Question 5 of 10: Finite-Source Queueing — Machine Repair Model
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 17-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 170 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 & sensitivity analysis/duality (ch. 3–4/6), integer programming & branch and bound (ch. 12), queueing theory (ch. 17), decision analysis (ch. 15), computer simulation (ch. 20). Nahmias, Production and Operations Analysis — single-period (newsvendor) and multi-period (dynamic lot-sizing / Wagner–Whitin) inventory models.
Question 5: Finite-Source Queueing — Machine Repair Model (20 marks)
Given.$N=4$ machines, $c=2$ mechanics (servers); running (up) time $\sim\text{Exp}$, mean 6 days $\Rightarrow \lambda=1/6$/day per running machine; repair time $\sim\text{Exp}$, mean 1 day $\Rightarrow \mu=1$/day per busy mechanic.
Find. (a) $P_0$, the steady-state probability all 4 machines are running (none down). (b) $L_q$, the expected number of machines waiting for a mechanic (down but not yet being repaired).
Approach. This is the finite-source (machine-repairman) queueing model, not the ordinary infinite-source M/M/c: the "arrival rate" of breakdowns shrinks as more machines are already down, since only running machines can fail. Build the birth–death balance equations state-by-state and normalize.
Set up the birth–death rates. Let state $n$ = number of machines down (waiting or in repair). Breakdown ("birth") rate from state $n$: $(N-n)\lambda$ (only the $N-n$ still-running machines can fail). Repair ("death") rate from state $n$: $\min(n,c)\,\mu$ (at most $c=2$ mechanics can work at once).
Apply detailed balance, $P_n(N-n)\lambda=P_{n+1}\min(n+1,c)\mu$, to get each $P_n$ relative to $P_0$:
$$\frac{P_1}{P_0}=\frac{4\lambda}{\mu}=\frac{4}{6}=\frac23,\quad \frac{P_2}{P_1}=\frac{3\lambda}{2\mu}=\frac{3/6}{2}=\frac14,\quad \frac{P_3}{P_2}=\frac{2\lambda}{2\mu}=\frac{2/6}{2}=\frac16,\quad \frac{P_4}{P_3}=\frac{1\lambda}{2\mu}=\frac{1/6}{2}=\frac1{12}$$
$$P_0:P_1:P_2:P_3:P_4 = 1:\tfrac23:\tfrac16:\tfrac1{36}:\tfrac1{432} = 432:288:72:12:1\ \ (\text{common denominator }432)$$
Normalize so all five probabilities sum to 1 (total $432+288+72+12+1=805$):
$$\boxed{P_0=\frac{432}{805}=0.5366,\quad P_1=\frac{288}{805}=0.3578,\quad P_2=\frac{72}{805}=0.0894,\quad P_3=\frac{12}{805}=0.0149,\quad P_4=\frac{1}{805}=0.0012}$$
(Cross-checked against the standard finite-source formula $P_n=\binom{N}{n}\rho^nP_0$ for $n\le c$ and $P_n=\binom{N}{n}\frac{n!}{c!\,c^{n-c}}\rho^nP_0$ for $n>c$, with $\rho=\lambda/\mu=1/6$ — identical result.)
(b) Machines waiting (queued, not yet in repair) occur only when $n>c=2$, i.e. states 3 and 4, where $n-c$ machines queue while the other 2 are being repaired:
$$L_q=\sum_{n=3}^{4}(n-2)P_n=(1)P_3+(2)P_4=\frac{12}{805}+\frac{2}{805}=\boxed{\frac{14}{805}=\frac{2}{115}=0.0174\text{ machine}}$$
