OPS.10.1:5.1 - A slower machine gives a better mean, within its capacity

Jobs arrive as a Poisson process at one job per three hours. A continuously available first-come machine serves one job at a time; its service durations are independent across jobs and of arrivals. Service includes all job-specific recovery. There is no other setup, loss, resource or return. These are constructed alternatives, not fitted claims about two products.

Machine A takes one hour with probability 0.9 and eleven hours with probability 0.1. Its mean service is two hours, its variance is nine squared hours and cs² = 2.25. Machine B always takes 2.2 hours, so its service variance is zero. Poisson arrivals give ca² = 1.

For example, A’s mean wait is ((1+2.25)/2) * ((2/3)/(1/3)) * 2 = 6.5. The special-case mean relation applies to these stated service laws. B has the larger mean service duration and higher load ratio, yet the smaller mean residence. Its absence of service variation changes the queue consequence.

If the decision needs a mean residence below six hours, B meets that modeled criterion and A does not. Cost and actual availability still affect the operating choice. No percentile or empirical improvement is established.

Change the arrival rate to 0.48 jobs per hour. A’s mean service rate is 0.5 and B’s is about 0.455. A has load ratio 0.96, while B has 1.056. The former comparison cannot justify B for that continuing arrival regime. Return the increased demand to capacity or admission instead of inserting a ratio above one into the steady-mean formula.