Question 3 of 7: Operator Fatigue, Fatigue Allowances, and Optimum Multiple-Machine Assignment

Question 3: Operator Fatigue, Fatigue Allowances, 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) Major Factors Affecting Operator Fatigue

Fatigue arises from several interacting sources. Physical/muscular factors include the magnitude and duration of the force exerted, the amount of static (postural-holding) versus dynamic muscular work, and awkward or constrained working postures that recruit muscle groups poorly suited to sustained effort. Environmental factors include heat and humidity, poor ventilation, noise, and inadequate or glare-producing illumination, all of which raise the physiological cost of doing the same physical work. Work-organization factors include the length of the working period without rest, the pace and monotony of the cycle, and the degree of mental attention or visual strain the task demands. Individual factors — general health, nutrition, sleep, age and level of training — also modulate how quickly a given workload produces fatigue in a specific worker. Methods engineering (Question 1) and good workplace design (Question 2(i)) attack the controllable half of this list directly; the fatigue allowance (part (ii)) compensates for what remains.

(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. Compute the theoretical break-even machine count $n'=(l+m)/(l+w)$; since it is not a whole number, price one full cycle at each of the two integers bracketing it (and check the trend on either side) to confirm which gives the lower unit cost.

1. Operator time per machine serviced. $l+w=2.00+0.12=\boxed{2.12\text{ min}}$.
2. Theoretical break-even machine count. $n'=\dfrac{l+m}{l+w}=\dfrac{2.00+6.00}{2.12}=\dfrac{8.00}{2.12}=\boxed{3.774}$. For an integer assignment $n\lt n'$ the operator gets round all $n$ machines before any needs attention again, so the machines set the pace (cycle $=l+m$, machine idle $=0$, the operator carries the idle time); for $n\gt n'$ the operator is the limiting resource (cycle $=n(l+w)$, operator idle $=0$, the machines carry the idle time). The candidates to price are therefore $n=3$ and $n=4$.
3. Cycle time and cost, $n=3$ (machine-limited). Since $3\lt 3.774$: cycle $T_c=l+m=8.00$ min, producing 3 finished pieces per cycle (operator busy $=3(2.12)=6.36$ min, idle $=1.64$ min per cycle). Cost per cycle $=\dfrac{8.00}{60}(8.00)+3\left(\dfrac{8.00}{60}\right)(24.00)=1.067+9.600=\boxed{\$10.667}$. Unit cost $=10.667/3=\boxed{\$3.556/\text{unit}}$.
4. Cycle time and cost, $n=4$ (operator-limited). Since $4\gt 3.774$: cycle $T_c=n(l+w)=4(2.12)=8.48$ min, producing 4 pieces per cycle (operator idle $=0$; each machine idle $=8.48-8.00=0.48$ 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 ($3.556 versus $3.675 for $n=4$); unit cost at $n=1,2$ is higher still ($4.267, $3.733) 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: $\boxed{n=3\text{ machines}}$.