Question
In: Economics
The municipality of a major city in Quebec is planning to build a new bridge to...
The municipality of a major city in Quebec is planning to build a new bridge to decrease the traffic load on the existing bridges connecting both sides of the city across the river. Construction is to start in 2020 and is expected to take four years at a cost of $25 million per year. After construction is completed, the cost of operation and maintenance is expected to be $2.5 million for the first year and to increase by 2.8% per year thereafter. The scrap/salvage value of the bridge at the end of year 2053 is estimated to be $5 million. Consider the present to be the end of 2018/beginning of 2019 and the interest rate to be 8%.
a) Draw a cash flow diagram for this project (from
present till end of year 2053).
b) Find the Present Worth of this project.
c) Find the Future Worth of this project.
Solutions
Expert Solution
a) Cash Flow Diagram 
B) & C) part using excel,
i = 8% = 0.08
For PV factor, we use formula 1/ (1+0.08)^(year - 2018)
Construction cost per year for four years = 25m
O&M cost for first year = 2.5m
Cash flow of O&M cost = 1.028 * cost of O&M in previous year
| Year | Construction cost | O&M Cost | Scrap value | Net cash flow | PV Factor | Present value |
| 2018 | 0.00 | 1 | 0.00 | |||
| 2019 | 0.00 | 0.925925926 | 0.00 | |||
| 2020 | -25000000 | -25000000.00 | 0.85733882 | -21433470.51 | ||
| 2021 | -25000000 | -25000000.00 | 0.793832241 | -19845806.03 | ||
| 2022 | -25000000 | -25000000.00 | 0.735029853 | -18375746.32 | ||
| 2023 | -25000000 | -25000000.00 | 0.680583197 | -17014579.93 | ||
| 2024 | -2500000 | -2500000.00 | 0.630169627 | -1575424.07 | ||
| 2025 | -2570000 | -2570000.00 | 0.583490395 | -1499570.32 | ||
| 2026 | -2641960 | -2641960.00 | 0.540268885 | -1427368.78 | ||
| 2027 | -2715935 | -2715934.88 | 0.500248967 | -1358643.62 | ||
| 2028 | -2791981 | -2791981.06 | 0.463193488 | -1293227.44 | ||
| 2029 | -2870157 | -2870156.53 | 0.428882859 | -1230960.94 | ||
| 2030 | -2950521 | -2950520.91 | 0.397113759 | -1171692.45 | ||
| 2031 | -3033135 | -3033135.49 | 0.367697925 | -1115277.63 | ||
| 2032 | -3118063 | -3118063.29 | 0.340461041 | -1061579.07 | ||
| 2033 | -3205369 | -3205369.06 | 0.315241705 | -1010466.01 | ||
| 2034 | -3295119 | -3295119.39 | 0.291890468 | -961813.94 | ||
| 2035 | -3387383 | -3387382.74 | 0.270268951 | -915504.38 | ||
| 2036 | -3482229 | -3482229.45 | 0.250249029 | -871424.54 | ||
| 2037 | -3579732 | -3579731.88 | 0.231712064 | -829467.06 | ||
| 2038 | -3679964 | -3679964.37 | 0.214548207 | -789529.76 | ||
| 2039 | -3783003 | -3783003.37 | 0.198655748 | -751515.36 | ||
| 2040 | -3888927 | -3888927.47 | 0.183940507 | -715331.29 | ||
| 2041 | -3997817 | -3997817.44 | 0.170315284 | -680889.41 | ||
| 2042 | -4109756 | -4109756.33 | 0.157699337 | -648105.85 | ||
| 2043 | -4224830 | -4224829.50 | 0.146017905 | -616900.75 | ||
| 2044 | -4343125 | -4343124.73 | 0.135201764 | -587198.12 | ||
| 2045 | -4464732 | -4464732.22 | 0.125186818 | -558925.62 | ||
| 2046 | -4589745 | -4589744.72 | 0.115913721 | -532014.39 | ||
| 2047 | -4718258 | -4718257.58 | 0.107327519 | -506398.88 | ||
| 2048 | -4850369 | -4850368.79 | 0.099377333 | -482016.71 | ||
| 2049 | -4986179 | -4986179.11 | 0.092016049 | -458808.50 | ||
| 2050 | -5125792 | -5125792.13 | 0.085200045 | -436717.72 | ||
| 2051 | -5269314 | -5269314.31 | 0.078888931 | -415690.57 | ||
| 2052 | -5416855 | -5416855.11 | 0.073045306 | -395675.84 | ||
| 2053 | -5568527 | 5000000 | -568527.05 | 0.067634543 | -38452.07 | |
| Net Present Value | -101606193.87 | |||||
| Future value = Present value *(1+i)^t | Future Value | -1502282558.87 | ||||
Pls comment if you require any further explanation or you think the solution is wrong
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