Module15HWSol - Module 15 Chapter 9 D5 The present worth of...

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Module 15 Chapter 9 D5 The present worth of the project is given by PW(i) = $1.25M + $1M(P/A, i, 10) $0.35M(P/A, i, 10) $35K(P/G, i, 10) + $10K(P/F, i, 10) Solving the equation above for different MARR values, we get MARR PW -20% $16,837,043.40 -10% $7,420,497.66 0% $3,685,000.00 10% $1,946,627.08 20% $1,025,674.07 30% $484,174.29 40% $138,229.55 50% -$97,803.69 60% -$267,461.18 and From the graph, we see that the project is rejected when MARR is greater than 45%. -$1,000,000.00 $0.00 $1,000,000.00 $2,000,000.00 $3,000,000.00 $4,000,000.00 $5,000,000.00 0% 20% 40% 60% 80% PW MARR
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D7 ±²³´µ ¶ ·¸¹ º » º ¸¼½¼¼¼²¾¿À½ ³´½ »¼µ ·¸¹¼½¼¼¼ Á ·»¼½¼¼¼²¾¿À½ ³´½ »¼µ  ·¹¼½¼¼¼²¾¿Ã½ ³´½ »¼µ ¶ ¸¼ÄÅ The investment is attractive. D8 a) PW(15%) = $100K + $23K(P/A, 15%, 8) = $3,208 => accept project b) FW(10%) = $100K(F/P, 10%, 8) + $23K(F/A, 10%, 8) = $48,667 => accept project c) AE(25%) = $100K(A/P, 25%, 8)
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Module15HWSol - Module 15 Chapter 9 D5 The present worth of...

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