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This is a problem that arose from reading "Introduction to Probability Models, Tenth Edition" Chapter 5 written by Ross M. Sheldon. Specifically question 16.

The situation is as follows:

There are three jobs that need to be processed, with the processing time of job $i$ being exponential with rate $u_{i}$.There are two processors are available, so job processing of the jobs can immediately start, with process on the final job to start when one of the inital ones have finished.

Let $T$ be the total time that process takes, find $E[T]$.

I will leave my attempt at the problem below - It does not match up with the provided solution.

My question is how do you do this problem.

Attempt:

$$ \textrm{Let } X_{i}\sim Exp(u_{i}), i\in \left \{ 1,2,3 \right \} \\ \textrm{Let } I = \textrm{ Time until first departure} \\ \textrm{Let A } = \textrm{ Additional time required} \\ \textrm{However, we see that T = I + A. So E[T] = E[I+A] = E[I] + E[A]} \\ $$

$$ E[I] = \frac{1}{u_{1}+u_{2}} \textrm{ (The time until the first completion is exponential with rate } u_{1}+u_{2} \textrm{)} \\ $$

$$ E[A] = E[A|X_{1}<X_{2}]P(X_{1}<X_{2})+E[A|X_{1}>X_{2}]P(X_{1}>X_{2}) \\ = E[max(X_{2},X_{3})]\frac{u_{1}}{u_{1}+u_{2}} + E[max(X_{1},X_{3})]\frac{u_{2}}{u_{1}+u_{2}} \\ = (\frac{1}{u_{2}}+\frac{1}{u_{3}}-\frac{1}{u_{2}+u_{3}})\frac{u_{1}}{u_{1}+u_{2}} + (\frac{1}{u_{1}}+\frac{1}{u_{3}}-\frac{1}{u_{1}+u_{3}})\frac{u_{2}}{u_{1}+u_{2}} \\ $$

$$\textrm{ Hence we see that E[T] is just the sum of the above two expressions} $$

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Giacomo Nassif Jack
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  • How do you obtain $$E[max[X_2,X_3]] = \frac{1}{\mu_2} + \frac{1}{\mu_3} - \frac{1}{\mu_2+\mu_3}$$ ? – Michael Jul 04 '19 at 07:23
  • The expectation of the maximum of two iid exponential random variables is shown well in this post here: https://math.stackexchange.com/questions/2470451/expectation-of-the-maximum-of-two-exponential-random-variables – Giacomo Nassif Jack Jul 04 '19 at 07:51
  • Oh, it turns out your expression for $E[max[X_2,X_3]]$ is correct because it is the same as $$E[first] + E[next] = \underbrace{\frac{1}{\mu_2 + \mu_3}}{E[first]} + \underbrace{\left[\frac{\mu_2}{\mu_2 + \mu_3}\right]\frac{1}{\mu_3} + \left[\frac{\mu_3}{\mu_2+\mu_3}\right]\frac{1}{\mu_2}}{E[next]}$$ Your previous calculations were conceptually similar to my version of the formula and so I expected you to finish with my formula above, it was unusual you jumped to that other expression without justification. – Michael Jul 04 '19 at 14:57
  • I was actually playing around with both formulas yesterday. I started that way and then tried to link it to the formula I gave above. This formula can be proven through the typical distribution of the maximum method. However, a much more interesting way is using the algebraic identity max(x,y) = x + y - min(x,y) – Giacomo Nassif Jack Jul 05 '19 at 00:45
  • Yes, $\max[X,Y]=X+Y-\min[X,Y]$ seems nicest for that problem, but it does not seem to help for the case of a max with three or more, whereas $P[\max[X_1, ...., X_n]\leq x] = P[X_1\leq x, ..., X_n\leq x]$ works well. – Michael Jul 05 '19 at 04:31
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    I think another interesting approach for the max[X,Y,Z] is: max[X,Y,Z] = X + Y+ Z - min[X,Y] - min[X,Z] - min[Y,Z] + min[X,Y,Z]. Seems to follow something that resembles the inclusion-exclusion principle. – Giacomo Nassif Jack Jul 07 '19 at 01:26

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