Tractor Inventory — EOQ and Dynamic-Demand Lot Sizing: 23-Ind-A1 Operations Research, December 2018, Q1

Question 1: Tractor Inventory — EOQ and Dynamic-Demand Lot Sizing (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.

(a) Constant demand — classic EOQ

Given. Fixed order (setup) cost $K=\$2{,}500$/order; holding cost $h=\$500$/tractor/month; constant demand $D=15$ tractors/month. (Purchase price $\$6{,}500$ and sale price $\$10{,}000$ only fix the per-tractor margin, not the ordering policy — they don't enter the EOQ trade-off between ordering and holding cost.)

Find. The order quantity $Q^*$ that minimizes total monthly holding+ordering cost, and that minimum cost.

Approach. This is the classic Economic Order Quantity (EOQ) model: balance ordering cost (falls as order size grows) against holding cost (rises with order size) to find the minimizing lot size.

Apply the EOQ formula. $$Q^*=\sqrt{\frac{2KD}{h}}=\sqrt{\frac{2(2500)(15)}{500}}=\sqrt{150}=\boxed{12.25\text{ tractors}}$$
Compute the resulting monthly cost using the EOQ minimum-cost identity (ordering cost equals holding cost at the optimum, so total cost is twice either): $$TC^*=\sqrt{2KDh}=\sqrt{2(2500)(15)(500)}=\sqrt{37{,}500{,}000}=\boxed{\$6{,}123.72/\text{month}}$$ Cross-check via $TC(Q^*)=\dfrac{DK}{Q^*}+\dfrac{Q^*h}{2}=\dfrac{15(2500)}{12.25}+\dfrac{12.25(500)}{2}=3{,}061.86+3{,}061.86=6{,}123.72$ — the two components are equal, confirming $Q^*$ is the true minimum.
State the ordering policy. Order $Q^*\approx12.25$ tractors every $Q^*/D=12.25/15=0.8165$ month (about every 24.5 days), i.e. roughly 1.22 orders/month.

Check: EOQ is derived as a continuous quantity; a dealership orders whole tractors, so the practical policy rounds to $Q=12$ per order (cost rises by only $1.28/month at $Q=12$ ($15(2500)/12+12(500)/2=\$6{,}125.00$) vs. the true optimum — EOQ's cost curve is flat near the minimum). The exam asks for "the optimal ordering policy," so the exact $Q^*=12.25$ is reported as the answer, with the integer caveat noted.

(b) Non-constant demand — Wagner–Whitin dynamic programming

Given. Same $K=\$2{,}500$/order and $h=\$500$/tractor/month; forecast demand $d_1=20,\,d_2=25,\,d_3=12,\,d_4=3$ over 4 months; no backorders, unmet demand each month must be covered from stock ordered in that or an earlier month.

Find. The ordering plan (which months to order in, and how much) that minimizes total ordering+holding cost over the 4 months, and that minimum cost.

Approach. With demand varying by period, EOQ no longer applies; use the Wagner–Whitin dynamic-programming recursion: let $f(t)$ be the minimum cost to satisfy demand through month $t$, and for each $t$ consider every "last order" month $i\le t$ that covers periods $i,\dots,t$ in one order.

Set up the cost of one order in period $i$ covering through period $j$: $$C(i,j)=K+h\sum_{k=i}^{j}(k-i)\,d_k$$ and the recursion $f(t)=\min_{1\le i\le t}\big[f(i-1)+C(i,t)\big]$, $f(0)=0$.

Stage-by-stage evaluation (comparing every feasible "last order" period at each stage):

Wagner–Whitin recursion — minimum cost to cover months 1–t
$t$	Best last-order period $i$	$f(t)$
1	1 (order covers month 1 only)	$2,500
2	2 (order covers month 2 only)	$5,000
3	3 (order covers month 3 only)	$7,500
4	3 (order covers months 3–4 together)	$9,000

At every stage, ordering separately each month (no carry-over) beats combining orders, EXCEPT for months 3&4: combining them into one order placed in month 3 costs $C(3,4)=2500+500(1)(3)=\$4{,}000$ against $f(2)=\$5{,}000$, i.e. $\$9{,}000$ total — cheaper than ordering separately in 3 and 4 ($f(3)+K=7{,}500+2{,}500=\$10{,}000$), because month 4's demand (3 units) is small enough that one month's holding cost ($500\times3=\$1{,}500$) is less than a second $2,500 order fee.

Read off the optimal plan. $$\boxed{\text{Order 20 in month 1; order 25 in month 2; order 15 (12+3) in month 3, carrying 3 units into month 4}}$$ $$\boxed{\text{Minimum total cost}=f(4)=\$9{,}000}$$ Breakdown: 3 orders $\times\ \$2{,}500=\$7{,}500$ ordering cost, plus $\$1{,}500$ holding cost on the 3 units carried from month 3 into month 4.

Final results — Question 1
Item	Value
(a) EOQ $Q^*$	12.25 tractors/order (≈1.22 orders/month)
(a) Minimum monthly cost	$6,123.72
(b) Ordering plan	20 (month 1); 25 (month 2); 15 (month 3, covers 3&4)
(b) Minimum 4-month cost	$9,000

Topic: Inventory lot-sizing — static EOQ vs. dynamic (Wagner–Whitin) demand.

Key relations: EOQ $Q^*=\sqrt{2KD/h}$, $TC^*=\sqrt{2KDh}$; Wagner–Whitin recursion $f(t)=\min_i[f(i-1)+C(i,t)]$, $C(i,j)=K+h\sum_{k=i}^{j}(k-i)d_k$.

Why this works: EOQ assumes constant, infinite-horizon demand, so a single repeating order size is optimal. Once demand varies period-to-period, the trade-off must be re-evaluated at every possible "order now vs. wait" decision point; the Wagner–Whitin recursion is an exact dynamic program over that decision tree (never approximate) because the optimal plan always orders exactly enough to reach the next order point (the zero-inventory-ordering property).

Common pitfall: Applying EOQ directly to the fluctuating forecast (it ignores that never carrying inventory across a demand spike is sometimes cheaper, and vice versa), or forgetting that an order always occurs with zero starting inventory (so only $C(i,j)$ combinations matching that property need be considered).

Source: Nahmias, Production and Operations Analysis — EOQ (single-item) and dynamic lot-sizing (Wagner–Whitin algorithm) chapters.

Question 1 of 10: Tractor Inventory — EOQ and Dynamic-Demand Lot Sizing

Question 1: Tractor Inventory — EOQ and Dynamic-Demand Lot Sizing (20 marks)

(a) Constant demand — classic EOQ

(b) Non-constant demand — Wagner–Whitin dynamic programming