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chapter7A

# chapter7A - Additional Solutions Chapter Seven Replacement...

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Additional Solutions: Chapter Seven: Replacement Decisions 7S.1 The \$26 000 cost of the old machine is a sunk cost, and thus irrelevant to the analysis. The present cost of switching to the new machine is PC = 11 000 – 8 000 – 1 000( P/F, 0.2,5) – 2 000( P/A ,0.2,5) = 3 000 – 1 000(0.4019) -2 000(2.9906) = -3 383 Thus, the company will save \$3 383 by replacing the old machine. So they should do it. *7S.2 Retiring the machine now has a present value of £10 000. Retiring it in a year’s time has a present value of: PV 1 = (5 000+6 000)( P/F, 0.15,1) = 11 000(0.8696) = £9 566 Retiring it in two years time has a present value of: PV 2 = 5 000( P/F ,0.15,1) + (4 000+4 000) (P/F, 0.15,2) = 5 000(0.8696) + 8 000(0.7561) = £10 397 Retiring it in three years has a present value of: PV 3 = 5 000( P/F, 0.15,1) + 4 000 (P/F, 0.15,2) + (3 000+2 000) ( P/F, 0.15,3) = 5 000(0.8696) + 4 000(0.7561) +5 000(0.6575) = £10 660 And waiting until the end of the fourth year has a present value of: PV 4 = 5 000( P/F, 0.15,1) + 4 000( P/F, 0.15,2) + 3 000( P/F, 0.15,3) + 2 000( P/F ,0.15,4) = 5 000(0.8696) + 4 000(0.7561) + 3 000(0.6575) + 2 000(0.5717) = £10 488 We conclude that the machine should be retired at the end of the third year.

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7S.3 The first thing to do is to see if it is worth replacing one truck now. The cost of
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chapter7A - Additional Solutions Chapter Seven Replacement...

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