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23-CS-1 Engineering Economics · December 2013

Question 5 of 5: Cargo Vans — Different Lives

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Notes on this paper

National Exams — December 2013 — 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 are given below; standard compound-interest factors are used and minor rounding is immaterial.

Question 5: Cargo Vans — 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.

Assumption: the down payment is the initial (capital) cost recovered against the $7,500 salvage; the annual instalment is treated as a recurring annual cost.

(a) Necessary Assumption

The alternatives must be assumed repeatable—each replaced identically at the end of its life (or compared over the least-common-multiple period, 60 years). Annual Worth builds this repeatability in automatically.

(b) Annual Worth (i = 8%)

Van A (10 yr): $CR = 11{,}000(A/P,8\%,10) - 7{,}500(A/F,8\%,10) = 11{,}000(0.14903)-7{,}500(0.06903)=1{,}121.6$. Maintenance $=150+110(A/G,8\%,10)=150+110(3.8713)=575.8$.

$$EAC_A = 1{,}121.6 + 1{,}000 + 3{,}000 + 575.8 = \boxed{\$5{,}697/\text{yr}}$$

Van B (12 yr): $CR = 12{,}000(A/P,8\%,12) - 7{,}500(A/F,8\%,12) = 12{,}000(0.13270)-7{,}500(0.05270)=1{,}197.1$. Maintenance $=160+100(A/G,8\%,12)=160+100(4.5957)=619.6$.

$$EAC_B = 1{,}197.1 + 900 + 2{,}800 + 619.6 = \boxed{\$5{,}517/\text{yr}}$$

Since $EAC_B < EAC_A$, Van B should be selected.

(c) Present Worth

Present worth must be taken over a common period. Under the repeatability assumption of part (a) that is the least common multiple of the lives, 60 years (six cycles of Van A, five of Van B). A repeated asset's present worth over that period is its equivalent annual cost times $(P/A,8\%,60) = 12.37655$:

$$PW_A = 5{,}697(12.37655) = \$70{,}509 \qquad PW_B = 5{,}517(12.37655) = \$68{,}281$$

Van B has the lower present cost, by about $2,228, so Van B is again preferred. This is the same decision as Annual Worth, because both figures are scaled by the same positive factor.

(d) Do PW and AW Always Agree?

Yes, provided the same MARR and a consistent study period are used; they are equivalent measures.

(e) Salvage of B for a 10-Year Study

Truncating B to 10 years with unknown salvage $S_B$: $CR = 12{,}000(A/P,8\%,10) - S_B(A/F,8\%,10) = 1{,}788.3 - 0.06903\,S_B$; maintenance (10 yr) $=160+100(A/G,8\%,10)=547.1$; instalment 900; running 2,800. Set $EAC_B(10) = EAC_A = 5{,}697$:

$$1{,}788.3 - 0.06903\,S_B + 900 + 2{,}800 + 547.1 = 5{,}697 \;\Rightarrow\; 0.06903\,S_B = 338.0 \;\Rightarrow\; S_B \approx \boxed{\$4{,}900}$$

Over a 10-year study period, a salvage value of about $4,900 or more for Van B would make it the better choice than Van A.

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