Workforce Sizing for a Fixed-Deadline Contract: 22-Mec-B4 Integrated Manufacturing Systems, May 2013, Q5

Question 5: Workforce Sizing for a Fixed-Deadline Contract (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 fixed-quantity, fixed-deadline contract to be run on straight time only.

Contract and process data
Item	Product A	Product B
Quantity contracted	1,000	2,000
Standard time per unit	20 h	5 h
Set-up time per unit	3 h	2 h
Number of batches	1	1
Standard set-up time per batch	16 h	10 h
Working days available	30, at 8 h per day (straight time)
Organisational efficiency incl. operator productivity	80 per cent

Find. (a) the straight-time head count that just completes the original contract in 30 working days, and (b) the additional head count needed over the last 15 days if 300 further A's, in their own batch, are added at the half-way point.

Approach. Convert the contract into standard hours, inflate those standard hours to actual clock hours by dividing by the organisational efficiency, divide by the hours one worker can offer in the available time, and round up — a fractional worker cannot be hired and the deadline is contractual.

Check: the data list both a set-up time per unit and a standard set-up time per batch, and the answer below treats these as two distinct charges — a per-unit preparation allowance incurred on every piece, plus a once-per-batch machine set-up. This is the reading that makes every line of the data table load-bearing, and it is stated as an assumption in accordance with Note 1 of the paper. If the per-unit set-up were instead read as an alternative statement of the batch set-up rather than an addition to it, the standard hours would fall to 20,016 and 10,010 and the answers would become 157 and 60 workers; the method is identical either way, and the assumption must be declared with the answer.

Part (a) — total the standard hours for product A. Each unit carries its run time plus its per-unit set-up allowance, and the batch carries one machine set-up: $$H_A = n_A\,(t_A + u_A) + b_A\,B_A = 1000\,(20 + 3) + 1\,(16) = 23{,}000 + 16 = 23{,}016\ \text{h}$$ where $n$ is the quantity, $t$ the standard time per unit, $u$ the set-up time per unit, $b$ the number of batches and $B$ the standard set-up per batch.
Total the standard hours for product B, and for the contract. By the same construction, $$H_B = 2000\,(5 + 2) + 1\,(10) = 14{,}000 + 10 = 14{,}010\ \text{h}$$ so the contract requires $$H = H_A + H_B = 23{,}016 + 14{,}010 = \boxed{37{,}026\ \text{standard hours}}$$ Note how the batch set-ups, 26 hours between them, are negligible here: with one batch of each product the contract is overwhelmingly run time, which is why the batching policy is not the lever on this problem.
Convert standard hours to actual clock hours. Standard hours are what the work should take; the organisational efficiency of 80 per cent states that a clock hour on site delivers only 0.80 standard hours, so $$H_{\text{actual}} = \frac{H}{E} = \frac{37{,}026}{0.80} = \boxed{46{,}282.5\ \text{clock hours}}$$ The efficiency divides rather than multiplies: dividing inflates the requirement, which is the correct direction, whereas multiplying would shrink it and is the single commonest error on this type of question.
Convert to a head count. On straight time each worker offers $$h_w = 30\ \text{days} \times 8\ \text{h/day} = 240\ \text{h}$$ so the requirement is $$N = \frac{H_{\text{actual}}}{h_w} = \frac{46{,}282.5}{240} = 192.84\ \text{workers}$$ Rounding up, because 192 workers deliver only $192 \times 240 = 46{,}080$ clock hours and would miss the contractual deadline, $$\boxed{N = 193\ \text{workers}}$$ The crew is tight: 193 workers offer 46,320 clock hours against 46,282.5 required, a spare capacity of just 37.5 hours over the whole 30 days.
Part (b) — cost the added work. Three hundred more A's, in a separate batch of A, add $$H_X = 300\,(20 + 3) + 1\,(16) = 6{,}900 + 16 = \boxed{6{,}916\ \text{standard hours}}$$ and at the same 80 per cent efficiency, $$H_{X,\text{actual}} = \frac{6{,}916}{0.80} = 8{,}645\ \text{clock hours}$$
Convert the added work to additional workers. Only the back half of the schedule remains, so each additional worker can offer $$h_{w,X} = 15\ \text{days} \times 8\ \text{h/day} = 120\ \text{h}$$ giving $$N_X = \frac{8{,}645}{120} = 72.04 \quad\Rightarrow\quad \boxed{N_X = 73\ \text{additional workers}}$$ The existing crew cannot absorb any of it: its entire spare capacity is 37.5 hours over 30 days, roughly 19 hours across the remaining 15, which is 0.16 of one worker. The manufacturing department therefore runs 266 workers over the last 15 days.
Sanity-check the answer against the deadline. Rounding up from 72.04 to 73 buys $73 \times 120 = 8{,}760$ clock hours against 8,645 required, a cushion of 115 hours or about 1.3 per cent — thin enough that management should note the risk. Two further points belong with the answer. The head count nearly doubles the department for half the contract, so the "pull workers off other projects instantly" assumption in the question is doing a great deal of work; in reality the learning curve on a new batch of A would depress effective efficiency below 80 per cent for the first units. And if the efficiency figure is itself soft, the sensitivity is direct and linear: at 75 per cent efficiency part (a) becomes 206 workers and part (b) 77.

Final results — Question 5
Quantity	Value
Standard hours, product A (1,000 units, 1 batch)	23,016 h
Standard hours, product B (2,000 units, 1 batch)	14,010 h
Total standard hours, original contract	37,026 h
Actual clock hours at 80 per cent efficiency	46,282.5 h
Hours available per worker (30 days × 8 h)	240 h
(a) Workers required	193 (192.84 exact)
Standard hours, added 300 A's (separate batch)	6,916 h
Actual clock hours for the addition	8,645 h
Hours available per worker (15 days × 8 h)	120 h
(b) Additional workers required	73 (72.04 exact); 266 in total over the last 15 days

Topic. Aggregate capacity planning: converting a contract into standard hours and a standard-hour requirement into a straight-time head count.

Key relations. Standard hours $H = \sum_p \left[n_p (t_p + u_p) + b_p B_p\right]$; actual hours $H_{\text{actual}} = H / E$ with $E$ the organisational efficiency; head count $N = \left\lceil H_{\text{actual}} / (D \times h) \right\rceil$ for $D$ working days of $h$ hours.

Why this works. Standard time is defined as the time a qualified worker takes at a defined rate of working under defined conditions, which deliberately excludes the real losses of an operating plant: unbalanced lines, waiting for material, changeovers, absence, rework and the learning curve. The organisational efficiency is the single factor that reinstates all of them, so it must divide the standard hours. Once the requirement is in clock hours the head count is a pure capacity division; the ceiling function is what makes it an engineering answer rather than an arithmetic one, because the deadline is contractual and a partial worker buys nothing.

Common pitfall. Multiplying by the efficiency instead of dividing, which understates the crew by 36 per cent here and would put the contract two weeks late. The second pitfall is sizing part (b) on the full 30-day day-count instead of the 15 days that remain, which halves the answer; the third is quietly dropping either the per-unit set-up allowance or the batch set-up without saying so, which is why the assumption is declared explicitly above.

Source. Chase, Jacobs & Aquilano, Operations and Supply Chain Management, ch. 19 (Sales and Operations Planning) for the aggregate capacity conversion, and ch. 7A for standard times, allowances and worker efficiency; Groover, Automation, Production Systems, and Computer-Integrated Manufacturing, ch. 2, for batch set-up versus run time in the production-rate model.

Question 5 of 6: Workforce Sizing for a Fixed-Deadline Contract

Question 5: Workforce Sizing for a Fixed-Deadline Contract (20 marks)