11-CS-1 Engineering Economics · December 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 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 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.
The two alternatives have unequal lives (10 years against 5), so they cannot be compared over their own lives without a common basis. The assumption is repeatability: each alternative is assumed to be replaced by an identical asset, at the same costs and in the same service, for as long as the service is required — equivalently, both are compared over the least common multiple of their lives (LCM = 10 years, i.e. the hoist repeated twice). An Annual Worth comparison builds this assumption in automatically, which is why AW is the natural method for unequal lives; a Present Worth comparison must be done explicitly over the LCM.
Robotic arm (10 yr). Capital recovery:
Maintenance is $3,000/yr for ten years plus an extra $2,000/yr in years 6–10, so its present worth is $3{,}000(P/A,7\%,10)+2{,}000(P/A,7\%,5)(P/F,7\%,5)=21{,}071+5{,}847=\$26{,}918$, which annualizes to $26{,}918(0.142378)=\$3{,}833$/yr. Labour is nil.
Electric hoist (5 yr). $CR = 9{,}000(A/P,7\%,5)-1{,}000(A/F,7\%,5)=9{,}000(0.243891)-1{,}000(0.173891)=2{,}195-174=\$2{,}021$/yr. Maintenance is $400/yr for five years plus an extra $300/yr in years 4–5: $400(P/A,7\%,5)+300(P/A,7\%,2)(P/F,7\%,3)=1{,}640+443=\$2{,}083$, annualizing to $2{,}083(0.243891)=\$508$/yr. Labour is $30,000/yr.
Since $EAC_{\text{robot}} < EAC_{\text{hoist}}$, the robotic arm is more economic — it is cheaper by $8,413/yr. Its far larger first cost is outweighed by eliminating the operator, whose $30,000/yr is by itself larger than the arm's entire annual cost.
Present Worth must be taken over a common period, so use the LCM of 10 years — the arm once, the hoist bought twice (at $t=0$ and again at $t=5$). Taking costs as positive:
One hoist cycle costs $9{,}000 + 2{,}083 + 30{,}000(P/A,7\%,5) - 1{,}000(P/F,7\%,5) = 9{,}000+2{,}083+123{,}006-713 = \$133{,}376$, and the second cycle is that amount shifted five years:
The robotic arm is again preferred, cheaper in present-worth terms by $59,095. That is the same answer as (b), as it must be — the unrounded annual-worth gap of $8,413.81/yr multiplied by $(P/A,7\%,10)=7.023582$ gives back $59,095.
The study period now fixes the horizon at 5 years, so the repeatability assumption is dropped and the arm is valued on what it is worth when the study ends:
Over five years the arm's maintenance is the flat $3,000/yr (the step up to $5,000 never arrives), and labour is nil, so $EAC_{\text{robot}} = 31{,}017+3{,}000 = \$34{,}017$/yr. The hoist's life already equals the study period, so it is unchanged at $32,529/yr. Now $EAC_{\text{hoist}} < EAC_{\text{robot}}$, so over a 5-year horizon the electric hoist is more economical, by $1,488/yr — the arm's large first cost cannot be recovered in only five years even with the much higher salvage. The decision has reversed.
Yes. For any alternative, $AW = PW\,(A/P,i,n)$, and $(A/P,i,n)$ is a strictly positive constant that depends only on the MARR and the study period — both of which are the same for every alternative being compared. Multiplying every alternative's present worth by the same positive number cannot change their ranking, so the two methods must select the same alternative. The proviso is the one part (d) illustrates: they agree only when both are applied to the same study period and the same MARR. What changed the answer in (d) was the horizon, not the method.
