Question 6 of 10: Finite-Source Queueing — Overhead Crane Servicing Ten Machines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2019 — 17-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 175 marks across 10 questions and only 100 marks are required, so a candidate would normally answer a subset — all ten are solved below for completeness.
Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear programming & the simplex method (ch. 3–4), duality & sensitivity analysis (ch. 6), dynamic programming (ch. 11), network optimization & CPM/PERT project crashing (ch. 9–10), queueing theory incl. finite-source (machine-repair) models (ch. 17), decision analysis & the value of information (ch. 15–16), Markov chains (ch. 16), Monte Carlo simulation (ch. 20). Nahmias, Production and Operations Analysis — deterministic EOQ inventory models with and without planned shortages.
Finite-source (machine-repairman) queue: state $n$ = number of machines currently needing the crane (waiting or being served). Each of the $N-n$ still-running machines can call at rate $\lambda$; the single crane serves at rate $\mu$.
Given.$N=10$ machines; each machine's run ("up") time is exponential, mean 30 min $\Rightarrow \lambda=1/30$/min; crane service time is exponential, mean 10 min $\Rightarrow \mu=1/10$/min; single server (one crane), finite calling population = the 10 machines.
Find. (a) $P_0$, the fraction of time the crane is idle. (b) $L_q$, the expected number of machines waiting for the crane (i.e. down but not yet being serviced).
Approach. This is the finite-source (machine-repairman) single-server queueing model, not the ordinary infinite-source $M/M/1$ model — the "arrival rate" shrinks as more machines are already down, since only the $N-n$ still-running machines can call.
State probabilities. With $r=\lambda/\mu=(1/30)/(1/10)=1/3$ and $N=10$:
$$P_0=\left[\sum_{n=0}^{N}\frac{N!}{(N-n)!}\,r^{\,n}\right]^{-1},\qquad P_n=\frac{N!}{(N-n)!}\,r^{\,n}P_0\ (n=1,\dots,N)$$
Evaluating the sum numerically (10 terms growing from $n!/{(10-n)!}(1/3)^n$):
$$\boxed{P_0\approx 0.00081\ (0.081\%)}$$
— the crane is idle barely a tenth of one percent of the time; with 10 machines each able to call at $\lambda$ and only one crane at $\mu=3\lambda$, the system is heavily loaded and the crane is very rarely caught up.
Expected number in system.$L=\sum_{n=0}^{N}nP_n$, evaluated from the same distribution:
$$\boxed{L\approx 7.00\text{ machines}}$$
(out of 10, on average 7 are down — either waiting or being serviced — at any instant).
(b) Expected number waiting (queue only, excluding the one in service). Since the crane is busy whenever any machine is down, the fraction of time it is BUSY is $1-P_0$, and that busy crane accounts for exactly one of the $L$ machines in the system on average:
$$L_q=L-(1-P_0)=7.00-(1-0.00081)$$
$$\boxed{L_q\approx 6.00\text{ machines waiting}}$$
