Question 9.6: Consider two competing projects, for which MARR = 16%:...
Consider two competing projects, for which MARR = 16%:
| C | A | n | |
| Project A | $100 000 | $23 000 | 9 |
| Project B | 100 000 | 35 000 | 4 |
Here, C is the initial investment, A is the annual net cash inflow, and n is the service life of the asset. The ROR method yields:
project A 0 = -$100 000 + $23 000 (P/A, i *, 9)
(A/P, i*, 9) = 0.23
i* = 17.7% (by interpolation in Appendix A)
project B 0 = -$100 000 + $35 000(P/A, i*, 4)
(A/P, i*, 4) = 0.35
i* ≈ 15%
Hence, according to the MARR, project A is acceptable and project B is not.
However, suppose that the cash flows can be reinvested at 25%, compounded annually. Thus, the $23 000 annual cash inflows from project A are actually equivalent to a future value
$23 000(F/A, 25%, 9) = $593.400
nine years hence, and the annual cash inflows from project B are actually equivalent to a future value
$35 000(F/A, 25%, 4) (F/P, 25%, 5) = $615 900
nine years hence. Thus (the initial investments being equal) project B is actually the preferred alternative.