Question 3 of 7: Rating and Allowances in Stopwatch Time Study, and Optimum Multiple-Machine Assignment

Question 3: Rating and Allowances in Stopwatch Time Study, and Optimum Multiple-Machine Assignment (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.

(i) Why Rating and Allowances Are Critical and Controversial, and How to Alleviate the Problems

Performance rating and allowances are critical because they are the two adjustments that convert a raw stopwatch reading — specific to one operator, on one day, under one set of conditions — into a standard intended to be fair and achievable for any qualified worker under normal conditions indefinitely. Without a correct rating, a standard timed on a fast operator becomes unattainable by an average worker, and one timed on a slow operator overpays for ordinary performance; without correct allowances, even a perfectly rated normal time ignores the unavoidable non-productive time (personal needs, unavoidable delays, fatigue) that a real shift always contains. Both are controversial precisely because both ultimately rest on a trained analyst's judgement rather than on a directly measured physical quantity: two analysts observing the identical performance can, in good faith, assign different ratings, and the "correct" size of a fatigue or delay allowance for a given job is rarely backed by a plant-specific measurement. Because pay (under incentive systems) or output expectations are riding on the resulting standard, small differences in judgement translate directly into disputes between labour and management over whether a standard is fair.

Industry alleviates these problems through several complementary approaches: (1) rating training and certification, using synchronized rating films or videos of known, independently verified paces so analysts practise until their individual ratings converge within an accepted tolerance, with periodic re-certification; (2) consensus/group rating, averaging two or more trained analysts' independent ratings of the same observation to dilute any one analyst's bias; (3) adopting predetermined motion-time systems (MTM, MOST — Question 5) that replace subjective rating altogether with published, motion-level time values derived once from a large independent film study; (4) determining allowances empirically via work sampling (Question 6) — observing actual personal, delay and fatigue time directly across many random instants — rather than applying a traditional blanket percentage; and (5) involving worker/union representatives in setting and auditing rating and allowance practice, which improves the perceived fairness of the resulting standards even where the underlying judgement cannot be made perfectly objective.

(ii) Factors for Which Fatigue Allowance Is Given

A stopwatch-derived normal time (Question 4(i)) assumes a sustainable, continuous pace with no recovery built in, so a fatigue allowance is added to convert it into an achievable standard. The factors the allowance recognizes fall into three groups. Physical/energy factors: the force or weight handled, working position (standing, stooping, cramped or awkward postures cost more than a normal seated/standing posture), and the amount of muscular tension involved. Environmental factors: poor atmospheric conditions (heat, humidity, fumes, dust), poor lighting, excessive noise, and vibration. Mental/visual factors: the degree of mental strain or close attention/concentration the task requires, eye strain from close or precise visual work, and monotony or tediousness of a highly repetitive cycle. Each factor is rated (e.g. against the ILO-style point tables reproduced in Niebel) and the ratings summed to a total fatigue allowance percentage specific to the job, exactly as the personal, delay and fatigue percentages are combined in Question 4(i).

(iii) Optimum Number of Machines per Operator

This is a deterministic multiple-machine (interference) assignment problem: the operator services each machine (load/unload, then walks to the next), after which the machine runs unattended under automatic power feed while the operator moves on. The optimum assignment balances the cost of operator idle time (too few machines) against the cost of machine idle time (too many machines).

Given.

Find. The number of machines $n$ that minimizes the expected unit cost of output.

Approach. Combine the loading/unloading and walking times into one servicing time $s$, find the theoretical break-even machine count $n'=(s+m)/s$, then price the two integers bracketing $n'$ (and check the trend on either side) to confirm which gives the lower unit cost.

1. Servicing time per machine. $s=l+w=2.00+0.12=\boxed{2.12\text{ min}}$.
2. Theoretical break-even machine count. $n'=\dfrac{s+m}{s}=\dfrac{2.12+6.00}{2.12}=\boxed{3.830}$. For an integer assignment $n\lt n'$ the operator is the limiting resource (cycle $=s+m$, machine idle $=0$); for $n\gt n'$ the machines are the limiting resource (cycle $=ns$, operator idle $=0$). The candidates to price are therefore $n=3$ (below $n'$) and $n=4$ (above $n'$).
3. Cycle time and cost, $n=3$. Since $3<3.830$: cycle $T_c=s+m=2.12+6.00=8.12$ min, producing 3 finished pieces per cycle (operator busy $=3(2.12)=6.36$ min, idle $=1.76$ min per cycle). Cost per cycle $=\dfrac{8.12}{60}(8.00)+3\left(\dfrac{8.12}{60}\right)(24.00)=1.083+9.744=\boxed{\$10.827}$. Unit cost $=10.827/3=\boxed{\$3.609/\text{unit}}$.
4. Cycle time and cost, $n=4$. Since $4>3.830$: cycle $T_c=ns=4(2.12)=8.48$ min, producing 4 pieces per cycle (operator idle $=0$; each machine idle $=8.48-8.12=0.36$ min). Cost per cycle $=\dfrac{8.48}{60}(8.00)+4\left(\dfrac{8.48}{60}\right)(24.00)=1.131+13.568=\boxed{\$14.699}$. Unit cost $=14.699/4=\boxed{\$3.675/\text{unit}}$.
5. Decision. $n=3$ gives the lower unit cost ($3.609 versus $3.675 for $n=4$); unit cost at $n=1,2$ is higher still ($4.331, $3.789) and continues rising for $n\ge5$, so $n=3$ is the global minimum over all integer assignments, not merely the better of the two neighbours of $n'$. Optimum assignment: $\boxed{n=3\text{ machines}}$.