Injection Presses — Different Lives: 11-CS-1 Engineering Economics, December 2015, Q5

Question 5: Injection Presses — 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 press is replaced identically at the end of its life (or compared over the LCM period). Annual Worth builds this in.

(b) Annual Worth (i = 9%)

Press A (20 yr): $CR = 300{,}000(A/P,9\%,20)-75{,}000(A/F,9\%,20)=32{,}863.94-1{,}465.99=31{,}397.96$; maintenance $=1{,}500+200(A/G,9\%,20)=1{,}500+200(6.76745)=2{,}853.49$.

EAC_A = 31{,}397.96 + 4{,}500 + 3{,}000 + 2{,}853.49 = \boxed{\$41{,}751/\text{yr}}

Press B (25 yr): $CR = 350{,}000(A/P,9\%,25)-75{,}000(A/F,9\%,25)=35{,}632.19-885.47=34{,}746.72$; maintenance $=1{,}000+150(A/G,9\%,25)=1{,}000+150(7.83160)=2{,}174.74$.

EAC_B = 34{,}746.72 + 3{,}500 + 2{,}000 + 2{,}174.74 = \boxed{\$42{,}421/\text{yr}}

Since $EAC_A < EAC_B$, select Press A.

(c) Present Worth

Present Worth needs a common study period. The lives are 20 and 25 years, whose least common multiple is 100 years: Press A is replaced five times over, Press B four times. Under the repeatability assumption of part (a) each repetition costs the same equivalent annual amount, so the 100-year present worth of either alternative is simply its EAC spread over 100 years, with $(P/A,9\%,100)=11.1091$:

PW_A = 41{,}751(11.1091) \approx \$463{,}800 \qquad PW_B = 42{,}421(11.1091) \approx \$471{,}300

Press A has the lower 100-year present cost, so Press A is again preferred. The ranking cannot differ from part (b), because both EACs are multiplied by the same factor.

(d) Do PW and AW Always Agree?

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

(e) Press B Salvage for a 20-Year Study

Truncating B to 20 years with unknown salvage $S_B$: $CR = 350{,}000(A/P,9\%,20)-S_B(A/F,9\%,20)=38{,}341-0.0195465\,S_B$; instalment 3,500; operating 2,000; 20-yr maintenance $=1{,}000+150(A/G,9\%,20)=2{,}015$. Setting $EAC_B(20)=EAC_A=41{,}751$:

45{,}856 - 0.0195465\,S_B = 41{,}751 \;\Rightarrow\; S_B = \frac{4{,}105}{0.0195465} \approx \boxed{\$210{,}000}

Over a 20-year study period Press B becomes the better choice only if its salvage value is about $210,000 or more. That is a demanding requirement, not an easy one: it is roughly 2.8 times the $75,000 salvage the company actually expects, and it asks a 20-year-old press to retain 60% of its $350,000 purchase price. On the stated estimate the answer to part (b) is therefore robust — Press A would remain the choice unless Press B held its resale value far better than assumed.

Concept & Theory: The Recurring Unequal-Life Template

This is the same template as the earlier press/pump/van comparisons: compute each alternative's equivalent annual cost via capital recovery plus annualized operating and (gradient) maintenance costs, then pick the lower. Annual Worth handles the 20- vs 25-year mismatch directly under the repeatability assumption, and PW over the LCM gives the identical ranking. Solving for the salvage that equalizes the two (part e) is the standard sensitivity check, and it is worth reading the answer rather than just reporting it: the break-even salvage of about $210,000 is roughly 2.8 times the salvage the company actually expects, so the margin between the two presses, though only $670/yr, rests on an assumption that would have to be badly wrong before the choice changed.