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:

g1(t) = (t)

and that of time to k failures is the distribution of the sum of 2k exponential random variables, i.e.

gk(t) = (t)2k - 1 /(2k - 1)!

Hence:

v(t) = fk(t) = (t)2k - 1 /(2k - 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)dt = t/2 - (1 - )/4