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23-CS-1 Engineering Economics · December 2018

Question 2 of 5: Two 3-D Printers — Unequal Lives

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

Notes on this paper

National Exams — December 2018 — 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 follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 2: Two 3-D Printers — Unequal 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

Comparing unequal lives by annual worth assumes repeatability: each alternative can be replaced identically (same costs, in real terms) at the end of its life, so the service continues indefinitely. Equivalently, the alternatives are compared over the least-common-multiple study period (60 years here).

(b) Annual-Worth Comparison (i = 8%)

Machine A (n = 10): capital recovery $=11{,}000(A/P,8\%,10)-7{,}500(A/F,8\%,10)=1{,}639.3-517.7=\$1{,}121.6$; plus running+service $=\$4{,}000$; plus maintenance $150+110(A/G,8\%,10)=150+110(3.871)=\$575.8$.

$$AW_A = 1{,}121.6 + 4{,}000 + 575.8 = \$5{,}697\ \text{/yr}$$

Machine B (n = 12): capital recovery $=12{,}000(A/P,8\%,12)-7{,}500(A/F,8\%,12)=1{,}592.3-395.2=\$1{,}197.1$; plus running+service $=\$3{,}700$; plus maintenance $160+100(A/G,8\%,12)=160+100(4.596)=\$619.6$.

$$AW_B = 1{,}197.1 + 3{,}700 + 619.6 = \$5{,}517\ \text{/yr}$$

$AW_B(\$5{,}517) < AW_A(\$5{,}697)$ → select Machine B.

(c) Present-Worth Comparison

Over the LCM of 60 years, $PW = AW(P/A,8\%,60)$ with $(P/A,8\%,60)=12.37655$; the factor is the same positive constant for both, so the ranking is unchanged: Machine B ($PW_B = 5{,}516.70\times 12.37655 = \$68{,}278$ cost, versus $PW_A = 5{,}697.45\times 12.37655 = \$70{,}515$).

(d) Do PW and AW Always Agree?

Yes, provided the same study period and MARR are used; since $PW = AW(P/A,i,N)$ with a common factor, they can never rank alternatives differently.

(e) Salvage on B for a 10-Year Study Period

Truncating B to 10 years, its annual worth must fall below $AW_A=\$5{,}697$. With salvage $S$: $AW_B(10) = 12{,}000(A/P,8\%,10) - S(A/F,8\%,10) + 3{,}700 + [160+100(A/G,8\%,10)] = 6{,}035.49 - 0.069029\,S$. Setting this below $AW_A = 5{,}697.45$:

$$6{,}035.49 - 0.069029\,S < 5{,}697.45 \;\Rightarrow\; S > \frac{6{,}035.49 - 5{,}697.45}{0.069029} = \boxed{\$4{,}897}$$

Machine B's salvage value at the end of year 10 would need to exceed $4,897 (about $4,900) to make it the better choice. That is below the $7,500 it is expected to fetch at the end of its full 12-year life, so for a machine two years younger it is a plausible resale value: under a 10-year study Machine B remains the better choice unless its year-10 resale falls below about $4,900.