Question 2 of 10: Finite-Source Queueing — Crane Servicing Ten Machines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 150 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/integer programming, network optimization (CPM), dynamic programming, decision analysis, Markov chains and queueing theory; Nahmias, Production and Operations Analysis (7th ed.) — EOQ and inventory-control models.
Given. $N=10$ machines; each running machine calls for service at rate $\lambda = 1/30$ per minute (exponential run time, mean 30 min); the crane (single server) serves at rate $\mu = 1/10$ per minute (exponential service, mean 10 min).
Find. (a) Fraction of time the crane is idle, $P_0$; (b) expected number of machines waiting (queued, not yet being served), $L_q$.
Approach. This is the finite-source (machine-repairman) queueing model — a single repairman (crane), a finite population of $N=10$ potential callers, each generating calls only while it is not already down. Use the finite-source birth–death steady-state distribution rather than the infinite-source M/M/1 formulas.
Steady-state distribution. With $r=\lambda/\mu$ and $n$ = number of machines currently down (in queue or in service), the finite-source result is
$$P_n = \binom{N}{n} r^{\,n} P_0,\qquad n=0,\dots,N,\qquad P_0=\left[\sum_{n=0}^{N}\binom{N}{n}r^{\,n}\right]^{-1}.$$
Here $r=\lambda/\mu=(1/30)/(1/10)=1/3$.
Part (a) — crane idle fraction. The crane is idle exactly when no machine is down, i.e. state $n=0$, so the idle fraction is $P_0$ itself. Summing $\binom{10}{n}(1/3)^n$ for $n=0,\dots,10$ (largest terms near $n=3$–$4$ since $r=1/3$) gives
$$P_0=\dfrac{59{,}049}{1{,}048{,}576} = \boxed{0.0563\ (5.63\%)}.$$
The crane is busy the great majority of the time — with ten machines each able to call at $1/30$/min against a $1/10$/min service rate, demand routinely outpaces the single crane.
Expected number down, $L$. $\displaystyle L=\sum_{n=0}^{10} n P_n$. Carrying out the weighted sum (Python-verified) gives the clean result $L=5/2=\boxed{2.5\text{ machines down, on average}}$.
Part (b) — expected number waiting. $L$ counts every down machine, including the one currently being unloaded/reloaded by the crane. The queue (waiting only) excludes that one machine whenever the crane is busy, i.e. subtract the crane's utilization $(1-P_0)$:
$$L_q = L-(1-P_0) = 2.5-(1-0.0563) = \boxed{1.56\text{ machines waiting, on average}}.$$
