Question 2 of 9: Finite-Capacity Queueing — Should the Jewellery Store Add a Second Parking Spot?
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 17-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 135 marks across 9 questions (all worth 15 marks) and only 100 marks are required, so a candidate would normally answer a subset — all nine 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), integer programming (ch. 12), network optimization & CPM/PERT (ch. 9–10), queueing theory (ch. 17), decision analysis (ch. 15–16), Markov chains (ch. 16), equipment replacement (ch. 11/19). Nahmias, Production and Operations Analysis — single-period (newsvendor) inventory models.
Question 2: Finite-Capacity Queueing — Should the Jewellery Store Add a Second Parking Spot? (15 marks)
the numbers below are independently recomputed here and agree with that solved paper.
Given. Service (session) time = 0.5 hr/customer $\Rightarrow\mu=2$/hr; current layout is a single-spot, no-waiting system ($M/M/1/1$) that serves 10 customers per 7-hr day; sale rate 15% per served customer at $1,000 profit/sale; 10% of turned-away customers never return (permanently lost); the second spot (making it $M/M/1/2$) costs $100/day.
Find. Whether the expected daily profit gained from fewer permanently-lost customers exceeds the $100/day cost of the second spot.
Birth–death diagram for the proposed two-spot system ($M/M/1/2$, states 0/1/2 = customers in the system). The current one-spot system is the same chain truncated at state 1 ($M/M/1/1$) — any arrival while in state 1 is turned away.
Approach. Back out the underlying arrival rate $\lambda$ from the stated throughput of the current single-spot ($M/M/1/1$) system, use it to find the blocking probability of the proposed two-spot ($M/M/1/2$) system, then compare the expected daily profit from fewer permanently-lost customers against the $100/day cost.
Recover $\lambda$ from the current ($M/M/1/1$) throughput. With $\mu=2$/hr, the current system serves $\lambda\mu/(\lambda+\mu)$ customers/hr, which must equal the given 10/7-hr day = 1.4286/hr:
$$\frac{2\lambda}{\lambda+2}=\frac{10}{7}\ \Longrightarrow\ 14\lambda=10\lambda+20\ \Longrightarrow\ \boxed{\lambda=5\text{ customers/hr}}$$
so $\rho=\lambda/\mu=2.5$ (a heavily loaded system — more customers arrive than the single server can absorb).
Current ($M/M/1/1$) blocking and daily turn-aways.$P_0=\mu/(\lambda+\mu)=2/7=0.2857$, $P_1=\lambda/(\lambda+\mu)=5/7=0.7143$ (= blocking probability, since capacity is 1):
$$\text{blocked/day}=\lambda P_1(7\text{ hr})=5(5/7)(7)=25.0\text{ customers/day}$$
(and served/day $=5(2/7)(7)=10.0$, matching the given data — confirms $\lambda=5$).
Proposed ($M/M/1/2$) blocking and daily turn-aways. With $\rho=2.5$: $P_0=1/(1+\rho+\rho^2)=1/9.75=0.1026$, $P_2=\rho^2P_0=6.25(0.1026)=0.6410$ (= blocking probability, capacity 2):
$$\text{blocked/day}=\lambda P_2(7)=5(0.6410)(7)=22.44\text{ customers/day}$$
so the second spot reduces daily turn-aways by $25.00-22.44=2.564$ customers/day.
Value of the fewer turn-aways — only the 10% who never return represent a true lost sale (the other 90% simply come back another day, so no revenue is actually lost from them):
$$\text{permanently-lost customers avoided/day}=0.10(2.564)=0.2564$$
$$\text{expected profit gained/day}=0.2564\times0.15\times\$1000=\$38.46$$
Compare to the $100/day cost of the spot.
$$\boxed{\$38.46/\text{day gained} \;<\; \$100.00/\text{day cost} \;\Rightarrow\; \text{do NOT add the second parking spot}}$$
Adding it would cost the store about $100-38.46=\$61.54$/day net.
