3-D Printers — Different Lives: 11-CS-1 Engineering Economics, May 2016, Q4

Question 4: 3-D Printers — 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.

(a) Necessary Assumption

Repeatability—each printer is replaced identically at the end of its life (or compared over the LCM period, 12 years). Annual Worth builds this in.

(b) Annual Worth (i = 7%)

Machine A (6 yr): $CR = 6{,}400(A/P,7\%,6)-400(A/F,7\%,6)=1{,}343-56=1{,}287$; maintenance $=600+100(A/G,7\%,6)=830$.

EAC_A = 1{,}287 + 1{,}200 + 830 = \boxed{\$3{,}317/\text{yr}}

Machine B (4 yr): $CR = 4{,}400(A/P,7\%,4)-400(A/F,7\%,4)=1{,}299-90=1{,}209$; maintenance $=400+140(A/G,7\%,4)=598$.

EAC_B = 1{,}209 + 1{,}400 + 598 = \boxed{\$3{,}207/\text{yr}}

Since $EAC_B < EAC_A$, Machine B is more economic.

(c) Present Worth

Over the common (LCM = 12-year) period, $PW = EAC\times(P/A,7\%,12)$ for each; the same factor applies, so Machine B is again preferred.

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

Truncating A to 4 years with unknown salvage $S_A$: $CR = 6{,}400(A/P,7\%,4)-S_A(A/F,7\%,4)=1{,}889-0.225228\,S_A$; running 1,200; 4-yr maintenance $=600+100(A/G,7\%,4)=742$. Setting $EAC_A(4)=EAC_B=3{,}207.1$:

3{,}831.0 - 0.225228\,S_A = 3{,}207.1 \;\Rightarrow\; S_A \approx \boxed{\$2{,}770}

Over a 4-year study period, a salvage value of about $2,770 or more for Machine A would make it the preferred choice—about 43% of its $6,400 purchase price after four of its six years of service, which is a demanding but not impossible resale.

(e) Do PW and AW Always Agree?

Yes, with the same MARR and a consistent study period.

Question 4 of 5: 3-D Printers — Different Lives