- 2065 Kartik · 16 marks
Repeatability, 4-year study, infinite life
Short answer
Repeatability: AW A = Rs10,268, AW B = Rs11,349; select B. For4-year study with B's salvage Rs2,50,000, AW B = -309; select A. Infinite life: capitalized worth A = Rs85,566, B = Rs94,571; select B.
Select the best project.
| Project A | Project B | |
|---|---|---|
| Initial investment (Rs.) | 3,50,000 | 5,00,000 |
| Annual revenues (Rs.) | 1,90,000 | 2,50,000 |
| Annual expenses (Rs.) | 64,500 | 1,38,000 |
| Useful life (years) | 4 | 8 |
| Salvage value at the end of useful life | 0 | 0 |
| Use | ||
| a) Repeatability assumption | ||
| b) Study period is 4 years | ||
| c) Infinite project life |
Net annual revenue = revenue − expenses: A = 1,90,000 − 64,500 = 1,25,500; B = 2,50,000 − 1,38,000 = 1,12,000. Salvage values are zero. MARR is not given; assume 12%.
a) Repeatability assumption (compare AW over own lives)
AW(A) = −3,50,000(A/P,12%,4) + 1,25,500
= −3,50,000(0.32923) + 1,25,500 = 10,268
AW(B) = −5,00,000(A/P,12%,8) + 1,12,000
= −5,00,000(0.20130) + 1,12,000 = 11,349
B has the higher AW. Select B.
b) Study period = 4 years
Only the first 4 years of B are used, so B must be sold at year 4. Its value then is not given; assume it falls straight-line from 5,00,000 to 0 over 8 years, i.e. Rs 2,50,000 at year 4.
AW(A, 4 yr) = 10,268
AW(B, 4 yr) = −5,00,000(A/P,12%,4) + 2,50,000(A/F,12%,4) + 1,12,000
= −164,617 + 52,309 + 1,12,000 = -309
With no resale value at all, AW(B) would be even lower (−52,617). Either way A is clearly better. Select A.
c) Infinite project life
Capitalized worth CW = AW / i (projects repeated for ever).
CW(A) = 10,268 / 0.12 = 85,566
CW(B) = 11,349 / 0.12 = 94,571
Select B.
| Basis | AW(A) | AW(B) | Choice |
|---|---|---|---|
| a) Repeatability | 10,268 | 11,349 | B |
| b) 4-year study | 10,268 | -309 | A |
| c) Infinite (CW) | 85,566 | 94,571 | B |
Answer: B under repeatability and for infinite life; A when only 4 years of service are needed.
✓ Answer checked by Claude Opus 5.5 (high effort)
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