Economic Order Quantity — With and Without Planned Shortages: 23-Ind-A1 Operations Research, December 2014, Q1

Question 1: Economic Order Quantity — With and Without Planned Shortages (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Annual demand $D=40{,}000$ units/yr; ordering cost $K=\$16$/order; holding cost $h=\$2$/unit/yr; (part b) backorder/shortage cost $p=\$4$/unit/yr.

Find. (a) $TC(Q)$ and $Q^*$ with no shortages; (b) $TC(Q,s)$ and $Q^*,s^*$ with planned backorders.

Approach. Build the classic EOQ cost function from ordering + holding (+ shortage) terms, then minimize by calculus (a single variable for part a, jointly in $Q$ and $s$ for part b).

Part (a) — cost function. With no shortages the inventory cycles between $Q$ and $0$, so average stock is $Q/2$. Ordering cost per year is $D/Q$ orders $\times K$, holding cost is $h\cdot Q/2$: $$TC(Q) = \dfrac{DK}{Q} + \dfrac{hQ}{2}.$$
Part (a) — optimize. Setting $dTC/dQ = -DK/Q^2 + h/2 = 0$ gives the standard EOQ: $$Q^* = \sqrt{\dfrac{2DK}{h}} = \sqrt{\dfrac{2(40{,}000)(16)}{2}} = \sqrt{640{,}000} = \boxed{800\text{ units}}.$$ Minimum yearly cost $TC(Q^*) = DK/Q^*+hQ^*/2 = 800+800 = \boxed{\$1{,}600/\text{yr}}$.
Part (b) — cost function with backorders. Allowing a maximum shortage $s$, the cycle carries positive stock $(Q-s)$ for a fraction $(Q-s)/Q$ of the cycle and is short by up to $s$ for the remaining fraction. Average positive inventory is $(Q-s)^2/(2Q)$ and average shortage is $s^2/(2Q)$: $$TC(Q,s) = \dfrac{DK}{Q} + \dfrac{h(Q-s)^2}{2Q} + \dfrac{p\,s^2}{2Q}.$$
Part (b) — optimize jointly. Solving $\partial TC/\partial Q = 0$ and $\partial TC/\partial s = 0$ simultaneously (standard planned-backorder EOQ result) gives $$Q^* = \sqrt{\dfrac{2DK}{h}\cdot\dfrac{h+p}{p}},\qquad s^* = Q^*\dfrac{h}{h+p}.$$ Substituting $D=40{,}000$, $K=16$, $h=2$, $p=4$: $$Q^* = \sqrt{640{,}000\times\dfrac{6}{4}} = \sqrt{960{,}000} = \boxed{979.8\text{ units}\ (\approx 980)},$$ $$s^* = 979.8\times\dfrac{2}{6} = \boxed{326.6\text{ units}\ (\approx 327)}.$$ Minimum yearly cost is lower than part (a) because backorders let the firm avoid some holding cost: $TC^*=\sqrt{2DKhp/(h+p)} = \boxed{\$1{,}306.4/\text{yr}}$.

Final results — Question 1
Quantity	No shortages	With planned shortages
Optimal order quantity $Q^*$	800 units	979.8 units (≈980)
Max planned shortage $s^*$	—	326.6 units (≈327)
Minimum total yearly cost	$1,600.00	$1,306.39

Topic: Deterministic EOQ with and without planned backorders.

Key relations: No-shortage EOQ: $Q^*=\sqrt{2DK/h}$, $TC^*=\sqrt{2DKh}$. Planned-backorder EOQ: $Q^*=\sqrt{(2DK/h)(h+p)/p}$, $s^*=Q^*h/(h+p)$, $TC^*=\sqrt{2DKhp/(h+p)}$.

Why this works: Both models trade an ordering cost that falls with $Q$ against a carrying/shortage cost that rises with $Q$; the convex sum has a unique interior minimum found by calculus. Allowing backorders adds a second decision variable $s$ that lets the firm substitute a cheaper shortage cost for some of the holding cost, which is why the optimal cycle quantity grows and the minimum cost falls whenever $p$ is finite.

Common pitfall: Forgetting the $(Q-s)$ and $s$ split when writing the holding/shortage terms (using $Q^2$ instead of $(Q-s)^2$), and reporting $s^*$ as if it were the shortage cost rather than the shortage quantity.

Source: Nahmias, Production and Operations Analysis (7th ed.), ch. 4 — deterministic EOQ models with and without backorders.

Question 1 of 10: Economic Order Quantity — With and Without Planned Shortages

Question 1: Economic Order Quantity — With and Without Planned Shortages (15 marks)