Question 1 of 10: EOQ — Special-Storage Inventory Item
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.
Find. (a) $TC(Q)$ and the cost-minimizing order quantity $Q^*$ with no shortages allowed. (b) $TC(Q,S)$ and the optimal $Q^*,S^*$ when planned shortages (backorders) are allowed.
Approach. This is the classical deterministic EOQ model: (a) balance ordering cost against holding cost; (b) allow a maximum backorder level $S$ and balance ordering, holding and shortage cost jointly, differentiating with respect to both $Q$ and $S$.
(a) Total yearly cost with no shortages. With $D/Q$ orders/yr and average on-hand inventory $Q/2$:
$$TC(Q)=\frac{DK}{Q}+\frac{hQ}{2}=\frac{40{,}000(16)}{Q}+\frac{2Q}{2}=\frac{640{,}000}{Q}+Q$$
Minimize: set $dTC/dQ=0$.
$$-\frac{DK}{Q^2}+\frac{h}{2}=0\ \Longrightarrow\ Q^*=\sqrt{\frac{2DK}{h}}=\sqrt{\frac{2(40{,}000)(16)}{2}}=\sqrt{640{,}000}$$
$$\boxed{Q^*=800\text{ units/order}}$$
Substituting back, $TC(Q^*)=640{,}000/800+800=800+800$, so $TC^*=\boxed{\$1{,}600.00/\text{yr}}$ (ordering and holding cost are equal at the optimum, the standard EOQ signature).
(b) Total yearly cost with planned shortages. With maximum shortage level $S$ (so peak on-hand inventory is $Q-S$), the average on-hand inventory is $(Q-S)^2/(2Q)$ and the average shortage is $S^2/(2Q)$:
$$TC(Q,S)=\frac{DK}{Q}+\frac{h(Q-S)^2}{2Q}+\frac{pS^2}{2Q}$$
Minimize jointly: setting $\partial TC/\partial Q=0$ and $\partial TC/\partial S=0$ and solving simultaneously gives the standard EOQ-with-backorders result:
$$Q^*=\sqrt{\frac{2DK}{h}\cdot\frac{h+p}{p}}=\sqrt{640{,}000\cdot\frac{2+4}{4}}=\sqrt{960{,}000}$$
$$\boxed{Q^*\approx 979.8\text{ units/order}}$$
$$S^*=Q^*\cdot\frac{h}{h+p}=979.8\times\frac{2}{6}$$
$$\boxed{S^*\approx 326.6\text{ units}}$$
Minimum total cost. Substituting back (equivalently $TC^*=\sqrt{2DKh\,p/(h+p)}$):
$$TC^*=\sqrt{2(40{,}000)(16)(2)(4)/6}$$
$$\boxed{TC^*\approx \$1{,}306.39/\text{yr}}$$
— lower than part (a)'s $1,600.00/yr, as expected: allowing planned shortages relaxes the trade-off and can only reduce (never increase) the minimum cost.
