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11-CS-1 Engineering Economics · May 2014

Question 4 of 5: CNC Machines — Different Lives

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

Notes on this paper

National Exams — May 2014 — 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 4: CNC Machines — 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 machine is assumed replaced identically at the end of its life (or compared over the least-common-multiple period, 60 years). Annual Worth builds this in.

(b) Annual Worth (i = 9%)

CNC plasma (15 yr): capital recovery $=90{,}000(A/P,9\%,15)-20{,}000(A/F,9\%,15)=11{,}165-681=10{,}484$. Maintenance (geometric, $A_1=1200,g=5\%$): present worth $=1200\frac{1-(1.05/1.09)^{15}}{0.04}=\$12{,}878$, annualized $\times(A/P,9\%,15)=\$1{,}598$.

$$EAC_{\text{plasma}} = 10{,}484 + 1{,}400 + 6{,}300 + 1{,}598 = \boxed{\$19{,}782/\text{yr}}$$

CNC laser (20 yr): capital recovery $=140{,}000(A/P,9\%,20)-50{,}000(A/F,9\%,20)=15{,}336-977=14{,}359$. Maintenance (geometric, $A_1=800,g=4\%$): PW $=800\frac{1-(1.04/1.09)^{20}}{0.05}=\$9{,}745$, annualized $=\$1{,}067$.

$$EAC_{\text{laser}} = 14{,}359 + 1{,}900 + 4{,}100 + 1{,}067 = \boxed{\$21{,}426/\text{yr}}$$

Since $EAC_{\text{plasma}} < EAC_{\text{laser}}$, select the CNC plasma.

(c) Present Worth

The lives are 15 and 20 years, so the common study period is their least common multiple, LCM = 60 years, over which each machine is assumed replaced identically. Over that period $PW = EAC\times(P/A,9\%,60)$, with $(P/A,9\%,60)=11.04799$:

$$PW_{\text{plasma}} = 19{,}782(11.04799) = \boxed{\$218{,}500}\qquad PW_{\text{laser}} = 21{,}426(11.04799) = \boxed{\$236{,}700}$$

The same factor multiplies both EACs, so the ranking cannot change: the CNC plasma is again preferred (lower present worth of cost)—the same decision as Annual Worth.

(d) Do PW and AW Always Agree?

Yes, given the same MARR and a consistent study period. The two measures differ only by the factor $(P/A,i,N)$, which is strictly positive, so multiplying every alternative's annual worth by it rescales all of them equally and can never change their order — only the units in which the answer is reported. Apparent disagreements between the two in practice always trace back to an inconsistent setup, most often comparing unequal lives on a present-worth basis without first extending both to a common study period (as was done here with the 60-year LCM), and not to the methods themselves.

(e) Laser Salvage for a 15-Year Study

Truncating the laser to 15 years with unknown salvage $S_L$: capital recovery $=140{,}000(A/P,9\%,15)-S_L(A/F,9\%,15)=17{,}368-0.034059\,S_L$; annual payment 1,900; operating 4,100; 15-yr maintenance annualized ≈ $1,004. Setting $EAC_{\text{laser}}(15)=EAC_{\text{plasma}}=19{,}782$:

$$24{,}372 - 0.034059\,S_L = 19{,}782 \;\Rightarrow\; S_L \approx \boxed{\$134{,}800}$$

A salvage of about $134,800 would be required—about 96% of the laser's own $140,000 down payment, and far above its stated $50,000 salvage. The laser would have to be worth almost as much after fifteen years of service as it cost new, which is not achievable for production machinery. Over a 15-year horizon the plasma therefore remains the better choice for any realistic laser salvage value.