--- title: "Replacement strategy for four-year need" subject: "Engineering Economics" chapter: 5 key: "semester_6/engineering_economics:replace-machine-aec-four-years" source: https://www.bctengineer.com/q/semester_6/engineering_economics/replace-machine-aec-four-years site: BCT Engineering --- # Replacement strategy for four-year need Engineering Economics · Chapter 5: Replacement Analysis - Asked in: 2072 Asoj, 4+4+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/replace-machine-aec-four-years - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5 ## Question An existing machine has market value of Rs. 10000 and decreases by Rs. 2000 per year. Its operating cost is Rs. 2500 in year 1 and increases by 20% each year for 4 years. New machine costs Rs. 20000 now and its market value will decrease by Rs. 20% per year for 4 years. Operating cost is Rs. 1500 in first year and increase by 30% each year. Calculate equivalent uniform annual cost of both existing and new machines. MARR = 15%. Formulate the best replacement strategy if we need the machine for four years only. ## Short answer Existing machine AEC for 4 years is Rs 6,352 (ESL 1 year, AEC Rs 6,000). New machine AEC for 4 years is Rs 7,582 (ESL 4 years). Best strategy: keep the existing machine for the entire 4 years, since its AEC is lower than any plan involving the new machine. ## Answer AEC(n) = [P + Σ O&M_k (P/F,15%,k) − MV_n (P/F,15%,n)] (A/P,15%,n), with P = 10,000 for the existing machine (its present market value) and 20,000 for the new machine. #### Existing machine (MV_n = 10,000 − 2,000n; O&M = 2,500(1.2)^(k−1)) | n | MV_n | O&M in year n | PW of cost | AEC(n) | |---|---|---|---|---| | 1 | 8,000 | 2,500 | 5,217 | 6,000 | | 2 | 6,000 | 3,000 | 9,905 | 6,093 | | 3 | 4,000 | 3,600 | 14,179 | 6,210 | | 4 | 2,000 | 4,320 | 18,136 | 6,352 | #### New machine (MV_n = 20,000(0.8)^n; O&M = 1,500(1.3)^(k−1)) | n | MV_n | O&M in year n | PW of cost | AEC(n) | |---|---|---|---|---| | 1 | 16,000 | 1,500 | 7,391 | 8,500 | | 2 | 12,800 | 1,950 | 13,100 | 8,058 | | 3 | 10,240 | 2,535 | 17,713 | 7,758 | | 4 | 8,192 | 3,296 | 21,646 | 7,582 | Example (existing, n = 2): PW = 10,000 + 2,500(0.8696) + 3,000(0.7561) − 6,000(0.7561) = 9,905; AEC = 9,905(A/P,15%,2 = 0.61512) = 6,093. - Existing machine: minimum AEC at n = 1, so ESL = 1 year (AEC = 6,000). - New machine: AEC keeps falling to n = 4, so ESL = 4 years (AEC = 7,582). #### Best replacement strategy for a 4-year need The need is for exactly 4 years, so compare "keep the existing machine for n years, then use the new machine for the remaining 4 − n years". Each plan is costed as one PW at 15% and then converted to a 4-year AEC. | Defender | Then | PW of cost | |---|---|---| | Keep defender 4 yr | - | 18,136 | | Keep defender 3 yr | new for 1 yr | 19,039 | | Keep defender 2 yr | new for 2 yr | 19,811 | | Keep defender 1 yr | new for 3 yr | 20,620 | | Replace now | new for 4 yr | 21,646 | Check, plan "keep 4 years": PW = 18,136; AEC = 6,352. **Answer: Keep the existing machine for the full 4 years and do not buy the new one.** Its AEC is Rs 6,352 per year for the 4 years, lower than every strategy that buys the new machine (the best of which costs Rs 6,669 per year and a new machine kept 4 years costs Rs 7,582 per year). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. AEC of the existing machine is 6,000; 6,093; 6,210; 6,352 for n = 1-4. What is its ESL? A. 1 year B. 2 years C. 3 years D. 4 years Answer: A. AEC is lowest at n = 1 and rises afterwards. (Key point: Existing machine ESL) 2. What P is used for the existing machine in AEC(n)? A. Its original purchase price B. Zero C. Its present market value of 10,000 D. The price of the new machine Answer: C. The defender is valued at its present market value; earlier cost is sunk. (Key point: Investment for existing machine) 3. How is the best strategy for exactly 4 years found? A. Compare each machine's minimum AEC only B. Choose the machine with lowest first cost C. Always replace now D. Cost each plan 'keep existing n years then new for 4-n years' as a PW and compare Answer: D. Because the horizon is fixed, all combinations are compared on equal 4-year PW; the best has the lowest PW. (Key point: Strategy for fixed 4-year need) 4. What is the recommended strategy? A. Keep the existing machine all 4 years (AEC 6,352) B. Replace now with the new machine C. Keep existing 1 year, then new D. Keep existing 2 years, then new Answer: A. PW of 18,136 is the lowest of all plans, which beats any plan buying the new machine. 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