- 2065 Chaitra · 10 marks
Nepal Airlines jet plane choice, 4 years and infinite
Short answer
For 4-year need: annual equivalent cost A = 8.68 crore, B (using6 crore salvage at year 4) = 9.62 crore; select A. For infinite period: capitalized cost A = 72.37 crore, B = 62.98 crore; select B.
Nepal Airlines is planning to purchase a jet plane. The estimates on two types of plane under consideration are:
| Project | Plane A | Plane B |
|---|---|---|
| First investment cost | 25,00,00,000 | 30,00,00,000 |
| Annual O & M | 1,50,00,000 | 1,00,00,000 |
| Useful Life | 4 years | 6 years |
| Salvage value | 5,00,00,000 | 6,00,00,000 |
| MARR = 12 % | ||
| Which plane is the best one if it is believed that the plane will be used for i) 4 years and ii) infinite period? |
All amounts in Rs crore (1 crore = 1,00,00,000). Both planes are cost-only alternatives, so the one with the lower equivalent annual cost (AEC) is chosen. MARR = 12%.
Factors: (A/P,12%,4) = 0.32923, (A/F,12%,4) = 0.20923; (A/P,12%,6) = 0.24323, (A/F,12%,6) = 0.12323.
(i) Service needed for 4 years
AEC(A) = 25(A/P,12%,4) − 5(A/F,12%,4) + 1.5
= 25(0.32923) − 5(0.20923) + 1.5
= 8.231 − 1.046 + 1.5 = 8.685 crore/yr
Plane B has a 6-year life, so for a 4-year need its value at year 4 is required. It is not given, so assume the quoted salvage value of Rs 6 crore is what it realises at year 4 (the only figure available):
AEC(B, 4 yr) = 30(A/P,12%,4) − 6(A/F,12%,4) + 1.0
= 9.877 − 1.255 + 1.0 = 9.622 crore/yr
AEC(A) = 8.68 < AEC(B) = 9.62, so Plane A is selected for 4 years.
Sensitivity of the assumption: B would win only if its year-4 market value exceeds about Rs 10.5 crore (for example, Rs 14 crore from straight-line decline of 30 → 6 gives AEC = 7.95). On the given data, A is chosen.
(ii) Service needed for an infinite period
Each plane is replaced by an identical one at the end of its life (repeatability), so capitalized equivalent cost CE = AEC / i.
AEC(B, 6 yr) = 30(A/P,12%,6) − 6(A/F,12%,6) + 1.0
= 7.297 − 0.739 + 1.0 = 7.557 crore/yr
CE(A) = 8.685 / 0.12 = 72.37 crore
CE(B) = 7.557 / 0.12 = 62.98 crore
| Plane | AEC (crore/yr) | Capitalized cost (crore) |
|---|---|---|
| A (4 yr) | 8.685 | 72.37 |
| B (6 yr) | 7.557 | 62.98 |
Answer: (i) For 4 years choose Plane A (AEC 8.68 vs 9.62 crore/yr). (ii) For an infinite period choose Plane B (AEC 7.56 vs 8.68 crore/yr; capitalized cost 63.0 vs 72.4 crore).
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- AW sensitivity to investment and net annual cash flow · Engineering Economics, chapter 6
- PW sensitivity to investment, income, life by ±25% · Engineering Economics, chapter 6
- Define ESL; find ESL of 50,000 asset · Engineering Economics, chapter 5
- Modified B/C ratio for 100,000 project · Engineering Economics, chapter 3
- Repeatability, 4-year study, infinite life · Engineering Economics, chapter 4
- IRR by trial and error; investment decision · Engineering Economics, chapter 3
- EVA; 60 days, 25% work, Rs 12,00,000 · Project Engineering, chapter 4
- EPS and book value; ABC project Rs 6,00,000 · Project Engineering, chapter 6
- Service reservoir of 30 m diameter, 1,00,000 population · Water Supply Engineering, chapter 7
- EVA; BCWS 9,50,000, BCWP 8,00,000, ACWP 10,00,000 · Project Engineering, chapter 4