Time to finish eight jobs (sum of exponentials)

Question

A server processes jobs one at a time. Each job's processing time is exponential with mean 0.5 seconds (rate $\lambda = 2$ per second), and the jobs are independent. Let $T$ be the time to finish 8 jobs.

What are the mean and variance of $T$ ?

Accepted Answer

A sum of n independent exponentials with the same rate is called an Erlang (a Gamma with integer shape). You do not need its full density to get the moments, because means and variances of independent pieces just add: - Mean = n 	imes (1/\lambda) = n/\lambda. - Variance = n 	imes (1/\lambda^2) = n/\lambda^2. Each job has mean 1/\lambda = 0.5 s and variance 1/\lambda^2 = 0.25. For 8 jobs: \mathbb{E}[T] = 8 	imes 0.5 = 4 	ext{ s}, \qquad \operatorname{Var}(T) = 8 	imes 0.25 = 2. Standard deviation \sqrt{2} \approx 1.41 s. Notice the total is less bursty in relative terms than a single job:…

Time to finish eight jobs (sum of exponentials)

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