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11-CS-1 Engineering Economics · May 2019

Question 5 of 5: Waterjet Cutting Machine — Replacement Analysis

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National Exams — May 2019 — 11-CS-1 Engineering Economics. Three hours; open book; any non-communicating calculator permitted. Any four of the five questions constitute a complete exam paper and each question is of equal value (25 marks). Fully worked solutions to all five follow.

Question 5: Waterjet Cutting Machine — Replacement Analysis (25 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.

Method. Keeping the old machine for $n$ more years means forgoing its $49,000 market value now, paying the operating and maintenance cost of each year kept, and recovering the salvage value $S_n$ at the end. Bring all of that to $t=0$ and spread it over the $n$ years:

$$EAC(n) = \left[\,49{,}000 + \sum_{k=1}^{n} OM_k\,(P/F,10\%,k) - S_n\,(P/F,10\%,n)\right](A/P,10\%,n)$$

The discount factors needed are $(P/F,10\%,k) = 0.909091,\;0.826446,\;0.751315,\;0.683013$ for $k=1\ldots4$.

(a) Equivalent Annual Cost for One to Four More Years

  1. One year. With a single year there is no discounting to do inside the bracket — capital recovery plus the year's operating cost is enough:
    $$EAC(1) = 49{,}000(1.10) - 31{,}500 + 17{,}000 = 53{,}900 - 31{,}500 + 17{,}000 = \$39{,}400$$
  2. Two years. Present worth of the cost of keeping, then annualise with $(A/P,10\%,2)=0.576190$:
    $$P(2) = 49{,}000 + 15{,}454.55 + 17{,}619.83 - 16{,}425.61 = \$65{,}648.77$$
    $$EAC(2) = 65{,}648.77(0.576190) = \$37{,}826$$
  3. Three years, with $(A/P,10\%,3)=0.402115$:
    $$P(3) = 49{,}000 + 15{,}454.55 + 17{,}619.83 + 20{,}139.75 - 11{,}762.59 = \$90{,}451.54$$
    $$EAC(3) = 90{,}451.54(0.402115) = \$36{,}372$$
  4. Four years, with $(A/P,10\%,4)=0.315471$:
    $$P(4) = 49{,}000 + 15{,}454.55 + 17{,}619.83 + 20{,}139.75 + 23{,}068.08 - 4{,}604.87 = \$120{,}677.34$$
    $$EAC(4) = 120{,}677.34(0.315471) = \$38{,}070$$
Years kept, $n$Salvage $S_n$O&M in year $n$Present worth of keepingEAC
1$31,500$17,000$35,818$39,400
2$19,875$21,320$65,649$37,826
3$15,656$26,806$90,452$36,372
4$6,742$33,774$120,677$38,070

(b) Remaining Economic Life of the Old Machine

The economic life is the number of years that minimises the equivalent annual cost. The series falls from $39,400 to $37,826 to $36,372 and then turns up to $38,070, so

$$n^{*} = \boxed{3\ \text{years},\quad EAC_{\min} = \$36{,}372}$$

The minimum is interior, which is the usual shape and reflects two opposing forces. Spreading the $49,000 of forgone market value over more years pulls the annual cost down; the rising operating and maintenance cost ($17,000 growing to $33,774) and the shrinking salvage push it up. Through year 3 the first effect dominates; in year 4 the second takes over, helped by the salvage collapsing from $15,656 to $6,742.

(c) Should the Plant Replace the Old Machine Now?

Yes — replace now. The challenger can deliver the same service for an equivalent annual cost of $35,000, whereas the best the defender can manage over any remaining service life is $36,372 per year, at its economic life of three years:

$$EAC_{\text{new}} = \$35{,}000 \;\lt\; EAC_{\min,\text{old}} = \$36{,}372$$

Keeping the old machine costs the plant about $1,372 per year more than the alternative, on the defender's own most favourable terms. Nor does the year-by-year marginal-cost test rescue it: the cost of keeping the old machine for just one more year is

$$MC_1 = \underbrace{17{,}000}_{\text{O\&M}} + \underbrace{(49{,}000-31{,}500)}_{\text{loss in market value}} + \underbrace{0.10(49{,}000)}_{\text{interest on capital tied up}} = \$39{,}400$$

which also exceeds $35,000. Both tests say the same thing, so the recommendation is firm.

(The marginal costs of the second and third years are $36,095 and $33,013, so they do not rise monotonically. That is precisely why the minimum-EAC comparison, not a single year's marginal cost, is the governing test here: a defender whose marginal cost dips later cannot be judged on year 1 alone, and on the correct test it still loses.)

(d) What Is the Sunk Cost?

The $130,000 paid for the old waterjet cutting machine seven years ago is the sunk cost. It is money already spent and unrecoverable; no decision taken today can change it, so it must not appear anywhere in the analysis.

Two points justify the answer:

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