measured in radians . We are going to examine the behaviour of this function as x gets close to 0. Note that x cannot equal 0, as this involves dividing by 0, which is not allowed. This need not concern us, as we are only interested in what happens as x gets close to 0. Some results for x...| approaching 0 'from the left' (i.e. using values of x that are more negative than 0 and hence to the left of 0 on the number line) are: x f ( x ) -1.0 0.841471 -0.9 0.870363 -0.8 0.896695 -0.7 0.920311 -0.6 0.941071 -0.5 0.958851 -0.4 0.973546 -0.3 0.985067 -0.2 0.993347 -0.1 0.998334 -0.01 0.999983 From this table we can see that as x gets closer to 0 from the left the function gets closer to 1. We write to denote the limit as x approaches 0 from the left. Since we approach x from one direction only (the left) we refer to this as a one-sided limit. We can write our discovery from the table as: , where 0 - implies approaching 0 from the left. The function sin( x ) should be familiar from basic trigonometry, and is the ratio of the opposite and the hypotenuse of a right angled triangle. Radians is a natural measure of an angle with 2 radians in a full circle.  |../MathematicsSession/Maths00020.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Definition of a Limit ¤ Similarly, some results for x approaching 0 from the right are: x f ( x ) 1 0.841471 0.9 0.870363 0.8 0.896695 0.7 0.920311 0.6 0.941071 0.5 0.958851 0.4 0.973546 0.3 0.985067 0.2 0.993347 0.1 0.998334 0.01 0.999983 From this table we can see that as x gets closer to 0 from the right the function again gets closer to 1. We...| write to denote the limit as x approaches 0 from the right. We can write our discovery from the table as: Hence both the one-sided limits from the left and the right are 1. Making the distinction between the direction in which we approach x = 0 is unimportant, as we get the same answer of 1. We therefore write: and say that the limit of f ( x ) as x tends towards 0 is 1. Lots of other similar results are possible. By producing tables like the ones we've looked at check that:  |../MathematicsSession/Maths00030.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|One-Sided Limits ¤ These have already been introduced but for all the previous examples the limits from the left and right have been the same. This is not always the case. For example, consider the function: where x is the absolute value or modulus of x , which is defined as x = x if x 0 and x = - x otherwise. Considering each of the one sided...| limits in turn we obtain: and . Hence the one sided limits are not the same and: does not exist.  |../MathematicsSession/Maths00040.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Properties of Limits ¤ Limits have many properties and these include: , where k is a constant. Click here for an example. . Click here for an example . . Click here for an example . , assuming that | |../MathematicsSession/Maths00050.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Understanding the Derivative ¤ An important idea in mathematics is the gradient (or steepness) of a function. The graphs of some functions are steeper than others and differentiation is a way to quantify this. The derivative is also a useful tool in finding the optimum value of a function, as will be shown later. In order to understand the...| definition of the derivative, click on the following to update the chart. Consider the problem of finding the gradient of the function at the value of x . We define the gradient of the function at x as being the gradient of the tangent to the function's graph at x . The tangent line just touches the curve at this value of x . We can approximate this tangent line by using a small step, h , along the vertical axis. Hence the gradient of the curve is approximately:  |../MathematicsSession/Maths00060.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Definition of the Derivative ¤ So, the gradient of the curve is approximately: Using smaller and smaller values of h gives more accurate estimates of the gradient. Of course, h cannot equal 0, as then there is no line to draw and we divide by 0 in the formula above. However, we can take the limit as h tends towards 0 and obtain the gradient as:...| Since the gradient depends on the value of x , this expression is also a function of x , and is derived from the function as shown. This function is denoted by f' ( x ) and is referred to as the derivative. Obtaining this function is called differentiation. Hence we write: This limit is the definition of the derivative. If we use the notation y = f ( x ), then the derivative is often written as .  |../MathematicsSession/Maths00070.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rules for Calculating Derivatives ¤ Using the definition of the derivative we can prove many important results. Here is a summary of the eight rules for calculating derivatives. Click on each to jump a new screen for the full details with examples and exercises. The derivative of a constant rule: if f ( x ) = c then f' ( x ) = 0. The powers of x...| rule: if f ( x ) = ax n then f' ( x ) = anx (n-1) . Trigonometric functions: if f ( x ) = sin( x ) then f' ( x ) = cos( x ) if f ( x ) = sec( x ) then f' ( x ) = sec( x )tan( x ) if f ( x ) = cos( x ) then f' ( x ) = -sin( x ) if f ( x ) = cot( x ) then f' ( x ) = -csc 2 ( x ) if f ( x ) = tan( x ) then f' ( x ) = sec 2 ( x ) if f ( x ) = csc( x ) then f' ( x ) = - csc( x )cot( x ) if f ( x ) = a sin( bx ) then f' ( x ) = ab cos( bx ) if f ( x ) = a cos( bx ) then f' ( x ) = -ab sin( bx ) Exponential and logarithmic functions: if f ( x ) = e x then f' ( x ) = e x if f ( x ) = ae bx then f' ( x ) = abe bx . if f ( x ) = ln ( x ) then f' ( x ) = 1/ x The sum rule: if f ( x ) = g ( x ) &plusmn; h ( x ) then f' ( x ) = g' ( x ) &plusmn; h' ( x ). The product rule: [ u ( x ) v ( x )]' = u ( x ) v' ( x ) + v ( x ) u' ( x ). The quotient rule: The chain rule: [ u ( v ( x ))]' = u' ( v ( x )) v' ( x ) .  |../MathematicsSession/Maths00080.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 1: The Derivative of a Constant ¤ If f ( x ) = c , where c is a constant, then the function remains at the same value for all x . It neither increases or decreases and therefore has a gradient of 0. Therefore f' ( x ) = 0. Example Find f' ( x ) when f ( x ) = 2. This is f ( x ) = c , where c = 2, so f' ( x ) = 0. So what do you think f' ( x )...| is when f ( x ) = 3 ? Type your answer in the textbox, and click the &quot;Check Answer&quot; button. f' ( x ) = Yes, that's right. 3 is a constant and therefore the differential is 0. No, that's not right. 3 is a constant and therefore the differential is 0. Return  |../MathematicsSession/Maths00080a.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Return Rule 2: The Powers of x ¤ If f ( x ) = ax n then f' ( x ) = anx (n-1) . Examples If f ( x ) = 2 x 5 then f' ( x ) = 10 x 4 . If then Have a go at these questions. Type your answer in the textboxes, and click the &quot;Check Answer&quot; button. 1. What is f' ( x ) when f ( x ) = 3 x 5 ? f' ( x ) = to the power of Yes, that's right, f '( x )...| = 15 x 4 . No, that's not right, f '( x ) = 15 x 4 . 2. What is f' ( x ) when f ( x ) = x 12 ? f' ( x ) = to the power of Yes, that's right, f '( x ) = 12 x 11 . No, that's not right, f '( x ) = 12 x 11 .  |../MathematicsSession/Maths00080b.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 3: Trigonometric Functions ¤ Assuming that the angle x is measured in radians and that , and we have: If f ( x ) = sin( x ) then f' ( x ) = cos( x ) If f ( x ) = cos( x ) then f' ( x ) = -sin( x ) If f ( x ) = tan( x ) then f' ( x ) = sec 2 ( x ) and: If f ( x ) = sec( x ) then f' ( x ) = sec( x )tan( x ) If f ( x ) = cot( x ) then f' ( x ) =...| -csc 2 ( x ) If f ( x ) = csc( x ) then f' ( x ) = -csc( x )cot( x ) Two generalisations of the first two rules are often helpful: If f ( x ) = a sin( bx ) then f' ( x ) = ab cos( bx ) If f ( x ) = a cos( bx ) then f' ( x ) = - ab sin( bx ) Click here for a examples and exercises on differentiating trigonometric functions . Return  |../MathematicsSession/Maths00080c.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 4: Exponential and Logarithmic Functions ¤ In order to produce elegant results for exponential and logarithm functions we need to define the constant e : For this value we have that: If f ( x ) = e x then f' ( x ) = e x More generally: If f ( x ) = ae bx then f' ( x ) = abe bx . Also, defining ln( x ) = log e ( x ), we have that: If f ( x ) =...| ln( x ) then . Examples If f ( x ) = 4 e -3 x then f' ( x ) = -12 e -3 x If f ( x ) = e 4 x then f' ( x ) = 4 e 4 x Have a go at these questions. 1. What is f' ( x ) when f ( x ) = 2 e -7 x ? f' ( x ) = to the power of Yes, that's right, f '( x ) = -14 e -7 x . No, that's not right, f '( x ) = -14 e -7 x . 2. What is f' ( x ) when f ( x ) = -2 e - x ? f' ( x ) = to the power of Yes, that's right, f '( x ) = 2 e - x . No, that's not right, f '( x ) = 2 e - x . Return  |../MathematicsSession/Maths00080d.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Return Rule 5: The Sum Rule ¤ This rule means that if we have several functions added together (or indeed taken away) then we can differentiate each one in turn and add (or take away) the derivatives: If f ( x ) = g ( x ) &plusmn; h ( x ) then f' ( x ) = g' ( x ) &plusmn; h' ( x ). Examples If f ( x ) = 2 x 3 - sin( x ) then f' ( x ) = 6 x 2 +...| cos( x ) If f ( x ) = x - ln( x ) then f' ( x ) = 1 - 1/ x Try these questions. 1. What is f' ( x ) when f ( x ) = 3 x 2 + 10? f' ( x ) = + Yes, that's right, f '( x ) = 6 x . No, that's not right, f '( x ) = 6 x . Don't forget that a constant differentiates to 0. 2. What is f' ( x ) when f ( x ) = 3sin(4 x ) + sin( x )? f' ( x ) = + Yes, that's right, f '( x ) = 12cos(4 x ) + cos( x ). No, that's not right. Using the if f ( x ) = a sin( bx ) then f' ( x ) = ab cos( bx ) rule as well, f '( x ) = 12cos(4 x ) + cos( x ).  |../MathematicsSession/Maths00080e.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 5: The Product Rule ¤ Unfortunately, things are not quite so easy when we have two functions multiplied together. It is not the case that: [ u ( x ) v ( x )]' = u' ( x ) v' ( x ). That is we cannot simply differentiate each function in turn and multiply the results together. The product rule says that: [ u ( x ) v ( x )]' = u ( x ) v' ( x ) +...| v ( x ) u' ( x ). Example If f ( x ) = (4 x 2 - 1)(7 x 3 + x ) find f' ( x ). Here we could multiply out the brackets and use earlier rules, instead use the product rule. Let u ( x ) = (4 x 2 - 1) and v ( x ) = (7 x 3 + x ). Therefore u' ( x ) = 8 x and v' ( x ) = (21 x 2 + 1). The product rule gives f' ( x ) = (4 x 2 - 1)(21 x 2 + 1) + (7 x 3 + x ) (8 x ). Multiplying out and simplifying this gives f' ( x ) = 140 x 4 - 9 x 2 - 1. Click here for a couple of exercises on using the product rule . Return  |../MathematicsSession/Maths00080f.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Return Rule 7: The Quotient Rule ¤ This rule enables us to differentiate a function divided by another function. The quotient rule says that: Example Find f' ( x ) when . Let u ( x ) = x 2 - 1 and v ( x ) = x 3 + x. Then u' ( x ) = 2 x and v' ( x ) = (3 x 2 + 1). The quotient rule gives . and simplifying the numerator gives: . Try these questions....| Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. What do you think the differential is when: f ( x ) = sin( x ) / x ? 2. What do you think the differential is when: f ( x ) = 3 x / (2 x +1)?  |../MathematicsSession/Maths00080g.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 8: The Chain Rule ¤ This rule enables us to differentiate 'functions of functions'. For example consider the function f ( x ) = cos( x 2 + 9) This is a function of a function as first we take the value of x and apply the 'square and add 1' function. Having done this we apply another function, the cos function. In general, we will call the...| first function to be applied v ( x ) and the second u ( x ). The chain rule says that: [ u ( v ( x ))]' = u' ( v ( x )) v' ( x ) Example For our example of f ( x ) = cos( x 2 + 9), we have v ( x ) = x 2 + 9 and u ( v ( x )) = cos( v ( x )) . We therefore have v' ( x ) = 2 x and u' ( v ( x )) = -sin( v ( x )) = -sin( x 2 + 9). The chain rule now gives us that: f' ( x ) = -sin( x 2 + 9) 2 x = -2 x sin( x 2 + 9) Click here for a couple of exercises on using the chain rule . Return  |../MathematicsSession/Maths00080h.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 6: Exercises using the Product Rule ¤ Try these questions. Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. What do you think the differential is when f ( x ) = x 2 tan( x )? f' ( x ) = x 2 sec 2 ( x ) + 2 x tan( x ) 2. What do you think the differential is when f ( x ) = x 3 sin( x )?...| f' ( x ) = 3 x 2 sin ( x ) + x 3 cos( x ) Return  |../MathematicsSession/Maths00080i.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 8: Exercises using the Chain Rule ¤ Try these questions. Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. What do you think the differential is when f ( x ) = ( x 3 + 7 x + 1) 35 ? f' ( x ) = 35( x 3 + 7 x + 1) 34 (3 x 2 + 7). This is derived because we have v ( x ) = x 3 + 7 x + 1 and u...| ( v ( x )) = ( v ( x )) 35 . So v' ( x ) = 3 x 2 + 7 and u' ( v ( x )) = 35 (v ( x )) 34 . 2. What do you think the differential is when ? This is derived because we have v ( x ) = 4 + 3 = 4 + 3 x 1/2 and similarly u ( v ( x )) = ( v ( x )) 1/2 . So v' ( x ) = 3/2 x -1/2 and u' ( v ( x )) = 1/2( v ( x )) -1/2 . Therefore, f' ( x ) = (3/4)(1/ x 1/2 )(1/(4 + 3 x 1/2 ) 1/2 ) Return  |../MathematicsSession/Maths00080j.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 3: Examples and Exercises using Trigonometric Functions ¤ Examples If f ( x ) = 3sin(7 x ) then f' ( x ) = 21cos(7 x ) If f ( x ) = 2cos( x ) then f' ( x ) = -2sin( x ) Have a go at these questions. 1. What is f' ( x ) when f ( x ) = 3cos(10 x )? f' ( x ) = Yes, that's right, f '( x ) = -30sin(10 x ). No, that's not right, f '( x ) =...| -30sin(10 x ). 2. What is f' ( x ) when f ( x ) = sin(4 x )? f' ( x ) = Yes, that's right, f '( x ) = 4cos(4 x ). No, that's not right, f '( x ) = 4cos(4 x ). Return  |../MathematicsSession/Maths00080k.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Maximum and Minimum Problems ¤ An important application of differentiation is finding maximum and minimum points of functions. When modelling real life situations, a maximum point may provide a maximum profit. A minimum point might indicate minimum loss. Hence such points are of interest. Assuming that a function is differentiable, local maximum...| and minimum points are located where the function's gradient is zero, as shown below. For all three points shown, f '( x ) = 0 and hence these points are called stationary points. Two things should be noted. Just because the gradient is zero does not mean that the point is a maximum or a minimum point. The maximum and minimum points shown are only relative maximum and minimum points. They are either greater than or less than the points around them but other parts of the graph are greater than the maximum and less than the minimum points.  |../MathematicsSession/Maths00090.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|The Second Differential Test ¤ Nonetheless finding relative maximum and minimum points is a useful technique and may well provide global maximum and minimum points in practice. If the function is twice differentiable then this second derivative test is useful. Suppose that the function f ( x ) is twice differentiable at a stationary point, x 0 ,...| to produce f ''( x ). If f ''( x ) &gt; 0 then there is a relative minimum at x 0 . If f ''( x ) &lt; 0 then there is a relative maximum at x 0 . For example, if f ( x ) = x 3 then f '( x ) = 3 x 2 and f ''( x ) = 6 x . Let's find any relative maximum and minimum points of f ( x ) = x 4 - 2 x 2 . f '( x ) = 4 x 3 - 4 x = 4 x ( x - 1)( x + 1) f ''( x ) = 12 x 2 - 4 Now, f '( x ) = 0 when x = -1, 0 or 1. Hence these are the possible relative maximum and minimum points. Applying the second derivative test to these points: f ''(-1) = 8 &gt; 0, f ''(0) = -4 &lt; 0, f ''(1) = 8 &gt; 0. This means that there is a relative maximum at x = 0, and relative minima at the points x = -1 and x = 1.  |../MathematicsSession/Maths00100.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Find the Relative Maximum and Minimum ¤ Have a go yourself at finding the relative maximum and minimum points for these functions. Type a value for the maximum and minimum points if you find them, or click no stationary points if you don't think there are any and click the &quot;Check Answer&quot; button when you have finished. 1. f ( x ) = x ( x...| - 1) 2 ? Stationary points: Minimum when x = , Maximum when x = No stationary points Yes, that's right, the relative minimum point occurs when x = 1 and the relative maximum point is when x = 1/3. With f '( x ) = 3 x 2 - 4 x + 1 = x (3 x - 4) + 1, values for x of 1 and 1/3 make f '( x ) = 0, and with f ''( x ) = 6 x - 4, you can see that f ''(1) > 0 and is therefore a minimum and that f ''(1/3) Yes, the relative minimum point occurs when x = 1. However the relative maximum point is when x = 1/3. This is derived from f ( x ) = x ( x 2 - 2 x + 1) and using the product rule, f '( x ) = 3 x 2 - 4 x + 1 = x (3 x - 4) + 1. Values for x of 1 and 1/3 make f '( x ) = 0, and with f ''( x ) = 6 x - 4, you can see that f ''(1) > 0 and is therefore a minimum and that f ''(1/3) Yes, the relative maximum point is when x = 1/3, however the relative minimum point occurs when x = 1. This is derived from f ( x ) = x ( x 2 - 2 x + 1) and using the product rule, f '( x ) = 3 x 2 - 4 x + 1 = x (3 x - 4) + 1. Values for x of 1 and 1/3 make f '( x ) = 0, and with f ''( x ) = 6 x - 4, you can see that f ''(1) > 0 and is therefore a minimum and that f ''(1/3) No, that's not right, the relative minimum point occurs when x = 1 and the relative maximum point is when x = 1/3. This is derived from f ( x ) = x ( x - 1) 2 = x ( x 2 - 2 x + 1) and using the product rule, f '( x ) = 3 x 2 - 4 x + 1 = x (3 x - 4) + 1. Values for x of 1 and 1/3 make f '( x ) = 0, and with f ''( x ) = 6 x - 4, you can see that f ''(1) > 0 and is therefore a minimum and that f ''(1/3) No, that's not right, there are 2 stationary points. The relative minimum point occurs when x = 1 and the relative maximum point is when x = 1/3. This is derived from f ( x ) = x ( x - 1) 2 = x ( x 2 - 2 x + 1) and using the product rule, f '( x ) = 3 x 2 - 4 x + 1 = x (3 x - 4) + 1. Values for x of 1 and 1/3 make f '( x ) = 0, and with f ''( x ) = 6 x - 4, you can see that f ''(1) > 0 and is therefore a minimum and that f ''(1/3) 2. f ( x ) = x 3 + 5 x - 2? Stationary points: Minimum when x = , Maximum when x = No stationary points Yes, that's right! There are no relative maximum or minimum points for this function, because f '( x ) = 3 x 2 + 5 is therefore greater than 0 for all values of x . No, that's not right, there aren't actually any relative maximum or minimum points for this function. This is because f '( x ) = 3 x 2 + 5 and is therefore greater than 0 for all values of x .  |../MathematicsSession/Maths00110.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|The Integral ¤ Integration is the inverse operation of differentiation. To illustrate the idea, imagine we know that f' ( x ) = x . What is f ( x )? There are lots of possibilities, for example f ( x ) = x 2 /2 as then f' ( x ) = x . However, f ( x ) = x 2 /2 + 3, f ( x ) = x 2 /2 - 5, and in fact f ( x ) = x 2 /2 + c where c is any constant all...| provide f' ( x ) = x . Hence when we differentiate the function f ( x ) = x 2 /2 + c , we get x . Integration is the inverse process to differentiation, so that when we differentiate and then integrate we should get back to where we started. We write this as . In general we have that: . Here c is referred to as the constant of integration. There are many rules for integration and these will be explained on the next screen.  |../MathematicsSession/Maths00120.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rules for Integration ¤ Here is a summary of the seven rules for integration. Click on each to jump a new screen for the full details with examples and exercises. Integral of a constant rule: k d x = kx + c . Integral of powers of x rule: where n is not equal to -1. Integrals of trigonometric functions: and . Integrals of exponential functions and...| 1/ x : and . Integration of a sum rule: ( f ( x ) &plusmn; g ( x )) d x = f ( x ) d x &plusmn; g ( x ) d x . Integration by substitution: Step 1: Make a choice for u , say u = g ( x ). Step 2: Obtain d u /d x = g '( x ). Step 3: Substitute u = g ( x ), d u = g' ( x ) d x . Now the entire integral should be in terms of u . Step 4: Evaluate the integral. Step 5: Write the final answer in terms of x . Integration by parts: u ( x ) v' ( x ) d x = u ( x ) v ( x ) - v ( x ) u' ( x ) d x.  |../MathematicsSession/Maths00130.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 1: The Integral of a Constant ¤ Since when we differentiate the function kx we get k , we have that k d x = kx + c . Example 5d x = 5 x + c Evaluate the following integrals. Type your answer in the textboxes without using spaces, and click the &quot;Check Answer&quot; button. 1. 3 d x . f ( x ) = Yes, that's right, 3 d x = 3 x + c . No,...| that's not right, 3 d x = 3 x + c . 2. -1 d x . f' ( x ) = Yes, that's right, -1 d x = - x + c . No, that's not right, -1 d x = - x + c . Return  |../MathematicsSession/Maths00130a.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 2: The Integral of Powers of x ¤ Since when we differentiate the function we get ax n , we have that . Note that this rule cannot work for n = -1, as this involves dividing by 0. Hence this rule does not apply for this value of n . Example 4 x 3 d x = x 4 + c Try evaluating the following integrals. Have a go at working out the answer on paper...| before you click on the &quot;Check Answer&quot; button. 1. 2 x 7 d x . f ( x ) = . 2. x 10 d x . f ( x ) = . Return  |../MathematicsSession/Maths00130b.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 3: Integrals of Trigonometric Functions ¤ There are many types of integrals containing trigonometric functions. Two of the most important are: Since when we differentiate we get a cos( bx ) we have that: . Also, since when we differentiate we get a sin( bx ) we have that: . Examples Try evaluating the following integrals. Have a go at working...| out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. 2sin(5 x ) d x . f ( x ) = . 2. 10cos(2 x ) d x . f ( x ) = 5sin(2 x ) + c . Return  |../MathematicsSession/Maths00130c.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 4: Integrals of Exponential Functions and 1/ x ¤ The rule for finding the integral of a power of x failed to find , as the powers of x rule does not apply to a power of -1. However since when we differentiate the function ln( x ) we get , we have that = ln( x ) + c . Also since when we differentiate the function we get ae bx , we have that ae...| bx d x = + c . Examples Try evaluating the following integrals. Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. 4 e 2 x d x . f ( x ) = 2 e 2x + c . 2. 10 e -x d x . f ( x ) = -10 e -x + c . Return  |../MathematicsSession/Maths00130d.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 5: Integration of a Sum ¤ Just as with differentiation, if we have several functions added together (or indeed taken away) then we can integrate each one in turn and add (or take away) the integrals: ( f ( x ) &plusmn; g ( x )) d x = f ( x ) d x &plusmn; g ( x ) d x . Example Try evaluating the following integrals. Have a go at working out...| the answer on paper before you click on the &quot;Check Answer&quot; button. 1. ( x + sin( x )) d x . f ( x ) = x 2 /2 - cos( x ) + c . 2. (10 - 5cos(2 x )) d x . f ( x ) = 10 x - 5/2sin( x ) + c . Return  |../MathematicsSession/Maths00130e.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 6: Integration by Substitution ¤ By making an appropriate substitution, integrals often become easier to evaluate. The method is probably best shown by an example first. Let's evaluate: ( x 2 + 1) 50 2 x d x Make the substitution u = u ( x ) = x 2 + 1. This means that u' ( x ) = 2 x . Using the alterative notation described on screen 7, we...| have . This provides us with d u = 2 x d x . Substituting u and d u into the integral results in u 50 d u. This is readily dealt with and is . In terms of the original variable x we have that the integral is: . Click here for a summary of the steps and a couple of exercises on using the integration by substitution rule . Return  |../MathematicsSession/Maths00130f.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 7: Integration by Parts ¤ The product rule when differentiating gives us that: [ u ( x ) v ( x )]' = u ( x ) v' ( x ) + v ( x ) u' ( x ) Integrating both sides gives us: u ( x ) v ( x ) = u ( x ) v' ( x ) d x + v ( x ) u' ( x ) d x + c Rearranging this gives the formula for integration by parts as: u ( x ) v' ( x ) d x = u ( x ) v ( x ) - v (...| x ) u' ( x ) d x . Here the constant of integration is omitted as it will be re-introduced when the integral on the right hand side is evaluated. Click here for an example and a couple of exercises on using the integration by parts rule . Return  |../MathematicsSession/Maths00130g.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 7: Examples and Exercises using the Integration by Parts Rule ¤ Example Find f ( x ) when f' ( x ) = xe x d x . Let u ( x ) = x and v' ( x ) = e x . We differentiate u ( x ) = x in order to get u' ( x ) = 1 and integrate v' ( x ) = e x in order to get v ( x ) = e x . The constant of integration is omitted as it cancels out. Substituting into...| the integration by parts formula gives: xe x d x = xe x - e x d x = xe x - e x + c. Try these questions. Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. What do you think the integral is when f' ( x ) = xe -x d x ? f ( x ) = -xe -x - e -x + c 2. What do you think the differential is when f' ( x ) = ln( x ) d x ? f ( x ) = x ln ( x ) - x + c Return  |../MathematicsSession/Maths00130h.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Rule 6: Exercises using Integration by Substitution ¤ The procedure for integration by substitution is carried out in practice like this in five steps: Make a choice for u , say u = g ( x ). Obtain d u /d x = g '( x ). Substitute u = g ( x ), d u = g' ( x ) d x . At this stage, the entire integral should be in terms of u . No x 's should remain....| If they do, then try a different substitution, or another method. Evaluate the integral. Write the final answer in terms of x . Try these questions. Have a go at working out the answer on paper before you click on the &quot;Check Answer&quot; button. 1. What do you think the integral is when f' ( x ) = sin 2 ( x )cos( x ) d x ? f ( x ) = . This is achieved by using u = sin( x ). 2. What do you think the differential is when f' ( x ) = ? f ( x ) = . This is achieved by using u = x 3 + 1. Return  |../MathematicsSession/Maths00130i.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Finite Integral of a Function From a to b ¤ An important problem in mathematics is finding areas under the curve of a function as shown in this figure. This area is denoted by: The fundamental theorem of Calculus is a remarkable result linking the problem of finding areas and differentiating. It says that if the function is continuous on the...| interval from a to b , and if F ( x ) is a function that differentiates to give f ( x ), namely , then the area under the curve is given by:  |../MathematicsSession/Maths00135.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Calculating Finite Integrals ¤ Let's evaluate . Here the function differentiates to give x . So this may be used. In addition to this, so may any function , where c is a constant, as all these functions differentiate to give x . Hence by integrating the function and neglecting to add c provides a suitable function to use for finding areas. The...| area is therefore . Can you calculate the following finite integrals? Type your answers into the textboxes in the fraction format indicated using the lowest common denominator and click the &quot;Check Answer&quot; button. = / = / = / Yes, that's right. It integrates to x 4 /4, and substituting for x for the range 2 to 3 gives 81 - 16/4 = 65/4 . No, that's not right. It integrates to x 4 /4, and substituting for x for the range 2 to 3 gives 3 4 /4 - 2 4 /4 = 81 - 16/4 = 65/4 . Yes, that's right. It integrates to -1/ x , and substituting for x for the range 1 to 3 gives -1/3 - -1/1 = 2/3 . No, that's not right. It integrates to -1/ x (since 1/x 2 = x -2 ), and substituting for x for the range 1 to 3 gives -1/3 - -1/1 = 2/3 . Yes, that's right. Multiplying out and integrating gives (5 x 2 + 2 x 5 )/10. Substituting for x for the range -1 to 2 gives (20 + 64)/10 - ((5 - 2)/10) = 81/10 . No, that's not right. Multiplying out and integrating gives (5 x 2 + 2 x 5 )/10. Substituting for x for the range -1 to 2 gives (20 + 64)/10 - ((5 - 2)/10) = 81/10 .  |../MathematicsSession/Maths00137.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Summary ¤ The main aim of this session has been to cover the topics of : differentiation. Click here to jump to a summary of differentiation on screen 8 , and integration. Click here to jump to the summary of integration on screen 13 . | |../MathematicsSession/Maths00140.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Questions on Differentiation ¤ Here are some questions on differentiation. Click on each one as you begin to work out the answer, and after 10 seconds, you will be able to click on the associated &quot;Check Answer&quot; button to validate your solution. Find f '( x ) for the following functions: Answers: f '( x ) = 28 x 6 . f '( x ) = -2sin( x )...| - 3cos( x ) f '( x ) = x 3 e x + 3 x 2 e x f '( x ) = 7 e 7 x f '( x ) = 7 e 7 x  |../MathematicsSession/Maths00150.htm",
    "Statistics for Maintenance Modelling: Mathematics Preliminary|Questions on Integration ¤ Here are some questions on integration. Click on each one as you begin to work out the answer, and after 10 seconds, you will be able to click on the associated &quot;Check Answer&quot; button to validate your solution. Evaluate the following integrals: Answers: 2sin( x ) - 4cos( x ) + c x + c | |../MathematicsSession/Maths00160.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Introduction ¤ Reliability is defined to be the probability that a component or system will perform a required function for a given period of time when used under stated operating conditions. In other words, it is the probability of a non-failure over time. The risk associated with an event (e.g. the failure of a component) is most often defined...| as the probability of that failure occurring and the impact of that event once it has occurred. For instance the probability of an outage of the size seen in North West America in August 2003 is very small, but the impact of such an event is very large. Knowing these definitions of reliability and risk, it is clear that in order to study the two quantities we need to understand some ideas and concepts of probability . Indeed, the concepts of probability form the foundation of all decision making. By using the concepts of probability one is able to deal with uncertainty because this uncertainty can be measured. Once the uncertainty has been measured we can use the information to come to decisions. There are two general approaches to modelling uncertainty by using probability concepts: the more basic method makes use of the idea of sample space and events defined over the sample space. The probabilities of these events and the probabilities of other related events are then formed using 'unions' and 'intersections' of the elementary events. the second method is based on the concept of a random variable and a probability distribution associated with the random variable. A random variable is a variable that takes on certain values in accordance with specified probabilities.  |../BasicProbabilitySession/Prob00005.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Objectives ¤ By the end of this session you will understand: the key ideas and concepts of probability how Venn diagrams are used to represent the ideas of probability what conditional probability is how Bayes' theorem is used to predict the probability of a certain outcome what random variables are the concept of probability distributions and how...| to make use of them what expectation and variance are and how to calculate them the usefulness of the histogram.  |../BasicProbabilitySession/Prob00010.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Key Definitions and Concepts ¤ In order to understand the subjects studied in later sessions it will be beneficial to get to grips with the basic concepts of probability. Click on the concepts for a definition and examples. An Experiment The Sample Space An Event The Union of Events Intersection of Events The Complement of an Event An Event...| contained in Another Event An activity for which the outcome is not predictable with certainty is an experiment . However, although the outcome of the experiment is uncertain, the set of all possible outcomes is known. The set of all possible outcomes of an experiment is known as the sample space of the experiment, and is denoted S . Some examples of sample spaces are: If the experiment consists of flipping a coin once, then the sample space may be written as: S = H , T where H and T stand for heads and tails. If the experiment consists of measuring the lifetime of a bulb (in hours) then the sample space may be written as: S = x : 0 x &lt; where x represents the useful lifetime of the bulb. If the experiment consists of tossing two dice simultaneously, then the sample space consists of the 36 points given by: S = ( i , j ) : i , j = 1,2,3,4,5,6 where the outcome ( i , j ) is said to occur if i appears on one of the die, and j on the other. Any subset of the sample space S is called an event , E . Some examples of events, for the sample spaces described earlier, are: In the 1st sample space S = H , T , if E = H , then E is the event that a head was thrown. In the 2nd sample space S = x : 0 x &lt; , if E = x : 0 x 5 , then E is the event that the light bulb does not last longer than 5 hours. In the 3rd sample space S = ( i , j ) : i , j = 1,2,3,4,5,6 , if E = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) , then E is the event that the sum of the dice is 7. For any two events E and F of a sample space S , we define the new event E F to consist of all outcomes that are either in E , or in F , or in both . In other words, the event E F will occur if either E or F occurs. In the 3rd sample space S = ( i , j ) : i , j = 1,2,3,4,5,6 , if E = (1,6), (2,5) and F = (3,4) , then E F = (1,6), (2,5), (3,4) . The event E F is called the union of event E and the event F . For any two events E and F the intersection of E and F , written E F , consists of all outcomes that are both in E and in F . In the 2nd sample space S = x : 0 x &lt; , if E = x : 0 x 6 and F = x : 4 x 15 , then E F = x : 4 x 6 . For an event E , the definition of the complement of E , written E is the set of all outcomes in the sample space not equal to E . In other words E will occur if E does not . For instance, in 3rd sample space S = ( i , j ) : i , j = 1,2,3,4,5,6 , if event E = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) , then E = the event that the dice do not sum to 7 . Since either the event or its complement must occur, we have P ( E ) + P ( E ) = 1 i.e. the probability of the event E occuring plus the probability of it not occuring is 1. Finally, if all the outcomes of event E are contained in event F , then we say that E is contained in F , written E F (or F E ). This means that if E F , the occurrence of E necessarily implies the occurrence of F .  |../BasicProbabilitySession/Prob00020.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Questions on Key Concepts ¤ Let event A = the failure of component 1, and event B = the failure of component 2. Question 1 From the following choices, can you pick which one best describes A B ? Click on &quot;Check Answer&quot; when you've chosen. the event that both components have failed the event that neither component failed the event that...| component 1 failed or component 2 failed Yes, that's right. As the intersection of events A and B , A B represents both events occurring. No, neither component failing would have been represented by A B . A B is the intersection of events A and B , and describes the outcome that both components have failed. No, component 1 or 2 failing is the union of A and B - A B . A B is the intersection of events A and B , and describes the outcome that both components have failed. Question 2 From the following choices, can you pick the notation which describes the event that component 1 does not fail and component 2 fails? Click on &quot;Check Answer&quot; when you've made a selection. A B A B A B Yes, that's right. The complement of event A with the intersection of event B describes the sitiuation where component 1 doesn't fail, and component 2 does . No, A B represents the outcome that component 1 doesn't fail or that component 2 fails. However, the intersection of these two events - A B - describes the outcome that component 1 doesn't fail and component 2 does. No, A B is the event that component 1 fails and component 2 doesn't fail. It needs to be the other way round - A B - to describe the outcome that component 1 doesn't fail and component 2 does .  |../BasicProbabilitySession/Prob00030.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Introduction to Venn Diagrams ¤ Sample spaces and events and the relationships between the two are often shown by means of a Venn Diagram. The sample space S is depicted by a rectangle, while the events are depicted as regions, often circles, within the rectangle. This example shows a Venn diagram depicting the event E (the shaded area). This Venn...| diagram shows the event defined by the union of events E and F , written ( E F ).  |../BasicProbabilitySession/Prob00040.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Interpreting the Shaded Area in a Venn Diagram ¤ Bearing in mind the definitions on the previous page, click on one of the notations below which you think describes the shaded area in this Venn diagram, then click on &quot;Check Answer&quot;. E F E E F E F Yes, the shaded area is the intersection of E and F, written E F. No, the shaded area...| represents the intersection of E and F, and so E F is the correct answer.  |../BasicProbabilitySession/Prob00050.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Matching Venn Diagrams With Events ¤ Have a go at these. Drag each event from the bottom, and drop it on the Venn diagram whose shaded area you think the event describes. When you've finished, you will find out how many you've got right. You can click the Reset Button to try again. Event: Event: Event: Event: Event: Drag events: E E F G E F (E F)...| G E G Well done! You correctly identified all the events that the shaded areas in the event diagrams represent. I'm afraid that you've matched 2 of the events incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you've matched 3 of the events incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you've matched 4 of the events incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you haven't matched any of the events correctly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results.  |../BasicProbabilitySession/Prob00060.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Definition of Probability ¤ When one talks of a probability, it is the probability of a particular event occurring that is of interest. One way of defining the probability of an event is in terms of its relative frequency. Suppose that an experiment, with sample space S , is repeatedly performed under exactly the same conditions. For each event E...| of the sample space, we define n ( E ) to be the number of times in the first n repetitions of the experiment that the event E occurs. Then P ( E ), the probability of the event E , is defined by:  |../BasicProbabilitySession/Prob00070.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Axioms of Probability ¤ In order for the definition of probability to be valid, certain assumptions or axioms must be held true. These axioms are more or less intuitive, and state that: Axiom 1: 0 P ( E ) 1 The probability of event E is some number between 0 and 1. Axiom 2: P ( S ) = 1 The outcome will be a point in the sample space with...| probability 1. Axiom 3 For any sequence of mutually exclusive events E 1, E 2 &#133; The probability of at least one of these mutually exclusive events occurring is just the sum of their respective probabilities. Click here for an example.  |../BasicProbabilitySession/Prob00080.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Question on Probability ¤ If a die is rolled and we suppose that all six sides are equally likely to appear, what is the probability of rolling a multiple of three? Click on one of the options below, then click on &quot;Check Answer&quot;. 1/2 1/3 1/6 Yes, the probability is 1/3, since there are two multiples of three on a die, 3 and 6. So, from...| axiom 3, the probability is: P ( 3 ) + P ( 6 ) = 1/6 + 1/6 = 1/3 No, the probability is actually 1/3. This is because there are two multiples of three on a die, 3 and 6. So, from axiom 3, the probability of rolling either of these is: P ( 3 ) + P ( 6 ) = 1/6 + 1/6 = 1/3  |../BasicProbabilitySession/Prob00090.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Propositions of Probability ¤ Proposition 1: P ( E ) = 1 - P ( E ) This states that the probability that an event does not occur is 1 minus the probability that the event does occur. This proposition is derived from the fact that E E = S . Proposition 2: If E F , then P ( E ) P ( F ) If event E is contained in event F , then the probability of E...| is no greater than the probability of event F . Proposition 3: P ( E F ) = P ( E ) + P ( F ) - P ( E F ) This states that the probability of event E or F or both occurring is the sum of their respective probabilities minus the probability of E and F occuring. This proposition can be proved using a Venn diagram where ( E F ) is divided into three mutually exclusive sections. Click here for the proof. Click here for an example.  |../BasicProbabilitySession/Prob00100.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Conditional Probability ¤ The concept of conditional probability is one of the most important in all of probability theory. It's based on the notion that probabilities can be updated when certain events are observed. For instance if we are monitoring two components operating in parallel which share a common load, if one component fails, the...| probability of the other component failing increases as a result of the increased load placed on it. The updated probability of event E given that event F has occurred is the conditional probability of E given F has occurred and is written P ( E F ). Independent and Dependent Events When a coin is tossed twice, the two events are said to be independent, since the second flip of the coin is not influenced by the first flip. In short, two events are independent if the occurrence of one event does not effect the probability of the occurrence of the other. Two events E and F that are not independent are said to be dependent. Events E and F are independent if and only if: P ( E F ) = P ( E ) assuming P(F) 0, or P ( F E ) = P ( F ) assuming P(E) 0, or P ( E F ) = P ( E ) P ( F ) Click here for an example.  |../BasicProbabilitySession/Prob00110.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Definition of Conditional Probability ¤ The updated probability of event E given that a dependent event F has occurred is the conditional probability of E given F . The conditional probability is denoted P ( E F ). The following definition holds for the conditional probability if P ( F ) &gt; 0. This says that the probability that event E occurs...| given event F has occurred is equal to the probability of event E or event F or both events E and F occurring divided by the probability of event F occurring. If a coin is flipped twice and we assume that all four points in the sample space S = ( H , H ), ( H , T ), ( T , H ), ( T , T ) are equally likely, what is the conditional probability that both flips result in heads, given that the first flip does? Click on one of the options below, then click on &quot;Check Answer&quot;. 1/4 1/2 3/4 Yes, the probability is 1/2. If E = ( H , H ) denoted the event that both flips land heads, and F = ( H , H ), ( H , T ) the event that the first flip lands heads, then the desired probability is given by: No, the probability is actually 1/2. If E = ( H , H ) denoted the event that both flips land heads, and F = ( H , H ), ( H , T ) the event that the first flip lands heads, then the desired probability is given by:  |../BasicProbabilitySession/Prob00120.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Introduction to Bayes' Theorem ¤ The definition of conditional probability can be taken a step further to give Bayes' Theorem. Bayes' Theorem is useful when calculating the probability of a certain outcome for an experiment when the outcome depends on various intermediate stages. To understand Bayes' Theorem it is useful to look at a few examples....| Suppose the completion of a building renovation may be delayed because of a strike. The probabilities are: 0.4 that there will be a strike 0.9 that the renovation will be completed on time if there is no strike 0.3 that the renovation will be completed on time if there is a strike. What is the probability that the renovation job will be completed on time? If A is the event that the renovation will be completed on time and B is the event that there will be a strike, we are given P ( B ) = 0.4, P ( A B ) = 0.9 and P ( A B ) = 0.3. We can write: P ( A ) = P (( A B ) ( A B )) ( click here for an explanation ) = P ( A B ) + ( A B ) (from the definition of a union) = P ( B ) P ( A B ) + P ( B ) P ( A B ) (from the definition of conditional probability ) Substituting, gives: P ( A ) = (0.4)(0.3) + (1 - 0.4) (0.9) = 0.66 If P ( F ) > 0, the definition of conditional probability is: . Therefore, P ( E F ) = P ( F ) P ( E F ).  |../BasicProbabilitySession/Prob00130.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Rule of Total Probability ¤ The use of conditional probability in the last example can be generalised for the situation where the intermediate stage permits k different alternatives (whose occurrence is denoted by B 1, B 2,&#133;, Bk ). It requires the following theorem, sometimes called the rule of total probability : Have a go at this question....| A tool manufacturer rents machines from three other companies: 60% from company 1, 30% from company 2 and 10% from company 3. If 9% of the machines from company 1 need a service, 20% of the machines from company 2 need a service and 6% of the machines from company 3 need a service, what is the probability that one of the machines will need servicing? We'll call this event A . Type your answer in the textbox, and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. P ( A ) = Yes, that's right. If B 1, B 2, and B 3 are the events that the machines come from company 1, 2, or 3 respectively, we can substitute the probabilities into the rule of total probability to give: P ( A ) = (0.0)(0.9) + (0.3)(0.2) + (0.1)(0.06) = 0.12 No, P ( A ) = 0.12 . If B 1, B 2, and B 3 are the events that the machines come from company 1, 2, or 3 respectively, we can say that P ( B 1 ) = 0.6, P ( B 2 ) = 0.3, P ( B 3 ) = 0.1, P ( A B 1 ) = 0.09, P ( A B 2 ) = 0.2, P ( A B 3 ) = 0.06. Using the rule of total probability we get: P ( A ) = (0.0)(0.9) + (0.3)(0.2) + (0.1)(0.06) = 0.12  |../BasicProbabilitySession/Prob00140.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Bayes' Theorem ¤ Return Referring to the previous question, suppose that we are interested to know that if a machine needs a service, what is the probability that it came from company 2? To answer this type of question we need Bayes' Theorem. If B 1, B 2,&#133;, Bk constitute a partition of the sample space S and P ( B i ) 0 for i = 1, 2, ..., k...| then for any event A in S such that P ( A ) 0: So in the previous example, if a machine needed a service ( P ( A )), what is the probability that it came from company 2? To recap: P ( B 1 ) = 0.6, P ( B 2 ) = 0.3, P ( B 3 ) = 0.1, P ( A B 1 ) = 0.09, P ( A B 2 ) = 0.2 and P ( A B 3 ) = 0.06. Type your answer in the textbox, and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. P ( B 2 A ) = Yes, that's right. Substituting the probabilities into Bayes' Theorem gives: P ( B 2 A ) = (0.3)(0.2) / ( (0.6)(0.9) + (0.3)(0.2) + (0.1)(0.06) ) = 0.06 / 0.12 = 0.5 No, P ( B 2 A ) = 0.5 . Substituting the probabilities into Bayes' Theorem gives: P ( B 2 A ) = (0.3)(0.2) / ( (0.6)(0.9) + (0.3)(0.2) + (0.1)(0.06) ) = 0.06 / 0.12 = 0.5  |../BasicProbabilitySession/Prob00150.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Random Variables and Probability Distributions ¤ Although the concept of a random event is a useful one in describing random processes, a more useful concept is that of the random variable, and this leads us to the second approach to modelling uncertainty using probability distributions. As we've seen, the outcome of an experiment need not be a...| number. For example, the outcome when a coin is tossed can be 'heads' or 'tails'. However, we often want to represent outcomes as numbers. A random variable is a function that associates a unique numerical value with every outcome of an experiment . The value of the random variable will vary from trial to trial as the experiment is repeated. There are two types of random variable - discrete and continuous. Discrete Random Variables A discrete random variable is one which may take on only a countable number of distinct values such as 0, 1, 2, 3, 4, ... Discrete random variables are usually (but not necessarily) counts.If a random variable can take only a finite number of distinct values, then it must be discrete. Examples of discrete random variables include: the number of children in a family the Friday night attendance at a cinema the number of patients in a doctor's surgery the number of defective light bulbs in a box of ten.  |../BasicProbabilitySession/Prob00160.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|The Probability Mass Function ¤ For a discrete random variable X , the probability mass function is defined as: P ( x ) = P X = x This function says that the probability mass function for x is equal to the probability that the value of the observed random variable X equals the value x . The probability mass function is positive or zero for all x ....| Since X must take on one of the values x i , we have: The chart on the right shows the probability mass function for the sum of two dice. The discrete random variable ( X ) is the sum of the dice, with possible values of x ranging from 2 to 12. P ( x i ) is calculated by summing the probabilities of the different roles of the die that result in x i .  |../BasicProbabilitySession/Prob00170.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Continuous Random Variables and the Probability Density Function ¤ A continuous random variable is one which takes an infinite number of possible values. Continuous random variables are usually measurements. Examples include height, weight, the amount of sugar in an orange and the time required to run a mile. The way in which probabilities are...| assigned to discrete random variables is not applicable for continuous random variables, and instead of using a probability mass function, we use a probability density function . Mathematically the probability density function, f ( x ) is given by the equation: Unlike the discrete case, f ( x ) does not give the probability of X = x . General properties of probability density functions are: f(x) 0 for - &lt; x &lt; These two properties come from the fact that: the random variable must take on a value from - to probabilities cannot be negative.  |../BasicProbabilitySession/Prob00180.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|The Normal Distribution ¤ An example of the probability density function for the well known Normal (Gaussian) Distribution is given by: The probability density function for the standard Normal Distribution with = 1 and = 0 is shown here. Click on the chart for an interactive version in Excel where you can experiment with different values of and ....| In statistics the Normal distribution is the usual starting point to model a random variable. It is said that the Normal distribution is the distribution of a random variable which is made up of many small disturbances or effects. Furthermore the Normal distribution is used extensively in statistical inference. With regards to reliability and maintainence the Normal distribution takes on a less important role since lifetime distributions do not allow negative values, whereas the Normal distribution does.  |../BasicProbabilitySession/Prob00190.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|The Cumulative Density Function ¤ Return The cumulative density function gives the probability that a random variable X will be less than some specified value x (e.g. the probability that an item fails before time t ). Mathematically it is written: where f ( t ) is the probability density function at t . For the standard (mean zero and variance 1)...| Normal Distribution this formula can be written: Often Tables (see for example the tables on the StatsoftInc website ) are given in statistical text books giving the values of the following function: This quantity is represented by the shaded area in the distribution shown here.  |../BasicProbabilitySession/Prob00200.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Using Statistical Tables to Find the Probability of a Random Variable Having a Certain Value ¤ Lets have a look at working out the probability that a random variable having a standard Normal Distribution will take on a value less than 1.72 shown graphically here. First use the table which gives the probabilities that a standardised variable...| selected at random will be greater than a value of Z and look up the figure for 1.72. Click on the look-up table to see that the probability of a value greater than 1.72 is 0.0427. Therefore, the probability that a value of less than 1.72 occurs is 1 - 0.0427 = 0.9573.  |../BasicProbabilitySession/Prob00210.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Find the Probability of a Random Variable Falling Within a Range ¤ Find the probability that a random variable having a standard Normal Distribution will take on a value between 1.30 and 1.75. Click on the look-up table to calculate your answer. Type it in the textbox to 4 decimal places when you have worked it out and click the &quot;Check...| Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. P ( 1.75 &gt; Z 1.3 ) = Yes, that's right. You correctly subtracted the probability of Z = 1.75 from the probability of Z = 1.3 to get the answer of 0.0567 No that's not right. The answer is 0.0567 . This is obtained by subtracting the probability of Z = 1.75 from the probability of Z = 1.3. P ( Z=1.3 ) - P ( Z = 1.75 ) = 0.0968 - 0.0401 = 0.0567  |../BasicProbabilitySession/Prob00220.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Expectation and Variance ¤ The expected value and variance are two properties of random variables that help define how the random variable is distributed. The expected value gives an indication of where about the random variable is located, and the variance gives an indication of how the random variable is most likely spread about the expected...| value. The formula for both the expected value and the variance of a random variable depends on whether the random variable is discrete or continuous. Expected Value for Discrete Random Variables For a discrete random variable with probability mass function p ( x ), the expected value is given by: This function says that the expected value of X is a weighted average of all the possible values ( n ) that X can take on, each value being weighted by the probability mass that X assumes it. See if you can work out the expected value of X , where X is the outcome when a die is rolled. Type your answer into the textbox and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. E [ X ] = Yes, that's right, E [ X ] = 3.5. This is because for all i , P ( x i ) = 1/6, so we get: E [ X ] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 7/2 No that's not right. E [ X ] = 3.5 . Since p ( 1 ) = p ( 2 ) = p ( 3 ) = p ( 4 ) = p ( 4 ) = p ( 6 ) = 1/6, we get: E [ X ] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 7/2  |../BasicProbabilitySession/Prob00230.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Expected Value for Continuous Random Variables ¤ If X is a continuous random variable having probability density function f ( x ), then the expected value of x is defined to be: For example, for a uniformly distribution random variable with probability density function given by: The expected value can be worked out as follows: | |../BasicProbabilitySession/Prob00240.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Variance for Random Variables ¤ The variance for discrete and continuous random variables is given as: Discrete random variable with a probability mass function p ( x ) Continuous variance of a random variable X , with expected value Alternatively, often the easiest way to calculate the variance for both discrete and continuous random variables is...| by the formula: Var ( X ) = E [ X 2 ] - ( E [ X ]) 2 Can you calculate Var ( X ) if X represents the outcome when a die is rolled? Recall that when you calculated E [ X ] for rolling a die, it was 7/2. Type your answer into the textboxes in the fraction format indicated and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. Var ( X ) = / Yes, that's right, Var ( X ) = 35/12 . Since we already knew E [ X ], we had to calculate E [ X 2 ] = 1 2 (1/6) + 2 2 (1/6) + 3 2 (1/6) + 4 2 (1/6) + 5 2 (1/6) + 6 2 (1/6) = 91/6 Var ( X ) = 91/6 - (7/2) 2 = 35/12 No, that's not right, Var ( X ) = 35/12 . Since we already knew E [ X ], we had to calculate E [ X 2 ] = 1 2 (1/6) + 2 2 (1/6) + 3 2 (1/6) + 4 2 (1/6) + 5 2 (1/6) + 6 2 (1/6) = 91/6 Var ( X ) = E [ X 2 ] - E [ X ] 2 = 91/6 - (7/2) 2 = 35/12  |../BasicProbabilitySession/Prob00250.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Variance for Continuous Random Variables ¤ The concept of variance can be visualised by looking at the Normal distribution with two different variances. The chart below shows the probability density functions for the Normal distribution with variances 2 = 0.5 and 4. One can see how the distribution about the expected value (zero) is much different...| for differing variances. The larger the variance, the wider the spread about the mean.  |../BasicProbabilitySession/Prob00260.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|The Histogram ¤ We have already seen that the expected value (mean) and variance can be used to summarize the essential properties of a random variable. The purpose of a histogram is to graphically summarise the distribution of a univariate data set. The histogram graphically shows the following: center (i.e. the location) of the data spread (i.e....| the scale) of the data skewness of the data presence of outliers presence of multiple modes in the data. These features provide strong indications of the proper distributional model for the data. This histogram plots a data set containing the width of a specialised drill bit i.e. continuous data. The specifications of the drill bit require that the bit have a diameter between 10.1mm and 10.3mm. Using a histogram, one can immediately see the distribution of the drill bit widths.  |../BasicProbabilitySession/Prob00270.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Constructing a Histogram ¤ The histogram of drill bit diameters was formed using the most common form of the histogram. The plot is obtained by splitting the range of the data into equal-sized bins (called classes). Then for each bin, the number of points from the data set that fall into each bin are counted. That is: Vertical axis: Frequency...| (i.e. counts for each bin) Horizontal axis: Response variable The classes can either be defined arbitrarily by the user or via some systematic rule. The histogram has a variant which is called the relative histogram, where the counts are replaced by normalized counts. There are two common methods used to normalize the counts. The first method is where the normalized count is the count in a class divided by the total number of observations. In this case the relative counts are normalized to sum to one (or 100 if a percentage scale is used). This is the intuitive case where the height of the histogram bar represents the proportion of the data in each class. The second method is where the normalized count is the count in the class divided by the number of observations times the class width. For this normalization, the area (or integral) under the histogram is equal to one. From a probabilistic point of view, this normalization results in a relative histogram that is most akin to the probability density function. If you want to overlay a probability density on top of the histogram, use this normalization. Although this normalization is less intuitive (relative frequencies greater than 1 are quite permissible), it is the appropriate normalization if you are using the histogram to model a probability density function.  |../BasicProbabilitySession/Prob00280.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Exercise to Construct a Histogram in Excel ¤ X Bins -0.432560 -1.665600 0.125330 0.287680 -1.146500 1.190900 1.189200 -0.037633 0.327290 0.174640 -0.186710 0.725790 -0.588320 2.183200 -0.136400 0.113930 1.066800 0.059281 -0.095648 -0.832350 -3 -2 -1 0 1 2 3 Try this exercise to create a histogram in Excel ( click here to launch Excel ) from the...| data range in this table using the bins we have suggested here. To do this: Copy and paste each column into Excel (for a better layout it is best to copy each one separately and paste them in side-by-side). Select Tools &gt; Data Analysis from the Excel menu and select Histogram from the dialog box and click OK . If the Data Analysis option isn't available, click Add-Ins on the Tools menu. In the Add-Ins available list, select the Analysis ToolPak box, and then click OK . This opens up the dialog window where you specify the data underlying the histogram. Select any pre-existing range in the Input Range field , and then select the data values in the X column . Select any pre-existing range in the Bin Range field , and then select the data values in the Bins column . Note that if you omit this field, the bins will be automatically generated, and that if you specify your own bins, they must be in ascending order. Click the Output Range radio button , and then click the cell in the Excel spreadsheet where you want the output display to start. Finally, click the Chart Output checkbox . Click here to see what the output should look like .  |../BasicProbabilitySession/Prob00290.htm",
    "Statistics for Maintenance Modelling: Basic Probability Concepts|Summary ¤ The main objectives of this session should be to re-familiarise yourself with some of the basic concepts of probability. Many of these concepts will be used later on in the course and a good understanding of them now will be of great benefit later on. You should know: What a sample space is and what is meant by an event. What the union,...| intersection and complements are of events. How to use Venn diagrams to represent events. What conditional/unconditional probability is and what the difference is between the two. Whether two events are independent or dependent. The concept of Bayes' Theorem. What a random variable is - continuous or discrete? What the probability mass function and the probability density functions are. What the cumulative distribution function is. Of the concepts of expectation and variance. How to produce a histogram for a data set.  |../BasicProbabilitySession/Prob00300.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Introduction ¤ The next two sessions introduce the distributions that are most commonly fitted to reliability lifetime data. The session after that then discusses how to carry out the fitting. As with all model fitting, we are not saying that the lifetime data exactly matches the chosen distribution, but rather that the distribution is a very good...| approximation to the data. Fitting a distribution to the data assists in making good Asset Management decisions as it smoothes some of the sampling noise and fills in gaps in the data coverage. Hence there are three main criteria involved in choosing a distribution: Consideration of any physical characteristics of the problem that link with a particular distribution, e.g. is there a reason to believe that the hazard rate is constant? However this kind of argument is often only appropriate for simple items. How well the distribution fits the observed data. How simple and easy to use the distribution is.  |../ContinuousLifetimeDist1Session/ContLifeDist100005.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Objectives ¤ By the end of this session you will know: what the basic concepts of the Reliability Function ( R ( t )) are what the basic concepts of the Hazard Function ( h ( t )) are what the Probability Density Function ( f ( t )) is what the properties and parameters of the Poisson Distribution are what the properties and parameters of the...| Weibull Distribution are what the properties and parameters of the Gumbel Distribution are what the R ( t ) of these distributions is what the h ( t ) of these distributions is how to graphically display these distributions.  |../ContinuousLifetimeDist1Session/ContLifeDist100010.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Reliability and Hazard Functions ¤ Let T be the time before an item fails and f ( T ) be the failure probability density function ( pdf ) of T , i.e. the probability of T being within a particular interval of values. For instance, the probability of an item failing between times a and b is . A domestic iron has the failure probability density...| function: where t is measured in years. The plot of this density function is shown on the right.  |../ContinuousLifetimeDist1Session/ContLifeDist100020.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Analysing the Reliability of an Iron ¤ The term is needed so that the probability density function integrates to 1 over the interval [0, ) i.e. the iron will eventually fail (and can only fail once!). A result from using calculus that is outside of this course gives the integral of f ( t ) as: when a, b [0, ). Hence the probability that an iron...| fails in its first year of use is: And the probability that an iron fails in its third year of use is:  |../ContinuousLifetimeDist1Session/ContLifeDist100030.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Reliability Function ¤ As we saw in the previous session, the cumulative density function ( cdf ) - which gives the probability that an item fails before time t - is given by: F ( t ) = P ( T t ) = . It follows that: . The probability that the item survives beyond time t is one minus the cdf . This is called the reliability function (also...| known as the survival function) and is denoted by R ( t ) (or S ( t )). Hence: R ( t ) = 1 - F ( t ) = . It follows that: . Return  |../ContinuousLifetimeDist1Session/ContLifeDist100040.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Combining the Probability Density Function and the Reliability Function ¤ So, if an item has lasted (survived) 2 years, how do you work out the probability that it fails in its third year? This is not as this ignores the fact that many items do not make it to the end of year 2. In fact only R (2) of the items make it past year 2 and so the number...| who fail in year 3 as a fraction of those that make it past year two is: .  |../ContinuousLifetimeDist1Session/ContLifeDist100045.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Hazard Function ¤ Following on from this, the rate at which items fail at the end of year 2 (start of year 3) is: . Generalising this gives the (conditional) failure rate function or the hazard function: The hazard function expresses the propensity for a part to fail shortly after time t given that it has survived to time t , i.e. the...| probability of failing in the time interval ( t , t + t ) assuming that the part has lasted until time t , is approximately t h ( t ). Click here for additional explanation to help understand the hazard function . The Exponential Form of the Reliability (Survival) Function Since from the previous screen, , we have . Solving this differential equation gives us: Return  |../ContinuousLifetimeDist1Session/ContLifeDist100050.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Determining the Probability Density Function From the Hazard Function ¤ Often it is easier to specify/approximate the hazard function than the probability density function, e.g. the hazard function might be specified to follow a piecewise linear bath tub curve, shown below. The probability density function is then determined from inversion of the...| hazard function. Note that: if R ( t ) 1, then h ( t ) f ( t ) the first step in finding the probability density function from the hazard function is to see that the hazard function is the negative of the derivative of the reliability function so that integrating minus the hazard function gives the log of the reliability function: , then the cummulative hazard function is H ( t ) = h ( u ) d u , and it is readily seen that R ( t ) = e - H ( t )  |../ContinuousLifetimeDist1Session/ContLifeDist100060.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Analysing the Hazard Function for an Iron ¤ The reliability function for the iron's failure probability density function of is: Hence the hazard function is: | |../ContinuousLifetimeDist1Session/ContLifeDist100070.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Mean Residual Life ¤ In a similar fashion to the way the hazard function gives the failure rate conditional on having survived so far, the mean residual life is the mean time to the failure conditional on having survived so far. It is defined to be: mrl ( x ) = E ( X - x x )) Note that it excludes the time already survived from the value, i.e. it...| deals with the residual life. Return  |../ContinuousLifetimeDist1Session/ContLifeDist100080.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Poisson Distribution ¤ The Poisson distribution is often used to model how many failures occur within a time interval where it is assumed that: the number of failures is proportional to the length of the interval the number of failures in one time interval does not affect the number of failures in another interval. A typical situation would be...| a complex system that is made up of many similar but not necessarily identical components. Although each component is reliable, their large number means that there is a chance that one or more might fail. As the number of failures within an interval can only take whole number (integer) values, the Poisson distribution is a discrete distribution. Its probability density function for the number of failures x in an interval of length t is: The Poisson distribution has the single parameter - it is called the average rate of the Poisson process . The mean of the distribution is t and the standard deviation is . Return  |../ContinuousLifetimeDist1Session/ContLifeDist100090.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Graphical Representation of the Poisson Distribution ¤ The effect of different values of for a time interval of length 1 are shown below, but you can click on any of the charts for an interactive version in Excel where you can experiment with different values of t yourself. Click here to find out how to calculate the Poisson distribution in...| Excel.  |../ContinuousLifetimeDist1Session/ContLifeDist100100.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Relating the Poisson Distribution to the Hazard Function ¤ The reliability (survivor) function is the probability that there are no failures in an interval of length t , i.e.: The failure probability density function is the derivative of minus R ( t ) with respect to time, so: (This is the exponential distribution. For more, see the Weibull...| distribution on screen 16.) Therefore the hazard rate is: Hence the hazard rate is constant, i.e. an item's age does not affect its chance of failing. We have actually been discussing what is referred to as a homogeneous Poisson process as time intervals of the same length are treated in the same way. Click here to learn about non-homogeneous Poisson processes . And click here to find out how the Poisson distribution can model the binomial distribution .  |../ContinuousLifetimeDist1Session/ContLifeDist100110.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Using the Poisson Distribution as a Model to Replace Lightbulbs ¤ The manufacturing of large specialised electrical/electronic products is often carried out in large project halls. Due to the ceiling height in these halls, changing failed light bulbs requires some effort so there is a tendency to wait for a number of bulbs to have failed before...| carrying out the replacement. So as to avoid affecting production, the replacement task is carried out at weekends. However for safety reasons there is a limit on how many bulbs can have failed before immediate remedial action is needed. Hence a balance needs to be struck between replacing bulbs when only a small number have failed and waiting until larger numbers have failed and running the risk that the safety limit will be exceeded and a more costly maintenance action will be needed. A natural model for the number of bulbs failing in a week is a Poisson distribution. Suppose that = 1.1 then the probability that there are: 0 failures in a week is 0.33 1 failure in a week is 0.37 2 failures in a week is 0.20 3 failures in a week is 0.07 Hence if the safety limit is 5 failures and at the weekend when the decision is made not to maintain this weekend, there is then 1 failure, the probability that an emergency maintenance is needed in the forthcoming week is: 1 - (0.33 + 0.37 + 0.20 + 0.07) = 0.03 (i.e. 1 - the probability of upto 4 failures in total)  |../ContinuousLifetimeDist1Session/ContLifeDist100120.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Weibull Distribution ¤ Return Weibull distributions are widely used for modelling lifetime distributions. One of the main reasons for this is the flexibility of the distribution shapes that can be approximated by varying the Weibull's two parameters. The pdf is given by: Here is the scale parameter and is the shape parameter (sometimes is...| referred to as the characteristic life as it is the time at which 63.2% of the population will have failed). Click here to see an alternative parameterisation for the Weibull distribution which is also used frequently throughout this module. If is 1, then we have the exponential distribution , and . The mean of the distribution is: and the standard deviation is: . Here is the gamma function defined by for t &gt; 0 ( click here for more on the gamma function ). Hence for the exponential distribution, the mean is (since (2) = 1).  |../ContinuousLifetimeDist1Session/ContLifeDist100130.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Graphical Representation of the Weibull Distribution ¤ The probability density function for the Weibull distribution is shown against time below with = 2 and = 2, but you can click on the chart for an interactive version in Excel where you can experiment with different values of alpha and beta yourself. Click here to find out how to calculate...| the Weibull distribution in Excel.  |../ContinuousLifetimeDist1Session/ContLifeDist100140.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Reliability Function for the Weibull Distribution ¤ Return The reliability (survival) function is . Hence, for the exponential distribution, . This is shown against time below with = 2 and = 2, but you can click on the chart for an interactive version in Excel where you can experiment with different values of alpha and beta yourself. Using...| Excel with the notation of this section, consider the Weibull distribution with = 3 and = 1. If an item has survived to year 2, what is the probability that it fails between years 3 and 4? Type your answer into the textbox to 3 decimal places and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. Click here to launch Excel. Yes, that's right. The correct expression for this is (F(4) - F(3)) / R(2). Using the Weibull function with our notation in Excel produces (WEIBULL(4,1,3,TRUE)-WEIBULL(3,1,3,TRUE))/(1-WEIBULL(2,1,3,TRUE)) = 0.203 . No, that's not right. The correct expression for this is (F(4) - F(3)) / R(2). Using the Weibull function with our notation in Excel produces (WEIBULL(4,1,3,TRUE)-WEIBULL(3,1,3,TRUE))/(1-WEIBULL(2,1,3,TRUE)) = 0.203 .  |../ContinuousLifetimeDist1Session/ContLifeDist100150.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Hazard Function for the Weibull Distribution ¤ Hence the hazard function is: . I.e. it is just a power of t . If is less than 1, then the hazard rate decreases with age. If = 1, then the hazard rate is constant and the Weibull degenerates into the Poisson/exponential case. If is greater than 1, then the hazard rate increases with age. In...| particular, if is 2, then the hazard function increases linearly with time . The hazard function against time is shown below with = 2 and = 2, but you can click on the chart for an interactive version in Excel where you can experiment with different values of alpha and beta yourself. Return  |../ContinuousLifetimeDist1Session/ContLifeDist100160.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Using the Weibull Distribution to Test Electrical Relays ¤ Fifteen electrical relays were tested to failure. The number of cycles ( 10 5 ) before failure occurred were: 2.1, 4.9, 6.1, 8.0, 8.5, 10.3, 10.6, 12.0, 14.5, 16.4, 16.5, 18.9, 22.6, 23.5 and 30.3. A Weibull distribution with = 16 10 5 and = 1.7 is a good model for this data as can be seen...| from this plot of the actual data and the Weibull reliability function for the electrical relays. This data is taken from O'Connor (1995) p. 79 where it was adjusted as the original data suggested that relays did not fail before 4.1 10 5 cycles.  |../ContinuousLifetimeDist1Session/ContLifeDist100165.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Gumbel (Extreme Value) Distribution ¤ In many reliability situations, it is the items with shorter lifetimes that dictate policy as failures can cause system interruptions and so they need to be kept within limits. Therefore distributions tailored to modelling these poorer items are often fitted to reliability data, particularly the extreme...| value (or Gumbel) distribution. If a large set of items have lifetimes given by the exponential distribution, then the extreme value distribution models the smallest of these lifetimes. The probability density function for the extreme value distribution is: Here is the scale parameter and is the location parameter. The mean of the distribution is - 0.577 . The reliability function is: The hazard function is: The extreme value distribution has no shape parameter - its shape is fixed. Click here to find out how to calculate the Gumbel distribution in Excel.  |../ContinuousLifetimeDist1Session/ContLifeDist100170.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|The Effects of Changing the Parameters on the Gumbel Distribution ¤ The effects of changing and on the Gumbel distribution are shown below, but you can click on any of the charts for an interactive version in Excel where you can experiment with different values of mu and sigma yourself. | |../ContinuousLifetimeDist1Session/ContLifeDist100180.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Using the Gumbel Distribution to Model Breaking Strengths of Wires ¤ The breaking strengths of 15 equal length wires are 48, 52, 58, 61, 62, 63, 64, 65, 69, 69, 72, 72.5, 75, 75 and 76 Newtons. Assigning a cumulative probability of 1/15 to 48, 2/15 to 52, etc., and then plotting the reliability function gives the solid line in this figure. The...| reliability function for a Gumbel distribution with = 69.5 and = 7.8 is shown as the dotted line. This data is taken from Practical Reliability Engineering by O'Connor (1995) p. 82 . Note: Assigning multiples of 1/15 for the cumulative probabilities is slightly inaccurate as we want to find the mid-points of the probability bands and these bands are narrower nearer the mean. More accurate cumulative probabilities are given by B&eacute;rnard's median rank value of , where i is the ranked order of occurrence (i.e. 48 is 1, 52 is 2, etc.) and n is the number of observations. Hence 48 would receive a value of 0.045, 52 a value of 0.110, etc.  |../ContinuousLifetimeDist1Session/ContLifeDist100185.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Using the Gumbel Probability Density Function in Excel ¤ Use Excel's Solver to find the value for which the extreme value's probability density function is greatest when is 5 and t is 3.0. Type your answer into the textbox to 2 decimal places and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working...| out the answer. Click here to launch Excel. Click here if you can't see the Solver option on the Tools menu. = Yes, that's right. Typing EXP((3-5)/ A 1)*EXP(-EXP((3-5)/ A 1))/ A 1 as f ( t ) and using the Solver function to find the value of which maximises this expression results in = 1.48 . No, = 1.48 . If EXP((3-5)/ A 1)*EXP(-EXP((3-5)/ A 1))/ A 1 is the probability density function in cell B1 with an initial value for in cell A1, and the Solver dialog is set up as shown then 1.48 is the value that maximises the probability density function.  |../ContinuousLifetimeDist1Session/ContLifeDist100188.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Summary ¤ Fitting a parametric model to reliability data summarises the data using just a few parameters but more importantly allows decisions to be made in regions where there are not huge amounts of data. Apart from the Poisson/exponential distribution which is normally used because the hazard rate is thought to be approximately constant, the...| emphasis for the other distributions introduced in this and the next session, is on choosing the most appropriate one based on the shape of their curves. If this selection is done by considering a set of failure data, then the curves are likely to be the probability density or cumulative distribution curves. However if the decision is made more on the basis of engineering modelling and experience, then it is likely that the hazard function curve will be used. The main characteristics of the distributions are: Exponential - it has a constant hazard rate. The probability of an individual (surviving) item failing does not increase with time. Weibull - it can take on a wide variety of unimodal shapes depending on the parameter settings while having the advantage of analytical expressions for the reliability and hazard functions. Depending on the parameter values, the hazard rate is either decreasing, increasing or constant. The latter case corresponds to the exponential function. Extreme Value (Gumbel) - it has a probability density function shape similar to the normal distribution's bell shape but with a longer tail to the left and a shorter tail to the right.  |../ContinuousLifetimeDist1Session/ContLifeDist100190.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 1 ¤ What is the probability of a 3 year old item failing in the next year? Click on one of the options below, then click on &quot;Check Answer&quot;. R (3) - R (4) h (3) None of these Yes, that's right. The fraction of all items that fail between the start of year 3 and the start of year 4 is = F (4) - F (3). However only R (3) items make...| it to the start of year 3, hence F (4) - F (3) has to be increased to take this into account. No, that's not right. The fraction of all items that fail between the start of year 3 and the start of year 4 is = F (4) - F (3). However only R (3) items make it to the start of year 3, hence F (4) - F (3) has to be increased to take this into account.  |../ContinuousLifetimeDist1Session/ContLifeDist100200.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 2 ¤ If f ( a ) &gt; 0, then what is the result of ? Click on one of the options below, then click on &quot;Check Answer&quot;. Always &lt; 1 Always equal to 1 Always &gt; 1 Sometimes &gt; 1 and sometimes &lt; 1 Yes, that's right, it's always greater than 1. This is because as and R ( x ) &lt; R ( a ) for x &gt; a , then . In fact, = . No,...| that's not right. It is always greater than 1. This is because as and R ( x ) &lt; R ( a ) for x &gt; a , then . In fact, = .  |../ContinuousLifetimeDist1Session/ContLifeDist100210.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 3 ¤ What is the probability density function for the hazard function h ( x ) = ( x )? Click on one of the options below, then click on &quot;Check Answer&quot;. Weibull with = and = 2. Poisson with average rate . Extreme value with = 0 and = . None of these. Yes, that's right. In the section on the Weibull distribution, it was mentioned...| that = 2 gave a straight line hazard function. No, that's not right. A hazard function of ( x ) produces a straight line, and if you remember in the section on the Weibull distribution, it was mentioned that = 2 gave a straight line hazard function.  |../ContinuousLifetimeDist1Session/ContLifeDist100220.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 4 ¤ What is the probability density function of R ( x ) = x 3 for 0 x 1? Click on one of the options below, then click on &quot;Check Answer&quot;. - x 4 /4. - 3 x 2 . 3 - 3 x 2 . None of these. That's right, it's none of these because it's not a valid hazard function and so does not have a probability density function. This is because x...| 3 is an increasing function so can't be a reliability function. No, that's not right. It's none of these because it's not a valid hazard function and so does not have a probability density function. This is because x 3 is an increasing function so can't be a reliability function.  |../ContinuousLifetimeDist1Session/ContLifeDist100230.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 5 ¤ Is h (2) &gt; h (3)? Click on one of the options below, then click on &quot;Check Answer&quot;. Sometimes. Always. Never. Yes, sometimes is correct because h ( x ) is a measure of how likely an item of age x is to fail in the near future. Therefore it can go up or down. No, that's not right. The answer is sometimes, because h ( x ) is...| a measure of how likely an item of age x is to fail in the near future and therefore it can go up or down.  |../ContinuousLifetimeDist1Session/ContLifeDist100240.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 6 ¤ Is f ( x ) &gt; h ( x )? Click on one of the options below, then click on &quot;Check Answer&quot;. Sometimes. Always. Never. Yes, f ( x ) will never be greater than h ( x ) since h ( x ) = f ( x ) / R ( x ) and 0 R ( x ) 1. No, f ( x ) will never be greater than h ( x ) since h ( x ) = f ( x ) / R ( x ) and 0 R ( x ) 1. | |../ContinuousLifetimeDist1Session/ContLifeDist100250.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 7 ¤ Is R ( x ) &gt; F ( x )? Click on one of the options below, then click on &quot;Check Answer&quot;. Sometimes. Always. Never. Yes, that's right. R ( x ) = 1 - F ( x ) and F ( x ) goes from 0 to 1. So for small values of x , R ( x ) &gt; F ( x ) but for large values of x , R ( x ) &lt; F ( x ). No, the answer is actually sometimes....| This is because R ( x ) = 1 - F ( x ) and F ( x ) goes from 0 to 1. So for small values of x , R ( x ) &gt; F ( x ) but for large values of x , R ( x ) &lt; F ( x ).  |../ContinuousLifetimeDist1Session/ContLifeDist100260.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 8 ¤ What is ? Click on one of the options below, then click on &quot;Check Answer&quot;. f ( x ). h ( x ). - f ( x ). h ( x ) f ( x ). 1 - f ( x ). Yes, that's right. R ( x ) = 1 - F ( x ) so d R/ d x = d l/ d x - d F/ d x = 0 - f ( x ). No, the answer is -f ( x ). This is because R ( x ) = 1 - F ( x ) so d R/ d x = d l/ d x - d F/ d x =...| 0 - f ( x ).  |../ContinuousLifetimeDist1Session/ContLifeDist100270.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 1|Question 9 ¤ Assuming x 0, a &gt; 0 and b &gt; 0, what is F ( x ) for ? Click on one of the options below, then click on &quot;Check Answer&quot;. There is no analytical solution for F ( x ) None of these Yes, that's right. Differentiating gives f ( t ). The constant of integration is 1 since F (0) = 0. No, that's not right. Differentiating gives...| f ( t ). The constant of integration is 1 since F (0) = 0.  |../ContinuousLifetimeDist1Session/ContLifeDist100280.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Introduction ¤ The previous session discussed the probability density function, the cumulative density function, the reliability function, the hazard function and several distributions that model lifetimes. This session continues the review of lifetime distributions by covering the commonly used: normal and lognormal distributions, before going on...| to the more specialised: gamma logistic loglogistic and Pareto distributions.  |../ContinuousLifetimeDist2Session/ContLifeDist200005.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Objectives ¤ By the end of this session you will know: what the basic properties and parameters of the Normal and Lognormal distributions are what the properties and parameters of Gamma distribution, Logistic and Log Logistic distributions are what the properties and parameters of Pareto distribution are what the R ( t ) of these distributions is...| what the h ( t ) of these distributions is how to graphically display these distributions.  |../ContinuousLifetimeDist2Session/ContLifeDist200010.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Normal Distribution ¤ The normal distribution plays a central role in statistics partly because of the central limit theorem which says that the distribution of any reasonably well behaved function of a sample mean tends to the normal distribution as the sample size increases to infinity. So it is perhaps not surprising that one of the most...| common distributions used for lifetime modelling is the normal distribution (sometimes referred to as the Gaussian distribution, particularly in engineering settings). The probability density function of the normal distribution is f ( t ) = . The mean is and the standard deviation is . Hence just alters the position and not the shape of the curve. The effects of changing can be seen in these graphs of f ( t ), but you can click on them for an interactive version in Excel where you can experiment with different values of and yourself.  |../ContinuousLifetimeDist2Session/ContLifeDist200020.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Cumulative and Reliability Functions for the Normal Distribution ¤ Unfortunately the cumulative density function does not have an analytical form, and so has to be approximated or solved numerically. Many statistics books have tables of this integral for the case = 0, = 1, i.e. the standard normal distribution, and ( t ) is defined as the...| resulting integral, i.e. ( t ) = . Hence in the general case F ( t ) = . As R ( t ) = 1 - F ( t ), the reliability function has the same problem. The shape of the reliability function is: As with the extreme value distribution, the normal distribution allows negative lifetimes. However again the probability of negative values is minimal if is taken to be large enough compared with .  |../ContinuousLifetimeDist2Session/ContLifeDist200030.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Normal Distribution ¤ Although there is no analytical expression for the hazard function resulting from a normal distribution, it is increasing (and its rate of change / gradient is increasing) as is illustrated here. Click here to find out how to calculate the Normal distribution in Excel. The probability density...| function stemming from a study of over four hundred 40 watt light bulbs is shown on the right. The normal distribution with mean 1065 and standard deviation 191 is also shown over the data. This data set is based on the one illustrated by Cox (1998) on page 16.  |../ContinuousLifetimeDist2Session/ContLifeDist200040.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Lognormal Distribution ¤ The pdf of the lognormal distribution is f ( t ) = if t 0 and zero otherwise. Here ln t is the natural logarithm of t . The lognormal distribution is often more appropriate for lifetime data than the normal distribution as it allows a wider variety of shapes to be modelled and it does not have the problem of negative...| lifetimes, as shown in these charts of f ( t ). As you can see, if is much larger than , then the lognormal distribution is approximately a normal distribution. Click on the charts for an interactive version in Excel where you can experiment with different values of and . The lognormal's mean is and its standard deviation is .  |../ContinuousLifetimeDist2Session/ContLifeDist200050.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Cumulative and Reliability Functions for the Lognormal Distribution ¤ The cumulative density function can be written as: F ( t ) = = . Hence ln t has a normal distribution, i.e. the log of the lifetimes are normally distributed - hence the distribution's name. The shape of the reliability function is: | |../ContinuousLifetimeDist2Session/ContLifeDist200060.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Lognormal Distribution ¤ Again there is no analytical expression for the hazard function. Furthermore the hazard function is no longer monotonic but is peaked as is shown below and is zero for t = 0 and t = . In practice it is unusual to meet a decreasing hazard function for large t , and so this is a drawback of the...| lognormal distribution. However it is sometimes used when large t values are of no interest, i.e. the region to the left of the hazard function's peak is all that is of interest. Click here to find out how to calculate the Lognormal distribution in Excel.  |../ContinuousLifetimeDist2Session/ContLifeDist200070.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Example of Fitting the Lognormal Distribution to Failure Times ¤ PET film is used for electrical insulation inside SF 6 gas insulated transformers. The failure times from 15 accelerated life-tests were 50.25, 87.75, 87.77, 92.90, 92.91, 95.96, 108.3, 108.3, 117.9, 123.9, 124.3, 129.7, 135.6, 135.6. Fitting a lognormal distribution with = 4.66 and...| = 0.23 to the cdf for the failure times, gives: The benefit of fitting a parameterised distribution can be seen in the smoothing of the jagged observation data. Note: This data comes from Wang (2000) . The graph suggests that the lognormal is a reasonably fit to the data, but the high to ratio suggests that the normal distribution would also be suitable.  |../ContinuousLifetimeDist2Session/ContLifeDist200080.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Gamma Distribution ¤ The gamma distribution generalises the exponential distribution. It can be regarded as the time to the N th failure if the time to failure is exponentially distributed. The probability density function of the gamma distribution is: f ( t ) = if t 0 and zero otherwise. Here is the gamma function (see The Summary of the...| Gamma Function screen in the previous session). When N = 1, then we get the exponential distribution f ( t ) = . However N does not have to be an integer. The mean is and the standard deviation is . The effects of changing lambda and N can be seen in these graphs of f ( t ), but you can click on them for an interactive version in Excel where you can experiment with different values yourself.  |../ContinuousLifetimeDist2Session/ContLifeDist200090.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Cumulative and Reliability Functions for the Gamma Distribution ¤ The gamma distribution gives rise to curves similar to those of the Weibull distribution. However the gamma distribution leads to more complicated equations, e.g. for the reliability/survival function. There is no simple analytical form for the cumulative distribution function -...| the simplest form involves an infinite series of ratios of gamma functions. Fortunately Excel provides functions for calculating the probability density function and the cumulative distribution function, as well as the inverse of the cumulative distribution function. Click here to find out how to calculate the Gamma distribution in Excel. The shape of the reliability function is shown on the right:  |../ContinuousLifetimeDist2Session/ContLifeDist200100.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Gamma Distribution ¤ There is no simple form for the hazard function. If N &gt; 1, then the hazard function is increasing. If N &lt; 1, then the hazard function is decreasing. For N = 1, we have the exponential case, i.e. a constant hazard function. The shape of the hazard function is shown on the right: | |../ContinuousLifetimeDist2Session/ContLifeDist200110.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question to Calculate the Hazard Function for the Gamma Distribution ¤ Using Excel, estimate the derivative of the hazard function at t = 4 for the parameter values N = 10 and lambda = 3. Click on one of the options below which approximates the answer, then click on &quot;Check Answer&quot;. 0.25. 0.36. 0.45. 0.55. None of these. Yes, that's...| right. Evaluating the hazard function at 4, and at a nearby point such as 4.00001, and dividing both by the difference 0.00001 is achieved in Excel by: ((GAMMADIST(4.00001, 10,1/3,FALSE)/(1-GAMMADIST(4.00001, 10,1/3,TRUE)))-(GAMMADIST(4.0, 10,1/3,FALSE)/ (1-GAMMADIST(4.0,10,1/3,TRUE))))/0.00001 0.36 . No, the answer is approximately 0.36 . You need to evaluate the hazard function at 4, and at a nearby point such as 4.00001, and divide both by the difference 0.00001. Since and R ( t ) = 1 - F ( t ), this is achieved in Excel by: ((GAMMADIST(4.00001, 10,1/3,FALSE)/(1-GAMMADIST(4.00001, 10,1/3,TRUE)))-(GAMMADIST(4.0, 10,1/3,FALSE)/ (1-GAMMADIST(4.0,10,1/3,TRUE))))/0.00001.  |../ContinuousLifetimeDist2Session/ContLifeDist200120.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Example of Fitting the Gamma Distribution to Reliability Data ¤ 22 patients with cancer were given drug and radiation therapy. Three patients left the treatment course - at days 79, 82 and 446. The uncensored survival (death) times were 30, 67, 95, 148, 170, 171, 176, 193, 200, 221, 243, 261, 262, 263, 399, 414, 446, 464 and 777 days. If the...| relatively small number of censored observations are ignored, then the chart shows the reliability (survival) function along with that of the gamma distribution with parameters N = 2.82 and = 0.01. This data is taken from Cox (1984) .  |../ContinuousLifetimeDist2Session/ContLifeDist200130.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Logistic Distribution ¤ The logistic function is . The shape of the distribution is: Click here to find out how the logistic function is used in regression analysis. | |../ContinuousLifetimeDist2Session/ContLifeDist200140.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Probability Density Function for the Logistic Distribution ¤ The logistic function could be used as a cumulative distribution function but it is preferable to add the parameters and to allow flexibility in the shape in a similar way to the how they are used to give the standard normal distribution greater flexibility. As a result, the...| probability density function of the logistic distribution is: f ( t ) = for &gt; 0. The mean is and the variance is . The effects of changing and can be seen in these graphs of f ( t ), but you can click on them for an interactive version in Excel where you can experiment with different values yourself.  |../ContinuousLifetimeDist2Session/ContLifeDist200150.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Cumulative and Reliability Functions for the Logistic Distribution ¤ The cumulative distribution function is: F ( t ) = The reliability function is: R ( t ) = The shape of the reliability function is shown here: | |../ContinuousLifetimeDist2Session/ContLifeDist200160.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Logistic Distribution ¤ The hazard function is: h ( t ) = . Hence as t , the limiting value is -1 . The shape of the hazard function is shown here: Click here to find out how to calculate the Logistic distribution in Excel. | |../ContinuousLifetimeDist2Session/ContLifeDist200170.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Logistic Distribution Fitted To the CDF of Lightbulb Failure Times ¤ The graph shows the logistic distribution fitted to the cumulative density function of the light bulb failure time data that was used to illustrate the normal distribution. The parameters used for the logistic distribution were = 1010.8 and = 88.6. The use of the logistic...| distribution to model this data was taken from &quot;The handbook of the logistic distribution&quot;, editor N. Balakrishnan Dekker, New York 1992, page 297.  |../ContinuousLifetimeDist2Session/ContLifeDist200180.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Loglogistic Distribution ¤ The loglogistic function gives rise to probability distributions, reliability functions and hazard functions that are similar to those that can be obtained from the Weibull distribution (and the lognormal distribution) but the formulae for R ( t ) and h ( t ) are simpler. Its probability density function is: f ( t )...| = when &gt; 0, &gt; 0 and t 0. That is if the (natural) logarithm of a variable X is distributed according to the logistic distribution, then X follows the loglogistic distribution. The effects of changing and can be seen in these graphs of f ( t ), but you can click on them for an interactive version in Excel where you can experiment with different values yourself.  |../ContinuousLifetimeDist2Session/ContLifeDist200190.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Reliability Function for the Loglogistic Distribution ¤ If &gt; 1, then the mean is: . (If 1, then the mean is undefined.) If &gt; 2, then the variance is: . (If 2, then the variance is undefined.) The reliability function is: R ( t ) = . The shape of the reliability function is shown on the right. | |../ContinuousLifetimeDist2Session/ContLifeDist200200.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Loglogistic Distribution ¤ The hazard function is h ( t ) = . Hence if 1, then the hazard function decreases, whereas if &gt; 1 it increases until t = then decreases to zero as shown in the shape of the hazard function shown here: Click here to find out how to calculate the Loglogistic distribution in Excel. | |../ContinuousLifetimeDist2Session/ContLifeDist200210.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Example of Fitting the Loglogistic Distribution to the Emergence of Teeth ¤ The chart shows the cumlative distribution function for the loglogistic distribution being used to model the emergence times of teeth in Flemish boys. The three curves show the fitted loglogistic distribution for different teeth. The data comes from LEROY, R., BOGAERTS,...| L., LESAFFRE, R. and DECCLERCK, D. (2003) The emergence of permanent teeth in Flemish children. Community Dentistry and Oral Epidemiology , 31 , 30-39.  |../ContinuousLifetimeDist2Session/ContLifeDist200220.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Pareto Distribution ¤ The Pareto distribution originated in the modelling of incomes where a distribution that replicated the long right tail (that is there was no upper cut off for incomes) was needed. Its probability density function is given by: f ( t ) = for &gt; 0, &gt; 0 and t . The basic reciprocal power shape can be seen in these...| graphs of f ( t ). Click on them for an interactive version in Excel where you can experiment with different values of and yourself.  |../ContinuousLifetimeDist2Session/ContLifeDist200230.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Cumulative and Reliability Functions for the Pareto Distribution ¤ If &gt; 1, then the mean is and when &gt; 2, the variance is . (When 1 the mean is undefined and when 2 the variance is undefined.) As the cumulative distribution function is F ( t ) = , the reliability function is also a reciprocal power R ( t ) = . The shape of the...| reliability function is illustrated here.  |../ContinuousLifetimeDist2Session/ContLifeDist200240.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question to Find the Mean Residual Life for the Pareto Distribution ¤ Find the mean residual life function, MRL ( x ) = , for the Pareto distribution (i.e. the expected life left for an item of age x ). Is it: None of these. Yes, that's right. Since = - , f ( t ) = and R ( t ) = 1 - F ( t ) = = we get and . Subtracting the former from the latter...| gives . No, that's not right. Since = - , f ( t ) = and R ( t ) = 1 - F ( t ) = = we get and . Subtracting the former from the latter gives .  |../ContinuousLifetimeDist2Session/ContLifeDist200250.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|The Hazard Function for the Pareto Distribution ¤ The hazard function is just the reciprocal function h ( t ) = . Hence the hazard rate decreases with t which means that the Pareto distribution is only appropriate where an item becomes more reliable with age. The Pareto distribution gives rise to curves similar to the tail of the lognormal...| distribution and has the advantage that the equations for the reliability and hazard functions are much simpler. However while the Pareto may fit the tail of the data well, the lognormal often gives a better overall fit as it can model an early peak as well as the tail, e.g. when modelling incomes. Click here to find out how to calculate the Pareto distribution in Excel.  |../ContinuousLifetimeDist2Session/ContLifeDist200260.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Example of Fitting the Pareto Distribution to Software Failures During Testing ¤ The table gives the interval between successive software failures recorded during factory testing. Pareto models have often been fitted to this type of data as the time to failure is expected to increase as more errors are found - and software does not wear out like...| mechanical items. Fitting a Pareto distribution with = 0.5 and = 0.29 gives: Time range (hours) Number of failures 0-1 99 01-Feb 54 02-Mar 15 03-Apr 7 04-May 0 05-Jun 1 06-Jul 2 Total 178 This data comes from MUSA, J .D. (1979) Validity of execution-time theory of software reliability. IEEE Transactions on Reliability , 28 (3) (August), 18 -191.  |../ContinuousLifetimeDist2Session/ContLifeDist200270.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Summary ¤ Although particular lifetime distributions are sometimes chosen on theoretical grounds, in the more uncertain environment away from the laboratory, the complex interactions of the various factors contributing to failure means that choosing a distribution based on its shape is usually more appropriate. In this case, the number of...| parameters a model has and the ease of manipulation of the probability density, reliability and hazard functions are also factors - although the support provided by Excel makes this less important than it was. The probability density functions for the different distributions considered in this and the previous session, often only differ by the size of the right hand tail - it is usually the hazard functions that are more distinctive. The main characteristics of the distributions considered in this session are: Normal - symmetric bell shaped probability density function, increasing hazard function (and increasing rate of change of the hazard function). Lognormal - the probability density function has a range of shapes from the normal's bell shape to a no left tail/long right tail shape. Gamma - models the situation where several individual failures have to occur before there is a system failure. For N &gt; 1, the hazard function is increasing but the rate of increase is decreasing. Logistic - this leads to an s shaped increasing hazard function. However the use of the logistic distribution for reliability modelling is relatively rare, more usually the loglogistic distribution is used. Loglogistic - the loglogistic's hazard function is either decreasing or peaked depending on whether 1 or not. Pareto - this has a reciprocal hazard function and is suitable for modelling situations when there is no hard right hand cut off.  |../ContinuousLifetimeDist2Session/ContLifeDist200280.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 1 ¤ The following distributions have similar hazard functions with the parameters: Loglogistic with = 0.1206, = 5.878 Pareto with = 0.1, = 0.05 Lognormal with = 0.0001, = 6.569. What is R (3) - R (2) for the lognormal? Click on one of the options below, then click on &quot;Check Answer&quot;. -0.012 -0.018 -0.024 -0.031 None of these Yes,...| that's right. R ( t ) = 1 - F ( t ) so in Excel you can calculate it with: =(1-LOGNORMDIST(3,0.0001,6.569))-(1-LOGNORMDIST(2,0.0001,6.569)) which gives the answer -0.0244. No, the answer is -0.024. This is because R ( t ) = 1 - F ( t ) so in Excel you can calculate it with: =(1-LOGNORMDIST(3,0.0001,6.569))-(1-LOGNORMDIST(2,0.0001,6.569)) which gives the answer -0.0244.  |../ContinuousLifetimeDist2Session/ContLifeDist200290.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 2 ¤ The following distributions have similar hazard functions with the parameters: Loglogistic with = 0.1206, = 5.878 Pareto with = 0.1, = 0.05 Lognormal with = 0.0001, = 6.569. What is R (3) - R (2) for the pareto? Click on one of the options below, then click on &quot;Check Answer&quot;. -0.028 -0.034 -0.047 -0.062 None of these Yes,...| that's right. R ( t ) = ( /t)^ so in Excel you can calculate it with: =(POWER(0.05/3,0.1) - POWER(0.05/2,0.1)) which gives the answer -0.0275. No, the answer is -0.028. This is because R ( t ) = ( /t)^ so in Excel you can calculate it with: =(POWER(0.05/3,0.1) - POWER(0.05/2,0.1)) which gives the answer -0.0275.  |../ContinuousLifetimeDist2Session/ContLifeDist200300.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 3 ¤ The following distributions have similar hazard functions with the parameters: Loglogistic with = 0.1206, = 5.878 Pareto with = 0.1, = 0.05 Lognormal with = 0.0001, = 6.569. What is R (3) - R (2) for the loglogistic? Click on one of the options below, then click on &quot;Check Answer&quot;. -0.0047 -0.0056 -0.0078 -0.0082 None of...| these Yes, that's right. R ( t ) = 1/(1 + t^ ) so in Excel you can calculate it with: =(1/(1+5.878*3^0.1206) - 1/(1+5.878*2^0.1206)) which gives the answer -0.0056. No, the answer is -0.0056. This is because R ( t ) = 1/(1 + t^ ) so in Excel you can calculate it with: =(1/(1+5.878*3^0.1206) - 1/(1+5.878*2^0.1206)) which gives the answer -0.0056. Note: Although the hazard functions can be similar as shown here, the reliabilities can be quite different. The extreme case is the Pareto where parameter does not appear in the hazard function.  |../ContinuousLifetimeDist2Session/ContLifeDist200310.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 4 ¤ For the logistic distribution, what is the constraint on / so that F (0) = 0.001? Click on one of the options below, then click on &quot;Check Answer&quot;. 1.37 4.73 5.81 6.91 None of these Yes, that's right. F ( t ) = so 0.001 = 1/(1 + e^( / )). Hence / = ln(999) = 6.91. No, the answer is 6.91. This is because F ( t ) = so 0.001 =...| 1/(1 + e^( / )). Rearranging gives e^( / ) = 999, hence / = ln(999) = 6.91.  |../ContinuousLifetimeDist2Session/ContLifeDist200320.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 5 ¤ For the normal distribution what is the constraint on / so that F (0) = 0.001? Click on one of the options below, then click on &quot;Check Answer&quot;. 2.35 2.65 3.01 3.09 None of these Yes, that's right / = 3.09. Using =NORMINV(0.001,$A1,1) in the target cell and using the Solver tool to find the value of zero by changing the cell...| $A1, you get 3.09, i.e. the mean when the standard deviation is 1. No, the answer is that / = 3.09. You can find this by using =NORMINV(0.001,$A1,1) in the target cell and using the Solver tool to find the value of zero by changing the cell $A1, you get 3.09, i.e. the mean when the standard deviation is 1.  |../ContinuousLifetimeDist2Session/ContLifeDist200330.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Question 6 ¤ Which of the following distributions model the situation where items wear out, i.e. the hazard rate increases as t ? For each distribution, click yes or no, and when you've finished click on &quot;Check Answer&quot;. Exponential Yes No That's right, the hazard rate is constant. No, it doesn't because the hazard rate is constant....| Weibull Yes No Yes that's right. It does, if &gt; 1. No, it does if &gt; 1. Extreme value Yes No Yes that's right. No, it doesn't. Normal Yes No Yes that's right. No, it doesn't. Lognormal Yes No That's right, because the hazard rate decreased to zero as t . No, it doesn't because the hazard rate decreased to zero as t . Gamma Yes No Yes that's right. It does if N &gt; 1. No, it does if N &gt; 1. Logistic Yes No That's right, the hazard function levels off to -1 for large t . No, it doesn't because the hazard function levels off to -1 for large t . Loglogistic Yes No That's right, because for a large enough t the hazard rate decreases. No, it doesn't because for a large enough t , the hazard rate decreases. Pareto Yes No That's right, the hazard rate decreases with t . No, it doesn't because the hazard rate decreases with t .  |../ContinuousLifetimeDist2Session/ContLifeDist200340.htm",
    "Statistics for Maintenance Modelling: Model Selection|Introduction ¤ This session describes the process of choosing candidate parametric models of lifetime data, given failure time data. The data may result from planned lifetime testing of components or equipment, or may have been culled from records of equipment failures. Such lifetime data has its own problems, particularly censoring, in that not...| all failure times are known. Sometimes a further analysis such as the determination of an optimum time for age-based replacement can be done non-parametrically, so avoiding the whole problem of model choice (see Baker, 1993 ). However, most analyses do require parametric models of equipment lifetime, and this session describes the preliminary work needed in choosing models. Bear in mind that most textbooks were written before the availability of current computing power. It was necessary for the practitioner to carry out a lot of exploratory work perhaps on Weibull graph paper, before resorting to the jealously-guarded mainframe computer to fit the model selected, or doing laborious calculations by hand. With current computing power, much of this is unnecessary. There are only a few worthwhile models for lifetime data, and there are good criteria nowadays for deciding which model is best, such as the corrected Akaike Information Criterion ( AIC ) (see Burnham and Anderson, 2002 ). This allows for the effect that models with more parameters are expected to fit the data better. Hence, one can progress quickly to the model fitting methods described in the next two sessions, fitting the few feasible models that exist and settling for the best, without the need for extensive exploratory work. These labour-saving and computationally expensive approaches must surely become standard practice with time, and in this case, the important topic to study in this session is the concept of censoring. If informative censoring is present, or censoring is wrongly treated or ignored, powerful computers will guide the user infallibly to the wrong conclusion. In general, the provenance of the data must be carefully studied, especially when dealing with data from maintenance records. There are many horror stories where errors and biases in data collection have been ignored, resulting in the 'garbage in, garbage out' effect.  |../ModelSelectionSession/ModSel00005.htm",
    "Statistics for Maintenance Modelling: Model Selection|Objectives ¤ By the end of this session, you will understand: Type I censoring and Type II censoring the non-parametric estimation of R ( t ) and h ( t ) the Kaplan - Meier Estimator how to use graphical methods to aid model selection. | |../ModelSelectionSession/ModSel00010.htm",
    "Statistics for Maintenance Modelling: Model Selection|Censoring ¤ Analysis of failure data would be much easier if every item put on test could be followed to fail. Unfortunately this is often not possible. When we do not the know the failure times of some items, we say that they are censored . There are several types of censoring, some of which complexify the statistical analysis of failure data but...| can be dealt with, and others which can not. Click on the types for a description. Type 1 censoring Type 2 censoring Random (noninformative) censoring Informative censoring Interval censoring Left censoring The simplest type of censoring is the so-called type 1 censoring , which occurs when several items are put on test, and the testing is terminated at some time t . We know then that an item has either failed at a known time before time t , or that it survived to time t . Type 2 censoring is similar to type 1, but here testing stops when a certain number of items have failed. This is less common than type 1 censoring. Type 1 and type 2 censoring result from a planned experiment, but very commonly censoring arises in an unplanned and random way. This is particularly true outside the maintenance context, for example in the study of survival following onset of some medical condition. Here patients may die from other causes, be lost to records through moving away from the area, and so on. Similarly, when collecting data on equipment failures from maintenance records, some items may have failed from other causes unrelated to the ones considered in one's study, or records may be missing. For example, a television set put on test for electronic defects is dropped and smashed. Random censoring is noninformative , in that we can only deduce that the item had not failed by the time when it was censored. This type of censoring can be dealt with easily. There is a crucial distinction between noninformative and informative censoring. Informative censoring occurs when the censoring time is related to the failure time. For example, a technician withdraws items from testing when he or she thinks that they are about to fail. Assuming that we only know that the item had not failed under test is here of course wrong-it was just about to fail. Similarly, sometimes censoring happens when nature plays the part of the too-conscientious technician. For example, patients with a serious medical condition taking a medication drop out of the study with a seemingly-unrelated problem, but which was in fact a side effect of the drug. Interval censoring means that failure is known to have occurred within some time interval, but the exact time of failure is not known. In left censoring , observation only starts at some time , and some items are found to have already failed. Return  |../ModelSelectionSession/ModSel00020.htm",
    "Statistics for Maintenance Modelling: Model Selection|Modelling with Censored Data ¤ Trying to model informative censoring is probably a waste of time, and the data will be irretrievably spoilt if much informative censoring has occurred. One must be on watch to be clear that any censoring that occurred is uninformative. Generally, nonparametric methods can only deal with simple types of censoring,...| but parametric model-building can cope with anything, except informative consoring. In summary, if more than a tiny proportion of failure times are censored, statistical procedures that are adapted to cope with this must be used. Setting censored failure times to the time of censoring is a crude device that will over-estimate the true failure rate, and dropping censored observations from the analysis will also over-estimate the failure rate.  |../ModelSelectionSession/ModSel00030.htm",
    "Statistics for Maintenance Modelling: Model Selection|Questions on Censored Data ¤ Question 1 Which of the following describes the situation where all items are put on test simultaneously, and testing stops at a fixed time with some items unfailed? Click on &quot;Check Answer&quot; when you've chosen. Type 1 censoring Type 2 censoring Neither of these Yes, that's right, this is type 1 censoring. No,...| it's type 1 censoring. Type 1 censoring occurs when several items are put on test, and the testing is terminated at some time t . We know then that an item has either failed at a known time before time t , or that it survived to time t . No, it's type 1 censoring. Type 1 censoring occurs when several items are put on test, and the testing is terminated at some time t . We know then that an item has either failed at a known time before time t , or that it survived to time t . Question 2 An item of equipment has two failure modes, and we are interested in one of them. Failures due to the other mode are regarded as censored observations. Is this a valid procedure? Click on &quot;Check Answer&quot; when you've made a selection. Always Only if the failure modes are independent Only if the failure modes are independent Yes, that's right. If the failure modes are independent then the censoring is non-informative and can be dealt with easily. No, the procedure is only valid if the failure modes are independent and therefore the censoring is non-informative. No, the procedure is only valid if the failure modes are independent. If the failure modes are related, then the censoring is informative and difficult to deal with, even with parametric models.  |../ModelSelectionSession/ModSel00040.htm",
    "Statistics for Maintenance Modelling: Model Selection|Nonparametric Estimates of the Survival Function - Kaplan-Meier Estimator ¤ Return When choosing models for lifetime data, estimates of the survival (reliability) function R ( t ) and particularly of the hazard function h ( t ) are very useful. The behaviour of the hazard function can at once rule out some lifetime models and suggest others. For...| such exploratory analysis we need a nonparametric method. In the absence of censoring, we would estimate R ( t ) from failure times t i of n items as: The Kaplan-Meier Estimator The Kaplan-Meier estimate reduces to this in the absence of censoring. It is: ( t ) = ( n i - d i ) / n i where d i items fail at time t i , out of n i items that had not failed before time t i . Here we have also allowed for tied failures, so that d i can be greater than 1. The figure on the right shows the survival function for a Weibull distribution with = 1, = 2.5, and the Kaplan-Meier estimate from a simulated sample of size 40.  |../ModelSelectionSession/ModSel00050.htm",
    "Statistics for Maintenance Modelling: Model Selection|Nonparametric Estimates of the Hazard Function ¤ From session 3 we have that the cumulative hazard function H ( t ) is given by H ( t ) = -ln( R ( t ))), so that we could estimate ( y ) = -ln( ( t )). In practice the so-called empirical cumulative hazard function: ( t ) = d i / n i is commonly used. The intuitive logic is that the proportion of...| items still on test failing at time t i , d i / n i , can be thought of as contributing to the integral of the hazard H ( t ) = h ( u ) d u . The figure shows the estimated cumulative hazard function for the Weibull sample. Unfortunately, it is asking too much of the data to try to plot what we would really like, an estimate of the hazard function h ( t ) rather than the cumulative hazard H ( t ). Simple formulae for the standard deviation of the reliability and hazard estimates are known, and may be found for example in Lawless (1982) .  |../ModelSelectionSession/ModSel00060.htm",
    "Statistics for Maintenance Modelling: Model Selection|Example of Calculating Kaplan-Meier Estimator ¤ If we have the following failure time data: 1, 2.5 (censored), 3, 4, what is the Kaplan-Meier estimator? Thus the sample size is n = 4. For each time interval, we need to calculate ( t ) = ( n i - d i ) / n i where d i items fail at time t i , out of n i items that had not failed before time t i ....| From t = 0 to t &lt; 1, no items have failed, so R = (4 - 0) / 4 = 1. From t = 1 to t &lt; 2.5, the first item failed at t = 1, therefore, R = (4 - 1) / 4 1 = 3/4. From t = 2.5 to t &lt; 3, R = 3/4. This is because there is no failure at t = 2.5. The only result of the censoring is to reduce the size of the risk set. From t = 3 to t &lt; 4, two items have not failed, at the point one item fails at t = 3. Therefore, R = (2 - 1) / 2 3/4 = 3/8. For t 4, R = (1 - 1) / 1 3/8 = 0.  |../ModelSelectionSession/ModSel00070.htm",
    "Statistics for Maintenance Modelling: Model Selection|Comparison of Distributions ¤ The aim is to use simple statistics derivable from the data to choose a suitable lifetime distribution to fit to the data. The most useful candidate distributions have already been discussed in the Continuous Lifetime sessions. The Weibull Distribution The Weibull distribution will give at least a passable fit to...| nearly all failure data, and this generalises the exponential through adding a shape parameter that differs from unity, so that R ( t ) = . When &gt; 1 the distribution has a hazard function increasing with age (a so-called IFR or increasing failure rate distribution), and when &lt; 1 the hazard function decreases with age ( DFR distribution). Hence the Weibull distribution can fit both failure data from mechanical devices that wear out with age, and from electronic devices that have 'teething problems' or 'infant mortality'. This is caused by defective items being progressively weeded out as time on test elapses. The Gamma Distribution Apart from the Weibull, which should be the first choice, the gamma distribution is often found useful. The gamma distribution also has two parameters, also generalises the exponential distribution, and can also accommodate IFR or DFR distributions. The difference between these two is that the hazard function at large t levels off to a constant for the gamma distribution, whereas it continues to change for the Weibull. When &gt; 1, the Weibull distribution is therefore short-tailed, whereas the gamma produces a long tail of failures at large times t . The gamma distribution in fact behaves like an exponential distribution in the tail. One can see from the graph of the empirical hazard function from data simulated from a Weibull distribution with = 2.5 that the hazard function is continually increasing, and hence that a Weibull rather than a gamma distribution would fit the data.  |../ModelSelectionSession/ModSel00080.htm",
    "Statistics for Maintenance Modelling: Model Selection|Comparison of Distributions (2) ¤ The Normal Distribution The normal distribution is not really satisfactory, as it has no lower limit for failure time. The Pareto Distribution The Pareto distribution can be derived as the failure time distribution of a population of items, each having an exponentially distributed failure time, but with a failure...| rate drawn as a random variable from a gamma distribution. This gives a very long-tailed distribution, with a hazard function that decreases progressively to zero at large times. Intuitively, as time passes, the items with high failure rates are progressively weeded out, leaving only the 'good' items to fail. The Lognormal and Log-logistic Distributions The lognormal and log-logistic distributions are occasionally used and are similar. Here the hazard function first increases, and later decreases. This latter effect gives of course a long tail of failures at large times. This behaviour is the exact opposite of the famous 'bathtub curve'. Again, it could arise when items have an increasing hazard of failure with age, but are drawn from a population where some have higher failure rates. These are progressively weeded out with time. This type of effect is common when modelling mortality after an operation among patients with a disease such as cancer, where the hazard of death first increases because the disease takes time to recur, and then decreases because the cancer has not spread in the remaining patients. Hence a lognormal distribution fits such data well.  |../ModelSelectionSession/ModSel00090.htm",
    "Statistics for Maintenance Modelling: Model Selection|Plotting Skewness Against the Coefficient of Variation ¤ The plot of cumulative hazard function is a guide as to whether the hazard is increasing with time or tending to a constant. Besides the hazard function, the moments of the distribution such as the variance and third moment (skewness) are also useful in distinguishing between candidate...| models. They are expressed in scale-independent units as the coefficient of variation and the sample skewness. For a random variable X , if the mean is and the variance is 2 , the coefficient of variation is / and the skewness is E ( X - ) 3 / 3 . If censoring is not an issue, one can plot the estimated skewness and coefficient of variation from the data as a point on the graph shown in Cox and Oakes (1984) . This chart shows skewness against the coefficient of varation for 5 lifetime distributions. Two-parameter distributions such as the Weibull, gamma, log-logistic and lognormal distributions describe curves as their parameters vary, and one can see which curve passes closest to the data point. The exponential distribution with one parameter is represented by the point (1, 2), where the Weibull and gamma curves cross. Long-tailed distributions have high skewness, so that the log-logistic and lognormal distribution lines lie at the top left of the graph. This is a very useful method of model selection, giving much insight into what might be the best model, and if any others might be worth trying. Simply feed the data into a statistics package like MINITAB, and calculate descriptive statistics.  |../ModelSelectionSession/ModSel00100.htm",
    "Statistics for Maintenance Modelling: Model Selection|Questions on the Hazard Function for Different Distributions ¤ Consider these questions before checking the answers. How does the hazard function normally behave for items that wear out with age, and which distributions can model this? How does the hazard function normally behave for electronic items that do not wear out with age, and which...| distributions can model this? What do you think the hazard function looks like for the mortality of human beings? Answer 1 The hazard of failure increases. The Weibull and gamma can model this, but not the exponential or Pareto distributions. The lognormal and log-logistic distributions give a hazard function that increases, and then decreases, so these also are probably not suitable. Answer 2 Failures caused by initial defects give a hazard function that decreases with age, and this can be modelled by the Weibull, Gamma and Pareto distributions. Answer 3 We are subject both to early (childhood) death from defects such as hole-in-the-heart, and also as a result of ageing. The human survival function is therefore the famous bathtub curve, which first decreases to around age 15 and then increases, with hazard exponentially increasing with age. There is also a blip from risky juvenile behaviour around age 20.  |../ModelSelectionSession/ModSel00110.htm",
    "Statistics for Maintenance Modelling: Model Selection|Other Graphical Methods of Model Selection ¤ Probably the most useful graphical method is that already discussed of plotting skewness against the coefficient of variation. Another class of graphical methods relies on plotting some statistic that falls on a straight line for the particular distribution being considered. For example, for an...| exponential distribution we have that ln R ( t ) = - t . Hence plotting the Kaplan-Meier estimate ln ( t ) against failure time should yield a straight line graph through the origin. This is of course a plot of the estimated cumulative hazard function. It will curve upwards for a Weibull distribution with &gt; 1 and downwards for a Weibull distribution with &lt; 1. To get a straight line graph for the Weibull distribution, we could plot: ln (-ln R ( t ))) = ln t + ln against ln t . The slope then equals , and the intercept is ln , so that and can be estimated. A problem here is that after taking logarithms twice, almost anything will be roughly linear! Lawless (1982) describes these and other plots in detail. Given the difficulty of deciding whether a scatter of data points is near-enough linear, and the ease of use of more decisive methodology such as fitting all likely models and choosing the one with the smallest corrected AIC (see screen 1), these are not discussed here in more detail.  |../ModelSelectionSession/ModSel00120.htm",
    "Statistics for Maintenance Modelling: Model Selection|Case Study: Time to Failure of Epoxy Electical Cable ¤ Some data from an MSc thesis by Stone (1978) is quoted in Lawless (1982) page 477. This is time to failure of epoxy electrical cable insulation specimens under voltage stress. The failure occurred in two stages, but here we only consider the time to the appearance of a microscopic defect....| Three of the 20 failure times (the three largest) were censored, as the cable had not failed. Testing was stopped after different times for these three. The figure shows the nonparametric estimate of the cumulative hazard function, from which one can see that the hazard appears to be progressively decreasing with time. This suggests the type of scenario discussed earlier, in which the more defective cables fail early on, leaving the good ones. The very long tail suggests that a very long tailed distribution such as the Pareto might be best. The coefficient of variation was 1.02 and the skewness (assuming that the 3 censored failure times were actual failures) was 1.4. From the plot of distribution types, this would suggest maybe a Weibull distribution with &lt; 1, or because the skewness has been underestimated, maybe a lognormal. One would progress to fit Weibull, lognormal and Pareto distributions, and in fact the Pareto distribution fits well. The exploratory analysis has shown that a long-tailed distribution that can model decreasing hazard is needed, and also that there are no obvious errors in the data. Having learnt that, it is best to progress immediately to model fitting.  |../ModelSelectionSession/ModSel00130.htm",
    "Statistics for Maintenance Modelling: Model Selection|Summary ¤ Initial analysis of failure data should consider possible errors and biases in the data. If censoring is present, this must be allowed for in subsequent analysis. If it is informative censoring, further analysis will be largely invalidated. Before model fitting, nonparametric estimates of the reliability (survival) function and...| cumulative hazard of failure can be plotted. This latter plot is crucial in understanding which distributions might fit the data. The key distinction is between distributions with increasing and those with decreasing hazard functions. As there are only a few distributions that might be tried, and computing is now fast and easy, prolonged exploratory analysis is unnecessary. Initial exploratory plots are still useful though, since they are easy to use and are more easily understood by non-statisticians than more mathematical approaches, but one should proceed to fit several survival models, using a program or package that copes with censoring. Choose the best model on the basis of a model-choice criterion, such as the corrected Akaike Information Criterion. Note that not all statistics packages will do everything that is required.  |../ModelSelectionSession/ModSel00140.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Objectives ¤ When studying certain behaviors and patterns in data we often want to model these characteristics in the data so as to make inferences about future events to occur in the particular process we are examining. Model selection deals with finding which model is the most appropriate for the data in hand. Model fitting deals with estimating...| the parameters of that model so as to fit the data in hand specifically. The previous session dealt with model selection. In this session we deal with the problem of parameter estimation and diagnostics once a model has been chosen. By the end of this session, you will understand: parameter estimation measures of appropriateness for parameter estimates how to obtain parameter estimates using maximum likelihood or the method of moments.  |../ModelFitting1Session/ModFit100010.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Parameter Estimation ¤ Once a parametric model has been decided upon, the model is fitted to the real data by estimating the unknown parameters, . For example, if it has been decided that the Weibull distribution is to be used to model the strength of some carbon fibres, then to fit the model we must estimate the two parameters and . The data are...| the observations which provide information on what values these parameters should take to best model the data. In parameter estimation the observed can be thought of as fixed points, and the model parameters are varied so they best explain the observations (data). Good methods of estimation have properties such as unbiasedness , minimum variance and consistency . A parameter estimate is said to be consistent if it approaches its true value as the sample size tends to infinity. We will look at the first two properties in more detail in the next few screens. The amount of data restricts how much information can be gained from it. More data implies more information. If more parameters are used in the model, the data will be spread more sparsely, resulting in an increased uncertainty as to the accuracy of the estimated parameters. There are two methods for estimating parameters: point estimate . A point estimate is a single number, and is simply a statistic calculated using the data that is used as an estimate of a population parameter. For example, we could say that the estimate for the population mean for time to failure of a batch of light bulbs is 410 hours - a single value. interval esimate . An interval estimate is not a single number, but two numbers, defining a range. For example, we could say that the estimate for the population mean for time to failure of the same batch of light bulbs is between 400 hours and 420 hours.  |../ModelFitting1Session/ModFit100020.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Bias of an Estimate ¤ Suppose we want to obtain an estimate of a population parameter . The estimator of will be denoted by (&quot;theta hat&quot;). To test this estimator, we would calculate several times using several samples from the population. The bias of an estimate refers to the distribution of the estimates around the true parameter value....| An unbiased estimator is one which has a mean (expected value) equal to the true parameter estimate. Mathematically this is written: Let be a point estimate for a parameter . Then is an unbiased estimator if E ( ) = . Otherwise is said to be biased . The bias of an estimator is given by: B ( ) = E ( ) - This chart shows the sampling distribution for a positively biased point estimator, i.e. E ( ) &gt; .  |../ModelFitting1Session/ModFit100030.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Examples of Some Common Unbiased Estimators ¤ Formal methods for deriving point estimators and their bias are beyond the scope of this session. It suffices here to give a couple of examples. For the Normal distribution, given by: Unbiased point estimates of the parameters and are given by: and Note that when estimating the variance 2 it seems more...| intuitive to divide by N rather than N - 1, however, it can be shown that dividing by N gives a biased estimate.  |../ModelFitting1Session/ModFit100040.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Measuring the Accuracy of Estimators ¤ Point estimates on their own do not tell us how close we might be to the true value of the parameter concerned. Furthermore, different samples from the same population will give different estimates of the same parameter. The variability in parameter estimates is characterised by the standard error . To...| calculate the standard error we first need the variance , given by: . The standard error (s.e.) is thus given by: .  |../ModelFitting1Session/ModFit100050.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Mean Squared Error ¤ Another quantity of interest is the mean squared error ( m.s.e. ) defined by: m.s.e. ( ) = E - 2 . The mean square error can help us understand the bias of an estimate. When the estimator is unbiased, i.e. E ( ) = , the variance and the mean squared error are the same. When the estimator is biased in some way, the mean squared...| error is the variance plus a contribution from the bias. It actually turns out that the contribution from the bias is simply the square of the bias, i.e: m.s.e. ( ) = V ( ) + B ( ) 2 To conclude, an estimate has two properties, accuracy and bias. In practice it is thought that a higher precision is more important than the bias, since bias can often be quantified and a suitable correction made to the original estimate.  |../ModelFitting1Session/ModFit100055.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Confidence Interval Estimates ¤ A confidence interval is one type of interval estimate that gives information on the probability of the true parameter value being in the given range. An interval estimate will hopefully have two properties. Firstly, the interval will contain the true value of , and second it will be relatively narrow. The two...| endpoints of the interval will be functions of the sample statistics and will vary from sample to sample. Confidence limits are the upper and lower endpoints of the confidence interval. They are calculated using a combination of the point estimate and the standard error of the estimate. Suppose that L and U are the lower and upper confidence limits respectively (these are random numbers since they can change with each sample of data taken). Then if: P ( L U ) = 1 - , the probability 1 - is the confidence coefficient . The resulting random interval [ L , U ] is called the two-sided confidence interval ( click here to learn why we used a two-sided interval, and how to calculate them from one-sided statistics tables ). In general, many estimates can be assumed to be normally distributed, provided the sample size is large . This comes from the central limit theorem . If this assumption holds, the confidence limits are of the form: where is a percentage point from the standard normal distribution. The central limit theorem says that the distribution of any reasonably well behaved function of a sample mean tends to the normal distribution as the sample size increases to infinity.  |../ModelFitting1Session/ModFit100060.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Finding Values at Percentage Points of Normal Distributions ¤ The percentage points ( ) of the Normal distribution can be found from the Percentage points of the T distribution for infinite degrees of freedom. The tabulation is for one-tail only, but can be used for two-tailed intervals. For example, a 95% two-sided confidence interval (two-sided...| means the true parameter value could actually be higher or lower than the estimate) is made up of the 2.5% and 97.5% points of the standard normal distribution. This gives us = 0.025 (= 1 - 0.975). From the snapshot of the t -table below we see that Z takes the value of 1.96 for a two-sided confidence interval at the 95% point. This value is worth remembering as it is often used in calculations. In reliability it is often the case that only the lower bound is of interest. This would mean that a 95% one-sided confidence interval was made up of the 5% point and the 100% point (trivial) of the standard normal distribution.  |../ModelFitting1Session/ModFit100070.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Calculating Confidence Limits ¤ For example, in a study, a parameter has been estimated to be 1.3947. The standard error has also been calculated to be 0.06368. What is the 99% confidence interval for the parameter estimate? We know that we need to look up the two-sided 99% point from the Normal tables. We only have the one-sided table to look at...| though. However, since the distribution is symmetric, we can say that our two-sided 99% interval is the same as a one-sided 99.5% interval, giving = 0.0005. As you can see in the t-table for infinite degrees of freedom, Z = 3.291. We can now calculate the 99% confidence interval as: 1.3947 3.291 0.06368 = [ L = 1.185, U = 1.604] Now have a go yourself and calculate the 90% confidence interval for the above example. Type your answer in the textbox, and click the &quot;Check Answer&quot; button. Lower confidence interval = Upper confidence interval = Yes, that's right. The 90% confidence interval corresponds to = 0.1 for the one-sided tables. We want a two-sided confidence interval. We therefore want to look at the = 0.05 point in the tables. The Z value is 1.645. This gives us our confidence interval as 1.3947 1.645 0.06368 = [ L = 1.290, U = 1.499] No that's not correct. The 90% confidence interval corresponds to = 0.1 for the one-sided tables. We want a two-sided confidence interval. We therefore want to look at the = 0.05 point in the tables. The Z value is 1.645. This gives us our confidence interval as 1.3947 1.645 0.06368 = [ L = 1.290, U = 1.499]  |../ModelFitting1Session/ModFit100080.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Maximum Likelihood Estimation ¤ Return Maximum Likelihood Estimation is the most widely used method of estimation in all of statistics. The reasons for this are two-fold. Firstly the reasoning behind it is intuitive, and second, it can be proved that in most cases the maximum likelihood estimate is the minimum variance, unbiased estimate of the...| particular parameter in question. The following example illustrates the logic behind the method of maximum likelihood. Consider three rooms containing a light bulb and a light switch. The light bulbs may be working or not working with equal probability, but we do not know how many are working and how many are broken. However, we are allowed to randomly sample two lights (i.e. enter two rooms and turn the light on!). If our random sample yields two working lights, what would be a good estimate of the total number of working lights? The third rooms' light can either be working or broken, i.e. the number of working lights is either two or three. If there are two working lights and one broken, the probability of randomly picking two working lights as we did is: On the other hand, if there are three working lights the probability of randomly picking two working lights is 3/3 2/2 = 1. As the latter is larger than the former, it is reasonable to assume that all three rooms have working lights as this estimate maximises the likelihood of obtaining the observed sample . It is possible for there to be one room with a broken light, but the observed outcome of our experiment gives a larger likelihood of there being no broken lights. This demonstrates the method for finding an estimator based on trying to maximise the likelihood of the observed outcome.  |../ModelFitting1Session/ModFit100090.htm",
    "Statistics for Maintenance Modelling: Model Fitting|The Likelihood Function ¤ Return Mathematically the definition of maximum likelihood is: Suppose that the likelihood function depends on k parameters 1 , 2 , ..., k . Choose as estimates those values of the parameters that maximise the likelihood: L ( X 1 , X 2 , ..., X n 1 , 2 , ..., k ). The likelihood function is a random function and is...| obtained by replacing the unknown random x 's with the data points X and regarding the probability density function as a function of , where is a vector of parameters. For a probability density function f x ( x ; ) and a random variable x , the likelihood function is: If X 1 , ..., X n are independent and identically distributed and the X i have densities f i ( x ), then the likelihood function is given by: L ( ; X ) = f 1 ( X 1 )... f n ( X n ) = f ( X i ) The value of that gives the maximum of the likelihood function maximises the probability of the observed data i.e. the likelihood function attains its maximum value when the parameters have their 'most likely' values.  |../ModelFitting1Session/ModFit100100.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Log-Likelihood and Maximum Likelihood Functions ¤ Log-Likelihood Function The values of that maximise the likelihood function are those that precisely maximise the log-likelihood function: The log-likelihood is often much easier to deal with, as it converts a product to a summation, and more used in practice. Likelihood functions are...| data-dependent, and are not measures of the fit of the model, or the plausibility of the parameter estimates. They summarise the information about model parameters in a data set, given a model. Accordingly, likelihood functions can only be compared within a data set for different models, not across different data sets. Maximum-Likelihood Function To obtain maximum likelihood estimates, a maximum must be found with respect to in the log-likelihood function. , the estimated parameter (the value of which maximises the likelihood function) can usually be found as the solution to: and by checking that so that corresponds to the maximum of the likelihood function. When a maximum cannot be found this way, numerical optimisation of the log-likelihood function must be used.  |../ModelFitting1Session/ModFit100110.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Example of the Log-Likelihood Function ¤ The likelihood (and therefore the log-likelihood) can be calculated for a range of values of the parameter to be estimated. If one plots the log-likelihood function, one can often see a clear maximum for a particular value of the parameter. The value of the parameter corresponding to this maximum is the...| maximum likelihood estimate. For example, the likelihood function for the exponential distribution is plotted here. 100 data points have been used as the observations. The maximum in the log-likelihood function occurs around the value of = 0.5. Hence the maximum likelihood estimate for this particular data set is = 0.5.  |../ModelFitting1Session/ModFit100120.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Finding the Maximum Likelihood Estimator for the Exponential Distribution ¤ If X 1 , X 2 , ..., X n are the values of a random sample from an Exponential population, what is the maximum likelihood estimator of its parameter ? The probability density function of an exponential random variable is given by: The likelihood function is therefore given...| by: The log-likelihood function is thus given by: Differentiating with respect to we get: Equating this derivative to zero and solving for , we get the maximum likelihood estimate: We should also check here that we have found a maximum not a minimum by checking that the second derivative is negative. This turns out to be the case. Click here for another example using the normal distribution .  |../ModelFitting1Session/ModFit100130.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Finding the Maximum Likelihood Estimator for the Normal Distribution ¤ Return If X 1 , X 2 , ..., X n constitute a random sample of size n from a Normal population with mean and variance 2 , what are the maximum likelihood estimates of these two parameters? The likelihood function is given by: This gives the log-likelihood function as:...| Differentiating with respect to we get: Differentiating with respect to 2 we get: Equating the first partial derivative to zero and solving for we get: and equating the second partial derivative to zero and solving for 2 we get:  |../ModelFitting1Session/ModFit100135.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Estimation Methods Used in Sample Moments ¤ The Method of Moments is one of the oldest methods for deriving point estimators. The k th moment of a random variable, taken about the origin, is: The corresponding k th sample moment is the average: The method of moments is based on the intuitively appealing idea that sample moments should provide good...| estimates for the corresponding population moments. Mathematically, the method of moments can be written as: Choose as estimates those values of the parameters that are solutions of the equations , for k = 1,2, ... t , where t is the number of parameters to be estimated. This will yield t equations in t unknowns (the parameters to be estimated).  |../ModelFitting1Session/ModFit100140.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Using the Method of Moments to Find One Unknown Parameter ¤ A random sample of n observations, X 1 , X 2 ,&#133;,X n , is selected from a population in which the X i possess a uniform probability density function over the interval (0, ) where is unknown. We would use the method of moments to estimate the parameter as follows: The value of for a...| uniform random variable is: The corresponding first sample moment is: Equating the population and sample moment, we get:  |../ModelFitting1Session/ModFit100150.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Using the Method of Moments to Find Two Unknown Parameters ¤ A random sample of n observations, X 1 , X 2 ,&#133;,X n , is selected from a population in which the X i possess a gamma probability density function with parameters and . The gamma probability density function is given by: We would use the method of moments to estimate the unknown...| parameters and as follows. Because we are looking to estimate two parameters we need two equations, i.e. we need to equate the first two population moments with the first two sample moments. The first two population moments of the gamma distribution can be looked up in a statistical text, and are equal to: = = and = 2 + 2 2 Equating these values with the corresponding sample moments we get: and Have a go at solving these equations to find the two parameters and . Click here for the solution.  |../ModelFitting1Session/ModFit100160.htm",
    "Statistics for Maintenance Modelling: Model Fitting|Summary ¤ In this session we have seen that there are two types of parameter estimates - point estimates and interval estimates. The basic properties of estimators have been discussed, namely, bias and variance. We have seen how to derive parameter estimates for a given model using the two most fundamental parameter estimation methods in...| statistics. The methods presented were maximum likelihood estimation and the method of moments. You should now be able to: Understand the concepts of parameter estimation Understand the ideas of bias and variance of a parameter estimate Be able to use maximum likelihood and the method of moments to derive estimates for parameters from common distributions Be able to plot likelihood functions. In the next session, Testing Model Fitness, we will see how these parameter estimates are used in practice, and how fitted models are assessed for their appropriateness.  |../ModelFitting1Session/ModFit100170.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Objectives ¤ In the previous session we discussed methods to fit models to data, for example we estimated the parameters of a model for a specific data set using maximum likelihood estimation. Once a model has been fitted to data it is advisable to check the appropriateness and the adequacy of the model. In this session we introduce formal...| statistical tests and graphical techniques for checking whether a model is adequate. By the end of this session, you will understand: what goodness of fit is how to test for goodness of fit: Pearson's test the Kolmogorov-Smirnov test the A2 test the W2 test tests for normality how to assess goodness of fit using graphical methods: the Q-Q plot the P-P plot  |../ModelFitting2Session/ModFit200010.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Goodness of Fit ¤ Once a parametric model has been fitted to a data set it is good practice to check (test) whether the model fits the data to an acceptable level. Such a check or test is appropriately named a 'goodness of fit' test. One can either use a formal statistical test, where a test statistic is calculated and compared to a theoretically...| derived statistic, or one can use a graphical procedure to assess the goodness of fit with the eye. Both methods are useful and both are presented in this session.  |../ModelFitting2Session/ModFit200020.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Pearson's Test ¤ We now introduce several quantitative statistical tests for goodness of fit. The first such test is called Pearson's test (the chi-square test). This test is used to examine whether a sample of data came from any given distribution. The test can be used in conjunction with any univariate probability distribution for which the...| cumulative distribution function can be derived. For the goodness of fit test, a test statistic is calculated from the given data and the proposed fitted model. The test statistic is defined as: where: O i is the o bserved frequency of the data, when the data has been allocated into k bins. For non-binned data, the values of O i can be found from a histogram which has been generated and/or a frequency table. The choice of the number of bins, k , does have an effect on the validity of the test, but as a general rule one can use a bin-width of 0.3 times the standard deviation for the sample, with the upper and lower bins starting at the sample mean plus and minus 6 times the standard deviation. E i is the e xpected frequency of the data in bin i , and is calculated by: E i = where n is the sample size, F is the cumulative distribution function of the proposed fitted distribution and Y ui and Y li are the upper and lower limits of bin i . Click here for a histogram that illustrates this.  |../ModelFitting2Session/ModFit200030.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Hypothesis Testing ¤ If the proposed model is appropriate for the sample data, the test statistic follows a distribution with ( k - c ) degrees of freedom, where c is the number of estimated parameters of the fitted distribution. In terms of hypothesis tests: the null hypothesis, H 0 is that the sample data follow the specified distribution, and H...| 1 is the sample data do not follow the specified distribution. H 0 is rejected if: &gt; where is the percentage point of the distribution with ( k - c ) degrees of freedom, and is the significance level at which the test is to be ascertained, e.g. 0.05 for 95%. The percentage points of the chi-square distribution can be found from the table.  |../ModelFitting2Session/ModFit200040.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Example of Calculating the Goodness of Fit Test ¤ The following example is taken from Wolstenholme (1995) . An experiment was run to test the 'weakest link' in twenty-four fibres. Each of twenty-four fibres, taken from one larger one, were mounted in a frame which allowed sub-lengths of fibre of 5, 12, 30 and 75 millimetres to be separately loaded...| with ever-increasing weights. The order in which the fibres failed was observed. Interpreting the weakest-link property would lead to the conclusion that the longer lengths generally fail first, because the more material there is the higher the number of potential failure sites. The different failure times are grouped into events E 1 , E 2 and E 3 , and the expected proportion of observations in each group calculated. It is proposed that the expected number of observations in each group is associated with the Weibull distribution - H 0 (the derivation of the expected number of observations per group is not needed here). The results of the experiment are: Failure event E 1 E 2 E 3 O i 6.00 16.00 2.00 E i 6.65 9.40 7.95 The value of the test statistic is thus: = (6 - 6.65) 2 /6.65 + (16 - 9.4) 2 /9.4 + (2 - 7.95) 2 /7.95 = 9.15 If we look at the upper 5% point of the distribution table with ( k - c ) degrees of freedom (i.e. 3 - 1 = 2), the value is 5.99. At less than , it is evidence that at 95% confidence, the weakest-link data do not follow a Weibull distribution.  |../ModelFitting2Session/ModFit200050.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Question Testing Goodness of Fit Test Using ¤ It is reported that on a production line of a certain model of car there are 25% red cars, 15% blue cars, 20% yellow cars and 40% white cars produced. A sample of 200 cars yields 40 red, 35 blue, 30 yellow and 95 white. Do these data conform to the reported production percentages? Use the distribution...| table at the 5% confidence level. Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. Yes No Yes, that's right. If the production is as reported, the expected number of cars of each colour are as follows: Failure event Red Blue Yellow White Observed 40 35 30 95 Expected 50 30 40 80 The test statistic is thus: = (40 - 50) 2 /50 + (35 - 30) 2 /30 + (30 - 40) 2 /40 + (95 - 80) 2 /80 = 8.15 At the 5% confidence level with degrees of freedom = (number of categories - 1) = (4 - 1) = 0, from the tables we see that 0.05, 3 = 7.815. Thus since 8.15 &gt; 7.815 we can say that the production is significantly different from that reported. No, the data doesn't conform to the reported production percentages. If the production is as reported, the expected number of cars of each colour are as follows: Failure event Red Blue Yellow White Observed 40 35 30 95 Expected 50 30 40 80 The test statistic is thus: = (40 - 50) 2 /50 + (35 - 30) 2 /30 + (30 - 40) 2 /40 + (95 - 80) 2 /80 = 8.15 At the 5% confidence level with degrees of freedom = (number of categories - 1) = (4 - 1) = 0, from the tables we see that 0.05, 3 = 7.815. Thus since 8.15 &gt; 7.815 we can say that the production is significantly different from that reported.  |../ModelFitting2Session/ModFit200055.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Kolmogorov-Smirnov Test ¤ The Kolmogorov-Smirnov test uses the difference between the empirical distribution of the observed data and the proposed distribution function. The test looks for the maximum 'distance' between the two and tests whether or not this distance is statistically different from zero. The test can be understood using the...| following scenario. A distribution has been fitted to historic data and parameters estimated. Consider a sample of n uncensored new observations. The Kolmogorov-Smirnov test statistic is: D n = max F n ( t ( i ) ) - F 0 ( t ( i ) ) where: F n is the sample distribution function F 0 is the theoretical distribution function t ( i ) are the ordered data. Clearly, the value of D n will be small for a good fit, and large for a poor fit. Theoretically one can derive 'critical values' of the test statistic D n based on the sample size, n , and the level of significance. If the test statistic exceeds the chosen critical value, then the model would be rejected as suitable. If the test statistic is smaller than the critical value, the proposed model is accepted at the chosen level of significance. Click here to see the table containing the critical values at various significance levels.  |../ModelFitting2Session/ModFit200060.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Kolmogorov-Smirnov Test in Practice ¤ With the help of computer based statistical packages, you may not need to perform the test yourself. It is noted however that you need to know of its existence, meaning and interpretation. Computer packages such as Minitab, SPSS and Stata will perform the test for you for a given set of probability...| distributions. Disadvantage of the Kolmogorov-Smirnov Test The basic assumption of the test is that the parameters must be known before the model is fitted to the data, i.e. the parameters must be pre-specified and not estimated on the data. If one fits the proposed distribution to the data and then tests for model fit, model-dependent factors contribute to violations of the assumptions in the derivation of the test. However, for certain scenarios this may be allowed for, see the A 2 and W 2 tests later.  |../ModelFitting2Session/ModFit200065.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Example Using the Kolmogorov-Smirnov Test ¤ Data on the useful lifetime of a certain component in a desktop computer were collected and are shown in the table in column t . i t F n ( t ) F 0 ( t ) F n ( t ) - F 0 ( t ) 1 2 0.067 0.020 0.048 2 19 0.163 0.173 -0.010 3 41 0.260 0.336 -0.077 4 52 0.356 0.405 -0.050 5 52 0.452 0.405 0.046 6 54 0.548...| 0.417 0.131 7 86 0.644 0.577 0.067 8 318 0.740 0.958 -0.218 9 428 0.837 0.986 -0.150 10 554 0.933 0.996 -0.063 It is believed that the exponential distribution is appropriate for modelling this data, and the parameter , is thought to be 0.01. Since the sample is so small, the F n ( t ) column was constructed using the formula given by Leitch (1995) , and is F n ( t ) = . The F 0 ( t ) is given by F 0 ( t ) = 1 - . Note the maximum value of F n ( t ) - F 0 ( t ) is -0.218. The distributions for F 0 ( t ) and F n ( t ) are shown on the next screen.  |../ModelFitting2Session/ModFit200070.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Example Using the Kolmogorov-Smirnov Test - Distributions ¤ The empirical and theoretical distribution functions are shown here. Using the table , for sample size n = 10 the critical values are 0.41 at a 5% significance level, and 0.49 at 1%. It is clear that the data are well within these boundaries, and it can therefore be concluded that the...| data fit the exponential curve acceptably well.  |../ModelFitting2Session/ModFit200080.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Question Testing Goodness of Fit Using the Kolmogorov-Smirnov Test ¤ The following table shows the repair times for a machine. Test the hypothesis using the Kolmogorov-Smirnov test that the repair times are normally distributed with mean 73.2 and standard deviation 7.041 at the 10% confidence level. Calculate the test statistic and compare it with...| the critical value in the statistical tables . Select &quot;Yes&quot; or &quot;No&quot; if you think the repair times are or are not normally distributed and then click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. i Time 1 61.6 2 63.4 3 65.1 4 65.5 5 70.0 6 72.3 7 72.5 8 72.7 9 73.0 10 75.3 11 77.1 12 78.4 13 83.2 14 83.5 15 84.3 i Time F n ( t ) F 0 ( t ) F n ( t ) - F 0 ( t ) 1 61.6 0.045455 0.049728 -0.004270 2 63.4 0.110390 0.081984 0.028405 3 65.1 0.175325 0.124989 0.050336 4 65.5 0.240260 0.137066 0.103193 5 70.0 0.305195 0.324741 -0.019550 6 72.3 0.370130 0.449145 -0.079001 7 72.5 0.435065 0.460403 -0.025340 8 72.7 0.500000 0.471694 0.028306 9 73.0 0.564935 0.488669 0.076266 10 75.3 0.629870 0.617245 0.012625 11 77.1 0.694805 0.710176 -0.015370 12 78.4 0.759740 0.769904 -0.010160 13 83.2 0.824675 0.922233 -0.097560 14 83.5 0.889610 0.928247 -0.038640 15 84.3 0.954545 0.942542 0.012003 Yes No Yes, that's right. Having made the calculations shown here, the maximum distance between F n ( t ) and F 0 ( t ) is 0.103, corresponding to the 4th observation. From the statistical tables, the critical value at the 10% confidence level is 0.304, and since this is greater than the test statistic, we can say that according to the Kolmogorov-Smirnov test, and at the 10% level of significance that the data does indeed come from a normal distribution with mean 73.2 and standard deviation 7.041. No, the data does indeed come from a normal distribution with mean 73.2 and standard deviation 7.041. Having made the calculations shown here, the maximum distance between F n ( t ) and F 0 ( t ) is 0.103, corresponding to the 4th observation. From the statistical tables, the critical value at the 10% confidence level is 0.304, and since this is greater than the test statistic, we can say that according to the Kolmogorov-Smirnov test, and at the 10% level of significance that the data does come from the specified normal distribution.  |../ModelFitting2Session/ModFit200085.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Tests for Normality ¤ So far we have introduced: the Pearson's test to test whether the observations are statistically different from what would have been expected, and the Kolmogorov-Smirnov test to test the maximum observed difference between the observed empirical distribution function and the proposed theoretical distribution function. Both of...| these tests can be harnessed to test for normality, i.e. to test for the fit of a normal distribution to a data set. It is worth noting that the Kolmogorov-Smirnov test does not allow for parameter estimation in the general case, but there is a specific modification that can be made for testing normality (for example, see Neave (1985) ). Again, it is not necessary for you to be able to perform these tests manually, rather that you are aware of their existence and meaning. Computer software packages are capable of performing these and other such tests.  |../ModelFitting2Session/ModFit200090.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Example of Testing for Normality in Minitab ¤ The following data are thought to be from a Normal distribution. To test this hypothesis, the data are entered into Minitab and the Kolmogorov-Smirnov test is performed. Click here for the detailed instructions . X i 0.020 0.337 1.981 -0.533 -1.217 0.246 0.966 1.462 -0.689 0.468 0.097 -0.948 -1.016...| -1.213 -1.672 0.332 -1.000 1.168 -0.377 -1.785 The graphical output that is generated is a plot of normal probabilities versus the data - a normal P-P plot (which we will discuss later). The Kolmogorov-Smirnov test's p-value indicates that, at levels greater than 0.15, there is evidence that the data do not follow a normal distribution.  |../ModelFitting2Session/ModFit200100.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|A 2 and W 2 Tests ¤ As mentioned previously, the Kolmogorov-Smirnov test has the disadvantage that the proposed distribution parameters must already be specified, and not estimated using the data to be tested. The A 2 test, of Anderson and Darling and the W 2 test of Cramer and von Mises can be used in such a scenario. Originally developed under...| the same assumptions of the Kolmogorov-Smirnov test, the A 2 and W 2 tests have been adapted for the case when the model parameters are estimated from the data (usually using maximum likelihood estimation - see screen 10 in the Model Fitting session) and a test is needed to see if this fitted model is acceptable. The Cramer-von Mises W 2 test statistic is given by: W 2 = The first quantity is simply a measure of the mean square deviation of the distance between the observed distribution (the data) and the proposed fitted distribution. The 1/12 n term is a correction based on the sample size, decreasing the value of the test statistic as n increases. The Anderson-Darling A 2 test statistic is given by: A 2 =  |../ModelFitting2Session/ModFit200110.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|A 2 and W 2 Tests to Test Critical Values ¤ As for Pearson's test and the Kolmogorov-Smirnov test, critical values of the test statistics, namely A 2 and W 2 can be derived for a given number of data and significance level. These derived critical values can be re-scaled to account for the situation when the model parameters have been estimated on...| the data in question. Stephens (1977) used Monte Carlo simulation to determine scaling factors for the case of assessing the goodness of fit of the Gumbel and Weibull distributions. Recall that the probability density function for the Weibull distribution is given by: where is a scale parameter and is a shape parameter. One can either estimate both parameters using the data or just one, i.e. only , or only . For these two scenarios the scaling factors for the test statistics are different. Similarly, the probability density function for the Gumbel distribution is given by: where is the scale parameter and is the location parameter. Again one can either test when both parameters have been estimated using the data or when just one parameter has been estimated. The scaling factors for the test statistic for these two scenarios are slightly different. The table shows the scaling factors for the scenarios discussed, along with critical values at the 1% level and the 5% level. Scaling 5% sig. 1% sig. W 2 1 + 0.2 / 0.124 0.175 1 + 0.16 / n 0.222 0.338 A 2 1 + 0.2 / 0.757 1.038 1 + 0.3 / n 1.321 1.959  |../ModelFitting2Session/ModFit200120.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Shapiro-Wilks Test for Normality ¤ An alternative to the above mentioned tests, and a sometimes more favoured test for normality is one introduced by Shapiro and Wilk (1965) . The test statistic is given by: W = where the a i are functions of the sample size and distributional characteristics of the normal distribution. Values of these...| coefficients can be found in, for example, Pearson and Hartley (1972) . Interestingly, in contrast to other tests for normality, it is small values of the test statistic that indicate departures from normality.  |../ModelFitting2Session/ModFit200130.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Graphical Methods ¤ So far we have discussed methods for assessing goodness of fit that use calculated test statistics to assess whether a proposed distribution fits a set of data acceptably well. In real life, it is often more practical and easier to visually assess the goodness of fit for a particular fitted distribution. We will look at the two...| most popular methods for doing so, Q-Q plots and P-P plots. Q-Q Plots A Q-Q plot (quantile-quantile) plots the quantiles of the data and the theoretical quantiles of the fitted distribution against each other. If the data does come from the proposed distribution a straight line relationship will be observed. Straight lines are easily identifiable meaning Q-Q plots provide a simple visual test of a distribution for modelling a data set, that comparison of the histogram with the density function does not. Departures from a straight line indicate departures from the specified distribution. P-P Plots The P-P plot is another graphical technique for assessing whether or not a data set follows a given distribution such as the Normal or Weibull. The ordered data are plotted against a theoretical distribution in such a way that the points should form approximately a straight line.  |../ModelFitting2Session/ModFit200140.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Quantiles ¤ In reliability the guaranteed or 'safe' life period is often of interest. The quantiles of lifetime distributions can be used to estimate such lifetime periods. The p th quantile of the random variable X is the value x such that: Pr X x = p In other words, the probability that the observed value X is less than or equal to some value x...| , is p . The chart on the right shows the 0.75 quantile for some distribution function. Using the Normal tables , find the 0.9 quantile of the normal distribution. Type it in the textbox to 2 decimal places when you have worked it out and click the &quot;Check Answer&quot; button. Click the &quot;Hint&quot; button for some help in working out the answer. z = Yes, that's right. The nearest value to (1 - 0.9) is 0.1003, which occurs when z = 1.28 . No that's not right. The answer is 1.28 . Since the table shows probabilities of a random variable occuring which is greater than z , we need to look for a probability to this in the table of (1 - 0.9), and then add up the row and column heading values to obtain the value of z . The nearest probability in the table is 0.1003, and this occurs when z = 1.28 .  |../ModelFitting2Session/ModFit200150.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Estimating a Quantile ¤ Once a certain reliability has been specified, say no more than 1% failure within the guaranteed lifetime, the actual length of the lifetime must be estimated. If the probability of failure within the guaranteed lifetime, t p , is p , and the lifetime has the reliability function R ( t ) then the value of t p is given by...| solving: 1 - R ( t p ) = p where the lifetime t p is actually the quantile . So in order to estimate the lifetime, we must be able to estimate the quantiles of interest. For example, if we have fitted the Weibull model to a set of failure data, then the guaranteed lifetime, with no more than 1% failure, is given by the following equation: 0.01 = 1 - R ( t p ) = 1 - (substituting the Reliability Function for the Weibull distribution from session 3) Let us say that and have been estimated using maximum likelihood to be 80.04 and 2.51 respectively. This leads to: = 0.99 Taking logs we get: t p = (-ln (0.99) 80.04) 1/2.51 = 9.18. Click here for another example of estimating a quantile .  |../ModelFitting2Session/ModFit200152.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Contructing Q-Q Plots Using Excel ¤ The following guidelines will help you to construct a Q-Q plot in Excel: Arrange the original data in ascending order, and label the new data set from 1 to n (smallest value first). In the next column calculate the theoretical quantiles, given by: y i = F -1 (( i - 0.5) / n ) F is the cumulative distribution...| function you are concerned with. For the normal distribution this is given by the function NORMINV in Excel. For other distributions you may have to derive the inverse of the cumulative distribution function: e.g. for the Exponential F ( t ) = 1 - 1 - F ( t ) = t = (-1/ ) ln(1 - F ( t )) where F ( t ) = ( i - 0.5) / n Using a scatter plot, graph the empirical quantiles on the x -axis and the theoretical quantiles on the y -axis. If you want to use a straight line as a reference for a perfect fit, add another series corresponding to the empirical quantiles on to the existing plot.  |../ModelFitting2Session/ModFit200155.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Contructing P-P Plots Using Excel ¤ As for the Q-Q plot, the n sample data are ordered from the smallest to largest: x (1) x (2) ... x ( n ) Number the ordered data from 1 to n (smallest first) and calculate ( i - 0.5)/ n . Next calculate, F ( x i ), where x i is the data, and F is the cumulative distribution function you are concerned with. For...| the normal distribution this is given by the function NORMINV in Excel. Other distributions are also available, e.g. the exponential (1 - ) or the Weibull. A plot of ( i - 0.5)/ n versus F ( x i ) will yield a straight line if the fitted distribution is appropriate. If the P-P-plot turns out to be a straight line, but not in the same position as the line, the parameter estimates for the scale and location parameters are incorrect. Using Q-Q and P-P Plots Q-Q and P-P-plots can be generated for several competing distributions to see which provides the best fit, and the P-P plot generating the highest correlation coefficient is the best choice since it generates the straightest P-P plot. In addition, a straight line can be fit to the points and added as a reference line. The further the points vary from this line, the greater the indication of a departure from the specified distribution.  |../ModelFitting2Session/ModFit200160.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Examples of Q-Q and P-P Plots ¤ The following table shows the data for the wearout of ball bearings (in thousands of hours of operating time). i Time 1 68.0 2 69.6 3 71.1 4 71.4 5 74.3 6 74.6 7 75.5 8 77.6 9 77.8 10 78.0 11 78.2 12 80.2 13 80.3 14 81.9 15 83.0 16 85.6 17 87.4 18 87.7 19 88.4 20 98.3 The Q-Q plot for a fitted normal distribution...| is: It can be seen that the fit yields a relatively straight line, which suggests the Normal distribution is indeed a good fit to the data. Click here to see a P-P plot for a fitted normal distribution . What about fitting a different distribution to this data set? Click here to see a P-P plot for a fitted Weibull distribution . Click on the chart to see the underlying calculations in Excel.  |../ModelFitting2Session/ModFit200170.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Question to Construct and Interpret Your Own Q-Q and P-P Plots ¤ The following table gives the failure times obtained from a reliability test of a microchip to be placed in a new home PC. i Time 1 3.3 2 4.2 3 12.9 4 13.8 5 14.3 6 14.8 7 18.5 8 22.8 9 27.1 10 29.7 11 32.0 12 39.5 13 41.3 14 41.6 15 51.1 16 61.7 17 92.2 18 106.6 19 148.8 20 198.1...| The exponential distribution with = 0.01832 is believed to fit the data. Plot the Q-Q and P-P plots for this, and deduce if the exponential distribution is a good fit to the data. Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button to see the plots. Click the &quot;Hint&quot; button for some help in working out the answer. Yes No Yes, that's right. The following plots indicate that the exponential distribution is indeed a good fit to the data since both plots yield approxinately straight lines. However, one may like to calculate the test statistics for complete confidence that the model is appropriate. Click here to see the calculations behind the plots in Excel . No, as you can see from the following plots, the exponential distribution is a good fit to the data since both plots yield approxinately straight lines. However, one may like to calculate the test statistics for complete confidence that the model is appropriate. Click here to see the calculations behind the plots in Excel .  |../ModelFitting2Session/ModFit200180.htm",
    "Statistics for Maintenance Modelling: Testing Model Fitness|Summary ¤ This session presented several methods for assessing the goodness of fit for a given model fitted to a given data set. You should now be aware that once a model has been fitted to data (see session 6), appropriate diagnostic tests should be performed so as to assess the goodness of fit for that model. The assessment methods presented...| fall in to two categories: Quantitative statistical tests: the Chi-square test the Kolmogorov-Smirnov test the A 2 and W 2 tests. Graphical methods for assessing goodness of fit: Q-Q and P-P plots. Such methods plot empirical values of some quantity, obtained from the original data versus theoretical values obtained from the fitted model. Straight lines in either a Q-Q or P-P plot indicate a good fit. However, although useful and easily transparent to the reader, graphical methods do not provide a quantitative measure, or test, of how appropriate a model is. It may sometimes be necessary to perform some of the quantitative methods for assessing goodness of fit to aid ones decision on choosing the most appropriate model.  |../ModelFitting2Session/ModFit200190.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Introduction ¤ In earlier sessions we have been concerned with modelling the time to failure, with no other variables involved. However, it is often the case that the lifetime depends on other variables, some of which it may be possible to control. One example of this is the way failure time is dependent on the load applied to the component - it...| is likely that increasing the load will cause the component to fail sooner. In statistics, the terminology used for this situation is as follows: the failure time is a dependent variable, and the load is the independent variable, or covariate . Regression Models for Modelling a Dependent Variable Throughout statistics the use of regression models provide a powerful tool for modelling a dependent variable as a function of one or more independent variables or covariates. From the simplest regression model, the normal linear regression model, to the more complex generalised linear models and non-linear models, regression is an invaluable methodology for any researcher. For an introduction to regression and the basic principles there are many good books, for example see Neter et. al. (1996) . In this session we present the regression models used in reliability and maintenance modelling, namely 'accelerated life models', 'the proportional hazards model' and 'the Weibull model'. It is worth remembering that we are just introducing the theories and concepts behind these models here. Computer software packages such as Minitab, Stata and SPSS can be used to fit these models. You need to understand the background of each model, as introduced here, in order to help with correct implementation.  |../RegressionModellingSession/RegMod00005.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Objectives ¤ By the end of this session you will know: what accelerated life models are what proportional hazards models are what proportional odds models are what the Weibull based regression models are. | |../RegressionModellingSession/RegMod00010.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Accelerated Life Models ¤ Accelerated life models make use of the survival function (see The Reliability Function screen in the Continuous Lifetime Distributions 1 session). The premise is that the rate to failure of a component is 'accelerated' by some multiplier, the multiplier being a function of the covariate, for example the load. Formally,...| the accelerated life model holds when: S ( t ; x ) = P T t x In other words, the survival function of t and x is equal to the probability of the time to failure being greater than t given the value of x , where x is the covariate, say load again. S ( t ; x ) = S 0 ( t x ) where S 0 is some baseline survival function and x is a positive function of x . Example Let T denote the time to breakdown of an electrical component subjected to constant voltage x , and suppose x = for some &gt; 0 and S 0 ( t ) = . Then, S ( t ; x ) = , so T is exponentially distributed with mean (since from R ( t ) = , = -1 , and is the mean. See The Weibull Distribution and The Reliability Function for the Weibull Distribution screens in the Continuous Lifetime Distributions 1 session). Note that 0 = 0 here. The choice of baseline function is made by the modeller, but experience can be used to make more appropriate decisions.  |../RegressionModellingSession/RegMod00020.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Assessing the Appropriateness of an Accelerated Life Model ¤ So far we have discussed the concept of the accelerated life model, but how do we know if the accelerated life model is appropriate for our data set? As an initial guide the following plot can be used: If the data are grouped, the j th group of n j observed failure times having x = j and...| n j being large enough for a plot of the sample survival j ( t ) function against ln t , then the plot should resemble horizontally shifted copies of the baseline plot of S 0 ( t ) against ln t . This follows from: S j ( t ) = S 0 ( t j ) = S 0 = s 0 ( y + ln j ), with y = ln t say. S 0 represents the survival function of the log failure time when j = 1. See the case study later in this session for an example of using this plot ( screen 11 ). The above plot applies to censored data as well as uncensored data, in which case, the Kaplan-Meier estimator (see the Nonparametric Estimates of the Survival Function screen in the Model Selection session) is used for . This plot, and variations of them, are also used when assessing the appropriateness of proportional odds models and proportional hazards models, which are discussed in the next few screens. Return  |../RegressionModellingSession/RegMod00030.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Proportional Hazards Models ¤ The second type of regression model commonly used in maintenance and reliability modelling is the proportional hazards models. In the proportional hazards models the hazard function satisfies: h ( t ; x ) = h 0 ( t x ) where: h ( t ; x ) is the hazard under the conditions x h 0 ( t ; x ) is some baseline hazard...| function x is some positive function of x . The name 'proportional hazards models' comes from the simple mathematical fact that since h ( t ; x 1 )/ h ( t ; x 2 ) = x 1 / x 2 is independent of t , the hazards at different values of x are in constant proportion over time. For this model, the survival function satisfies: S ( t ; x ) = S 0 ( t ; x ) where S 0 is the baseline survival function corresponding to h 0 . This is simple to prove since: ln S ( t ; x ) = - h ( t ; x ) d t = - x h 0 d t = x ln S 0 ( t )  |../RegressionModellingSession/RegMod00040.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Satisfying an Accelerated Life and a Proportional Hazards Models ¤ One can actually make S ( t ; x ) satisfy both an accelerated life model and a proportional hazards model, so that: S ( t ; x ) = S a0 ( t ax ) = S p0 ( t ) with the baseline functions standardised so that a 0 = p 0 = 1. Then: S a0 ( t ) = S p0 ( t ) = for some &gt; 0 (see The...| Reliability Function for the Weibull Distribution screen in the Continuous Lifetime Distributions 1 session) and px = ax . A proof of this property can be found in Cox and Oakes (1984) , section 5.3.  |../RegressionModellingSession/RegMod00045.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Assessing the Appropriateness of a Proportional Hazards Model ¤ To identify whether a data set should be modelled using the proportional hazards model it is easiest to examine an appropriate plot of the data. In a variation of how we assessed the appropriateness of the accelerated life model on screen 4 , if the data are grouped, the j th group of...| n j observed failure times having x = j and n j being large, then the plots of -ln j ( t ) against t for the different groups should resemble multiples of the baseline curve -ln S 0 ( t ) against t . This is because: if S j ( t ) = S 0 ( t ) then -ln S t ( t ) = - j ln S 0 ( t ). Alternatively, the plots ln -ln j ( t ) of against t should resemble vertically shifted copies of ln -ln S 0 ( t ) against t . To yield a possibly neater plot, the t axis may be transformed by taking logs. This will not change the pattern of the vertically parallel curves. See the case study later in this session for an example of using this plot ( screen 12 ). Return  |../RegressionModellingSession/RegMod00050.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Proportional Odds Models ¤ In probability theory, the odds of a particular event, T &lt; t occurring is given by the ratio P T &lt; t / P T t . In the proportional odds model the survival function satisfies: = x The proportional odds model says that the odds under explanatory variables x is x times the baseline odds (first used by Bennett, 1983 )....| As x increases so does the probability of a shorter lifetime. By differentiating the above equation we can obtain an equation which holds for the hazard function in the proportional odds model. Differentiating with respect to t : = x Rearranging, the hazard function ratio thus satisfies: = x and then, using the original equation we get: = As S (0, x ) = S 0 (0), at t = 0 the hazard ratio is x and tends to 1 as t . I.e. x has less effect on the hazard as time goes on.  |../RegressionModellingSession/RegMod00060.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Assessing the Appropriateness of a Proportional Odds Model ¤ Can you think of two explanations for x having a diminishing effect on the hazard as time goes by? Click on the &quot;Check Answer&quot; button to find out. Either: the system adjusts to the factors imposed on it and the system reaches some stable state or, the factors operate only in...| the earlier stages. Testing Whether to Use a Proportional Odds Model In the same way as for the models already discussed, to help identify whether a data set should be modelled using the proportional odds model one can plot the data. If the data are grouped, the j th group of n j observed failure times having x = j and n j being large, plots of (1 - j ( t )) / j ( t ) against t for different groups should resemble multiples of the corresponding baseline curve, and hence multiples of each other. Again, plotting the logarithm of each axis may yield tidier curves, while merely shifting the baseline curve vertically. See the case study later in this session for an example of using this plot ( screen 13 ). Return  |../RegressionModellingSession/RegMod00065.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Models Based on the Weibull Distribution ¤ The three models introduced so far, namely the accelerated life model, the proportional hazards model and the proportional odds model, can be adapted and generalised. First described by Lawless (1986) , these generalisations will not be discussed here, although it is useful to be aware that they exist,...| and where to get more information about them from. In the broadest sense, regression based on the normal distribution serves as the most useful model in all of statistics. However, as a consequence of the nature of reliability data, regression models based on the Weibull distribution are of particular interest here. The general premise for models based on the Weibull distribution is to let the distribution parameters vary according to some relationship with a covariate, x . For instance, the Weibull survivor function is given by: S ( t ; x ) = where is the Weibull shape parameter (independent of x ), and x is a parameter whose exact value is dependent on the covariate x , for example, ln x = x T , where is a vector of regression coefficients. If we let S 0 ( t ) = then we can write S ( t ; x ) = S 0 ( t ax ) where ax = 1/ x and we have an accelerated life model. Alternatively, if we let S ( t ; x ) = S 0 ( t ) where px = x - , we have a proportional hazards model. Clearly the models based on the Weibull distribution provide a wide ranging and powerful tool for modelling reliability data. The models can be further generalised by letting the shape parameter also depend on a covariate. This may be appropriate when initial exploratory analysis of the data suggests that the Weibull shape parameter is not the same everywhere.  |../RegressionModellingSession/RegMod00070.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Case Study: Breaking Strengths of Carbon Fibres - Accelerated Life Model ¤ The models and concepts discussed so far in this session are by no means easy to understand, so we'll now examine a case study. The data were collected from experiments done at The University of Surrey. This analysis closely follows that of Watson and Smith (1985) and...| Lindgren and Rootzen (1987) , which was reproduced in Crowder et al. (2000) . We have four different lengths of carbon fibres, 1mm, 10mm, 20mm and 50mm. This is the plot of survival probability against the log strength for lengths. Click here for the data and calculations of failure stresses for each length of fibre . The plot is constructed as follows: Take the natural logarithm of the failure stresses for each fibre. Calculate the empirical cumulative distribution function using: ( t ) = (a formula given by Leitch, 1995 ). Then calculate the estimated survival probability from: ( t ) = 1 - ( t ). If the accelerated life model is appropriate for the data, the lines in the chart should look like horizontally shifted copies of one another. At this point we may decide this is observed here, but experience will tell us to check the other plots, and indeed, that other models are a more appropriate fit to the data. Return  |../RegressionModellingSession/RegMod00080.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Failure Stresses for Four Lengths of Carbon Fibres ¤ Failure stresses for four different lengths of carbon fibres, 1mm, 10mm, 20mm and 50mm. 1mm 10mm 20mm 50mm 2.247 1.901 1.312 1.339 2.640 2.132 1.314 1.434 2.842 2.203 1.479 1.549 2.908 2.228 1.552 1.574 3.099 2.257 1.700 1.589 3.126 2.350 1.803 1.613 3.245 2.361 1.861 1.746 3.328 2.396 1.865...| 1.753 3.355 2.397 1.944 1.764 3.383 2.445 1.958 1.807 3.572 2.454 1.966 1.812 3.581 2.474 1.997 1.840 3.681 2.518 2.006 852.000 3.726 2.525 2.021 1.852 3.727 2.532 2.027 1.862 3.728 2.575 2.055 1.864 3.783 2.614 2.063 1.931 3.785 2.616 2.098 1.952 3.786 2.618 2.140 1.974 3.896 2.624 2.179 2.019 3.912 2.659 2.224 2.051 3.964 2.675 2.240 2.055 4.050 2.738 2.253 2.058 4.063 2.740 2.270 2.088 4.082 2.856 2.272 2.125 4.111 2.917 2.274 2.162 4.118 2.928 2.301 2.171 4.141 2.937 2.301 2.172 4.216 2.937 2.339 2.180 4.251 2.977 2.359 2.194 4.262 2.996 2.382 2.211 4.326 3.030 2.382 2.270 4.402 3.125 2.426 2.272 4.457 3.139 2.434 2.280 4.466 3.145 2.435 2.299 4.519 3.220 2.478 2.308 4.542 3.223 2.490 2.335 4.555 3.235 2.511 2.349 4.614 3.243 2.514 2.356 4.632 3.264 2.535 2.386 4.634 3.272 2.554 2.390 4.636 3.294 2.566 2.410 4.678 3.332 2.570 2.430 4.738 3.346 2.586 2.431 4.832 3.377 2.629 2.458 4.924 3.408 2.633 2.471 5.043 3.435 2.642 2.497 5.099 3.493 2.648 2.514 5.134 3.501 2.684 2.558 5.359 3.537 2.697 2.577 5.473 3.554 2.726 2.593 5.571 3.562 2.770 2.601 5.684 3.628 2.773 2.604 5.721 3.852 2.800 2.620 5.998 3.871 2.809 2.633 6.060 3.886 2.818 2.670 - 3.971 2.821 2.682 - 4.024 2.848 2.699 - 4.027 2.88 2.705 - 4.225 2.954 2.735 - 4.395 3.012 2.785 - 5.02 3.067 2.785 - - 3.084 3.02 - - 3.09 3.042 - - 3.096 3.116 - - 3.128 3.174 - - 3.233 - - - 3.433 - - - 3.585 - - - 3.585 - riate when initial exploratory analysis of the data suggests that the Weibull shape parameter is not the same everywhere.  |../RegressionModellingSession/RegMod00085.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Case Study: Breaking Strengths of Carbon Fibres - Proportional Hazards Model To test whether the proportional hazards model is appropriate we can plot the ln-ln survivor function against ln( t ), as described on screen 7 . If the proportional hazards model is appropriate the plot should consist of vertically shifted copies of the same curve....| Further, if these curves are in fact parallel straight lines then a Weibull model is appropriate as described on screen 10. Return  |../RegressionModellingSession/RegMod00090.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Case Study: Breaking Strengths of Carbon Fibres - Proportional Odds Model ¤ The proportional odds model may be tested by plotting the log odds against ln( t ) as described on screen 9 . These four curves are supposed, if indeed the proportional odds model is appropriate, to be vertically shifted copies of the same curve. This is not the case, for...| example, the curves for lengths 1mm and 10mm appear to be concave and convex respectively. Return  |../RegressionModellingSession/RegMod00100.htm",
    "Statistics for Maintenance Modelling: Regression Modelling|Summary ¤ In this session we have seen how regression models can be used to model data where it is thought that lifetime depends on another variable or variables. Three key regression models were introduced that are often used in reliability modelling. Accelerated life models; we saw how the rate of failure is &#145;accelerated&#146; by some...| covariate, for example, the load. Proportional hazards models, where the hazards at different values of the covariate, say load again, are in constant proportion over time. Proportional odds models, where the odds at different values of some covariate, say load, are in constant proportion over time. Several plots were presented in order to aid identification of these models, i.e. which one would be appropriate for a particular data set. Note that it is not necessary for you to be able to reproduce the theoretical underpinnings of these models, or plots, rather you understand the basic concepts of each model and plot, and when it is appropriate to use them. After all computers do fit these models for us.  |../RegressionModellingSession/RegMod00110.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Introduction ¤ It is frequently the case that time to failure is dependent on other random variables --- characteristics which are perhaps subject to natural variation, but they may also be under a certain amount of control. These explanatory variables, sometimes called covariates, influence the life-time directly, or model it through the...| reliability function , and thus through the hazard function . Click here to jump to The Hazard Function screen in the Continuous Lifetime Distributions 1 session if you need to recap. The covariates may be time-dependent and they may be influenced over time by the unit itself (e.g. level of corrosion), or may not be directly involved in the failure process (e.g. applied stress). In this session we discuss the analysis of maintenance and reliability data using the proportional hazards and accelerated life models. The proportional hazards model has been briefly introduced in the previous session on Regression Modelling. These two models are also called regression models since the covariates are sometimes also called regression variables. R ( t ) = 1 - F ( t ) =  |../PropHazardSession/PropHaz00005.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Objectives ¤ By the end of this session you will know: what the Semiparametric Proportional Hazards model is how to estimate the baseline Hazard and Survival functions parametrically and non-parametrically how to check if a Proportional Hazards model is appropriate what the Accelerated Life model is. | |../PropHazardSession/PropHaz00010.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|The Proportional Hazards Model ¤ The proportional hazards model was first proposed by Cox (1972) . The vast majority of reported uses of the proportional hazards model have been in the areas of survival analysis of medical data. It is noted that however, the proportional hazards model has been increasingly used in reliability and maintenance...| modelling research. The proportional hazards ( PH ) model introduced in the previous session on Regression Modelling, models the effects on the hazard function of certain explanatory covariates or factors associated with the life-time of equipment. The PH model assumes that, under conditions given by a vector y = ( y 1 , y 2 , ... , y n ) of covariates, the hazard function is of the form: h ( t; y ) = h 0 ( t ) g ( y ) where h 0 ( t ) is the baseline hazard function and g ( y ) is some function of y . Thus the combined effect of y is simply to scale the hazard function up or down. Since: is independent of t , the hazards at different y values are in constant proportion over time. The function g ( y ) may take a variety of forms and the most commonly used one is which is the form that shall be used in this session where = ( 1 , 2 , ... n ).  |../PropHazardSession/PropHaz00020.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Example of the Proportional Hazards Model ¤ Suppose the baseline hazard is , and the parameters are = 0.01, = 2, then at time t = 100: h 0 ( t ) = 0.02. If we take into account covariates where 1 = 0.5, n = 1 and y 1 = 1, then at time t = 100 we have: = 0.032. Similarly if y 1 = -1, we have: h ( t; y ) = 0.012 at t = 100. This demonstrates how the...| hazard will change if the influence of covariates is considered.  |../PropHazardSession/PropHaz00030.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Analysis of the Semi-Parametric Proportional Hazards Model ¤ In reliability and maintenance analysis, the function for h 0 ( t ) is usually specified or has to be specified. However, in other applications the functional form h 0 ( t ) does not need to be completely specified, such as in medical statistics. If h 0 ( t ) is not specified, or the...| application is only focused on the meaning of , a semi-parametric form of h ( t; y ) can be analysed and estimated. Suppose that a random sample of n items yields r distinct lifetimes and n - r censored times (see the Censoring screen in the Model Selection session). Let the ordered life times be t (1) , ..., t ( j ) , ..., t ( r ) where t (1) is the shortest life time and t ( r ) is the longest among these r items. Let R j be the set of items with t t ( j ) , where t may be either an observed lifetime or a censored time. Here R j is the risk set at time t ( j ) , that is those items whose lifetimes are longer than t ( j ) and could yet fail. Estimation of may then be based on the 'likelihood function': L ( ) where y i ( j ) is the i th observed covariate associated with the item observed to fail at time t ( j ) , and y il is the i th observed covariate associated with item l among the risk set at time t ( j ) .  |../PropHazardSession/PropHaz00040.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Justifying Using the Likelihood Function ¤ There are several alternative ways of justifying the use of the likelihood function on the previous screen. The simplest argument is due to Cox (1972) . Consider the set of R j of items at risk immediately prior to t ( j ) . The conditional probability that the item failed at t ( j ) given items in R j...| that could fail at t ( j ) is: = (since P (fail at t ( j ) ) hazard t if t is small) = (since h ( t ( j ) y ( j ) ) = h 0 ( t ( j ) ) ) Treating the function on the previous screen as an ordinary likelihood, the general methods of maximum likelihood estimation discussed in the Model Fitting session may be used to estimate .  |../PropHazardSession/PropHaz00050.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Estimation of the Survival and Hazard Functions ¤ There are two ways to estimate the baseline hazard function. You can estimate = ( 1 , 2 , ... n ) and the parametric form of h 0 ( t ) together using the maximum likelihood method provided you have sufficient life data. This is done by writing down the log-likelihood function based upon r distinct...| lifetimes and n - r censored times, that is: Ln L ( ) = Constant + + = Constant + + (since f ( t i ; y ) = h ( t i ; y ) S ( t i ; y )) = Constant + + + (since S ( t i ; y ) = ) = Constant + - (since ln( e x ) = x , and combining the two sets) where: D is the set of complete life data C is the set of censored data f ( t i ; y ) is the probability density function of time to failure given y S ( t i ; y ) is the survival (or reliability) function given y h ( t i ; y ) is the hazard function given y . Maximising the log-likelihood function given on the previous screen in terms of unknown parameters of , one should have the estimated parameter values. If the baseline hazard function depends on parameters , these can be estimated in this way.  |../PropHazardSession/PropHaz00060.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Nonparametric Estimation of the Survival and Hazard Functions ¤ Alternatively, it is possible to obtain a nonparametric estimate of h 0 ( t ) once = ( 1 , 2 , ... n ) have been estimated. The next formula is a generalisation of the Kaplan-Meier (see the Nonparametric Estimates of the Survival Function screen in the Model Selection session)...| estimator described in the Preliminary Model Selection session: where = the item hasn't failed at time t ( j ) given failure times t (1) , ..., t ( r ) . A plot of the empirical survival function 0 ( t ) may suggest a suitable parametric form for 0 ( t ) and an estimate of the baseline cumulative hazard function is 0 ( t ) = -ln 0 ( t ).  |../PropHazardSession/PropHaz00070.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Model Checking ¤ Return Let us note the effect of y on the survival function in the form of the equation h ( t; y ) = h 0 ( t ) g ( y ) where g ( y ) = . Since S ( t ) = (from The Hazard Function screen in the Continuous Lifetime Distributions 1 session): S ( t; y ) = = = = S 0 ( t ) g ( y ) , taking logs of S ( t; y ) gives: ln S ( t; y ) = - g (...| y ) h 0 ( u ) du . Taking logs again of -ln S ( t; y ), we finally have: ln(-ln S ( t; y )) = ln( g ( y )) + ln(-ln( S 0 ( t ))) where S 0 ( t ) = is the baseline survival function, therefore ln S 0 ( t ) = - h 0 ( u ) du . In other words, the ln(-ln) survival functions given y are parallel for different covariates. This suggests that if there are many observations for each y , a check on the proportional hazards assumption can be made by computing the Kaplan-Meier estimator of the survival function i ( t ) = ( t ; y i ) from the data at each y i . Plots of ln(-ln i ( t )) versus t should be roughly parallel if a proportional hazards model is appropriate .  |../PropHazardSession/PropHaz00080.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Accelerated Life Model ¤ The accelerated life model is another important class of regression models. Suppose we observe failure times for a system subject to a range of loads. One model that is easy to understand is that the failure time of the unit is the product of some function of the load. This is known as an accelerated life model because of...| the interpretation that increasing the load has the effect of accelerating the rate of failure by a deterministic constant. More formally, the accelerated life model holds when the survival function S ( t; y ) = P ( T t y ) is of the form: S ( t; y ) = S 0 t &#149; g ( y )) where S 0 is the baseline survival (reliability) function and g ( y ) is a positive function of y . The interpretation is that the life time is scaled down or up by a factor g ( y ) .  |../PropHazardSession/PropHaz00090.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Example using a Weibull Distributed Baseline Survival Function ¤ Suppose for some and S 0 ( t ) = , then since S ( t; y ) = S 0 ( t ) g ( y ) (from screen 9), S ( t; y ) = , which is also a proportional hazard model with a Weibull distributed baseline survival function using (see The Reliability Function for the Weibull Distribution screen in the...| Continuous Lifetime Distributions 1 session). This shows that if the baseline survival function is Weibull, then we can produce the same result either by an accelerated life model or by a proportional hazards model . In the accelerated life model, the scaled life time t ' = t &#149; g ( y ) has survival function S 0 . Equivalently, the log lifetime may be expressed as ln t = ln t ' - ln g ( y ) where t ' has survival function S 0 ( t '). Thus the distribution of ln t is just that of ln t ' subject to a location shift of -ln g ( y ), and is some times referred to as location-scale modelling. As an initial guide to whether an accelerated life model is appropriate, a plot of observed log times against y can provide information on the form of g ( y ), and can yield rough estimates of its parameters. Since ln t = ln t ' - ln g ( y ), a plot of ln t against y is effectively a plot of -ln g ( y ) against y , ignoring the &quot;error&quot; term ln t '.  |../PropHazardSession/PropHaz00100.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Case Study 1 - Failure of Insulating Fluid Subject to Voltage Stress ¤ Nelson (1972) describes the result of a life test experiment in which specimens of a type of electrical insulating fluid were subject to a constant voltage stress. The length of time until each specimen failed, or &quot;broke down&quot; was observed. These are the results for...| three groups of specimens, tested at voltages of 28, 30 and 32 kilovolts. The chart shows the product-limit (Kaplan-Meier) estimates of the ln(-ln( i ( t )) against time to failure. It can be seen that the curves are roughly parallel, which indicates a potential proportional hazards model. M i n u t e s 28 kv 30 kv 32 kv 68.85 17.05 194.9 0.4 3.91 426.07 22.66 47.3 82.85 0.27 110.29 21.02 7.74 9.88 0.69 108.29 175.88 89.29 100.58 1067.6 139.07 215.1 27.8 144.12 2.75 13.95 20.46 0.79 53.24 43.4 15.93  |../PropHazardSession/PropHaz00110.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Case Study 2 - Survival Data for Cancer Patients ¤ The table on the next screen presents survival data on 40 advanced lung cancer patients, taken from a large body of data discussed by Prentice (1973) . The main purpose of the study was to compare the effects of two chemotherapy treatments in prolonging survival time. All patients in the group...| represented here received prior therapy and were then randomly assigned to one of the two treatments, termed &quot;standard&quot; and &quot;test&quot;. Survival times t from the start of the treatment for each patient are recorded. Censored observations are marked with an asterisk and correspond to patients who were still alive at the time the data was collected. A number of covariates were thought to be important in relation to survival time, and these are also shown for each patient. Different patients can have different types of tumours. Here tumours have been classified into four broad types (squamous, small, adeno and large). Each patient is given a performance status at the time of diagnosis. This is a measure of the general medical status on a scale of 0 to 100: 10, 20, and 30 mean that the patient is completely hospitalised; 40, 50, and 60, mean that they are partially confined to hospital; and 70, 80, and 90, mean they are able to care for themselves. The age of the patient The number of months from diagnosis of lung cancer to entry into the study.  |../PropHazardSession/PropHaz00120.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Case Study 2 - Table of Survival Data for Cancer Patients ¤ Squamous Tumor Small Tumor Adeno Tumor Large Tumor t Survival days y 1 Perf. status y 2 Age in years y 3 Months in study t Survival days y 1 Perf. status y 2 Age in years y 3 Months in study t Survival days y 1 Perf. status y 2 Age in years y 3 Months in study t Survival days y 1 Perf....| status y 2 Age in years y 3 Months in study &quot;Standard&quot; Treatment 411 70 64 5 54 80 63 4 8 20 61 19 177 50 66 16 126 60 63 9 153 60 63 14 12 50 63 4 12 40 68 12 118 70 65 11 16 30 53 4 200 80 41 12 82 40 69 10 56 80 43 12 250 70 53 8 8 40 63 58 21 40 55 2 100 60 37 13 25* 70 48 9 287 60 66 25 1 70 48 11 10 40 67 23 &quot;Test&quot; Treatment 999 90 54 12 103* 70 36 22 18 40 69 5 164 70 68 15 231* 50 52 8 2 40 44 36 90 60 50 22 19 30 39 4 991 70 50 7 20 30 54 9 84 80 62 4 43 60 49 11 1 20 65 21 51 30 59 87 340 80 64 10 201 80 52 28 231 70 67 18 44 60 70 13 15 50 40 13  |../PropHazardSession/PropHaz00130.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Case Study 2 - Analysis of Survival Data for Cancer Patients ¤ The main objective is to compare the effects of the two treatments upon survival time, but in doing so one must take into account other factors that may also influence survival. What actually needs to be studied here is the joint effect of the various factors (i.e. treatment and...| covariates) in the experiment on survival time. As a first step in model fitting, a simple proportional hazards model was used, namely h ( t; y ) = h 0 ( t ) g ( y ) = . This is called an exponential PH model where the baseline hazard is a constant of (see The Hazard Function for the Weibull Distribution screen in the Continuous Lifetime Distributions 1 session). We start with a model involving all of the covariates. This can be done by associating a vector y = ( y 1 , y 2 , ... , y 7 ) with each individual, where: y 1 = Performance status y 2 = Age y 3 = Months since entry into study y 4 = 1 if tumor type is squamus, 0 otherwise y 5 = 1 if tumor type is small, 0 otherwise y 6 = 1 if tumor type is adeno, 0 otherwise y 7 = 0 if treatment is test, 1 if it is standard  |../PropHazardSession/PropHaz00140.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Case Study 2 - Estimation of Survival Data Parameters ¤ It is wise to centre some of the covariates to show the variation against the mean more clearly, and here just y 1 , y 2 and y 3 are centred since they are the dominant covariates. The log likelihood (from screen 7) is then: Ln L ( ) = Constant + r (ln ) - where r is the number of complete...| life times. Maximum likelihood estimates for = ( 1 , 2 , ... 7 ) and (the baseline constant hazard rate) are readily obtained by some numerical iteration algorithms, and are listed in this table. 1 2 3 4 5 6 7 4.722 0.054 0.0085 0.0034 0.3448 -0.121 -0.863 -0.277 This suggests that y 1 (performance status) is important in terms of confined impact of 1 y 1 to the hazard function, but that y 2 and y 3 are not. In addition, once performamce status has been accounted for, treatment and tumor type do not appear to have sizeable effects, though there is a suggestion that the Adeno tumor type is important. Ln L ( ) = Constant + +  |../PropHazardSession/PropHaz00150.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Summary ¤ In this session, we have discussed the analysis of reliability data using the proportional hazards and accelerated life models. The merit of the proportional hazard and accelerated life models is the ability to incorporate covariates into the reliability or maintenance model where the time to failure is not only dependent upon the time...| the item has survived but also upon the levels of covariates. Also, the vast majority of reported cases of using the PH or accelerated life models have been in the analysis of survival data in health care, though the semi-parametric hazard-based model is beginning to have more of an impact in the reliability and maintenance modelling field. The full PH or accelerated life model has a role to play in condition based maintenance modelling since the observed condition monitoring information may be regarded as the covariates. However, care must be taken when fitting the model to data since you must distinguish the covariates which may influence the state of the item monitored and the covariates which may be a function of the underlying state. The former can be modelled by the PH model such as speed, stress, loading, or temperature, and the latter should be modelled through a different mechanism such as the filtering model introduced in Condition Based Maintenance sessions of the Maintenance Modelling module.  |../PropHazardSession/PropHaz00160.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Exercise 1 ¤ Let t denote the time to failure of an electrical insulator subject to constant voltage y , and suppose for some and S 0 ( t ) = e -t . Prove that both proportional hazards and accelerated life models can produce the same S ( t; y ) = . Prove that t is exponentially distributed with a scale parameter g ( y ). What is the mean of t ?...| Answer 1 The survival function of the exponential distribution is usually of the form of (from The Reliability Function for the Weibull Distribution screen in the Continuous Lifetime Distributions 1 session) with the hazard function of h 0 ( t ) = . In this case, since S 0 ( t ) = e -t , = 1. If we use a PH model with = 1, then we have h ( t; y ) = h 0 ( t ) g ( y ) = and S ( t; y ) = = (from screen 9 ). Using an accelerated life model we have S ( t; y ) = S 0 t &#149; g ( y )) = S ( t ). Substituting t for the parameter in the baseline survival function we get S 0 ( t ) = , so the result is the same. Answer 2 If S ( t; y ) = , then the pdf (from The Reliability Function screen in the Continuous Lifetime Distributions 1 session) is , so t is exponentially distributed with scale parameter . Answer 3 If f ( t ) is in the form , then the mean is 1/ . Since we've just proved that = g ( y ), the mean is 1/ g ( y ). If f ( x ) = ae bx then f' ( x ) = abe bx .  |../PropHazardSession/PropHaz00170.htm",
    "Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models|Exercise 2 ¤ This table shows the failure times for steel specimens at four stress levels: Stress 0.870 0.990 1.090 1.180 Failure times 1.670 0.800 0.012 0.073 2.200 1.000 0.180 0.089 2.510 1.370 0.200 0.117 2.900 2.250 0.240 0.135 3.000 2.950 0.260 0.175 4.700 3.700 0.320 0.262 7.530 6.070 0.320 0.270 14.700 6.650 0.420 0.350 27.900 7.050 0.440...| 0.386 37.400 7.350 0.880 0.456 Plot the product-limit (Kaplan-Meier) estimates of the ln(-ln( i ( t )) against the times to failure and assess whether a proportional hazards model is appropriate. Engineering background suggests that at stress level y , failure times should have approximately a Weibull distribution with a scale parameter related to stress s by a power law relationship = cs p and with a shape parameter which is independent of s . Assess this continuous model graphically and against the discrete alternative that failure times at the i th stress level have a Weibull distribution with parameters i and , ( i = 1,2,3 and 4). Are you satisfied that should be considered the same for all four stresses? Click here for the charts and the answers to the questions.  |../PropHazardSession/PropHaz00180.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Introduction ¤ A repairable system is one which after failing to function properly may be restored to satisfactory working order by replacing or repairing certain components except replacement of the entire system. The system will, throughout its lifetime, undergo alternating in-service times and repair/maintenance times. In practice many systems...| are of this type. Examples featured in this session are: motor vehicles air-conditioning units in an aircraft main propulsion diesel engines of USS Grampus some software during the production and testing stages. Of interest are the times of failures and the costs incurred, both of which may be plotted against time to reveal patterns and any unusual features of the data. Repair and maintenance times may also be of interest, but it will be assumed initially that these are small relative to the in-service times.  |../RepairableSystemsSession/RepSys100005.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Objectives ¤ By the end of this session you will know: what the MCNF is what the MCCF is what the ROCOF is what the HPP is what the NHPP is how to estimate the parameters of a chosen NHPP what repair time is what maintainability is what availability is how to calculate MCNF and MCCF . | |../RepairableSystemsSession/RepSys100010.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Framework ¤ Let T 1 , T 2 , T 3 , ... be the random variables of the times of failure of the system, X i , i = 1, 2,&#133; be the time between the ( i - 1)th and i th failure. We may take 'time' to be operating time, or real time if the 'down' times are considered to be negligible. Observing the system over time, we are looking for any change in...| the nature of the X i . There will be no change if the X i are independent and identically distributed. Departure from this condition may constitute trend and the detection and estimation of trends is of major interest. Even if there is no long-term trend there may be some other structure of practical importance, for example inadequate repairs may produce a run of short failure times among longer times. So there may be dependence or autocorrelation between the X i . In repairable systems one generally has the times of successive failures of a single system, where the intervals between failures may be neither independent nor identically distributed. This is a key feature which results in the use of different statistical methods for repairable systems from those needed in component reliability. The Nature of Time In the above discussion we have used time in its most general sense. In any particular application, time needs to be defined carefully and appropriately. For a continuously operating refrigerator in reasonably constant conditions real time may be appropriate. More commonly, however, operating time may be more appropriate than real time. On the other hand, some non-negative quantity other than time itself may be suitable. For instance, the mileage of a motor vehicle may be more useful for assessing reliability than real time and easier to record than operating time.  |../RepairableSystemsSession/RepSys100020.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Graphical Methods ¤ Three functions of prime interest relating to repairable systems are: the Mean Cumulative Number of Failures ( MCNF ) the Mean Cumulative Cost Function ( MCCF ) the Rate of Occurrence of Failures ( ROCOF ). Plots of the sample equivalents of these measures often reveal simple underlying power laws describing the evolution of...| the measures with time. The MCNF and MCCF may be plotted directly against time, or perhaps more usefully, the ln MCNF or ln MCCF against ln t . Then an underlying power law may be estimated by drawing a straight line through the points. If more than one identical system is considered, then the number of failures is averaged over the systems in operation.  |../RepairableSystemsSession/RepSys100030.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|MCNF and MCCF ¤ An equation which includes a variable raised to a power. MCNF Let N ( t ) be the number of failures in the time interval (0, t ), then: MCNF = E ( N ( t )). If this can be written in power law form as: E ( N ( t )) = , we have: ln MCNF = ln A + ln t A plot of ln MCNF against ln t will reveal whether a power law is a suitable model...| for the MCNF . It also provides an estimate for via the slope of the fitted line. Signs of non-linearity may indicate an inadequate model, or features such as a 'kink' in the line may indicate a change in the system due, for example, to an effective quality improvement programme. MCCF If C ( t ) is the cumulative cost at time t , then similarly we might be able to write: MCCF = E ( C ( t )) = and ln MCCF = ln B + ln t .  |../RepairableSystemsSession/RepSys100040.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|ROCOF ¤ Under the power-law assumption, the ROCOF , or intensity function, is defined by: ( t ) = E ( N ( t )) = A Thus if: 0 &lt; &lt; 1, then the ROCOF decreases = 1, the ROCOF is constant &gt; 1, then the ROCOF increases. Ascher and Feingold (1984) coined the terms of &quot; happy system &quot; and &quot; sad system &quot; where the ROCOF is...| decreasing and increasing, respectively. The existence of trend is therefore characterised by linear plots which have slopes significantly different from 1. An approximate measure of the possible departure from 1 can be gained via a confidence interval estimate of the slope. The ROCOF should not be confused with the hazard function. The hazard describes the conditional probability of failure within t + t of a non-repairable component given survived to t . It is possible for each X i to have an increasing hazard but for the system to have a constant ROCOF , or indeed any other combination of behaviours. A battery may have Weibull distribution life time with shape parameter greater than 2, implying an increasing hazard, but a system in which the battery is replaced on failure by a similar battery may display constant ROCOF .  |../RepairableSystemsSession/RepSys100050.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Example of MCNF and MCCF for Motor Vehicles ¤ The following data concern repairs to 954 motor vehicles in a warranty period. The values have been averaged over the 954 vehicles and are months in service ( MIS ), cumulative number of failures per unit and cumulative cost per unit. MIS 1 2 3 4 5 6 7 8 MCNF 0.397 0.547 0.688 0.807 0.95 1.064 1.173...| 1.305 MCCF 11.84 16.44 20.84 24.68 29.73 32.76 36.82 41.14 ln MIS 0 0.30103 0.477121 0.60206 0.69897 0.778151 0.845098 0.90309 ln MCNF -0.40121 -0.26201 -0.16241 -0.09313 -0.02228 0.026942 0.069298 0.115611 ln MCCF 1.073352 1.215902 1.318898 1.392345 1.473195 1.515344 1.566084 1.614264 The chart shows ln MCNF and ln MCCF plotted against ln MIS . Both plots approximate to a straight line with coincidentally similar slopes. The power estimated by the slope is approximately 0.6, so both number of failures and costs are showing a decreasing rate of occurrence over the period.  |../RepairableSystemsSession/RepSys100060.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Example of MCNF and MCCF for Six Repairable Systems ¤ This example concerns repair and cost data for six repairable systems, all commissioned at time zero but each subject to retirement over time. The data cover a period of 42 months and is illustrated below, each event shown as a failure ( ) or retirement ( ). The cost of repair (in hundred...| dollars) and the calculation of the MCNF and MCCF is shown in the table on the right. Event Time Cost System in service Mean no failures MCNF Mean cost MCCF 1 2 1 6 1/6=0.17 0.17 1/6=0.17 0.17 2 5 3 6 1/6=0.17 0.33 3/6=0.2 0.67 3 8 1 6 1/6=0.17 0.50 1/6=0.17 0.83 4 8 2 6 1/6=0.17 0.67 2/6=0.33 1.17 5 12 1 6 1/6-0.17 0.83 1/6=0.17 1.33 6 12 7 14 1 5 1/5=0.2 1.03 1/5=0.2 1.53 8 16 9 17 2 4 1/4=0.25 1.28 2/4=0.5 2.03 10 18 3 4 1/4=0.25 1.53 3/4=0.75 2.78 11 19 2 4 1/4=0.25 1.78 2/4=0.5 3.28 12 20 13 26 1 3 1/3=0.33 2.11 1/3=0.33 3.61 14 29 15 33 16 39 2 1 1/1=1 3.11 2/1=2 5.61 17 42  |../RepairableSystemsSession/RepSys100070.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Logs Plotted for Six Repairable Systems ¤ This chart shows ln time plotted against ln MCNF and ln MCCF . Both plots are linear, indicating that a power law model is appropriate for both MCNF and MCCF . The slopes of the lines indicate how the rate of failures and costs are varying over time. In the case of MCNF , the gradient is approximately one...| and therefore the failure rate is fairly constant. The costs show a rate of change of 1.16 which indicates increasing costs, but given the length of time over which the data were collected, this factor may be purely due to inflation. This latter point shows that it is particularly important when costs are considered to measure them in the most appropriate way. Over a long period of time it may be necessary to use the concept of discounting or other different measures.  |../RepairableSystemsSession/RepSys100080.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Homogeneous Poisson Process Models ( HPP ) ¤ A stochastic point process may be regarded as a sequence of events distributed in a continuum according to some probability mechanism. If we take the continuum to be time and the events as failures, it is a frame for repairable systems. There are a variety of stochastic point process models for the...| failures of repairable systems, but we will focus on the Poisson process (see The Poisson Process screen in the Continuous Lifetime Distributions session). This process is conceptually simple and it can model happy and sad systems as well as systems with constant ROCOF . The relevant statistical methodology is well developed and easy to apply. For an HPP , natural assumptions, which may or may not be satisfied in any particular example, are: failures occurring in disjoint time intervals are independent the ROCOF is constant. If both assumptions are satisfied, then it is called a homogeneous Poisson process. Two important properties of a HPP are: the number of failures N in a time interval of length t has a Poisson distribution with mean t , so that: P ( N = k ) = ( t ) k / k ! the times between successive failures are independent random variables each having an exponential density with scale parameter , so that: P ( t &gt; x ) =  |../RepairableSystemsSession/RepSys100090.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Non-homogeneous Poisson Process Models ( NHPP ) ¤ The mean time between failures is 1/ . A system for which we expect the HPP to be a good model is one in which there are many components liable to fail, but the probability of failure of any one of them is small. In applications, assumption 2 (on the previous screen) is critical. Many systems are...| in practice either improving or deteriorating with time. In this case we need more general models, and the NHPP is one of them. The assumptions for a NHPP are as for the HPP except that the ROCOF varies with time. Consider a NHPP with a time-dependent ROCOF , ( t ), then the number of failures in the interval ( t 1 , t 2 ] has a Poisson distribution with mean: E ( N ( t 1 , t 2 )) = ( t ) d t Thus the probability of having k failures in ( t 1 , t 2 ] is: P ( N = k ) = ( ( t ) d t ) k / k ! When k = 0, we have the probability of no failures in ( t 1 , t 2 ] given by: P ( N = 0) = By choosing a suitable parametric form for ( t ), one can obtain a flexible model for the failures of a repairable system in a 'minimal repair' set-up, i.e. when only a small proportion of the component parts of the system are replaced on repair.  |../RepairableSystemsSession/RepSys100100.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Parametric Forms of Lambda( t ) for NHPP Models ¤ Two particular forms for ( t ) are: 1 ( t ) = - gives a simple model to describe a happy system ( 1 &lt; 0) or a sad system ( 1 &gt; 0). If 1 = 0 we have a HPP . 2 ( t ) = - referred to as the Weibull or power law process (with A = ). The name Weibull process is a misnomer here since only the time...| to the first failure has a Weibull distribution. Once again this model may represent a sad or happy system as discussed above. The parameters within ( t ) can be estimated using the maximum likelihood method. Suppose we observe a system over the time interval (0 , t 1 ] with failures observed at t 1 , t 2 , ... t n . The likelihood may be obtained as follows. The probability of observing no failures in (0, t 1 ), one failure in ( t 1 , t 1 + t 1 ), no failures in ( t 1 + t 1 , t 2 ), one failure in ( t 2 , t 2 + t 2 ) and so on up to no failures in ( t n + t n , t 0 ) is, for small t 1 , t 2 ,..., t n : L = ( t 1 ) t 1 ( t 2 ) t 2 ... Dividing through by t 1 , t 2 ,..., t n gives: L = t i and the log-likelihood is: ln L = ln ( t i ) - ( t ) d t. Maximising this with respect to the parameters within ( t ) estimates their values.  |../RepairableSystemsSession/RepSys100110.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Return Fitting a NHPP Model and Estimating Parameters Using Lambda 1 ( t ) ¤ Let's look at some likelihood-based statistical methods for fitting a NHPP with ( t ) = 1 ( t ) = or 2 ( t ) = to some sets of repairable systems data. Estimating Parameters by Substituting Lambda 1 ( t ) into the Log-Likelihood Putting ( t ) = 1 ( t ) = into the...| log-likelihood equation on the previous screen, the log-likelihood for a repairable system observed over the time interval (0, t 0 ] with failure times at t 1 , t 2 , ... t n is: ln L = ln ( t i ) - ( t ) d t = ln ( ) - d t = n 0 + 1 t i - where n is the n th failure. To obtain the maximum likelihood estimates of 0 and 1 , one can maximise this equation using any optimisation software such as Excel Solver or Mathcad. Alternatively, one can differentiate it with respect to 0 and 1 and then set the derivatives to zero. We shall look at this next.  |../RepairableSystemsSession/RepSys100120.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Estimating the Maximum Likelihood Parameters from Lambda 1 ( t ) ¤ Differentiating the log-likelihood with respect to 0 and 1 and setting the derivatives to zero: ln L = n - = 0 ln L = t i - = 0 After some algebra, 0 may be expressed explicitly in terms of 1 , that is: = and finally we have: t i + n 1 -1 n t 0 (1 - ) -1 = 0 From this equation we...| can find 1 and therefore 0 = ln ( n 1 ( - 1) -1 ). The programme involved in solving this equation is modest. A simple interactive search or repeated bisection will suffice.  |../RepairableSystemsSession/RepSys100130.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Fitting a NHPP Model and Estimating Parameters Using Lambda 2 ( t ) ¤ Return When ( t ) = 2 ( t ) = , substituting ( t i ) by 2 ( t i ) in the log-likelihood, ln L = ln ( t i ) - ( t ) d t , we have: ln L = ln ( ) - d t = n (ln + ln ) + ( - 1) ln t i - Differentiating the log-likelihood with respect to and and setting the derivatives to zero, we...| have: ln L = n -1 - = 0 ln L = n -1 + n ln + t i - ln ( t 0 ) = 0 Solving the above equations simultaneously we have the closed form maximum likelihood estimators for and as: = and = .  |../RepairableSystemsSession/RepSys100140.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Example of Fitting a NHPP Model to Software Failures and Estimating Parameters ¤ Return The following are the intervals in days between successive failures of a piece of software developed as part of a large data system: 9, 12, 11, 4, 7, 2, 5, 8, 5, 7, 1, 6, 1, 9, 4, 1, 2, 2, 6, 1, 11, 33, 7, 91, 2, 1, 87, 47, 12, 9, 135, 258, 16, 35. In the...| software reliability context, we want to know if there is evidence of reliability growth, i.e. is 1 &lt; 0? Observation of the software was made for 850 days, resulting in 34 observed failures. Therefore, n = 34, t 0 = 850, and t i = 6980. Estimating Parameters by Substituting Lambda 1 ( t ) into the Log-Likelihood Using Excel's solver to solve the equation on screen 14, t i + n 1 -1 n t 0 (1 - ) -1 = 0, we have 1 = -0.004 which suggests that there is evidence of reliability growth ( 1 modifies t in 1 ( t )). 0 = ln ( n 1 / ( - 1) -1 ), which is -1.96. Estimating Parameters by Substituting Lambda 2 ( t ) into the Log-Likelihood Applying this model to the software failure data, we have = 0.0007 and = 0.1542. This also confirms that reliability is improving since &lt; 1. Which Model of Lambda ( t ) is Best? To determine which model is better in this case, one has to use other criteria such as assessing the goodness of fit, (discussed throughout the Testing Model Fitness session), or simply comparing the values of ln L . The one with the larger likelihood is the better of the two - which in this case is 2 ( t ).  |../RepairableSystemsSession/RepSys100150.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Using Excel's Solver ¤ Using the Solver facility (accessed from the Tools menu), if you've typed the expression for t i + n 1 -1 n t 0 (1 - ) -1 in B1 and a nominal value for 1 in B2, then: the Target Cell for solver should be B1 select the Value of radio button Equal To 0 set the By Changing Cells entry to B2 Then click the Solve button. When the...| Solver Results dialog appears, click the Keep Solver Solution radio button. Click here to open an Excel spreadsheet with the data from the example and the above expression for you to solve . If the Solver option isn't available: Click Add-Ins on the Tools menu. In the Add-Ins available list, select the Solver Add-in box, and then click OK . Close  |../RepairableSystemsSession/RepSys100150a.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Example of Fitting a NHPP Model to Failure Repairs and Estimating Parameters ¤ The following data are a subset of the times in operating hours of unscheduled maintenance actions (failure repairs) for a diesel engine of USS Grampus: 860, 1258, 1317, 1442, 1897, 2011, 2122, 2439, 3203, 3298, 3902, 3910, 4000, 4247, 4411, 4456, 4517, 4899, 4910,...| 5676, 5755, 6137, 6221, 6311, 6613, 6975, 7335, 8158, 8498, 8690, 9042, 9330, 9394, 9426, 9872, 10191, 11511, 11575, 12100, 12126, 12368, 12681, 12795, 13399, 13668, 13780, 13877, 14007, 14028, 14035, 14173, 14176, 14449, 14587, 14610, 15070. The plot of ln MCNF against ln t , shown below, looks roughly linear, suggesting a power law model may be suitable. You can see from the plot that the slope of the line is roughly 1 so a constant ROCOF may be suggested. We can fit ( t ) = 2 ( t ) = to this data set using the equations on the previous screen to estimate and . This gives = 4.93 E -07 and = 1.2178 which suggests that ROCOF is slightly increasing. However, is extremely small which may mean it's not a good model choice (this is explored in exercise 3 on screen 27 at the end of this session). Return  |../RepairableSystemsSession/RepSys100160.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Testing for a Trend - The Laplace Test ¤ The process in which the ROCOF varies with time may be described as a NHPP with ( t ). It is important to test whether ( t ) is time dependent or not. Under the assumption of constant ( t ), the times of failures (not between failures), T 1 , T 2 , T 3 , ... T n , are ordered uniform variables over (0, T 0...| ] with mean T 0 /2 and variance T 0 2 /12. Let t 1 , t 2 , t 3 , ... t n and t 0 denote their realisations and by the central limit theorem , the statistic t i / n has an approximate normal distribution with mean t 0 /2 and variance t 0 2 /(12 n ), so the statistic: U = = is approximately N (0,1), and may be used to test the null hypothesis of no trend, that is 1 = 0 or 1 = 1 in 1 ( t ) and 2 ( t ) respectively. The equation we've generated here is commonly called the Laplace test . The central limit theorem says that the distribution of any reasonably well behaved function of a sample mean tends to the normal distribution as the sample size increases to infinity.  |../RepairableSystemsSession/RepSys100170.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Examples of Testing for a Trend ¤ Example 1 Failures occur at times 15, 42, 74, 117, 168, 233, 410. Using the equation for the Laplace test : U = ((15 + 42 + 74 + 117 + 168 + 233) - 6 410 / 2) / (410 ) = - 2.0 For testing H 0 : no trend, against H 1 : decreasing ROCOF , the critical value for 2.5% significance level is -1.96 (from the normal...| distribution table ) which &gt; U, so H 0 is rejected at this level and it is concluded that the incidence of failure is tending to reduce. Example 2 Taking the data of software failures used earlier on screen 15 , we will test whether there is a decreasing trend. Since n = 34, t 0 = 850, and t i = 6980, we have from the equation for the Laplace test that: U = (7015 - 1/2 34 850) / (850 ) = -5.2209 Since H 0 : no trend, and H 1 : decreasing trend, if we choose 1% as the significance level, the critical value is -2.33, which is greater than -5.2209, and strongly suggests we can accept H 1 . U =  |../RepairableSystemsSession/RepSys100180.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Extending the Laplace Test ¤ Strictly speaking, the Laplace test is based on the null hypothesis that the process is a homogeneous Poisson process. However, the test may be used for the more general hypothesis that the process is a renewal process (see the next session), that is, the inter-event times are identically and independently distributed...| ( IID ). This can be done by the U statistic being divided by the coefficient of variation of the inter-event times, which is estimated by the sample standard deviation divided by the sample mean. The exponential distribution has a coefficient of variation equal to 1, so this modification embraces the U of the equation on screen 18. The formulation of the Laplace test rests on the central limit theorem, which would imply that the test is only good when the value of n is large. However, the test proves to be fairly robust against departure from normality and may be applied even when n is modest. U =  |../RepairableSystemsSession/RepSys100190.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Repair Time ¤ One way of describing the occurrence of failures is to state the mean time between failures, often known as MTBF . The main difference between the MTBF and mean time to failure, known as the MTTF , is in the usage. The MTTF may apply to first failure or a non-repairable failure, whereas the MTBF is used in the context of repairable...| systems. Note that the MTBF is usually the mean operating time between failures. There is, however, considerable variation in the use of these terms and the definitions should be checked within any given text. Repair actions taken after system or component failures are corrective maintenance actions. Actions taken before failure are called preventive maintenance actions, which may include regular servicing and replacement of parts before significant wear or total failure has occurred. The modelling of repair time is of considerable interest as this affects the necessary human resources, spares provisioning and operational effectiveness. Some of the issues are considered in the second part of this course, namely maintenance modelling.  |../RepairableSystemsSession/RepSys100200.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Maintainability ¤ Maintainability is defined to be the probability that a component or system that has failed will be restored to full working order within a specified period of time, t . Times are frequently assumed to be exponentially distributed, and therefore the rate of repair is given by the reciprocal of mean time to repair ( MTTR ) ( click...| here for a summary of the exponential distribution ). In this case, the maintainability, M ( t ), may be expressed as: M ( t ) = Prob (repair completed before t ) = P ( X t ) ( X is exponentially distributed with mean = MTTR ) = 1 -  |../RepairableSystemsSession/RepSys100210.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Availability ¤ Reliability and maintainability are combined in the term availability which is the proportion of 'up' time . The long-term or steady-state availability, , is given by: = . It should be noted that provides no information about the number of repairs. It is most likely that 'down' time involves more than the time to repair. Factors...| might include the time to diagnose, spare part location and testing. For example, the expression for availability therefore often involves mean downtime ( MDT ) rather than the MTTR , so is revised as: = The unavailability, is given by 1 - .  |../RepairableSystemsSession/RepSys100220.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Summary ¤ This session introduced: the concept of repairable systems a few graphical methods for presenting: the Mean Cumulative Number of Failures ( MCNF ) the Mean Cumulative Cost Function ( MCCF ) the Rate of Occurrence of Failures ( ROCOF ). Attention was paid to the ROCOF where Poisson processes are involved. HPP and NHPP - two specific ROCOF...| for the NHPP were given in detail the likelihood formulation for estimating the unknown parameters within each ROCOF testing the trend of the ROCOF and the Laplace test the importance of repair time in maintenance modelling and the concepts of maintainability and availability.  |../RepairableSystemsSession/RepSys100230.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Exercise 1 ¤ This data set gives 13 observed intervals in operating hours between failures of the air-conditioning equipment in a Boeing 720 aircraft: 413, 14, 58, 37, 100, 65, 9, 169, 447, 184, 36, 201, 118. Carry out a Laplace test at a 5% significance level to test the trend. With the null hypothesis H 0 : no trend and H 1 : decreasing trend,...| can we reject the null hypothesis? Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button when you've finished. Click here for the normal distribution table . Yes No Yes, that's right, the null hypothesis cannot be rejected. t 0 = 1851, n = 13, and t i = 12641, which gives U = (12641 - (13 1851) / 2) / (1851 (13 / 12) 1/2 ) = 0.3163. At a 5% significance level, the critical value is -1.64 (we're looking for a decreasing trend). At less than 0.3163, it suggests that the null hypothesis cannot be rejected. No, in fact the null hypothesis cannot be rejected. t 0 = 1851, n = 13, and t i = 12641, which gives U = (12641 - (13 1851) / 2) / (1851 (13 / 12) 1/2 ) = 0.3163. At a 5% significance level, the critical value is -1.64 (we're looking for a decreasing trend). At less than 0.3163, it suggests that the null hypothesis cannot be rejected. Plot ln MCNF against ln t (cumulative failure time). Is there a straightforward model for MCNF available? Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button when you've finished. Yes No Yes, there isn't a straightforward model available. As you can see from the plot, the linearity is not good and there is no straightforward model for the failure function. There is some suggestion of cycles in the data, but more observations would be needed to identify any firm pattern. No, actually there isn't a straightforward model available. As you can see from the plot, the linearity is not good and there is no straightforward model for the failure function. There is some suggestion of cycles in the data, but more observations would be needed to identify any firm pattern.  |../RepairableSystemsSession/RepSys100240.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Exercise 2 ¤ Following on from question 1, the next 10 failures were observed after the equipment had a major overhaul after the 13th failure. The intervals were 34, 31, 18, 18, 67, 57, 62, 7, 22, 34. Plot ln MCNF against ln t using the data before and after the overhaul. Does there seem to be a change in the failure behaviour after the overhaul?...| Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button when you've finished. Yes No Yes, the plot again hints at some cyclic behaviour but also clearly shows the effect of the overhaul - it may have had an adverse effect on the failure behaviour of the unit since there appears to be an increase in the rate of occurrence of failures. It may be of interest to know whether the ROCOF has moved to a new level but is still constant. This can be done by resetting the time origin to zero at the 13th failure and redoing the chart and Laplace test. No, actually the plot hints at some cyclic behaviour but also clearly shows the effect of the overhaul - it may have had an adverse effect on the failure behaviour of the unit since there appears to be an increase in the rate of occurrence of failures. It may be of interest to know whether the ROCOF has moved to a new level but is still constant. This can be done by resetting the time origin to zero at the 13th failure and redoing the chart and Laplace test. Carry out a Laplace test at a 5% significance level to test the trend using all the data. With the null hypothesis H 0 : no trend and H 1 : increasing trend, can we reject the null hypothesis? Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button when you've finished. Click here for the normal distribution table . Yes No Yes, that's right, the null hypothesis is rejected. t 0 = 2201, n = 23, and t i = 33074, which gives U = (33074 - (23 2201) / 2) / (2201 (23 / 12) 1/2 ) = 2.5474. At a 5% significance level, the critical value is 1.64. At less than 2.5474, it suggests that H 0 can be rejected (since we're looking for an increasing trend). No, in fact the null hypothesis can be rejected. t 0 = 2201, n = 23, and t i = 33074, which gives U = (33074 - (23 2201) / 2) / (2201 (23 / 12) 1/2 ) = 2.5474. At a 5% significance level, the critical value is 1.64. At less than 2.5474, it suggests that H 0 can be rejected (since we're looking for an increasing trend).  |../RepairableSystemsSession/RepSys100250.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Exercise 3 ¤ Carry out a trend test on the data set from screen 17 of unscheduled maintenance actions for a diesel engine of USS Grampus carried out at times: 860, 1258, 1317, 1442, 1897, 2011, 2122, 2439, 3203, 3298, 3902, 3910, 4000, 4247, 4411, 4456, 4517, 4899, 4910, 5676, 5755, 6137, 6221, 6311, 6613, 6975, 7335, 8158, 8498, 8690, 9042, 9330,...| 9394, 9426, 9872, 10191, 11511, 11575, 12100, 12126, 12368, 12681, 12795, 13399, 13668, 13780, 13877, 14007, 14028, 14035, 14173, 14176, 14449, 14587, 14610, 15070 at a 2.5% significance level. With the null hypothesis H 0 : no trend and H 1 : increasing trend, can we reject the null hypothesis? Select &quot;Yes&quot; or &quot;No&quot; and click the &quot;Check Answer&quot; button when you've finished. Click here for the normal distribution table . Yes No Yes, that's right, the null hypothesis cannot be rejected. t 0 = 15070, n = 56, and t i = 461738, which gives U = (461738 - (56 15070) / 2) / (15070 (56 / 12) 1/2 ) = 1.2218. At a 2.5% significance level, the critical value is 1.96. Since this is greater than 1.2218, it suggests that the null hypothesis cannot be rejected. No, in fact the null hypothesis cannot be rejected. t 0 = 15070, n = 56, and t i = 461738, which gives U = (461738 - (56 15070) / 2) / (15070 (56 / 12) 1/2 ) = 1.2218. At a 2.5% significance level, the critical value is 1.96. Since this is greater than 1.2218, it suggests that the null hypothesis cannot be rejected. Return  |../RepairableSystemsSession/RepSys100260.htm",
    "Statistics for Maintenance Modelling: Repairable Systems|Exercise 4 ¤ Use the data set from screen 16 of successive failures of a piece of software developed as part of a large data system: 9, 12, 11, 4, 7, 2, 5, 8, 5, 7, 1, 6, 1, 9, 4, 1, 2, 2, 6, 1, 11, 33, 7, 91, 2, 1, 87, 47, 12, 9, 135, 258, 16, 35. Compute Ln L by substituting 1 ( t ) = and 2 ( t ) = into the log likelihood as discussed on screens...| 13 and 15 to give: resp. Which model gives a better result? Select the one you think, and then click the &quot;Check Answer&quot; button when you've finished. ln L = n 0 + 1 t i - ln L = n (ln + ln ) + ( - 1) ln t i - Yes, that's right, 2 ( t ) is better choice. For 1 ( t ) the log likelihood is -728, and for 2 ( t ) the log likelihood is -163.34. Since 2 ( t ) creates a ln L with a higher value, this choice is better. No, in fact 2 ( t ) is better choice. For 1 ( t ) the log likelihood is -728, and for 2 ( t ) the log likelihood is -163.34. Since 2 ( t ) creates a ln L with a higher value, this choice is better.  |../RepairableSystemsSession/RepSys100270.htm",
    "Statistics for Maintenance Modelling: Continuous Lifetime Distributions 2|Introduction ¤ The previous session discussed the probability density function, the cumulative density function, the reliability function, the hazard function and several distributions that model lifetimes. This session continues the review of lifetime distributions by covering the commonly used: normal and lognormal distributions, before going on...| to the more specialised: gamma logistic loglogistic and Pareto distributions.  |../RepairableSystems2Session/RepSys200005.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Objectives ¤ By the end of this session you will know: what renewal theory is what the renewal counting process ( N ( t )) is what the renewal function ( M ( t )) is how to calculate the renewal function analytically how to calculate the renewal function numerically what Laplace transforms are what the asymptotic behaviour of the renewal function...| is what alternating renewal processes are what discrete renewal processes are.  |../RepairableSystems2Session/RepSys200010.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|What is a Renewal Process? ¤ Imagine a light bulb, which has a probability density function of f ( x ), lasting for time x before it fails. When it fails, it is immediately replaced with a new one. The successive intervals between failures are random variables X 1 ,..., X n , and these form a renewal process. The candles shown here illustrate a...| similar scenario. The main application of renewal processes is in reliability and replacement. In addition, they crop up constantly when considering other more complex stochastic processes. The recognition of an embedded renewal process and the application of the relevant theory is then an important step in solving the complex problem. In this session, you will learn what a renewal process is, how the renewal function can be calculated, and how it is used in simple replacement models. Can you think of any other examples of renewal processes in daily life? Click here to see some of our ideas.  |../RepairableSystems2Session/RepSys200020.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Definition of the Renewal Process ¤ Consider a renewal process (e.g. of light bulbs). Then we can define: the renewal counting process as N ( t ), t 0 , where N ( t ) is the number of events (e.g. failures) by time t , where intervals between events are i.i.d. (independent and identically distributed) random variates with pdf. f ( x ),...| distribution function F ( x ), survival function S ( x ). the intervals between failure times X 1 ,..., X n where n 1 the waiting time to the n th event as the quantity W n = X 1 + ... + X n , where n 1 Then W n , n 0 and N ( t ), t 0 are called the renewal process . This is illustrated on the right for the renewal counting process N ( t ), failure times X and waiting times W .  |../RepairableSystems2Session/RepSys200030.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|The Renewal Function and Renewal Density ¤ The expected number of events by time t , M ( t ) = E N ( t ) , is called the renewal function . Its derivative v ( t ) = d M ( t )/d t is called the renewal density . As examples of typical renewal functions and densities, the charts below show the renewal function and renewal density for an Erlang(4)...| distribution (the theoretical calculation will be shown later). The pdf of the Erlang( n ) is:  |../RepairableSystems2Session/RepSys200040.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Renewal Process Replacement Costs ¤ The expected cost of operating a system for time t , that is replaced on failure at cost c and has no running costs is cM ( t ). The cost of issuing a warranty is an example. Block replacement of components is another. Here components (such as light bulbs that are hard to access) are replaced on failure, and all...| bulbs are replaced as planned at time t . If: the cost of replacing a failed bulb is c f the cost of a block replacement per bulb is c b then: the expected accumulated cost for a bulb under this strategy up to time t is c f M ( t ) + c b the cost per unit time is ( c f M ( t ) + c b )/ t . Minimising Costs The obvious best policy is to choose the replacement age t to minimise total cost per unit time. A Non-Time Based Application Inventory is another (rather strange!) application. Here demand for a product per week (say) is the random variate (rather than a time), and rather than mean number of failures by time we seek the mean number of weeks for total demand to exceed stock level.  |../RepairableSystems2Session/RepSys200050.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Why Renewal Theory? ¤ The main interest in renewal theory is the calculation of the renewal function, because this is what is needed for applications. We are interested in calculating it analytically where we can, i.e. in getting a mathematical expression in closed form in terms of elementary functions. Later on we shall obtain such a result, and...| shall be at pains to replace an infinite series expansion with an expression involving sines and cosines. This exemplifies what is meant by 'closed form' and 'elementary functions'. Such calculations give insight into renewal theory and no mathematician can resist them! We are also interested in calculating the renewal function numerically. After all, this is what is needed in practice. Finally, its asymptotic behaviour for large times t is of interest. It is of course also possible to calculate higher moments than the renewal function (the mean number of failures). This is touched on here only very briefly. See Cox (1962) for more details.  |../RepairableSystems2Session/RepSys200060.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Questions on the Conditions for Block Replacement ¤ What conditions do we require for block replacement to be appropriate? Click on one of the options below, then click on &quot;Check Answer&quot;. c f &lt; c b c f &gt; c b c f = c b Yes, that's right. The cost of block replacement must be cheaper, otherwise failure replacement is better. No, the...| cost of block replacement must be cheaper otherwise failure replacement is better. Is the scheme appropriate for any type of failure-time distribution? Click on &quot;Check Answer&quot; when you've made a selection. Yes No Yes, that's right. The failure time must be IFR - increasing failure rate. I.e. the hazard rate increases with age. No that's not right. The failure time must be IFR - increasing failure rate. I.e. the hazard rate increases with age.  |../RepairableSystems2Session/RepSys200070.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Calculating the Renewal Function Analytically: Exponential Distribution ¤ The simplest case is that of the exponential distribution for the time between failures. Here the result is trivial. We have f ( x ) = (from The Weibull Distribution screen in the Continuous Lifetime Distributions 1 session), where the mean is E ( X ) = = 1/ . The...| calculation of the renewal function M ( t ) relies on the generalism that if at least events k occur by time t , this is the same as saying that the time to the k th event must be no later than t . Mathematically, this becomes: N ( t ) k iff W k t. The distribution of W k is Erlang, because W k is the sum of k exponential random variables, so the pdf . is: f k ( x ) = ( x ) k - 1 /( k - 1)! Now the probability that the k th item fails before time t is the cdf (see The Cumulative Density Function screen in the Basic Probability Concepts session), so: Prob ( N ( t ) k ) = Prob ( W k t ) = F k ( t ) = f k ( x ) d x = 1/( k - 1)! (substituting the pdf. above) Integrating progressively by parts we obtain: Prob ( N ( t ) k ) = 1 - Another way of looking at this is that the number of renewals in the interval from 0 to t is a Poisson random variate with mean t where is the rate of occurance (see The Poisson Distribution screen in the Continuous Lifetime Distributions 1 session).  |../RepairableSystems2Session/RepSys200080.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Calculating the Renewal Function Analytically: Exponential Distribution (2) ¤ Now: Prob ( N ( t ) = k ) = Prob ( N ( t ) k ) - Prob ( N ( t ) ( k + 1)) = ( t ) k / k ! (using the equation on the previous screen) Since , M ( t ) = = ( t ) = t , as the sum = (since e x = 1 + x + x 2 /2! + ... x n / n ! = ) This tells us that the expected number of...| events in time t is just the interval t times the rate of occurrence, .  |../RepairableSystems2Session/RepSys200090.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Calculating the Renewal Function Analytically: Exponential Distribution (3) ¤ This method of finding M ( t ) can be modified to make the calculations easier. In general, the following identity is useful: Since Prob( N ( t ) = k ) = F k ( t ) - F k +1 ( t ) (because F k is the probability that k or more events have occurred), M ( t ) = k ( F k ( t...| ) - F k +1 ( t )) = F 1 - F 2 + 2( F 2 - F 3 ) + ... = F 1 + F 2 + F 3 + ... = F k ( t ) So in general, the renewal density v ( t ) = d M ( t )/d t , or v ( t ) = f k ( t ) is sometimes a good way in to calculate M ( t ). Click here for an intuitive proof of M ( t ). Return  |../RepairableSystems2Session/RepSys200100.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Calculating the Renewal Function Analytically: Erlang (2) Distribution ¤ Let's take the case where the failure of the system requires 2 exponential failures. This would apply if, for example, a system comprised two identical components, where one is a spare (not in use). The system fails when both components have failed. The pdf. of time to...| failure is: g 1 ( t ) = ( t ) and that of time to k failures is the distribution of the sum of 2 k exponential random variables, i.e. g k ( t ) = ( t ) 2 k - 1 /(2 k - 1)! Hence: v ( t ) = f k ( t ) = ( t ) 2 k - 1 /(2 k - 1)! = ( t )/1! + ( t ) 3 /3! + ( t ) 5 /5!... These are the odd terms of an exponential expansion, and so: v ( t ) = - /2 = ( /2) 1 - from which: M ( t ) = v ( t )d t = t /2 - (1 - )/4  |../RepairableSystems2Session/RepSys200110.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Calculating the Renewal Function Analytically: Erlang (2) Distribution (2) ¤ Note that M ( t ): increases initially as M ( t ) = 1/2( t ) 2 for small t (quadratically) and then increases linearly as t equals t /2 - 1/4 for large t . Since for the interfailure time distribution, the mean = 2/ , and the variance 2 = 2/ 2 , for large t when 0, M ( t...| ) t /u + ( 2 - 2 )/(2 2 ) This formula also holds for the exponential distribution, where = . We shall see that this asymptotic formula is true generally for 'non-arithmetic' distributions with finite variance. Some harder exact calculations are deferred until after this asymptotic formula has been discussed in more detail.  |../RepairableSystems2Session/RepSys200120.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Laplace Tranforms ¤ We define the Laplace transform of f ( x ) as: ( s ) = f ( x ) d x , where s &gt; 0. The inverse transform is made either by using a table of transforms of common functions, or by doing a contour integration. The pdf. of the sum of two random variables is given by the convolution of the two pdfs , and a very useful feature of...| Laplace transforms is that the transform of a convolution is the product of transforms of the two convolved functions. This means that, on taking transforms, a pdf. that is a k -fold convolution is replaced by the k th power of the Laplace transform of the failure time. Mathematically, convolution means that if X 1 and X 2 are random variates, the pdf f ( t ) of X + X 1 + X 2 is given by f ( t ) = f 1 ( y ) f 2 ( t - y ) d y , and the Laplace transforms obey ( s ) = 1 ( s ) 2 ( s ) Laplace transforms are therefore very useful in renewal theory, because the series of convolutions of probabilities that crop up become simple geometric series when expressed as Laplace transforms. You can however follow the essentials of renewal theory without bothering about these transforms, or come back to them later. For a fuller account, and a table of important Laplace transforms see the Laplace Transforms page on the Mathworld website .  |../RepairableSystems2Session/RepSys200130.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Questions on Laplace Tranforms ¤ Match the following functions with its Laplace transform. Drag each function and drop it on its Laplace transform - there are a few extra possibilities in there just to make you think! When you've dropped the last function, you will find out how many you've got right. The table of important Laplace transforms on...| the Laplace Transforms page on the Mathworld website will help you. You can click the Reset Button to try again. is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: is the Laplace transform of: Functions: 1 x x 2 Well done! You correctly identified all the Laplace transforms. I'm afraid that you've matched 1 of the Laplace transforms incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you've matched 2 of the Laplace transforms incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you've matched 3 of the Laplace transforms incorrectly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results. I'm afraid that you haven't matched any of the Laplace transforms correctly. Click &quot;Reset&quot; to have another go, or &quot;Show Answer&quot; to see the correct results.  |../RepairableSystems2Session/RepSys200140.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Asymptotic Formulae ¤ The behaviour of the renewal function at large time t values is its asymptotic behaviour as t . Studying this gives useful insight, but it is not directly related to calculations needed for replacement modelling, and this more theoretical section can be skipped on first reading. We start with the result we proved on screen 10...| : M ( t ) = F k ( t ) where F k is the k -fold convolution of the distribution function (the distribution function corresponding to the k -fold convolution of the pdf. , or the distribution function of the sum of k random variables). This is the probability that at least k events have occurred (the k th event has occurred) by time t . Let the Laplace transform of the pdf f be ( s ). Then the transform of F k ( t ) is ( s ) k / s . Taking the Laplace transform of the equation for M ( t ), and summing the geometric series, we obtain: ( s ) = (1/ s ) ( s ) k = (1/ s ) ( s ) k = This looks a lot simpler than the original expression, and will yield the asymptotic formula fairly easily (but by a non-rigorous argument).  |../RepairableSystems2Session/RepSys200150.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Asymptotic Formulae (2) ¤ We use the expansion: ( s ) = f ( t ) d t = 1 - s + (1/2) s 2 ( 2 + 2 ) + ..., which gives the approximation: ( s ) = (1/ s ) , or ( s ) = (1/ s 2 ) or ( s ) = 1/( s 2 ) + (1/2) on expanding the denominator as a power series in s . The inverse of this transform is: M ( t ) = t / + (1/2) which is the asymptotic result. | |../RepairableSystems2Session/RepSys200160.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Question on Asymptotic Formulae ¤ What is the result for M ( t ) when 2 = 0 and when 2 = 2 (the exponential distribution)? Can you make sense of these results? When 2 = 0, we have simply a lifetime that is always , and in this limiting case, the number of renewals by time t is t / - (1/2). By starting with a new component, rather than one halfway...| through its life, we have half a renewal less to make. With large 2 , we have on average more replacements to make, because some can fail very quickly. For the exponential distribution, 2 = 2 and and M ( t ) = t / , so the asymptotic formula is exactly correct. We have M ( t ) = t / because the exponential distribution is memoryless - the failure rate does not depend on when the previous failure occurred, and there is a constant hazard of failure of 1/ . This point is quite deep. A renewal process for the exponential distribution is identical to an asynchronous renewal process.  |../RepairableSystems2Session/RepSys200170.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|The Integral Equation ¤ The equation: M ( t ) = F k ( t ) can be re-expressed as an integral equation: M ( t ) = F ( t ) + f ( u ) M ( t - u ) d u and this can also be Laplace-transformed to yield: ( s ) = (which was derived earlier by a different route) In fact, there are many starting points for deriving these three formulae. An illuminating one...| is to derive M ( t ) = F ( t ) + f ( u ) M ( t - u ) d u directly. We start by assuming that the first renewal occurs at time u . The renewal function M ( t ) is then 1 + M ( t - u ), the renewal at time u plus the expected number of renewals after that. This occurs with pdf. f ( u ), so relaxing the assumption (the first renewal can occur at any time from 0 to t ), we have: M ( t ) = f ( u ) d u + f ( u ) M ( t - u ) d u It is possible to use this integral equation to derive the asymptotic formula for M ( t ) directly without taking Laplace transforms.  |../RepairableSystems2Session/RepSys200180.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Numerical Calculation ¤ Often we can not get an analytic expression for M ( t ). For small t , M ( t ) = F ( t ) (when the probability of 2 or more replacements is small). For large t , we have the asymptotic formula. Finding M ( t ) in between is the problem! The renewal integral equation is the basis of the most popular method of calculating M (...| t ). In Xie's method, it is discretized, and solved recursively for M ( ), M (2 ), etc. It is also possible to use the Laplace transform of this equation. If we can find the transform of the pdf. , we can invert the transform numerically, using some method such as Talbot's method. Talbot's method is described by Murli and Rizzardi (1990) , and compared favourably with other methods by Duffy (1993) . If we cannot find the transform of f ( t ) analytically (as we cannot for the Weibull distribution), we can perhaps approximate f ( t ) so that we can find an expression for its transform.  |../RepairableSystems2Session/RepSys200190.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|The Monte-Carlo Numerical Calculation ¤ A much more direct method that is very easy to program is to generate random variates from the distribution f ( t ), and to then simply count the number of renewals by time t . Repeating this many times gives the expectation M ( t ). This Monte-Carlo method will be slow, but with a modern computer, it...| doesn't matter. This method (using a cunning algorithm to make it run very fast) was developed by Baker (1993) as a method of computing nonparametric estimates of the renewal function from observed data on times to failure. The Monte-Carlo method would be to arrange the sample of n failure times in all n ! possible ways, and so to compute M ( t ). A recursive algorithm makes this calculation very fast indeed. This approach removes the need for fitting a parametric form to f ( t ) and then computing M ( t ) from that. Return  |../RepairableSystems2Session/RepSys200200.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Discrete Renewal Processes ¤ When time is discrete, it is possible to solve the renewal equation very simply. It becomes: M n = F n + M n - k p k where p k is the probability that failure occurs at time k F n is the distribution function of the probabilities p k at time n M n and is the renewal function at time n . Then and we solve recursively...| for M 0 , M 1 , ..., M n . Usually, p 0 = 0, when we have the slight simplification: M n = F n + M n - k p k In discrete-time block replacement, failures are replaced at cost c f at times 1.. n - 1, and a planned replacement is done at time n at cost c b whether a failure has occurred at time n or not. The cost per unit time is therefore: ( c b + c f M n - 1 )/ n . It is simple to find the time n that minimises cost per unit time.  |../RepairableSystems2Session/RepSys200210.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Question on Discrete Renewal Processes ¤ A long-life light bulb has the following properties: probabilities of failure in years 1 through 3 of use are p 1 = 0.1, p 2 = 0.4, p 3 = 0.5 cost c b of block (scheduled) replacement is &pound;2 cost c f of failure replacement is &pound;3. Solve the discrete renewal equation for years n = 1 to n = 10 and...| find the optimal cost block replacement period. Optimal time to replace at n = Yes, that's right. F 1 = p 1 = 0.1, F 2 = p 1 + p 2 = 0.5, F 3 = p 1 + p 2 + p 3 = 1. Since by year 3 the bulb is certain to fail, F 3 = F 4 = F 5 etc. The recursive calculation of the renewal function gives: M 0 = 0, M 1 = F 1 = 0.1, M 2 = F 2 + p 1 M 1 = 0.5 + 0.1 0.1 = 0.51, M 3 = F 3 + p 1 M 2 + p 2 M 1 = 1 + 0.1 0.51 + 0.4 0.1 = 1.091, M 4 = F 4 + p 1 M 3 + p 2 M 2 + p 3 M 1 = 1 + 0.1 1.091 + 0.4 0.51 + 0.5 0.1 = 1.3631 etc. The costs per unit time are then: Year 1 = ( c b + c f M 0 ) = 3 0 + 2 = 2, Year 2 = ( c b + c f M 1 )/2 = (3 0.1 + 2)/2 = 1.15, Year 3 = ( c b + c f M 2 )/3 = (3 0.51 + 2)/3 = 1.1767, Year 4 = 1.3182, Year 5 = 1.2179, Year 6 = 1.2472, Year 7 = 1.2601, Year 8 = 1.2400, Year 9 = 1.2513 and Year 10 = 1.2504. Hence you can see that year 2 produces the minimum replacement cost figure. No, the optimal replacement time is 2 . The distribution functions are F 1 = p 1 = 0.1, F 2 = p 1 + p 2 = 0.5, F 3 = p 1 + p 2 + p 3 = 1. Since by year 3 the bulb is certain to fail, F 3 = F 4 = F 5 etc. The recursive calculation of the renewal function gives: M 0 = 0, M 1 = F 1 = 0.1, M 2 = F 2 + p 1 M 1 = 0.5 + 0.1 0.1 = 0.51, M 3 = F 3 + p 1 M 2 + p 2 M 1 = 1 + 0.1 0.51 + 0.4 0.1 = 1.091, M 4 = F 4 + p 1 M 3 + p 2 M 2 + p 3 M 1 = 1 + 0.1 1.091 + 0.4 0.51 + 0.5 0.1 = 1.3631 etc. The costs per unit time are then:> Year 1 = ( c b + c f M 0 ) = 3 0 + 2 = 2, Year 2 = ( c b + c f M 1 )/2 = (3 0.1 + 2)/2 = 1.15, Year 3 = ( c b + c f M 2 )/3 = (3 0.51 + 2)/3 = 1.1767, Year 4 = 1.3182, Year 5 = 1.2179, Year 6 = 1.2472, Year 7 = 1.2601, Year 8 = 1.2400, Year 9 = 1.2513 and Year 10 = 1.2504. Hence you can see that year 2 produces the minimum replacement cost figure.  |../RepairableSystems2Session/RepSys200220.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Alternating Renewal Processes ¤ Here for example a machine might be alternately working or being repaired after failure. The successive times to failure and repair times form a renewal process. Successive idle and busy periods of the server of a queue form an alternating renewal process. The main result we want to derive is the probability that...| the system is working at time ( t ), and its asymptotic value. Let the pdfs of time to failure and repair be f a ( t ), f b ( t ) respectively. The time from insertion of a component in a socket to its renewal (end of its repair period) has pdf. f ( t ) = f a ( u ) f b ( t - u ) d u , and these periods constitute an ordinary renewal process. In terms of Laplace transforms, = a b and ( s ) = ( s )/( s (1 - ( s ))). The probability satisfies the equation: ( t ) = S a ( t ) + v ( u ) S a ( t - u ) d u , where S a is the survival function of time to failure, and v is the renewal density. Recall that v ( t ) d t is the probability that any failure (1st, 2nd, 3rd, etc) occurs in time d t , as v ( t ) = . The equation for ( t ) says that the probability that the system is working at time t is the probability that it never failed, or that it last failed at some time u and has not failed since.  |../RepairableSystems2Session/RepSys200230.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Evaluating Pie( t ) ¤ We could use this equation to evaluate ( t ) numerically or analytically, after calculating v . Taking Laplace transforms: ( s ) = (1 - ( s ) a )(1 + ( s ))/ s where = a b /(1 - a b ), and so: We can find ( t ) by inverting the Laplace transform. Asymptotically, expanding a = 1 - a s , b = 1 - b s , ( s ) = a /( s ( a + b ))...| and so ( t ) a /( a + b ), which is clearly the long-run proportion of time that the system is 'up'.  |../RepairableSystems2Session/RepSys200240.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Other Types of Renewal Process ¤ Asynchronous Renewal Process In an asynchronous renewal process, where the first time to renewal has pdf. S / . In general, a modified renewal process has an arbitrary first time to renewal, with pdf. (say) f a , which is different from the distribution of the remaining times. The renewal equation is modified: M (...| t ) = F a ( t ) + M ( t - u ) f ( u ) d u . Such processes are sometimes used to model situations where a component is supposed to have a different time to failure after its first repair. General Alternating Renewal Process Alternating renewal processes generalise to the case where there are n possible states, not just two, and there is a probability P ji of jumping to j state from state i . Times spent in each state have pdfs f i ( t ). Reward-Renewal Process Renewal-reward processes are also sometimes useful. Here we may think of a system where failure or operation has an associated cost Y , where Y is a random variable that may be correlated with X . We may want to find the expected cost to time t .  |../RepairableSystems2Session/RepSys200250.htm",
    "Statistics for Maintenance Modelling: Renewal Processes|Summary ¤ A renewal process is best exemplified by the successive failures of a light bulb --- a series of failures where the interfailure period is a random variable from some distribution. An understanding of renewal processes is needed for simple replacement models, such as block replacement. The key concept is the renewal function : the...| expected number of renewals to some time t . This session has explained how to do several calculations with renewal functions, but the most practical is the numerical calculation of the renewal function using the integral equation. For those who like discrete time approximations, a maintenance problem example is given where the problem is solved in discrete time. It is probably best for study to at least read the introduction to this session, but not to bother too much about the details, and to then come back to it later if and when more detail is needed to solve particular maintenance problems.  |../RepairableSystems2Session/RepSys200260.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Introduction ¤ Before studying the content of this session, you should be familiar with sessions 1-11 in this module. In particular, you should be familiar with elementary algebra, calculus, probability and statistics, including parameter estimation and hypothesis testing. Recommended reading books related to this module are Bayesian Theory by...| Bernardo and Smith, Kendall's Advanced Theory of Statistics: Bayesian Inference by O'Hagan and Bayesian Statistics: an Introduction by Lee. Other relevant texts that might be of interest include Theory of Probability by Jeffreys, Statistical Inference: an Integrated Approach by Migon and Gamerman, Statistics: a Bayesian Perspective by Berry, Bayesian Statistics - Principles, Models and Applications by Press and Bayesian Data Analysi s by Gelman, Carlin, Stern and Rubin. The full details of all of these can be found in the References . This session introduces you to the theory and practice of Bayesian methods to improve the accuracy of analysis and inference when compared with the simpler na&iuml;ve approach and the common frequentist approach (such as the methods for parameter fitting discussed in the Model Fitting session). It discusses and demonstrates the advantages of the Bayesian approach and describes techniques for resolving the computational difficulties encountered. Having studied this module, you will able to apply the principles of Bayesian analysis to problems arising in reliability and maintenance applications.  |../BayesianApproachSession/Bayesian00005.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Objectives ¤ By the end of this session you will know: what Bayes' theorem is what a prior distribution is what a posterior distribution is what a predictive distribution is how to specify reference and subjective priors what Bayes factors are how to apply Bayesian approach to reliability data. | |../BayesianApproachSession/Bayesian00010.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Bayes' Theorem ¤ In 1763, the Reverend Thomas Bayes published some research on probability in Philosophical Transactions of the Royal Society . This contained an incredibly important statement of what we now refer to as Bayes' theorem . In its simplest form, Bayes' theorem states that for two events and A and B , the conditional probability B of...| given that A has occurred can be expressed as: where it is sometimes useful to evaluate the probability of A using the law of total probability : P ( A ) = P ( A B ) P ( B ) + P ( A B c ) P ( B c ) where B c is the complement of the event B , that is the event that B does not occur. A more extensive form of this theorem is presented on the Bayes' Theorem screen in the Basic Probability Concepts session.  |../BayesianApproachSession/Bayesian00020.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example Using The Rule of Total Probability and Bayes' Theorem ¤ An aircraft warning light comes on if the landing gear on either side is faulty. Defining events W = &quot;warning light comes on&quot; and L = &quot;landing gear faulty&quot;, suppose we know that P ( W L ) = 0.999, P ( W L c ) = 0.005 and P ( L ) = 0.004. Our aim is to calculate...| the probability that the landing gear is faulty if the warning light comeson i.e. P ( L W ). Using the law of total probability: P ( W ) = P ( W L ) P ( L ) + P ( W L c ) P ( L c ) = 0.999 0.004 + 0.005 0.996 = 0.008976 Bayes' theorem gives: P ( L W ) = P ( W L ) P ( L ) / P ( W ) = 0.999 0.004 / 0.008976 = 0.45 (to 2 decimal places) So most (55%) of these warning lights are false alarms!  |../BayesianApproachSession/Bayesian00030.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Applying Bayes' Theorem to Random Variables ¤ Only in the mid-twentieth century were the real benefits of Bayes' theorem appreciated though. Not only does it apply to probabilities, but also to random variables. For example, suppose X is a discrete random variable and Y is a continuous random variable. Then the conditional probability density...| function of Y given X can be determined using Bayes' theorem, if we know the marginal distributions of X and Y , and the conditional distribution of X given Y : This rule for &quot;transposing the conditionals&quot; has proven to be crucial in a variety of important applications, including industrial quality control, medical screening and diagnosis, and criminal trials.  |../BayesianApproachSession/Bayesian00040.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Inferring Unknown Model Parameters ¤ Suppose that a continuous random variable X has a probability distribution that depends on an unknown parameter . For example, X might represent the fire-breach time of a door in minutes and it might have an exponential distribution with unknown mean = 1/ . Non-Bayesian Methods A na&iuml;ve approach to...| statistical inference would simply replace by a good guess based on expert opinions. However, this is inherently inaccurate and can lead to poor decisions. A better method is the frequentist approach to inference, whereby we evaluate an estimate for the unknown parameter based on a set of observed data D , typically consisting of a random sample of actual fire-breach times x 1 , x 2 ,..., x n . Subsequent analyses generally invoke the approximation , which can again lead to poor decisions. Even more important than applying Bayes' theorem to random variables, we can apply Bayes' theorem to unknown model parameters. This is the foundation of the Bayesian approach to statistical inference and has had an enormous and profound impact on the subject over the last few decades.  |../BayesianApproachSession/Bayesian00050.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Applying Bayes' Theorem to Unknown Model Parameters with the Posterior PDF ¤ In contrast, the Bayesian approach does not involve any guesses or estimates of unknown parameters in the model. Rather, it uses Bayes' theorem to update our prior beliefs about in response to the observed data D : so that we can use our posterior beliefs about for any...| subsequent inference involving X . This function is the posterior probability density function . The price that we pay for obtaining exact answers and avoiding approximations in this way comes in two parts: the need to assume a prior distribution for the increase in algebraic complexity. This session shows how to resolve these issues.  |../BayesianApproachSession/Bayesian00060.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|An Example of Applying Bayes' Theorem to Unknown Model Parameters ¤ Return Suppose the unknown parameter represents the proportion of car batteries that fail within two years and our prior beliefs about can be expressed in terms of the probability density function: g ( ) = 2(1 - ); 0 &lt; &lt; 1. Suppose also that we observe three car batteries,...| one of which fails within two years and two of which do not. Then we can express the likelihood of these data using the binomial probability mass function: p ( D ) = 3 (1 - ) 2 which is the discrete equivalent to the probability density function f ( D ) referred to on the previous screen. As p ( D ), the discrete equivalent of f ( D ), does not depend on , an application of Bayes' theorem from the previous screen gives: g ( D ) p ( D ) g ( ) (1 - ) 3 and so our posterior beliefs about the unknown parameter can be expressed as a beta distribution D Be (2,4). We elaborate on this process further when we look at prior and posterior distributions.  |../BayesianApproachSession/Bayesian00065.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Prior and Posterior Distributions ¤ The previous screen derived an equation for updating a prior probability density function g ( ) for an unknown parameter , based on some observed data D to give a posterior probability density function g ( D ). The term f ( D ) is proportional to the likelihood function, (introduced on The Likelihood Function...| screen in Model Fitting session) when considering maximum likelihood estimation. If the data set D consists of a random sample of observations x 1 , x 2 ,..., x n of a continuous random variable X with probability density function f ( x ), then the likelihood function becomes: L ( ; D ) f ( X i ). As the term f ( D ) does not depend on , we can therefore write: g ( D ) L ( ; D ) g ( ) in other words, posterior is proportional to likelihood times prior . This is the fundamental rule for Bayesian inference.  |../BayesianApproachSession/Bayesian00070.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example of Using the Fundamental Rule ¤ On screen 8 , we considered an example of using Bayes' theorem for unknown model parameters, in the context of a binomial random variable with probability of success . This is an illustration of how the fundamental rule &quot; posterior is proportional to likelihood times prior &quot; can be applied. To...| clarify this demonstration, the likelihood function for that example takes the form: L ( ; D ) p ( X i ) = (1 - ) 2 where the probability mass function p ( X i ) corresponds with a Bernoulli distribution. Consequently, the posterior probability density function of given the data D has the form: g ( D ) L ( ; D ) g ( ) (1 - ) 3 which agrees with the result we obtained previously.  |../BayesianApproachSession/Bayesian00075.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Finding the Posterior Mode, Median or Mean ¤ Having evaluated a posterior distribution using this rule, we can evaluate the posterior mode such that: g ( D ) g ( D ) by solving the equation: L ( ; D ) g ( ) = 0. The Constant of Proportionality However, to find the median or mean, and to use this posterior density to make any further inference, we...| need to determine the constant of proportionality in the fundamental Bayesian rule on the previous screen. In standard situations, we can recognise the functional form of L ( ; D ) g ( ) and hence quote published work on probability distributions to determine this constant of proportionality and so derive g ( D ) explicitly. In non-standard situations, we determine this constant of proportionality using numerical quadrature or simulation, both of which we discuss later.  |../BayesianApproachSession/Bayesian00080.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Reference Prior Distributions ¤ There are two main types of prior distribution, as represented by reference priors and subjective priors . We use a reference prior if no prior information is available. This offers a default Bayesian analysis that is not dependent upon any personal prior knowledge. The simplest reference prior is proposed by the...| Bayes-Laplace postulate and simply recommends the use of a uniform or locally-uniform prior g ( ) 1 for all in the region of support . However, different parameterisations can lead to different inferences with this approach. Jeffreys' Invariant Prior To avoid this inconsistency, the standard univariate reference prior that analysts now adopt is Jeffreys' invariant prior , defined by: g ( ) ; where I ( ) = - is Fisher's expected information. An extension exists for the case of a parameter vector , though we usually assume the components of are independent, so g ( ) is just the product of the univariate invariant priors.  |../BayesianApproachSession/Bayesian00090.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example using Jeffreys' Invariant Prior ¤ For the exponential distribution with parameter , f ( x ) = ; x &gt; 0 so I ( ) = = = and the invariant prior is: g ( ) = ; &gt; 0 which is improper (i.e. the distribution doesn't integrate to 1), though this is unimportant as the corresponding posterior distribution is usually proper. Books on Bayesian...| methods present tables of invariant prior and posterior distributions for common models.  |../BayesianApproachSession/Bayesian00100.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Subjective Prior Distributions ¤ Subjective prior distributions should be used if prior information is available. They represent the best available knowledge about unknown parameters and can be specified using smoothed histograms, relative likelihoods or parametric families. The first two are arbitrary and computationally awkward so we now...| investigate the last of these. A family of priors C is closed under sampling if: g ( ) C g ( D ) C , so that the posterior density has the same functional form as the prior density. This property is particularly appealing, as our prior knowledge can be regarded as posterior to some previous information. Again, we tend to suppose that components in multi-parameter problems are independent, so that their joint prior density is the product of corresponding univariate marginal priors. Natural Conjugate Priors For sampling distributions f ( x ) that belong to the exponential family, which includes Bernoulli, binomial, negative binomial, Poisson, exponential, normal and gamma models, such closed priors exist and are called natural conjugate priors. Books on Bayesian methods present tables of the conjugate prior and posterior distributions for common models. However, many applications that arise in reliability and maintenance are not amenable to such simple analyses. For example, the Weibull distribution is not a member of the exponential family. As a result of this, the constant of proportionality in the expression: g ( D ) L ( ; D ) g ( ) can rarely be evaluated analytically and numerical computation is usually required.  |../BayesianApproachSession/Bayesian00110.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Strategy for Using Priors ¤ Subjective Priors As some prior knowledge is always available, a subjective prior should be used whenever possible. To avoid the inconsistency of using natural conjugate priors when they exist and other forms of subjective prior when they do not, the following recommendations provide a simple, consistent and...| comprehensive strategy for general use: infinite range - &lt; &lt; , use a normal prior distribution for semi-infinite range 0 &lt; &lt; , use a gamma prior distribution for finite range 0 &lt; &lt; 1, use a beta prior distribution for . These priors match with the natural conjugate priors for simple models and extend to deal with more complicated models. Reference Priors However, reference priors are useful in these circumstances: for an objective analysis with no specific personal inputs for comparison with similar analyses by other investigators as baselines to assess the sensitivity of results to choice of prior.  |../BayesianApproachSession/Bayesian00120.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Prior - Posterior Analysis for an Exponential Distribution ¤ Return For the exponential distribution with parameter &gt; 0, the sampling distribution is f ( x ) = ; x &gt; 0 Prior Density We use a gamma prior density for , of the form: g ( ) = a - 1 ; &gt; 0. Here, a and b are specified constants known as hyperparameters and chosen so that g ( )...| represents genuine prior knowledge about . Posterior Density If we observe a random sample of lifetimes D = x 1 , x 2 ,..., x n , the posterior density for also has a gamma form: g ( D ) = a + n - 1 ; &gt; 0. This now allows us to make accurate inference and decisions involving X and .  |../BayesianApproachSession/Bayesian00130.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Predictive Distributions ¤ One of the most important and useful features of the Bayesian approach arises when we wish to make predictions about future values of the random variable X where f ( x ) is specified. Prior Predictive Probability Density Function If is unknown, the prior predictive probability density function of X is: f ( x ) = f ( x )...| g ( ) d Posterior Predictive Probability Density Function If data D are observed, the posterior predictive probability density function of X is: f ( x D ) = f ( x ) g ( D ) d f ( x ) L ( ; D ) g ( ) d In contrast, a frequentist approach either uses the approximation f ( x D ) f ( x ) or gives a point prediction = with a prediction interval if available.  |../BayesianApproachSession/Bayesian00140.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Prior Predictive Density for an Exponential Model ¤ For the exponential sampling distribution with Jeffreys' invariant prior, the prior predictive density is given by: f ( x ) 1/ d 1/ x ; x &gt; 0 which is improper. However, this does provide information about the relative likelihoods for different values of X . For example, the ratio of...| probabilities that X lies in the intervals (5, 10) and (10, 20) is given by: = = = 1 So X is equally likely to lie in these two intervals.  |../BayesianApproachSession/Bayesian00150.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Posterior Predictive Density for an Exponential Model ¤ Even if we subsequently observe a random sample of lifetimes D = x 1 , x 2 ,..., x n the posterior predictive density: f ( x D ) n - 1 d ( click here for an explanation ) = ; x &gt; 0 ( click here for an explanation ) is still improper, though we can evaluate relative likelihoods as we did...| for the prior predictive density. In contrast, a frequentist approach would merely generate the approximation X D Ex (1/ ) and could do no better than guess a value for X before observing any data. The exponential distribution of 1/the sample mean  |../BayesianApproachSession/Bayesian00160.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example Using Prior and Posterior Predictive Densities with a Gamma Prior ¤ For the exponential sampling distribution with a gamma prior, the prior predictive density is given by: f ( x ) = a - 1 d = ; x &gt; 0 which corresponds with a special form of gamma-gamma distribution. If we subsequently observe a random sample of lifetimes D = x 1 , x 2...| ,..., x n the posterior predictive density is given by: f ( x D ) a + n - 1 d = which again corresponds with a gamma-gamma distribution. As before, a frequentist approach would merely yield the approximation X D Ex (1/ ).  |../BayesianApproachSession/Bayesian00170.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Parameterising Subjective Prior Distributions ¤ We now consider the difficult problem of assigning values to the hyperparameters of subjective prior distributions. Suppose we have a model for a continuous random variable X with probability density function f ( x ) and corresponding subjective prior density g ( ). Typically, this prior consists of...| two unknown hyperparameters, which we can call a and b . We set fixed values for these hyperparameters, to reflect our prior knowledge about . For two hyperparameters, we need two distinct pieces of information such as the upper and lower tertiles (33 and 66 percentiles) of . Unfortunately, parameters are not observable and we cannot make accurate statements about them directly. However, we can do this indirectly by considering the prior predictive distribution introduced on screen 16 , which is also a function of a and b . We saw on screen 20 that the prior predictive density for the exponential sampling distribution (perhaps for a lifetime X ) with a gamma prior is given by: f ( x ) = ; x &gt; 0 Hence the prior predictive cumulative distribution function is: F ( x ) = d x = 1 - ; x &gt; 0 If an expert specifies tertiles L and U , such that F x ( L ) = and F x ( U ) = , then we can solve these two nonlinear simultaneous equations numerically for a and b . These can then be substituted into our prior density f ( x ).  |../BayesianApproachSession/Bayesian00190.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example of Parameterising Subjective Prior Distributions ¤ In the above example, the exponentially distributed random variable X might represent the lifetime of a new type of light bulb in operating hours. Suppose that, based on subjective knowledge of similar light bulbs, we believe that one third of the new type will fail within 2,500 operating...| hours and one third will last for at least 7,500 operating hours. This implies that we believe that the remaining third will fail between these two values. Then L = 2,500 and U = 7,500, so we need two solve the nonlinear equations: and numerically for a and b . There are many algorithms for solving these simultaneous equations and several computer packages that contain these algorithms. Mathcad gives the values a = 3.5240 and b = 20,502, so the prior distribution for the exponential parameter is specified completely as Ga (3.5240, 20,502).  |../BayesianApproachSession/Bayesian00195.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Bayes Factors ¤ Suppose we have two competing hypotheses H 0 and H 1 that the sampling distribution of the random variable X is given by the probability density functions f 0 ( x 0 ) and f 1 ( x 1 ) respectively. For instance, these might be Weibull and gamma lifetime distributions. We wish to choose between them, based on prior knowledge about...| the unknown parameters 0 and 1 and observed data such as the random sample D = x 1 , x 2 ,..., x n . The prior odds are defined by: and represent our prior beliefs about the relative likelihoods of the two hypotheses. The posterior odds : include the observed data and so represent our updated, posterior beliefs about the relative likelihoods of the two hypotheses. The Bayes factor is then defined as the ratio of the posterior odds to the prior odds, and using Bayes' theorem we have: BF = = So the posterior odds equals Bayes factor times prior odds , and measures the weight of evidence objectively in terms of the hypotheses. People with different prior odds for H 0 and H 1 can calculate their own posterior odds. The derivation of the R.H.S is beyond the scope of this session.  |../BayesianApproachSession/Bayesian00200.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example Using Bayes Factors ¤ Suppose H 0 represents the event that a bearing is critically worn and H 1 is its complement, that the bearing is not critically worn. Some evidence D is available to support H 0 , perhaps based on acoustic or vibration measurements. If the evidence is a hundred more times likely to arise from a bearing that is...| critically worn than from a bearing that is not, it can be summarised by means of the Bayes factor: = 100 Then different experts can assign their own prior odds to these events, such as: = 1/2 and then calculate their own corresponding posterior odds, in this case: = 100 1/2 = 50 so this particular expert believes that the bearing is fifty times more likely to be critically worn than not, given the available evidence.  |../BayesianApproachSession/Bayesian00210.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Bayesian Decision Theory ¤ Consider a set of possible decisions d with associated utility function u ( d , ), which depends on an unknown parameter . The best decision is that which maximizes the expected utility : u ( d , ) = u ( d , ) g ( ) d with respect to the prior g ( ). Alternatively, we can minimize the expected loss : l ( d , ) = l ( d ,...| ) g ( ) d for some loss function l ( ). If we observe exchangeable data D = x 1 , x 2 ,..., x n from the sampling density f ( x ), the criterion to maximize (minimize) is the posterior expected utility (loss) defined by: u ( d , ) = u ( d , ) g ( D ) d where g ( D ) L ( ; D ) g ( ) is the posterior density.  |../BayesianApproachSession/Bayesian00220.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Example Using Bayesian Decision Theory ¤ Which of two alarm systems should we buy if they cost c i units and fail at times X i where: X i i Ex ( i ) for i = 1, 2 respectively? Assuming replacements on failure for an infinite horizon, the elementary renewal theorem gives the expected cost per unit time for action i as the loss function: l ( i , i )...| = c i i. Then l ( i , i ) = c i E ( i ) and we choose system i which minimises this expected loss, where E ( i ) is the prior mean.  |../BayesianApproachSession/Bayesian00230.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Numerical Computation ¤ In Bayesian analysis, we often encounter intractable integrals such as: posterior densities g ( D ) L ( ; D ) g ( ) d posterior predictive densities f ( x D ) = f ( x ) g ( ) d posterior expected utilities = u ( d , ) = u ( d , ) g ( D ) d . To find out what you can do in this situation, click on the following methods:...| Monte Carlo simulation The Inversion Method The Rejection Method Monte Carlo simulation can be used to approximate any integral of this form by generating many pseudo-random numbers 1 , 2 ,..., n from the prior or posterior density in the integrand, ( h ( ) below), and evaluating the unbiased estimator: s ( ) h ( ) d 1/ n s ( i ). This is easy but computationally expensive. The inversion method converts readily available U (0, 1) pseudo-random numbers into values using the inverse cumulative distribution function: i H -1 ( u i ). The rejection method is more versatile but less efficient. In its simplest form, this method involves enclosing the target probability density function within a rectangle and generating a pair of U (0, 1) pseudo-random numbers to identify a point within this rectangle. If the probability density at the same x value as that chosen exceeds the corresponding y value, we accept this x value as a pseudo-random observation of U (0, 1). Otherwise, we reject this value and select another pair of pseudo-random numbers.  |../BayesianApproachSession/Bayesian00240.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Generating Pseudo-Random Numbers From Joint Distributions ¤ To generate pseudo-random numbers from joint distributions, work with products of univariate distributions instead, such as h ( 1 , 2 , 3 ) = h ( 1 ) h ( 2 1 ) h ( 3 1 , 2 ): generate 1 from h ( 1 ) generate 2 from h ( 2 1 ) generate 3 from h ( 3 1 , 2 ). Sometimes, we need to generate...| pseudo-random numbers from joint distributions when only the conditional distributions are available or convenient. This requires a Markov chain Monte Carlo (MCMC) method , and when this course material was created, WinBUGS software is freely available from the winbugs website for performing MCMC calculations. The most common MCMC method is Gibbs sampling , such as h ( 1 , 2 , 3 ) with initial guesses 1 (0) , 2 (0) and 3 (0) : generate 1 (1) from h ( 1 2 (0) , 3 (0) ) generate 2 (1) from h ( 2 1 (1) , 3 (0) ) generate 3 (1) from h ( 3 1 (1) , 2 (1) ) generate 1 (2) from h ( 1 2 (1) , 3 (1) ) and so on. The pseudo-random numbers are 1 = 1 ( m ) , 2 = 2 ( m ) and 3 = 3 ( m ) for some large value m .  |../BayesianApproachSession/Bayesian00250.htm",
    "Statistics for Maintenance Modelling: The Bayesian Approach|Summary ¤ Bayesian inference represents a methodology for mathematical modelling and statistical analysis of random variables and unknown parameters. It provides an excellent alternative to the frequentist approach which gained immense popularity throughout the twentieth century. Whereas the frequentist approach is based upon estimation,...| hypothesis testing and approximate inference, the Bayesian approach is based upon probability theory and provides complete solutions to practical problems. The Bayesian approach is regarded as infinitely superior by its advocates. However, it does depend upon the existence and specification of subjective probability to represent individual beliefs, whereas the frequentist approach is almost completely objective. It also involves more computational complexity, though this is not much of a hindrance with modern computing power. We have covered: the concepts beginning with Bayes' theorem, which is of great importance in its own right the construction of prior and posterior distributions the role of predictive distributions techniques for setting the hyperparameters of prior distributions the ability of the Bayesian approach to improve decision analysis - particularly in relation to practical problems in reliability and maintenance, as described when we looked at Bayes factors and Bayesian decision theory techniques for resolving the computational difficulties mentioned above.  |../BayesianApproachSession/Bayesian00260.htm"
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