Question 8 of 10: Tractor Inventory — Single-Period EOQ and Dynamic-Programming Lot Sizing

Question 8: Tractor Inventory — Single-Period EOQ and Dynamic-Programming Lot Sizing (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. Purchase cost $6,500/tractor, sale price $10,000/tractor; ordering (setup) cost $K=\$2{,}500$/order regardless of size; holding cost $h=\$500$/tractor/month. (a) constant demand $D=15$/month. (b) monthly forecast $d=(20,25,12,3)$ over 4 months.

Find. (a) EOQ ordering policy and its monthly holding+ordering cost; (b) the dynamic-programming (Wagner–Whitin) optimal order schedule and its minimum total cost over the 4 months.

Approach. Part (a) is a standard single-period EOQ (purchase/sale price is a margin distractor that does not affect the ordering-cost tradeoff). Part (b) has demand that varies sharply month to month, so instead of a constant EOQ, use Wagner–Whitin dynamic programming: for every "order in month $i$, cover through month $j$" span, compute the setup-plus-holding cost, then find the cheapest chain of spans covering all 4 months.

1. Part (a) — EOQ. With constant monthly demand $D=15$, ordering cost $K=2500$, holding cost $h=500$: $$Q^*=\sqrt{\dfrac{2DK}{h}}=\sqrt{\dfrac{2(15)(2500)}{500}}=\sqrt{150}=\boxed{12.2\text{ tractors/order}\ (\approx12\text{--}13)}.$$ Monthly cost $=DK/Q^*+hQ^*/2=\sqrt{2DKh}=\sqrt{2(15)(2500)(500)}=\boxed{\$6{,}123.72/\text{month}}$. The purchase/resale prices ($6,500 buy, $10,000 sell) set the per-unit margin but do not enter this minimization — the number of tractors bought each month is fixed by demand regardless of order size, so only the ordering pattern (K and h) affects cost.
2. Part (b) — span costs. Define $c(i,j)=K+h\sum_{t=i}^{j}(t-i)d_t$, the cost of one order in month $i$ that covers demand through month $j$ (holding each later month's units for the number of months until they are used). With $d=(20,25,12,3)$: e.g. $c(3,4)=2500+500(0)(12)+500(1)(3)=2500+1500=\$4{,}000$, versus ordering separately $c(3,3)+c(4,4)=2500+2500=\$5{,}000$ — combining the last two months saves $\$1{,}000$ because holding only 3 units for one month ($\$1{,}500$) is cheaper than a second $\$2{,}500$ setup.
3. Part (b) — DP recursion. Let $f(t)=$ minimum cost to cover months $1,\dots,t$: $f(0)=0$, $f(t)=\min_{1\le i\le t}\big[f(i-1)+c(i,t)\big]$. Evaluating all spans (Python-verified) gives $f(1)=2{,}500$ (order 1 covers month 1 alone), $f(2)=5{,}000$ (order 2 covers month 2 alone — combining 1–2 would cost $K+500(25)=15{,}000$, far worse), $f(3)=7{,}500$, $f(4)=9{,}000$ via combining months 3–4.
4. Part (b) — optimal schedule. Backtracking the DP gives three orders: month 1 (20 units), month 2 (25 units), month 3 (12+3=15 units, covering months 3–4): $$\boxed{f(4)=\$9{,}000\text{ minimum total cost, orders of }20,\ 25,\ 15\text{ in months }1,\ 2,\ 3.}$$