- 2072 Asoj · 6 marks
Choose machine under 5% inflation
Short answer
Compute inflation-free rate i' = (15%-5%)/1.05 = 9.524%. Present worth of costs: Machine X Rs 2,877,913, Machine Y Rs 3,077,730. Select X since it has lower cost.
Choose the best project from the following alternatives.
| Project | Machine X | Machine Y |
|---|---|---|
| First Cost | 15,00000 | 20,00000 |
| Life | 7 years | 7 years |
| Salvage Value | 200000 | 300000 |
| Annual operating and maintenance cost | 300000 | 250000 |
| Assume an average inflation of 5% for the next five years and interest rate is 15% / year. |
Assumptions: costs and salvage values are given in today's (constant) dollars; inflation of 5% a year applies throughout the 7-year life (the "5 years" in the question is taken as the whole planning period); 15% is the market interest rate. The machines give the same service, so the one with the lower cost is chosen.
Inflation-free rate: i' = (i − f)/(1 + f) = (0.15 − 0.05)/1.05 = 9.524%. Evaluating constant-dollar costs at i' gives the same result as converting to actual dollars and discounting at 15%.
Factors at i' = 9.524%: (P/A,i',7) = 4.9457; (P/F,i',7) = 0.5290.
PW(X) = 15,00,000 + 3,00,000(P/A,i',7) − 2,00,000(P/F,i',7)
= 15,00,000 + 3,00,000(4.9457) − 2,00,000(0.5290)
= 2,877,913
PW(Y) = 20,00,000 + 2,50,000(P/A,i',7) − 3,00,000(P/F,i',7)
= 20,00,000 + 2,50,000(4.9457) − 3,00,000(0.5290)
= 3,077,730
| Machine | PW of cost (Rs) | AEC at i' (Rs/yr) |
|---|---|---|
| X | 2,877,913 | 581,902 |
| Y | 3,077,730 | 622,305 |
(Without inflation, discounting at 15% gives PW(X) = 2,672,939 and PW(Y) = 2,927,324, so the ranking is unchanged.)
Answer: Select Machine X. It has the lower present worth of cost (Rs 2,877,913 against Rs 3,077,730).
✓ Answer checked by Claude Opus 5.5 (high effort)
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