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ENGR 120 · Engineering Economy
Problem Set 3 — Graded
Sample Student
83 / 100
83%
| Problem | Score |
|---|---|
| P1 · Simple vs. compound interest | 20 / 20 |
| P2 · Equivalent uniform annual cost | 16 / 20 |
| P3 · Internal rate of return | 25 / 25 |
| P4 · Present-worth comparison | 8 / 10 |
| P5 · Benefit–cost & sensitivity | 14 / 25 |
P4 · Present-worth comparison
8 / 10
Rubric
G1
Places the initial cost, annual savings, and terminal salvage with the
correct sign and timing 4 / 4
G2
Builds the present-worth comparison from the
correct cash flows, but
discounts the annual series one payment at a time 4 / 6
How to get full credit
Use (P/A, 8%, 8) for the annual series, then add the salvage with (P/F, 8%, 8).
- See Engineering Economy, Chapter 5.
- The method is explained in Class video 7.
Your work
PW = −28,000 + 3,200(P/F,8%,1) + 3,200(P/F,8%,2) + …
↑ each year discounted separately (should be P/A)
+ 12,000(P/F,8%,8)
PW ≈ $1,090
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