Tractor Inventory — EOQ and Dynamic-Programming Lot Sizing: 23-Ind-A1 Operations Research, December 2017, Q3

Question 3: Tractor Inventory — EOQ and Dynamic-Programming 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.

Check: the $6,500 purchase cost and $10,000 sale price set the per-tractor margin but do not enter the ordering-policy math (they are constant regardless of $Q$ or the order schedule, so they cancel out of the minimization) — both parts below minimize only the order + holding cost, as the question's own phrase "monthly inventory (i.e. holding and ordering) cost" confirms.

Given. Fixed order cost $K=\$2{,}500$/order (regardless of size); holding cost $h=\$500$/tractor/month. (a) constant demand $D=15$ tractors/month. (b) forecast demand for months 1–4: $d=(20,25,12,3)$ tractors.

Find. (a) The optimal ordering policy (order quantity, cycle length) and resulting monthly order+holding cost under constant demand. (b) Using dynamic programming (Wagner–Whitin lot sizing), the optimal order schedule and minimum total order+holding cost over the 4-month horizon.

Approach. Part (a) is a standard monthly-rate EOQ. Part (b) has time-varying demand, so apply the Wagner–Whitin DP: let $c(t,j)$ be the cost of a single order placed in month $t$ that covers demand through month $j$ (one fixed charge $K$ plus the holding cost of carrying each month's units from $t$ until they are used), and let $F(j)$ be the minimum cost to cover months $1,\dots,j$; then $F(j)=\min_{1\le t\le j}\{F(t-1)+c(t,j)\}$.

Part (a) — monthly EOQ. Treating the month as the time unit ($D=15$/month): $$Q^*=\sqrt{\frac{2KD}{h}}=\sqrt{\frac{2(2500)(15)}{500}}=\sqrt{150}$$ $$\boxed{Q^*_{(a)}=12.25\text{ tractors per order}}$$ with minimum monthly cost $TC^*=\sqrt{2KDh}=\sqrt{2(2500)(15)(500)}=\$6{,}123.72$/month, reordering every $T^*=Q^*/D=0.8165$ month (≈ 24.5 days, or about 5.4 orders per year).
Part (b) — build the order-block cost table $c(t,j)=K+h\sum_{i=t}^{j}(i-t)d_i$ for every possible single-order block covering months $t$ through $j$ (the item ordered for month $t$ itself is not held; the item for month $i>t$ is held $i-t$ months):
Wagner–Whitin block costs $c(t,j)$ ($), demand $d=(20,25,12,3)$
order in month $t$ ↓ / covers through $j$ → 1 2 3 4

1 2,500 15,000 27,000 31,500

2 — 2,500 8,500 11,500

3 — — 2,500 4,000

4 — — — 2,500

E.g. $c(1,2)=2500+500[(0)(20)+(1)(25)]=2500+12{,}500=15{,}000$; $c(3,4)=2500+500[(0)(12)+(1)(3)]=2500+1500=4000$.

Wagner–Whitin block costs $c(t,j)$ ($), demand $d=(20,25,12,3)$
order in month $t$ ↓ / covers through $j$ →	1	2	3	4
1	2,500	15,000	27,000	31,500
2	—	2,500	8,500	11,500
3	—	—	2,500	4,000
4	—	—	—	2,500

Forward DP recursion $F(j)=\min_{1\le t\le j}\{F(t{-}1)+c(t,j)\}$, $F(0)=0$:

DP recursion — minimum cost to cover months 1…j
$j$	candidates $F(t{-}1)+c(t,j)$	$F(j)$	best $t$
1	$F(0)+c(1,1)=0+2500$	2,500	1
2	$F(0)+c(1,2)=15{,}000;\ F(1)+c(2,2)=2500+2500=5000$	5,000	2
3	$F(0)+c(1,3)=27{,}000;\ F(1)+c(2,3)=11{,}000;\ F(2)+c(3,3)=5000+2500=7500$	7,500	3
4	$F(1)+c(2,4)=14{,}000;\ F(2)+c(3,4)=5000+4000=9000;\ F(3)+c(4,4)=10{,}000$	9,000	3

The winning candidate at $j=4$ is $F(2)+c(3,4)=9000$: ordering months 3 and 4 together ($12+3=15$ tractors, held 1 month for the 3) beats ordering month 4 alone, because the extra holding cost ($500\times1\times3=1500$) is cheaper than a whole extra order ($2,500).

Backtrack the optimal order schedule from $F(4)$ through its chosen predecessors: $$\boxed{F(4)=\$9{,}000\text{: order 20 in month 1, order 25 in month 2, order 15 (12+3) in month 3, no order in month 4}}$$

Final results — Question 3
Item	Value
(a) $Q^*$	12.25 tractors/order
(a) monthly order+holding cost	$6,123.72
(a) reorder interval	0.8165 month (≈24.5 days)
(b) optimal order schedule	Month 1: 20; Month 2: 25; Month 3: 15 (covers 3&4); Month 4: 0
(b) minimum 4-month cost	$9,000.00

Topic: Lot sizing under constant demand (EOQ) vs. time-varying demand (Wagner–Whitin dynamic programming).

Key relations: Constant-rate EOQ $Q^*=\sqrt{2KD/h}$; Wagner–Whitin recursion $F(j)=\min_{t\le j}\{F(t-1)+K+h\sum_{i=t}^{j}(i-t)d_i\}$, exploiting the "zero-inventory-ordering" property (an optimal policy never orders when inventory is already positive, so every order exactly matches a contiguous block of future demand).

Why this works: With time-varying demand the EOQ square-root formula no longer applies because the "rate" $D$ is not constant; the DP instead searches, exactly, over every way of partitioning the finite horizon into contiguous order-blocks, which is guaranteed optimal because the zero-inventory-ordering property collapses the search space from all real $Q$ to a small finite set of block boundaries.

Common pitfall: Reusing the part-(a) EOQ quantity as a fixed reorder rule in part (b) — a constant $Q\approx12$ ignores that ordering 15 units once in month 3 to also cover month 4's small demand of 3 is cheaper than two separate orders, which only the full DP search reveals.

Source: Nahmias, Production and Operations Analysis (7th ed.), ch. 4 (EOQ, deterministic time-varying demand — Wagner–Whitin); Hillier & Lieberman, ch. 11 (deterministic dynamic programming).

Question 3 of 9: Tractor Inventory — EOQ and Dynamic-Programming Lot Sizing

Question 3: Tractor Inventory — EOQ and Dynamic-Programming Lot Sizing (20 marks)