Question 3 of 7: Fatigue Factors, Fatigue Allowance, and Multiple-Machine Assignment

Question 3: Fatigue Factors, Fatigue Allowance, and 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 has both a physical and a psychological component, and a complete answer must address both: (1) physical/muscular effort — force exerted, weight handled, and especially the proportion of the cycle spent in static muscular exertion (holding rather than moving), which is disproportionately fatiguing for a given energy expenditure; (2) working posture — standing vs. sitting, reach above shoulder or below knee height, and twisted or stooped trunk postures (Class 5 motions, Question 2(ii)) all raise the physiological cost of a task; (3) environmental conditions — heat and humidity (impairing the body’s ability to dissipate metabolic heat), poor lighting, noise, and inadequate ventilation, each adding to physiological load independent of the task itself; (4) monotony and repetitiveness — a highly repetitive, low-variety cycle produces mental/psychological fatigue even when the physical load is light; (5) duration of continuous work without a break — fatigue accumulates across an unbroken work period and is only partly recoverable within the same period; and (6) individual factors — general health, physical conditioning, nutrition and sleep govern how quickly a given worker fatigues under an identical task.

(ii) Factors for Which Fatigue Allowance Is Given

A stopwatch-study fatigue allowance is normally built up from a small base allowance (covering the unavoidable minimum fatigue of even light, seated work) plus variable additions keyed to specific conditions present in the job being studied: standing vs. sitting position; abnormal position (stooping, reaching, twisted posture); use of force or muscular energy (weight lifted or force exerted, and whether it is applied statically or dynamically); poor lighting relative to the visual demand of the task; atmospheric conditions (heat, humidity, poor ventilation, fumes or dust); close attention or visual strain required by fine or exacting work; noise level; mental strain from the complexity of the process or the consequence of an error; and monotony/tediousness of a highly repetitive cycle. Each present factor adds a percentage to the base allowance, so a job with several adverse conditions accumulates a materially larger total fatigue allowance than the single flat 5% figure sometimes applied loosely in practice (as in Question 4(i)'s stated company policy) — that stated 5% should itself be understood as a company-wide simplification of this more detailed, condition-by-condition buildup.

(iii) Optimum Number of Machines Assigned to One Operator

Given.

Find. The optimum (minimum-cost) number of machines to assign to the operator.

Approach. Compute the theoretical break-even machine count $n'=(l+m)/(l+w)$ (Niebel & Freivalds: each machine needs attention every $l+m$ minutes — load/unload plus run — while the operator spends $l+w$ minutes per machine, since walking is operator time only); since it is not a whole number, price one full cycle at each of the two integers bracketing it and take the lower unit cost.

1. Operator servicing time per machine. $a=l+w=2.00+0.12=\boxed{2.12\text{ min}}$ (load/unload plus the walk to the next machine, once per machine per cycle).
2. Theoretical break-even machine count. $n'=\dfrac{l+m}{l+w}=\dfrac{2.00+6.00}{2.12}=\dfrac{8.00}{2.12}=\boxed{3.77\text{ machines}}$. For $n\le n'$ the operator can service every assigned machine before it needs attention again, so the MACHINES are the limiting resource (each machine’s own load-plus-run cycle, $l+m$, sets the pace and the operator carries idle time); for $n\ge n'$ the OPERATOR becomes the limiting resource (cycle $=na$, and the machines instead carry idle time). Since $3.77$ falls between 3 and 4, both integers must be priced.
3. Cycle time and cost at $n=3$ machines (machine-limited, $3\lt n'$). Cycle time $T_c=l+m=2.00+6.00=8.00$ min, producing 3 finished pieces per cycle; the operator’s loop takes only $3(2.12)=6.36$ min, so the operator is idle $1.64$ min per cycle and the machines are idle 0. Operator cost per cycle $=\dfrac{T_c}{60}R_o=\dfrac{8.00}{60}(8.00)=\$1.0667$; machine cost per cycle $=n\dfrac{T_c}{60}R_m=3\left(\dfrac{8.00}{60}\right)(24.00)=\$9.600$. Total cost per cycle $=1.0667+9.600=\$10.667$, so unit cost $=\dfrac{10.667}{3}=\boxed{\$3.556/\text{piece}}$.
4. Cycle time and cost at $n=4$ machines (operator-limited, $4\gt n'$). Cycle time $T_c=n(l+w)=4(2.12)=8.48$ min, producing 4 finished pieces per cycle; each machine needs only $8.00$ min, so each sits idle $0.48$ min per cycle. Operator cost per cycle $=\dfrac{8.48}{60}(8.00)=\$1.1307$; machine cost per cycle $=4\left(\dfrac{8.48}{60}\right)(24.00)=\$13.568$. Total cost per cycle $=1.1307+13.568=\$14.699$, so unit cost $=\dfrac{14.699}{4}=\boxed{\$3.675/\text{piece}}$.
5. Compare and select. $\$3.556/\text{piece}$ at $n=3$ is lower than $\$3.675/\text{piece}$ at $n=4$, so the optimum assignment is $\boxed{n=3\text{ machines per operator}}$.