23-CS-1 Engineering Economics · December 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2017 — 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.
Assumptions (permitted by NOTE 1): the labourer's $32,000/yr (salary $30,000 + benefits $2,000) is paid at each year-end over the 5-year horizon. The robot's $100,000 (purchase $90,000 + installation and first-time programming $10,000) falls at $t=0$; the first-time programming is already in that $100,000, so a $8,000 re-programming is charged at the end of every year that is a multiple of the re-programming interval and still inside the horizon — $t=1,2,3,4,5$ in part (a) and $t=2,4$ in part (b). This assumption is load-bearing: the part-(a) margin is only about $1,600, so dropping the year-5 re-programming would reverse the decision.
Labour: $PW_{\text{labour}} = 32{,}000(P/A,7\%,5) = 32{,}000(4.100197) = \$131{,}206$. Robot: $100,000 initial plus $8,000 reprogramming each of the 5 years:
The robot ($132,802) costs slightly more than the labourer ($131,206), so with yearly reprogramming it is not economic to replace (by a narrow ~$1,600 margin).
Reprogramming now occurs only at years 2 and 4: $PW_{\text{reprog}} = 8{,}000[(P/F,7\%,2)+(P/F,7\%,4)] = 8{,}000(0.87344+0.76290)=\$13{,}091$.
Now the robot is clearly cheaper—replacement is economic, saving about $18,100 in present worth. Less-frequent reprogramming is what tips the decision.
Labour: $EAC = \$32{,}000$/yr. Robot: $EAC = 100{,}000(A/P,7\%,5) + 8{,}000 = 24{,}389 + 8{,}000 = \$32{,}389$/yr.
The robot's $32,389/yr exceeds the labourer's $32,000/yr, so it is not economic—the same conclusion as part (a).
Yes, provided both methods use the same MARR and the same study period. Annual worth is simply present worth multiplied by the capital-recovery factor $(A/P,i,n)$, a strictly positive constant for a given $i$ and $n$; multiplying every alternative's present worth by the same positive number cannot change their ranking or the sign of their difference. Parts (a) and (c) show it directly: $132{,}802\times(A/P,7\%,5)=32{,}389$ and $131{,}206\times(A/P,7\%,5)=32{,}000$, so both methods say "do not replace".
The one case that needs care is alternatives with unequal lives. Annual worth can then be compared directly (it implicitly assumes each alternative is repeated on the same terms), whereas present worth must first be put on a common horizon — the least common multiple of the lives, or an explicit study period with salvage values. If present worth were computed over each alternative's own, different life, the two methods could appear to disagree; that is an error in the comparison, not a real difference between the methods.
