Question

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Question 1 0.6 pts The city of Mersey is planning to construct a bridge across the Cavendashim river. The first cost of the bridge will be $6,565,835. Annual maintenance and repairs will be $25,725 for the first 5 years, and then will increase to $36,476 for each of the next 10 years. For the final 5 years, the annual maintenance and repairs will increase again to $53,981 per year. A major overhaul of $465,724 will be performed at the end of year 9. Using an interest rate of 2%, what is the equivalent uniform annual cost for the 20-year period? Hint: Since this is an equivalent uniform annual cost, do you need to indicate that as negative? See example 6-3 and 6-4 in the book. Enter your answer as 1234 Round your answer. Do not use a dollar sign ("$"), any commas (";"), or a decimal point (":").

Answer 1

For calculating the Equivalent Uniform Annual Cost we need to convert the net present value to annual terms which we can do by treating the net present value obtained as the present value of an ordinary annuity of duration equal to the life of project.

Equivalent Uniform Annual Cost =NPV × r/(1 − (1 + r)^-n )

Where:

NPV = net present value

r = interest rate

n = life of project

We calculate the NPV in excel.

A B C D E 1। 2 Interest Rate | 2% Life of project 20 in years | 3 4 Year Initial Cost Annual cos Overhaul cost Total Cost Present value of each _ 6565435| _ 6565435 65,65,835 ]=E5 25725 _ 25725| 25,221]=E5/(1+SBS2)^A6 25725| _ 25725| | 24,726 ]=E7/(1+5B52)^A7 2572s| 25725 24,241]=E8/(1+5B12)^A৪ 25725| 25725] 23,766 ]=E9/(1+5B12)^A9 25725| 25725 23,300]=E10/(1+5B52)^A10 36476 | 36476 32,390 ]=E11/(1+5B12)^A11 36475 | 36476 31,755 ]=E12/(1+5B12)^A12 36476 | 36476 31,132]=E13/(1+5B12)^A13 36475| _ 465724] 502200| _ ,20,218 ]=E14/(1+5B52)^A14 1ol 36475| 36475| _ 29,923 ]=E15/(1+SB52)^A15 36476| 36475| 29,336 ]=E16/(1+SB52)^A16 36476 | 36476| 28,761 ]=E17/(1+5B52)^A17 3647s 28, 197]=E18/(1+5B12)^A14 36476 36476 | 27,644 ]=E19/(1+5B12)^A19 36476 36476| 27,102]=E20/(1+5BS2)^A20. 53931] 53991 39,322]=E21/(1+5B12)^A21 53991] 53991 33,551]=E22/(1+5B52)^A22 539811 53991] 37,795 ]=E23/(1+SB52)^A23 53941] 37,054 ]=E24/(1+SBS2)^A24 20] _53941] | | 53941[____ 36,328 ]=E25/(1+5B52)^A25 | 26 NPC | | T 75,62,598 ]=SUM(F5:F25) |||||||||||||||||||| 36475 462504]=F26 (B2/(1-(1+B2)^-E2)) 28 EUAC _29 NPVxr/(1- (1+r)-n)

This gives us the EUAC as 462504