Question

We have always assumed that components in a system fail independently. In life, this might not be the case. What effect does positive correlation between the lifetimes of two parallel components have on the system lifetime?

Assume that the two parallel components have lifetimes that are both described by normal distributions with means of 2,000 hours and standard deviations of 150 hours. Determine the distribution of system lifetime when the two component lifetimes are independent and again when they have a correlation of +0.8.

Figure out how to generate bivariate normal data in Minitab and create two worksheets: one with independent component lifetimes and the consequent system lifetimes and one with correlated lifetimes and the resulting system lifetimes. Use 3,000 replications in each worksheet. Include a screen shot showing both worksheets tiled in one data window

Answer 1

a) The rate for the first is lambda and for the second is mu, so the probability that the first occurs first is lambda/(lambda+mu)

b) The probability that the second occurs first the first 3 times, given the memoryless nature of the process, is (mu/(lambda+mu))^3

c) Each interval, the expected time until the next event is 1/(lambda + mu)

The probability that it is the first component failing in any interval is lambda/(lambda+mu)

In the n'th interval, either component failing leads to failure.

Thus, the probability of failure in the i'th interval, i < n, is, (as for i-1 intervals, the second component had to fail), (mu/(lambda+mu))^i-1 * lambda/(lambda+mu)

Thus, we have the sum i=1 to n-1 i/(lambda+mu)*(mu/(lambda+mu))^i-1 * lambda/(lambda+mu) +n*(mu/(lambda+mu))^n-1/(lambda+mu)

To calculate the sum i=1 to n-1 i/(lambda+mu)*(mu/(lambda+mu))^i-1 * lambda/(lambda+mu), calculate this as

sum i=1 to inf i/(lambda+mu)*(mu/(lambda+mu))^i-1 * lambda/(lambda+mu) -

sum i=n to inf i/(lambda+mu)*(mu/(lambda+mu))^i-1 * lambda/(lambda+mu)

The first sum is (consider this as the mean of the geometric distribution with parameter lambda/(lambda+mu) * 1/(lambda+mu)

1/(lambda/(lambda+mu)) * 1/(lambda+mu) = 1/lambda

sum i=n to inf i/(lambda+mu)*(mu/(lambda+mu))^i-1 * lambda/(lambda+mu)

Note that this is the geometric progression after n-1 intervals, so the expected value of the geometric, conditioned on reaching this, is n-1 + 1/(lambda/(lambda+mu))

Thus, we multiply this times (mu/(lambda+mu))^n-1*1/(lambda+mu)

Thus, the expected time is 1/lambda -

(n-1 + 1/(lambda/(lambda+mu)))*(mu/(lambda+mu))^n-1*1/(lambda+mu) +

n*(mu/(lambda+mu))^n-1/(lambda+mu) = (noting that the n terms cancel each other and remembering that we are subtracting the one term)

1/lambda + (mu/(lambda+mu))^n-1*1/(lambda+mu) -1/lambda*(mu/(lambda+mu))^n-1 or

1/lambda + (mu/(lambda+mu))^n-1 * (1/(lambda+mu) -1/lambda))

As a check, note that, when n = 1, this becomes

1/lambda + 1/(lambda+mu) - 1/lambda = 1/(lambda+mu), which is this answer