NivaarExam PrepOfficial exam papers ↗

11-CS-1 Engineering Economics · December 2013

Question 5 of 5: Cargo Vans — Different Lives

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

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 $=160+100(A/G,8\%,12)=160+100(4.5958)=619.6$.

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

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

(c) Present Worth

Over the common (LCM = 60-year) period, $PW = EAC\times(P/A,8\%,60)$; both scale by the same factor, so Van B is again preferred—the same decision as Annual Worth.

(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.

Back to the paper →