Professional Practice • Engineering Economics
Comparing Alternatives: PW, AW & Rate of Return
6 min read
Worked: two lighting retrofits
A parking structure can install retrofit A for $20,000 with $3,000 a year in energy and maintenance, or retrofit B for $28,000 with $1,500 a year. Both last 10 years with no salvage, and the owner’s MARR is 8%. (P/A, 8%, 10) = 6.710.
| A | B | |
|---|---|---|
| PW of costs | 20,000 + 3,000 × 6.710 = $40,130 | 28,000 + 1,500 × 6.710 = $38,065 |
| AW (A/P = 0.1490) | $5,981 per year | $5,673 per year |
B costs less on either measure. The increment agrees: $8,000 more up front buys $1,500 a year, so (P/A, i, 10) = 5.33, which is an incremental return of about 13.4% — above 8%.
Check it stuck: a set from Engineering Economics, the knowledge area this unit teaches.
Quiz this area0 of 8 units read in this chapter