Question 3 of 7: Performance Rating and Allowances, Fatigue Allowance Factors, and Optimum Multiple-Machine Assignment

Question 3: Performance Rating and Allowances, Fatigue Allowance Factors, 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 Performance Rating and Allowances Are Critical and Controversial, and How to Alleviate the Problems

A stopwatch records only the OBSERVED time of whichever operator happens to be on the job at the moment of the study, at whatever pace that individual actually worked — a mixture of true task content and the observed operator’s personal speed, skill, effort and any watch-consciousness distortion. Performance rating is the mechanism that removes this operator-specific component: the analyst judges the observed pace against an internalized concept of “100% normal” and multiplies the observed time by that rating to obtain a normal time that is, in principle, independent of which individual was timed. It is controversial precisely because it is a subjective judgement call, made by one analyst in the moment, with real pay consequences: a standard set from a fast operator would be unachievable by an average worker, and one set from a slow operator would pay a premium for ordinary performance. Allowances are controversial for a related reason — the personal, delay and fatigue percentages (part (ii)) are typically set once, by policy or negotiation, and then applied uniformly to jobs whose actual non-productive-time needs can differ substantially, so a blanket allowance can be simultaneously too generous on an easy job and too tight on a demanding one.

Industry has developed several complementary approaches to control this error rather than eliminate the judgement entirely: (1) rating training and certification — analysts practice on rating films of known, independently verified paces until their individual ratings converge within a tight tolerance of the accepted value, and are periodically re-certified; (2) group/consensus rating — averaging the independent ratings of two or more trained analysts on the same observation reduces the effect of any one analyst’s bias; (3) predetermined motion-time systems (MTM, MOST — Question 5) bypass subjective rating altogether by assigning a fixed time to each basic motion from motion-picture studies of many experienced operators; (4) statistical control of rating consistency — plotting an analyst’s ratings over time against a reference standard to detect drift toward looseness before it contaminates new standards; and (5) for allowances specifically, replacing a single blanket percentage with empirically measured allowances from work sampling (Question 6), which observes actual personal/delay/fatigue time directly across many random instants rather than assuming a fixed traditional percentage.

(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 to the next machine. 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 $3\lt3.830$: 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 $4\gt3.830$: 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: $\boxed{n=3\text{ machines}}$.