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hw5 sol

# hw5 sol - CE 167 Engineering Economy Assignment 5 Solution...

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CE 167: Engineering Economy Assignment 5 Solution Set Spring 2008 1) 0 1 2 3 4 5 7 8 \$180 \$ 100 \$90 \$ 100 \$ 100 6 \$90 \$ 100 \$ 100 a) without discounting year cashflow Account balance 0 -\$180 -\$180 1 -\$90 -\$270 2 \$0 -\$270 3 \$100 -\$170 4 \$100 -\$70 5 \$100 \$30 6 -\$90 -\$60 7 \$100 \$40 8 \$100 \$140 Payback period covers the time span until the account balance hits zero first. This happens between years 4 and 5. Using interpolation payback period = 4.7 years b. To solve for the payback period with discounted cash flows, repeat the above calculation, but discount each of the cash flows, at 5%,to their present value (P). year Cashflow (P/F, i%, n) P=F(P/F, i%, n) Account balance 0 -\$180 1 -180.00 -\$180.00 1 -\$90 0.9524 -85.72 -\$265.72 2 \$0 0.907 0.00 -\$265.72 3 \$100 0.8638 86.38 -\$179.34 4 \$100 0.8227 82.27 -\$97.07 5 \$100 0.7835 78.35 -\$18.72 6 -\$90 0.7462 -67.16 -\$85.87 7 \$100 0.7107 71.07 -\$14.80 8 \$100 0.6768 67.68 \$52.88 Discounted payback period is between years 7 and 8. Using interpolation discounted payback period = 7.2 years

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hw5 sol - CE 167 Engineering Economy Assignment 5 Solution...

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