Question 3 of 7: Lot sizing against a requirements schedule

Question 3: Lot sizing against a requirements schedule (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.

Average requirements $\bar{R} = 116.7$ units per week; preparation (set-up) cost $c_p = $$400 per lot; holding cost $c_H = $$4 per unit per week; planning horizon 12 weeks.

Find. The week-by-week inventory record (orders released and stock on hand) and the total incremental cost — set-up plus carrying — under each of the three lot-size policies, and hence which policy is cheapest over this schedule.

Approach. Size the lot three ways — a fixed economic quantity, a fixed economic time interval, and a variable lot balanced on part-periods — then run each policy week by week through the tabulated requirements, accumulating set-up cost per order released and carrying cost on the week-ending balance.

1. Part (a) — compute the economic lot size. The classical square-root formula balances the annualised (here, weekly) set-up cost against the carrying cost of the average cycle stock: $$Q^{*}=\sqrt{\frac{2\,\bar{R}\,c_p}{c_H}} =\sqrt{\frac{2(116.7)(400)}{4}}=\sqrt{23{,}340}=152.8 \;\Longrightarrow\;\boxed{Q^{*}=153\ \text{units per lot}}$$ The formula assumes demand is uniform, which this lumpy schedule is not; that mismatch is exactly what parts (b) and (c) are designed to expose.
2. Part (a) — run the fixed-quantity policy through the schedule. Whenever the balance on hand cannot cover the week's requirement, a whole lot of 153 is released — two lots at once in week 6, three in week 7, where a single lot is not enough. Carrying cost is charged on the week-ending balance at $4 per unit per week:

3. Part (a) — cost the policy. Ten lots are released, and the week-ending balances total 1,030 unit-weeks: $$\text{set-up} = 10 \times 400 = 4{,}000 \qquad \text{carrying} = 1{,}030 \times 4 = 4{,}120$$ $$\boxed{\text{total incremental cost (a)} = 4{,}000 + 4{,}120 = 8{,}120}$$ in dollars. Note how much stock this policy carries: because the lot is fixed while demand is not, the leftover from a large release sits through the quiet weeks 9 to 12.
4. Part (b) — compute the economic order interval. The periodic model fixes the time between orders instead of the quantity: $$T^{*}=\sqrt{\frac{2\,c_p}{c_H\,\bar{R}}}=\sqrt{\frac{2(400)}{4(116.7)}} =\sqrt{1.714}=\boxed{1.31\ \text{weeks}}$$ This is the same policy expressed differently — $Q^{*}/\bar{R} = 152.8/116.7 = 1.31$ weeks confirms it — but ordering can only happen on week boundaries, so $T$ must be rounded to 1 or to 2 weeks. Because the total-cost curve is flat near the optimum, the choice is settled by costing both, not by the rounding rule.
5. Part (b) — cost both roundings. With $T = 1$ week each order covers exactly that week's requirement, so nothing is ever carried; the two weeks with zero requirement need no order, giving ten set-ups: $$\text{total}_{T=1} = 10(400) + 0 = 4{,}000$$ With $T = 2$ weeks the orders become 55, 200, 525, 500, 100, 100 in weeks 1, 3, 5, 7, 9 and 11, the week-ending balances are 30, 125, 325, 100, 100, 100 (odd weeks) and zero in the even weeks, so $$\text{total}_{T=2} = 6(400) + 780(4) = 2{,}400 + 3{,}120 = 5{,}520$$ The one-week interval is cheaper by $1,520, so the economic periodic policy is $$\boxed{T = 1\ \text{week},\ \text{total incremental cost} = 4{,}000}$$ in dollars. It is lot-for-lot ordering in all but name, which is what an EOI of 1.31 weeks against a weekly bucket must collapse to.

6. Part (c) — establish the economic part-period. Part-period balancing lets the lot size vary, extending each order's coverage until the part-periods accumulated by the stock being carried come closest to the part-periods that one set-up buys: $$\text{EPP}=\frac{c_p}{c_H}=\frac{400}{4}=\boxed{100\ \text{part-periods (unit-weeks)}}$$ A part-period is one unit carried for one week; a requirement of $R_k$ placed $j$ weeks before it is needed contributes $R_k\,j$ part-periods.
7. Part (c) — build the lots. Starting at week 1: covering week 1 alone costs 0 part-periods, adding week 2 costs $30 \times 1 = 30$, and adding week 3 would cost a further $75 \times 2 = 150$ for a running total of 180, which overshoots 100 by more than 30 undershoots it. The first lot therefore covers weeks 1–2, i.e. 55 units. The same test applied from week 3 gives a lot of 200 covering weeks 3–4 (125 part-periods against 0 for week 3 alone), then week 5 alone (200 units, because week 6 would add $325 \times 1 = 325$), week 6 alone, and weeks 7–8 together (500 units, 100 part-periods — an exact hit on EPP). Weeks 9 and 11 have no requirement at all, so no lot starts there; the remaining orders are 100 in week 10 and 100 in week 12.
8. Part (c) — cost the policy. Seven lots are released and the week-ending balances total 255 unit-weeks: $$\text{set-up} = 7 \times 400 = 2{,}800 \qquad \text{carrying} = 255 \times 4 = 1{,}020$$ $$\boxed{\text{total incremental cost (c)} = 2{,}800 + 1{,}020 = 3{,}820}$$ in dollars. The week 7–8 tie is worth checking the other way: combining weeks 10 to 12 into one order would save a set-up (−$400) but carry 100 units through weeks 10 and 11 (+$800), so the split shown is correct. An exact part-period tie must always be settled by costing both branches rather than by the inequality.

Ranking the three policies makes the lesson explicit: part-period balancing at $3,820 beats the periodic model at $4,000, which in turn beats the fixed economic lot size at $8,120. The fixed quantity is more than twice as expensive as the best policy because a constant lot against a strongly varying requirement leaves residual stock sitting through every quiet week.