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- StudyBlue
- Tennessee
- University of Tennessee - Knoxville
- Industrial Engineering
- Industrial Engineering 405
- Dr. Dai
- Homework 1 Solution

Matt E.

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HW#1 Solution 4-18 (a) Cash flow in year k = ($14/ton)(10,000 tons)(1.05)k-1 $187,613 $178,679 i = 10% / yr $162,068 0 1 2 3 4 5 6 7 (b) Solution based on Section 4.8: F7 = $140,000(F/P,15%,6) + $147,000(F/P,15%,5) + $154,350(F/P,15%,4) + $162,068(F/P,15%,3) + $170,171(F/P,15%,2) + $178,679(F/P,15%,1) + $187,613 = $140,000(2.3131) + $147,000(2.0114) + $154,350(1.7490) + $162,068(1.5209) + $170,171(1.3225) + $178,679(1.15) + $187,613 = $1,754,102.16 Solution based on Section 4.14: F7 = P0 (F/P,15%,7) = EMBED Equation.3 (F/P,15%,7) = EMBED Equation.3 (2.6600) = $1,754,268.81 The difference in F7 amounts is due to rounding the interest factor values. 4-21 A = $22,000 0 1 2 3 4 5 i = 15% / yr P = $22,000 (P/A, 15%, 5) = $22,000 (3.3522) = $73,748.40 The company can justify spending up to $73,748.40 for this piece of equipment. F12 = $24,000 (F/P, ½%, 12) = $24,000 (1.0617) = $25,480.80 4-35 P = $3,000 (P/A, 20%, 10) + $7,000 (P/A, 20%, 5) = $3,000 (4.1925) + $7,000 (2.9906) = $33,511.70 You can afford to pay as much as $33,511.70 for the equipment. 4-40 Original Payments = A1 = $1,000 (A/P, 10%, 10) = $1,000 (0.1627) = $162.70 Balance at EOY 3 = $162.70 (P/A, 10%, 7) + $500 = $162.70 (4.8684) + $500 = $1,292.09 4-49 Equivalent cash inflows = Equivalent cash outflows Using time 0 as the equivalence point and N = total life of the system: $2,000(P/F,18%,1) + $4,000(P/F,18%,2) + $5,000(P/A,18%,N-2)(P/F,18%,2) = $20,000 $2,000(0.8475) + $4,000(0.7182) + $5,000(P/A, 18%, N-2)(0.7182) = $20,000 $5,000 (P/A,18%,N-2)(0.7182) = $15432.20 (P/A, 18%, N-2) = 4.297 From Table C-17, (P/A,18%,8) = 4.078 and (P/A,18%,9) = 4.303 Thus, (N-2) = 9 and N = 11 years . Note that if the system lasts for only 10 years, the present equivalent of the cash inflows < present equivalent of the cash outflows.

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