Question 3 of 7: Break-Even Justification of Numerically Controlled and Transfer Machines

Question 3: Break-Even Justification of Numerically Controlled and Transfer 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. Two independent make-or-buy-the-automation problems. Part (a) compares one lot on two machines; part (b) compares three production methods over a year.

Find. (a) the lot size above which the numerically controlled jig borer costs less than the standard machine, and (b) the ranges of annual quantity over which each of the three methods is the cheapest.

Approach. Both parts are linear break-even comparisons: write each alternative as a fixed charge plus a variable cost per piece, equate the pairs, and read off the crossover quantities; then identify which line forms the lower envelope in each range.

1. Part (a) — build the cost equation for the standard jig borer. The set-up is done on the machine, so it is charged at the shop rate: $$C_{\text{std}}(N) \;=\; t_s r_{\text{std}} \;+\; N\,t_{\text{run}} r_{\text{std}} \;=\; (1)(24) \;+\; N(6)(24) \;=\; 24 + 144N$$ where $N$ is the number of pieces in the lot.
2. Build the cost equation for the tape-controlled jig borer. Here the preparation is programming done in the office, so it is charged at the office rate, while the machining is charged at the higher NC shop rate: $$C_{\text{NC}}(N) \;=\; t_p r_{\text{office}} \;+\; N\,t_{\text{run}} r_{\text{NC}} \;=\; (2)(18) \;+\; N(5)(28) \;=\; 36 + 140N$$ Note the pattern that makes NC justification work at all: the fixed charge rises from $24 to $36, while the variable charge falls from $144 to $140 per piece.
3. Equate the two and solve for the break-even lot size. Setting $C_{\text{std}}(N) = C_{\text{NC}}(N)$, $$24 + 144N \;=\; 36 + 140N \quad\Longrightarrow\quad 4N \;=\; 12$$ $$\boxed{N^{*} = 3 \text{ pieces, at a cost of } \$456 \text{ either way}}$$
4. State the decision rule and check it either side. Below the crossover the standard machine is cheaper; above it the NC machine is. At $N = 2$ the standard machine costs $312 against $316; at $N = 4$ it costs $600 against $596. So the numerically controlled jig borer is justified for lots of four pieces or more, and every additional piece beyond the crossover saves $4, the difference $144 - 140$ between the two per-piece rates. Because the saving per piece is so small relative to the cost per piece, this is a marginal justification on cost alone — the real case for the NC machine on a job like this rests on repeatability, on the programme being available for the next order without a second set-up, and on freeing a skilled operator.
5. Part (b) — check the operation count and build the NC cost equation. The stated operations sum correctly, $50 + 12 + 38 + 36 + 16 = 152$, which confirms the one-hour cycle refers to all 152 operations on a machining centre. With negligible programming and set-up, $$C_{\text{NC}}(N) \;=\; N (1\ \text{h})(40) \;=\; 40N$$ This method has no fixed charge at all, which is precisely why it wins at low volume.
6. Build the conventional-machine cost equation. The $300,000 tooling is capital; what enters an annual cost comparison is the stated annual charge against it, $130,000. Adding one hour per piece at $24/h, $$C_{\text{conv}}(N) \;=\; 130{,}000 \;+\; N (1\ \text{h})(24) \;=\; 130{,}000 + 24N$$
7. Build the transfer-machine cost equation. The transfer machine is rated in pieces per hour, so its labour and overhead per piece is the hourly rate divided by the output rate: $$c_{\text{tr}} \;=\; \frac{16}{188} \;=\; \$0.0851 \text{ per piece}$$ $$C_{\text{tr}}(N) \;=\; 800{,}000 \;+\; 0.0851N$$ The variable cost is almost negligible — the machine is essentially all fixed cost, which is the signature of hard automation.
8. Locate the first crossover: NC centres against conventional machines. $$40N = 130{,}000 + 24N \quad\Longrightarrow\quad 16N = 130{,}000$$ $$\boxed{N_1 = 8{,}125 \text{ pieces per year, at } \$325{,}000}$$
9. Locate the second crossover: conventional machines against the transfer machine. $$130{,}000 + 24N \;=\; 800{,}000 + 0.0851N$$ $$23.9149N = 670{,}000 \quad\Longrightarrow\quad N_2 = 28{,}016 \text{ pieces per year}$$ $$\boxed{N_2 \approx 28{,}000 \text{ pieces per year, at } \$802{,}400}$$ The NC and transfer lines also cross, at 20,043 pieces, but conventional machining is already cheaper than both at that quantity, so that crossing does not appear on the lower envelope and is not a decision point.
10. Assemble the decision rule and test the ranking. Taking the minimum of the three costs at representative quantities confirms the envelope: at 5,000 pieces the NC centres cost $200,000 against $250,000 conventional; at 20,000 the conventional route costs $610,000 against $800,000 NC; at 50,000 the transfer machine costs $804,300 against $1,330,000 conventional. Hence $$\boxed{\text{NC centres } N < 8{,}125;\quad \text{conventional } 8{,}125 < N < 28{,}016;\quad \text{transfer } N > 28{,}016}$$
11. Confirm the transfer machine can physically make the quantity that justifies it. At 188 pieces per hour over a single 8-hour shift for 240 working days the transfer machine can produce 360,960 pieces per year, which is an order of magnitude above its break-even quantity. The machine is therefore never capacity-limited at the volumes where it is economic — it will run a small fraction of the year, and the decision is driven entirely by its annual fixed charge, not by its speed.