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OPS.10.1:5 - Archetypal Grounding

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.

ResultAB
Mean service, hours22.2
Offered-load ratio2/311/15
Mean queue wait, hours6.53.025
Mean arrival-to-completion time, hours8.55.225

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.

OPS.10.1:5.2 - Four hours on average, with a missed-deadline risk

Two independent jobs are ready at time zero. One server processes them in fixed order without interruption. Each service takes one or three hours with probability one half, independently of the other service. Both jobs must be complete by hour four.

Service durations, hoursLast completion, hourProbabilityBoth complete by four?
1, 121/4Yes
1, 341/4Yes
3, 141/4Yes
3, 361/4No

Mean last completion is four hours; the probability of meeting the deadline is three quarters. Replacing both durations by their mean of two hours would produce a single on-time schedule and discard that risk.

An additional independent server capable of the same work would allow both to start at zero and finish by three in every listed case. That is a conditional alternative; obtaining it is another operating action. Without it, a deadline of six covers all cases in this bounded model.

Now keep both marginal service distributions but make their durations equal, perhaps because a shared job condition affects both. The only outcomes are (1,1) and (3,3), each with probability one half. Mean last completion remains four; deadline probability falls to one half. The changed dependence reopens the probability result without changing the two means.

OPS.10.1:5.3 - An aggregate curve and a completed job

A controller represents remaining workload q in job-equivalents, starting at one with no new arrivals. Its assumed output rate is k*q, with k = 1 per hour. The balance gives q(t) = exp(-t): after one hour, about 0.368 job-equivalents remain. This model can support an aggregate regulation question where that outflow law fits.

A different operating account says one indivisible job takes exactly one uninterrupted hour on the available resource. Its completion event occurs at hour one. A constant-rate fluid balance, stopped at zero, also gives q(t) = max(1-t,0) in job-equivalents, but its intermediate fractions do not make the actual job partially delivered.

Even the exponential curve permits another interpretation under different premises. For one exponentially distributed service duration with mean one hour, it is the expected number of unfinished jobs; completion by one hour then has probability about 0.632. It is not a deterministic promise.

The practitioner chooses the account by the receiving question and the operating service law, not by whether the display uses a curve or discrete events. For the fixed one-hour deadline, use the completion event. For aggregate feedback, establish the outflow relation and how a commanded rate change can be realized. A new capacity setting without a corresponding operating mechanism leaves the proposed intervention unsupported.