hw3 sol - CE 167: Engineering Economy Assignment 3 Solution...

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CE 167: Engineering Economy Assignment 3 Solution Set Spring 2008 1) Call I the amount of money your grandparents will set aside two years from now. 0 1 2 3 4 5 7 8 I 15,000 Using NPV at year 0: 0 = - I (P/F, 5%, 2) + $15,000 (P/A, 5%, 5) (P/F, 5%, 2) = - I (0.9070) +$ 15,000 (4.329) (0.9070) I = $64,935 Alternatively, you could have used the NPV at year 2: 0 = - I + $15,000 (P/A, 5%, 5) 2.1) 240 MW = 110 MW (1 + 0.1) n 2.1818 = (1.10) n ln 2.1818 = n ln 1.10 n = ln 2.1818/ ln 1.10 n = 8.19 year The new capacity will thus be needed at n= 8 year. We round down 8.19 years to 8 year because of the ‘end-of-year convention’ (which means we lump arrows together and show them at the start or end of a period) but in this case we round to the start of the year because it is a capacity we must provide to meet demand. If we were to round up to 9 year, our customers would experience a blackout! 2.2) F = 110 MW (1+ 0.1) 13 = 379.7 MW = 380 MW F = 380 MW (total demand) - 240 MW (present capacity) (needed) 140 MW
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3.1) Find A/P A = ($25,000- $2,500) (A/P, 2%, 60) = $22,500 (0.02877) A = $647.28/mo. 3.2) Loan: $ 22,500 Duration of Loan: 60 months Interest rate: 2% per month Monthly payment: $ 647.28 End of Mo x Monthly Payment Interest Paid (2%) Principal Balance to be paid End of Mo x Monthly Payment Interest Paid (2%) Principal Balance to be paid 0 22500 30 647.280 296.941 350.339 14496.729 1 647.28 450.00 197.28 22302.72 31 647.28 289.93 357.35 14139.38 2 647.28 446.05 201.23 22101.49 32 647.28 282.79 364.49 13774.89 3 647.28 442.03 205.25 21896.24 33 647.28 275.50 371.78 13403.11 4 647.28 437.92 209.36 21686.89
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This note was uploaded on 02/27/2011 for the course CE 167 taught by Professor Horvath during the Fall '08 term at University of California, Berkeley.

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hw3 sol - CE 167: Engineering Economy Assignment 3 Solution...

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