Expected wait for the first of six services

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Six independent service counters each finish a job in an exponential time with a mean of 4 minutes. What is the expected time until the very first counter finishes its job?

Accepted Answer

The first finish is the minimum of the six service times. The minimum of independent exponentials is exponential with the rates added together. Each counter has rate \lambda = \frac{1}{4} per minute, so the first finish has rate 6\lambda = \frac{6}{4} and mean E[	ext{first finish}] = \frac{1}{6\lambda} = \frac{4}{6} = \frac{2}{3} \approx 0.667 	ext{ minutes}. Six counters running together deliver the first completion six times faster than a single counter would. Takeaway. More parallel exponential servers shrink the first-completion time in direct proportion: mean \mu/n for n servers.

Expected wait for the first of six services

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