ENG106W12_HW5-solution

ENG106W12_HW5-solution - ENG 106 Homework #5 Solutions...

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ENG 106 Homework #5 Solutions Winter 2012 Note: Unless otherwise stated, cash flows are in actual dollars and i and MARR are market rates. 5.28a&c: A newly constructed bridge costs $10,000,000. The same bridge is estimated to need renovation every 10 years at a cost of $1,000,000. Annual repairs and maintenance are estimated to be $100,000 per year. a) If the interest rate is 5%, determine the capitalized equivalent cost of the bridge. c) Repeat the analysis with an interest rate of 10%. Interest rate is 5%. Construction = Pc = 10,000,000 Renovation = Pr = A/ ι = 1,000,000 *(A/F, 5%, 10)/ ι = 1,000,000 *()/.05 = 1,590,000 (alternatively, could calculate effective interest rate for 10 yrs : ι e = (1.05)10 – 1 = 63% and then calculate Pr = A/ ι e = 1,590,000, as above) Maintenance = PM = A/ ι = 100,000/ ι = 100,000/.05 = 2,000,000 CE(5%) = Pc+Pr +PM = $13,590,000 Repeat the analysis with an interest rate of 10%. Pc = 10,000,000 Pr = A/ ι = 1,000,000 *(A/F, 10%, 10)/ ι = 1,000,000 *()/.1 = 627,000 PM = A/ ι = 100,000/ ι = 100,000/.1 = 1,000,000 CE(10%) = Pc+Pr +PM = $11,627,000 (As interest rate increases, CE decreases. But note that market ι and inflation rate are coupled, so the actual-dollar amounts on the time line would also probably be larger if the market rate was higher due only to inflation.) 5.39 Consider the following two mutually exclusive investment projects: A B n Cash Flow Salvage Value Cash Flow Salvage Value 0 -$12,000 -$10,000 1 -$2,000 $6,000 -$2,100 $6000 2 -$2,000 $4,000 -$2,100 $3,000 3 -$2,000 $3,000 -$2,100 $1,000 4 -$2,000 $2,000 5 -$2,000 $2,000 Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the end of year listed. Both projects will be available (and can be repeated) with the same costs and salvage values for an indefinite
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ENG106W12_HW5-solution - ENG 106 Homework #5 Solutions...

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