22-Mec-B4 Integrated Manufacturing Systems · May 2015

Question 1 of 7: Lot Sizing Against a Requirements Schedule

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: 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. A twelve-week requirements schedule for one purchased motor drive unit, together with the average weekly requirement used by the square-root formulas, the cost of preparing a lot and the cost of carrying one unit for one week.

Average requirement as printed on the paper, $\bar{R}=116.7$ units per week; lot preparation cost $c_{p}=$ $400 per lot; carrying cost $c_{h}=$ $4 per unit per week; planning horizon twelve weeks.

Find. The complete inventory record — order releases and week-ending balances — and the total incremental cost (lot preparation plus carrying) for each of the three lot-sizing policies, and hence the policy that should be adopted.

Approach. Establish the two square-root parameters from the average requirement, then simulate each policy week by week against the tabulated discrete requirements — a lot-size rule is only meaningful once it has been run through the actual schedule — and cost each record as (number of lots)×$c_{p}$ plus $c_{h}$ times the sum of the week-ending balances.

1. Reconcile the three statements of total demand before computing anything. The tabulated weekly requirements sum to $\sum R_{t}=1{,}480$ units, the summary line beneath Table 1 states 1,390 units, and the printed average implies $12\bar{R}=12(116.7)=1{,}400.4$ units. The three cannot all be right. Only the tabulated row can drive an inventory record, so the record is built from it; $\bar{R}$ is used exactly as the paper directs, inside the two square-root formulas.
2. (a) Economic lot size. With a preparation charge per lot and a linear carrying charge, the classical lot size balances the two:$$Q^{*}=\sqrt{\frac{2\bar{R}\,c_{p}}{c_{h}}}=\sqrt{\frac{2(116.7)(400)}{4}}=\sqrt{23{,}340}=152.8\ \text{units}$$so the policy orders in fixed lots of $\boxed{Q^{*}=153\ \text{units}}$, releasing as many whole lots as are needed to cover each week.

Running that fixed lot against the schedule gives the record below. Weeks 6 and 7 each need more than one lot, which is the first symptom of a policy that ignores the timing of demand.

Ten lots are released and the week-ending balances total 1,030 unit-weeks, so the two cost elements and their sum are$$TC_{a}=mc_{p}+c_{h}\sum_{t}I_{t}=10(\$400)+4(1{,}030)=\$4{,}000+\$4{,}120=\boxed{\$8{,}120}$$This is the dearest of the three policies, and the record shows why: the fixed lot is far too large for the opening weeks and far too small for weeks 6 and 7, so it carries stock at both ends of the horizon.

3. (b) Economic periodic reorder model. The same balance expressed as a time between orders rather than a quantity gives the economic order interval$$T^{*}=\sqrt{\frac{2c_{p}}{c_{h}\bar{R}}}=\sqrt{\frac{2(400)}{4(116.7)}}=\sqrt{1.7138}=1.31\ \text{weeks}$$identical information to $Q^{*}$, since $T^{*}=Q^{*}/\bar{R}$. Because orders can only be placed on week boundaries, $\boxed{T^{*}=1.31\ \text{wk}\rightarrow T=1\ \text{week}}$, and both roundings must be costed before that is accepted.

At $T=1$ the policy is lot-for-lot: each week's requirement is ordered in that week, nothing is ever carried into the following week, and the two zero-requirement weeks generate no order at all.

Ten orders are placed — weeks 9 and 11 need none — and carrying cost is zero:$$TC_{b}=10(\$400)+4(0)=\boxed{\$4{,}000}$$Costing the other rounding confirms the choice: at $T=2$ the twelve weeks are covered by six orders ($2,400) but 780 unit-weeks are carried ($3,120) for a total of $5,520, which is $1,520 dearer. The total-cost curve is flat near its minimum, so the rounding genuinely has to be evaluated rather than assumed.

4. (c) Part-period total cost balancing. The economic part period is the number of unit-weeks of stock whose carrying charge equals one lot preparation:$$EPP=\frac{c_{p}}{c_{h}}=\frac{400}{4}=\boxed{100\ \text{unit-weeks}}$$Starting in the first week with a live requirement, successive weeks are added to the lot while the accumulated part periods $\sum R_{t}(t-t_{0})$ stay nearest to 100; the lot is closed at whichever cover period leaves the accumulation closest to $EPP$.
5. Accumulate the part periods, lot by lot. Lot 1 opens in week 1: adding week 2 accrues $30(1)=30$ part periods, adding week 3 would accrue a further $75(2)=150$ for 180, and $|100-30|=70$ beats $|100-180|=80$, so the lot closes at week 2 with 55 units. Lot 2 opens in week 3: adding week 4 accrues $125(1)=125$, and $|100-125|=25$ beats $|100-0|=100$, so the lot covers weeks 3–4, 200 units. Lots 3 and 4 are single-week lots, because the next week's requirement alone (325 and 400 units) blows straight past 100 part periods. Lot 5 opens in week 7 and picks up weeks 8 and 9 — $100(1)+0(2)=100$ part periods exactly — for 500 units.
6. Settle the week-10 tie by cost, not by the inequality. Covering weeks 10–12 from one order accrues $100(2)=200$ part periods, and $|100-200|$ equals $|100-0|$. The tie is broken by costing both branches: one order saves a preparation ($400) but carries 100 units through weeks 10 and 11 ($800), a net loss of $400. Two orders it is.

Seven lots are prepared and only 255 unit-weeks are carried, so$$TC_{c}=7(\$400)+4(255)=\$2{,}800+\$1{,}020=\boxed{\$3{,}820}$$Part-period balancing wins because it is the only one of the three rules that looks at when demand occurs: it consolidates the two small early weeks and the flat tail, and refuses to consolidate the week-6 and week-7 peaks, which are exactly the decisions the other two rules get wrong.