Module15HWSol

# 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 Â ·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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