EOQ for Plastic Fasteners: 23-Ind-A4 Production Management, December 2015, Q2

Question 2: EOQ for Plastic Fasteners (20 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.

Quantity	Value
Usage rate	6,000 fasteners/week (52 operating weeks/yr) $\Rightarrow D=312{,}000$/yr
Unit cost $c$	0.002 cents $=\$0.00002$ each
Order cost $S$	$\$12$/order
Holding cost rate	25% of unit cost, annualized $\Rightarrow H=0.25c=\$0.000005$/unit-yr

Find. (a) Order interval and EOQ; (b) total annual ordering+holding cost at the EOQ; (c) the cost of continuing to order the part-(a) quantity if true demand is 12,000/week; (d) the true optimal cost at 12,000/week, and the difference.

Approach. Apply the classical EOQ model $Q^*=\sqrt{2DS/H}$ to get the order quantity and interval, then evaluate the total ordering+holding cost function $TC(Q)=\frac{D}{Q}S+\frac{Q}{2}H$ at the part-(a) quantity under both the original and the corrected demand, and separately at the true optimum for the corrected demand, to isolate the cost of having used the wrong order quantity.

Economic order quantity (part a). $$Q^*=\sqrt{\frac{2DS}{H}}=\sqrt{\frac{2(312{,}000)(12)}{0.000005}}=\boxed{Q^*\approx1{,}223{,}765\ \text{fasteners}}.$$ Orders per year $=D/Q^*=312{,}000/1{,}223{,}765\approx0.255$/yr, i.e. an order interval of $52/0.255\approx\boxed{204\ \text{weeks}\approx3.9\ \text{years}}$.
Total annual ordering + holding cost at $Q^*$ (part b). $$TC(Q^*)=\frac{D}{Q^*}S+\frac{Q^*}{2}H=\frac{312{,}000}{1{,}223{,}765}(12)+\frac{1{,}223{,}765}{2}(0.000005)=\boxed{\$6.12/\text{yr}}.$$
Cost if actual demand is 12,000/week but still ordering $Q^*$ from (a) (part c). True annual demand is now $D_2=12{,}000\times52=624{,}000$/yr, double the forecast, but the order quantity is still the part-(a) value: $$TC_{\text{wrong }Q}=\frac{D_2}{Q^*}S+\frac{Q^*}{2}H=\frac{624{,}000}{1{,}223{,}765}(12)+\frac{1{,}223{,}765}{2}(0.000005)=\boxed{\$9.18/\text{yr}}.$$
True optimal cost at 12,000/week (part d). Re-solving the EOQ formula at the corrected demand: $$Q_2^*=\sqrt{\frac{2D_2S}{H}}=\sqrt{\frac{2(624{,}000)(12)}{0.000005}}=1{,}730{,}665\ \text{fasteners},\qquad TC(Q_2^*)=\boxed{\$8.65/\text{yr}}.$$ Ordering the wrong (part-a) quantity instead of the true optimum costs $9.18-8.65=\boxed{\$0.52/\text{yr}}$ more, a $\boxed{6.1\%}$ penalty.

Quantity	Value
EOQ (part a), demand as forecast	1,223,765 fasteners, every ≈204 weeks (≈3.9 yr)
Total ordering+holding cost at EOQ (part b)	\$6.12/yr
Cost using part-a quantity at true 12,000/wk demand (part c)	\$9.18/yr
True optimal EOQ and cost at 12,000/wk (part d)	1,730,665 fasteners; \$8.65/yr
Penalty for having used the wrong EOQ	\$0.52/yr (6.1% above true optimum)

Check

The computed EOQ (≈1.22 million fasteners, a ≈3.9-year supply) is mathematically correct given the stated data but is an extreme result driven entirely by how cheap the fastener is (\$0.00002 each, so annualized holding cost is only \$0.000005/unit) — the EOQ formula has no awareness of practical limits such as warehouse space, shelf-life/plastic degradation, or minimum-order rules that a real purchasing department would also have to respect; those are not part of the given data and are flagged rather than silently overridden.

Concept & Theory

Topic. Classical Economic Order Quantity (EOQ) model and the cost of using an order quantity based on an incorrect demand forecast.

Key relations. $Q^*=\sqrt{2DS/H}$; total ordering+holding cost $TC(Q)=\frac{D}{Q}S+\frac{Q}{2}H$; holding cost per unit $H=(\text{holding rate})\times c$.

Why this works. $TC(Q)$ trades off ordering cost (which falls as $Q$ grows, since fewer orders are placed) against holding cost (which rises linearly with $Q$, since average on-hand inventory is $Q/2$); the EOQ is the quantity where these two opposing costs are exactly equal, which is also where their sum is minimized (setting $dTC/dQ=0$ gives $-DS/Q^2+H/2=0\Rightarrow Q^*=\sqrt{2DS/H}$). Because $TC(Q)$ is a shallow, symmetric curve near its minimum, moderate errors in the order quantity translate into only a small percentage cost penalty — exactly the pattern seen in part (d), where a 2× demand error cost only 6.1% extra.

Common pitfall. Forgetting that the EOQ formula answers "how much to order," not "when to reorder" — reorder timing/point requires lead-time demand and safety stock, not covered by this question; also assuming a large percentage forecast error must produce a proportionally large cost penalty, when the EOQ cost curve's flatness near its minimum means the opposite is often true.

Source. Nahmias & Olsen, Production and Operations Analysis, Ch. 4 (deterministic inventory models, EOQ).

Question 2 of 7: EOQ for Plastic Fasteners

Question 2: EOQ for Plastic Fasteners (20 marks)