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23-CS-1 Engineering Economics · May 2016

Question 4 of 5: 3-D Printers — Different Lives

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2016 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any four of the five questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all five questions follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 4: 3-D Printers — Different Lives (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.

(a) Necessary Assumption

The necessary assumption is repeatability: the service is needed for at least the least common multiple (LCM) of the two lives, here 12 years, and at the end of each life the machine is replaced by an identical machine with identical costs, so each machine's cash-flow cycle simply repeats (A twice, B three times) over that common period. Only then are the two alternatives compared over equal service. The alternative assumption is a common study period with an explicit salvage value for any machine cut short, which is what part (d) does. Annual Worth computed over each machine's own life builds the repeatability assumption in automatically.

(b) Annual Worth (i = 7%)

Machine A (6 yr): $CR = 6{,}400(A/P,7\%,6)-400(A/F,7\%,6)=1{,}343-56=1{,}287$; maintenance $=600+100(A/G,7\%,6)=830$.

$$EAC_A = 1{,}287 + 1{,}200 + 830 = \boxed{\$3{,}317/\text{yr}}$$

Machine B (4 yr): $CR = 4{,}400(A/P,7\%,4)-400(A/F,7\%,4)=1{,}299-90=1{,}209$; maintenance $=400+140(A/G,7\%,4)=598$.

$$EAC_B = 1{,}209 + 1{,}400 + 598 = \boxed{\$3{,}207/\text{yr}}$$

Since $EAC_B < EAC_A$, Machine B is more economic.

(c) Present Worth

Under the repeatability assumption the comparison period is the LCM of 6 and 4 years, i.e. 12 years (two cycles of A, three of B). Because every cycle repeats identically, the present worth of costs over 12 years is each machine's EAC capitalised at $(P/A,7\%,12)=7.94269$:

$$PW_A = 3{,}317(7.94269) = \boxed{\$26{,}346};\qquad PW_B = 3{,}207(7.94269) = \boxed{\$25{,}472}$$

Machine B has the lower present worth of costs (by about $874), so Machine B is again preferred. It cannot be otherwise: both EACs are multiplied by the same positive factor, so PW over the common period preserves the AW ranking.

(d) Salvage of A for a 4-Year Study

Truncating A to 4 years with unknown salvage $S_A$: $CR = 6{,}400(A/P,7\%,4)-S_A(A/F,7\%,4)=1{,}889-0.225228\,S_A$; running 1,200; 4-yr maintenance $=600+100(A/G,7\%,4)=742$. Setting $EAC_A(4)=EAC_B=3{,}207.1$:

$$3{,}831.0 - 0.225228\,S_A = 3{,}207.1 \;\Rightarrow\; S_A \approx \boxed{\$2{,}770}$$

Over a 4-year study period, a salvage value of about $2,770 or more for Machine A would make it the preferred choice—about 43% of its $6,400 purchase price after four of its six years of service, which is a demanding but not impossible resale.

(e) Do PW and AW Always Agree?

Yes—provided both are applied correctly, i.e. at the same MARR and over the same (or consistently repeated) study period. Then $PW = AW\times(P/A,i,n)$ with a strictly positive factor, so the two rank alternatives identically, as (b) and (c) show here. They appear to disagree only when misapplied—most commonly by computing each machine's PW over its own unequal life (A over 6 years, B over 4), which compares unequal service and is not a valid comparison.