Question 6 of 7: Choosing an Economic Order Quantity Approach in Three Awkward Cases

Question 6: Choosing an Economic Order Quantity Approach in Three Awkward Cases (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.

What the question is really testing. The classical Wilson EOQ, $Q^{*}=\sqrt{2DS/H}$, rests on four assumptions: demand is known and constant, each item's ordering cost is independent of every other item's, the item is held at one stocking point, and replenishment is instantaneous. Each of the three cases violates a different one of those assumptions, and the marks lie in naming the violation and in selecting the model that repairs it. Applying the plain formula item-by-item in any of the three would give a defensible-looking but wrong answer.

Part (a) — fifty items shipped weekly to one branch: joint replenishment. The assumption that fails here is cost independence. The dominant cost of a replenishment is not the paperwork for an individual item but the shipment: the truck, the consolidation, the receiving. That cost is incurred once per shipment no matter how many of the fifty items travel on it, so computing fifty separate EOQs would charge the shipping cost fifty times and produce fifty different, uncoordinated order cycles.

The correct approach is a joint (coordinated) replenishment model, expressed as a common order interval rather than as fifty quantities. Split the set-up cost into a major cost $S_0$ incurred once per shipment and a minor cost $s_i$ incurred for each item included, then choose the common review interval $T$ that minimises the total:

and order $Q_i = D_i T^{*}$ of each item, where $D_i$ is item $i$'s annual demand and $H_i$ its annual holding cost per unit. If the fifty items differ widely in value or usage, refine this to a power-of-two policy: compute each item's own economic interval $T_i = \sqrt{2s_i/(D_iH_i)}$, then include item $i$ on every $n_i$-th shipment where $n_i$ is the power of two nearest $T_i/T^{*}$. Fast movers then ship every week, slow movers every second, fourth or eighth week, and every shipment is still a single consolidated load. This nested structure is provably within about two per cent of the optimum and is far easier to operate than fifty independent cycles.

Two practical points complete the answer. Because a weekly shipment is already fixed, the real decision is not when to ship but which items to include and in what quantity, which makes this a periodic-review order-up-to (T, S) system in operation: for each item hold an order-up-to level $S_i = D_i(T + L) + z\sigma_i\sqrt{T+L}$ and each week order the difference between that level and the stock on hand. And the truck's capacity is a genuine constraint, so the joint solution should be checked against cube and weight limits and adjusted with a Lagrange multiplier on the binding one.

Part (b) — a highly seasonal item: dynamic lot sizing, or the newsvendor. The assumption that fails is constant demand, and it fails badly: an EOQ computed from average annual demand will be far too large in the off-season and far too small in the peak, and its implied constant order interval is meaningless when demand is concentrated in a few months.

Which model replaces it depends on whether the item can be carried between seasons. If it can, treat demand as a time-phased requirements schedule — period-by-period forecast quantities — and use a dynamic lot-sizing procedure that decides how many periods of requirement each order should cover:

• Wagner–Whitin dynamic programming gives the provably optimal schedule over the horizon and is the benchmark; it is now cheap enough to run but is sensitive to the horizon chosen.
• Silver–Meal extends the cover period while the average cost per period keeps falling — simple, and usually within a few per cent of optimal.
• Part-period balancing extends the cover period until accumulated part-periods most nearly equal the economic part-period $S/H$; intuitive to explain to a planner.
• Least unit cost extends while cost per unit falls; the weakest of the four but the easiest to compute by hand.

All four are driven by the same trade-off as EOQ — one more set-up against the stock a longer cover period creates — but they respect the lumpiness of the requirements instead of averaging it away. Where the seasonality is smooth rather than lumpy, an acceptable middle course is a period-by-period EOQ: recompute $Q^{*}$ from the demand rate prevailing in each month, which tracks the season without abandoning the formula.

If instead the item cannot be carried over — a fashion good, a dated product, a perishable — the problem is not lot sizing at all but a single-period (newsvendor) decision: order the quantity whose cumulative demand probability equals the critical ratio $C_u/(C_u + C_o)$, where $C_u$ is the margin lost by understocking and $C_o$ the loss on each unit left over. Two further devices belong in a complete answer: level the season by building ahead where holding cost is lower than overtime cost, which is an aggregate-planning decision rather than an inventory one; and where the peak is a single promotion, order to the promotion forecast and let the EOQ govern only the baseline demand.

Part (c) — a part passing through three stocking points: multi-stage (echelon) lot sizing. The assumption that fails is that the item is held at a single stocking point. Here one part number occupies four inventories in series — raw castings, semi-finished after chucking, finished components after milling and grinding, and the assembly line's consumption — and the lot size chosen at any stage becomes the demand pattern seen by the stage upstream of it. Computing an independent EOQ at each stage ignores that coupling and produces order quantities that do not nest, so stock accumulates at the interfaces and the raw-material stage is whipsawed by the finishing stage's ordering pattern.

The correct approach is a multi-stage serial inventory model using echelon stock and echelon holding costs. Three ideas carry the answer:

• Echelon stock. The echelon inventory at a stage is all the stock at that stage plus everything downstream of it. Decisions are taken on echelon stock, not local stock, because a casting that has been machined has not left the system — it has only changed form.
• Echelon holding cost. Charge each stage only the value it adds: $h_i^{e} = h_i - h_{i-1}$. The casting carries the purchase price; the chucking stage carries only the machining value added; the finishing stage only the milling and grinding value added. Using the full unit cost at every stage counts the same money three times and drives every lot size far too small.
• Nested, integer-multiple lot sizes. For a serial system the optimal policy is nested: each upstream stage orders in an integer multiple of the lot size of the stage below it, $Q_{i} = n_i Q_{i+1}$ with $n_i$ a positive integer (restricting the $n_i$ to powers of two costs at most a few per cent and makes the schedule repeat cleanly). Start from the final stage, whose demand is the known assembly-line usage, and work upstream.

Two refinements matter in this particular case. The chucking machine is automatic and therefore has a finite production rate, so its stage uses the production-lot form $Q(1-d/p)$ of Question 5 rather than the instantaneous EOQ. And because the assembly line uses the part continuously, the whole chain is a strong candidate for a pull system: rather than solving the multi-stage problem analytically every time the data change, size a fixed number of containers between stages by $k = D_L(1+\alpha)/C$ — lead-time demand plus a safety factor, divided by container size — and let each stage replenish only what the stage below consumed. The economic decision then shifts from lot sizing to set-up reduction, because every stage's lot size falls as the square root of its set-up cost, and reducing set-ups at the chucking machine shrinks all three inventories at once.