--- title: "Nepal Airlines jet plane choice, 4 years and infinite" subject: "Engineering Economics" chapter: 4 key: "semester_6/engineering_economics:nepal-airlines-jet-plane-4yr-infinite" source: https://www.bctengineer.com/q/semester_6/engineering_economics/nepal-airlines-jet-plane-4yr-infinite site: BCT Engineering --- # Nepal Airlines jet plane choice, 4 years and infinite Engineering Economics · Chapter 4: Comparative Analysis of Alternatives - Asked in: 2065 Chaitra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/nepal-airlines-jet-plane-4yr-infinite - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4 ## Question 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? ## 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. ## Answer 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 ```text 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): ```text 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. ```text 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which criterion is used to choose between the planes? A. Highest annual revenue B. Highest salvage value C. Lowest annual equivalent cost (AEC) D. Longest useful life Answer: C. Both planes are cost-only, so the lower AEC is selected. (Key point: Criterion for cost alternatives) 2. Which is the correct AEC for plane A over 4 years (crore)? A. 25(A/P,12%,4) - 5(A/F,12%,4) + 1.5 B. 25(A/F,12%,4) - 5(A/P,12%,4) + 1.5 C. 25(A/P,12%,4) + 5(A/F,12%,4) + 1.5 D. 25(A/P,12%,4) - 5(A/F,12%,4) - 1.5 Answer: A. Capital recovery less salvage annuity, plus annual O&M, gives 8.231 - 1.046 + 1.5 = 8.685. (Key point: AEC formula) 3. For an infinite period, how is the comparison made and what is the result? A. Capitalized cost = AEC / i; B is better (62.98 vs 72.37 crore) B. Capitalized cost = AEC x i; A is better C. Capitalized cost = AEC / i; A is better D. Only first-cost is compared; A is better Answer: A. With repeated replacement, CE = AEC/i. B's AEC of 7.557 beats A's 8.685 over its own life, so B is chosen. 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