Question 7 of 7: Optimum ratio of men to semiautomatic machines

Question 7: Optimum ratio of men to semiautomatic machines (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.

Operator wage rate $4.90 per hour; machine burden rate $18.00 per hour; a large number of identical machines, so any integer assignment is available.

Find. The number of machines one operator should tend, and the resulting lowest cost per unit produced.

Approach. Split the operator's work into the part that stops the machine and the part that does not, compute the theoretical number of machines that exactly fills the operator's time, then cost the two neighbouring integers — because the machines and the operator are paid at very different rates, the cheapest integer is not always the nearest one.

1. Classify the time-study elements. Three quantities govern a machine-coupling problem: the servicing time during which the machine is stopped, the operator's independent work plus walking, and the machine's own running time. $$l = 3.1 + 0.6 = 3.7\ \text{min},\qquad w = 2.4 + 1.9 + 0.4 = 4.7\ \text{min},\qquad t = 41.3\ \text{min}$$ The operator therefore spends $l + w = 8.4$ min on each machine he visits, while each machine is occupied for $l + t = 45.0$ min per piece.
2. Compute the theoretical number of machines. One operator can just keep $n$ machines busy when the time he spends going round them all equals the time a machine needs to complete a piece, $n(l+w) = l+t$: $$n' = \frac{l+t}{l+w} = \frac{3.7+41.3}{3.7+4.7} = \frac{45.0}{8.4} = \boxed{5.36\ \text{machines per operator}}$$ Since $n$ must be an integer, the answer is 5 or 6, and the two cases behave quite differently. Below $n'$ the system is machine-limited — every machine runs a full 45.0 min cycle and the operator has idle time. Above $n'$ it is operator-limited — the operator works continuously and every machine waits for him.
3. Write the cost per unit. Over one cycle the plant pays for one operator and $n$ machines, and produces $n$ units: $$\text{cost per unit}=\frac{T_c}{60}\cdot\frac{K_{\text{op}}+n\,K_{\text{mc}}}{n} \qquad\text{where}\quad T_c=\begin{cases} l+t & n \le n' \\[2pt] n(l+w) & n > n' \end{cases}$$ with $K_{\text{op}} = $$4.90 per hour and $K_{\text{mc}} = $$18.00 per hour. Both terms matter: adding a machine spreads the operator's wage over more units but adds a full machine burden, and once past $n'$ it also stretches the cycle.
4. Cost the machine-limited case, n = 5. Here $5(8.4) = 42.0 \le 45.0$, so the cycle is the machine cycle of 45.0 min = 0.75 h and the operator carries 3.0 min of idle time per cycle: $$\text{cost per cycle}=0.75\,(4.90 + 5 \times 18.00)=0.75(94.90)=71.175$$ $$\text{cost per unit}=\frac{71.175}{5}=\boxed{14.24\ \text{per unit}}$$ in dollars, at an output of $5(60)/45.0 = 6.67$ units per hour.
5. Cost the operator-limited case, n = 6. Now $6(8.4) = 50.4 > 45.0$, so the operator sets the pace and the cycle stretches to 50.4 min = 0.84 h, leaving each machine idle for 5.4 min waiting to be served: $$\text{cost per cycle}=0.84\,(4.90+6 \times 18.00)=0.84(112.90)=94.836 \qquad\Rightarrow\qquad \frac{94.836}{6}=15.81\ \text{per unit}$$ in dollars — 11 per cent dearer than five machines, because the sixth machine's burden is paid in full while the cycle it must share has grown by 12 per cent.
6. Check the other neighbour and conclude. Four machines give the same 45.0 min cycle but spread the operator over fewer units, $0.75(4.90+4 \times 18.00)/4 = 76.90(0.75)/4 = 14.42$ dollars per unit, so the cost curve has its minimum at five: $$\boxed{n = 5\ \text{machines per operator, at }14.24\ \text{dollars per unit}}$$ The result is the general one for machine coupling with a cheap operator and expensive machines: round the theoretical ratio down, because idle machine time costs $18.00 an hour and idle operator time only $4.90.
7. Test the alternative reading of the time study. If instead every handling element stops the machine, $l = 3.1+0.6+2.4+1.9 = 8.0$ min and $w = 0.4$ min, so $n' = (8.0+41.3)/8.4 = 5.87$. Costing the neighbours, $n = 5$ gives a 49.3 min cycle and $49.3(94.90)/(60 \times 5) = 15.60$ dollars per unit while $n = 6$ gives $50.4(112.90)/(60 \times 6) = 15.81$. The optimum is again five machines, so the ambiguity in the time study changes the cost figure by about 9 per cent but not the answer to the question actually asked.