Question 5 of 7: Design of an Inventory Control System for a New Product

Question 5: Design of an Inventory Control System for a New Product (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. A make-to-stock item replenished by internal production, so stock builds up gradually rather than arriving all at once: Every one of the seven data items (a) to (g) supplied by the question is used below.

Find. A complete operating policy — annual demand, run frequency and length, cycle length, maximum and average inventory, safety stock, reorder point, the implied set-up cost, and the annual cost of the system — sufficient for a stock clerk to run the item without further analysis.

Approach. Because supply and consumption overlap, the governing model is the production-lot (economic production quantity) model rather than the ordinary EOQ: build the inventory profile from the net accumulation rate $p-d$, derive the timing and cost figures from it, then set the reorder point from the demand during the maximum lead time.

1. Establish annual demand and the run frequency. Usage is uniform over the working year, so $$D \;=\; d \times (\text{days/yr}) \;=\; 20 \times 240 \;=\; 4{,}800 \text{ units per year}$$ and with a lot size of 1,000 units the number of production runs is $$m \;=\; \frac{D}{Q} \;=\; \frac{4{,}800}{1{,}000} \;=\; 4.8 \text{ runs per year}$$ — in practice five runs in some years and four in others, or a standing schedule of one run every 50 working days.
2. Determine the length of a run and of a cycle. The lot takes $$t_p \;=\; \frac{Q}{p} \;=\; \frac{1{,}000}{50} \;=\; 20 \text{ working days to produce}$$ and the lot lasts $$t_c \;=\; \frac{Q}{d} \;=\; \frac{1{,}000}{20} \;=\; 50 \text{ working days}$$ so each cycle consists of 20 days of simultaneous production and consumption followed by 30 days of consumption alone. Note $240/50 = 4.8$ cycles per year, which agrees with the run frequency above.
3. Compute the maximum and average inventory. While the run is under way stock accumulates at the net rate $p - d = 50 - 20 = 30$ units per day, so the peak reached at the end of the run is $$I_{\max} \;=\; (p-d)\,t_p \;=\; Q\left(1 - \frac{d}{p}\right) \;=\; 1{,}000\left(1 - \frac{20}{50}\right) \;=\; 600 \text{ units}$$ $$\boxed{I_{\max} = 600 \text{ units},\qquad \bar{I} = \tfrac{1}{2}I_{\max} = 300 \text{ units}}$$ The peak is only 600, not the full 1,000, because 400 units are consumed while they are being made — the single most important difference between this model and ordinary EOQ.
4. Cost the cycle stock. Carrying the average cycle inventory for a year costs $$C_{\text{hold}} \;=\; \bar{I}H \;=\; 300 \times 5.00 \;=\; \$1{,}500 \text{ per year}$$
5. Recover the set-up cost implied by the given lot size, and use it as a consistency check. The question states the lot size is economic, which means it satisfies the production-lot formula $Q = \sqrt{2DS/[H(1-d/p)]}$. Inverting for the set-up cost $S$, $$S \;=\; \frac{Q^{2}H\left(1-\dfrac{d}{p}\right)}{2D} \;=\; \frac{(1{,}000)^{2}(5.00)(0.60)}{2(4{,}800)} \;=\; \$312.50 \text{ per run}$$ Substituting $312.50 back into the formula reproduces $Q = 1{,}000$ exactly, and the annual set-up cost $mS = 4.8 \times 312.50$ comes to $1,500, which equals the annual holding cost — the balance that always holds at an economic lot size. The data are therefore internally consistent, and the shop now knows what a set-up must cost for the 1,000-unit lot to remain correct: if set-ups can be reduced below $312.50 the lot size should be cut.
6. Set the safety stock and the reorder point. The lead time from placing the order to the start of production is 10 days but may be as long as 15. Demand over the expected lead time is $$d L \;=\; 20 \times 10 \;=\; 200 \text{ units}$$ and the extra five days of possible delay must be covered by safety stock, $$SS \;=\; d \,\Delta L \;=\; 20 \times 5 \;=\; 100 \text{ units}$$ so the order is released when the on-hand balance falls to $$\boxed{ROP \;=\; d(L + \Delta L) \;=\; 20 \times 15 \;=\; 300 \text{ units}}$$ With the safety stock in place the profile oscillates between 100 and $600 + 100 = 700$ units, and the balance passes the reorder point 20 days after the end of each run.
7. Cost the safety stock and assemble the annual cost of the system. The safety stock is held permanently, so it costs $SS \times H = 100 \times 5.00$, that is $500 per year. The full annual cost is then $$TC \;=\; Dc \;+\; mS \;+\; \bar{I}H \;+\; SS\cdot H$$ $$TC \;=\; 72{,}000 \;+\; 1{,}500 \;+\; 1{,}500 \;+\; 500 \;=\; \$75{,}500 \text{ per year}$$ $$\boxed{TC = \$75{,}500/\text{yr}, \text{ of which } \$3{,}500 \text{ is inventory-related}}$$ Inventory management therefore controls under five per cent of the total cost of the item, which is worth stating: the leverage in this product lies in the $15 production cost, not in the stockroom.
8. State the operating policy in the form the stockroom will use. The system is a continuous-review, fixed-order-quantity (Q, ROP) system: review the balance after every issue; when the on-hand balance falls to 300 units, release a production order for 1,000 units; the run will start within 10 to 15 days and will take 20 working days to complete; the balance will bottom out near 100 units and peak near 700. Supporting figures for the stock record: 4.8 runs per year, one every 50 working days; peak inventory investment $10,500 (700 units at $15); inventory turnover $D/(\bar{I}+SS) = 4{,}800/400 = 12$ turns per year. Because the item is new, the policy should be reviewed after two or three cycles against actual usage, since the whole structure rests on the assumed 20 units/day.