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11-CS-1 Engineering Economics · May 2013

Question 4 of 6: Photocopiers — Different Lives

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Notes on this paper

National Exams — May 2013 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any five of the six questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all six questions are given below. Standard compound-interest factors are used throughout; minor rounding differences are immaterial.

Question 4: Photocopiers — Different Lives (20 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

To compare alternatives with unequal lives, one assumes the service is repeated (replaced) identically—each alternative is renewed under the same costs indefinitely (or over a common study period, the least common multiple of lives). The Annual-Worth method builds in this repeatability assumption automatically.

(b) Annual Worth (Equivalent Annual Cost, i = 7%)

System A (6 yr): capital recovery $= 3200(A/P,7\%,6) - 200(A/F,7\%,6) = 3200(0.20980) - 200(0.13980) = 643.4$. Maintenance $= 300 + 50(A/G,7\%,6) = 300 + 50(2.303) = 415.2$.

$$EAC_A = 643.4 + 600 + 415.2 = \boxed{\$1{,}658.6/\text{yr}}$$

System B (4 yr): capital recovery $= 2200(A/P,7\%,4) - 200(A/F,7\%,4) = 2200(0.29523) - 200(0.22523) = 604.5$. Maintenance $= 200 + 70(A/G,7\%,4) = 200 + 70(1.4155) = 299.1$.

$$EAC_B = 604.5 + 700 + 299.1 = \boxed{\$1{,}603.5/\text{yr}}$$

Since $EAC_B < EAC_A$, System B is preferable.

(c) Present Worth (common study period, LCM = 12 yr)

Repeating each over 12 years, $PW = EAC\times(P/A,7\%,12) = EAC\times7.9427$: $PW_A = \$13{,}173$ vs $PW_B = \$12{,}736$ (both costs). System B is again preferred—the same decision as Annual Worth.

(d) Salvage of A for a 4-Year Study

Over 4 years, A's EAC with an unknown salvage $S$: capital recovery $= 3200(A/P,7\%,4) - S(A/F,7\%,4) = 944.7 - 0.22523S$; maintenance (4 yr) $= 300 + 50(A/G,7\%,4) = 370.8$; running $600$. Set equal to $EAC_B = 1603.5$:

$$944.7 - 0.22523S + 600 + 370.8 = 1603.5 \;\Rightarrow\; 0.22523S = 311.9 \;\Rightarrow\; S \approx \boxed{\$1{,}385}$$

A four-year salvage value of about $1,385 or more for Photocopier A would make it the preferred choice.

(e) Do PW and AW Always Agree?

Yes—provided a consistent study period and the same MARR are used, Present Worth and Annual Worth are equivalent transformations and always give the same decision.