In: Economics
A state highway department is trying to decide whether it should “hot-patch” a short section of an existing country road or resurface it. If the hot-patch method is used, approximately 300 cubic meters of material would be required at a cost of $700 per cubic meter (in place). Additionally, the shoulders will have to be improved at the same time at a cost of $24,000. These improvements will last 2 years, at which time they will have to be redone. The annual cost of routine maintenance on the patched up road would be $5000. Alternatively, the state can resurface the road at a cost of $850,000. The surface will last 10 years if the road is maintained at a cost of $2000 per year beginning 4 years from now. No matter which alternative is selected, the road will be completely rebuilt in 10 years. At an interest rate of 9% per year,
a. make a recommendation on the best alternative based on the annual worth analysis.
b. make a recommendation based on the present worth method
The option1 requires material for 300 cubic meters and the rate
is $700 per cubic meter and also needs shoulder improvement.
( 300 * 700 ) + 24000 = 234000
This is valid for 2 years and the end of 2nd year, the same cost
will be again incurred.
The annual maintenance is $5000 per year.
Option B has an initial cost of $850000 and annual maintenance cost of $2000 per year.
PV = PMT / (1+Discount Rate)Duration
a) Option 1
PV of year 0 = 234000
PV of year 1 = 5000 / ( 1.09 ) = 4587.16
So we can create a table of this cash flow.
Year | Option 1 | PW @ 9% | Option 2 | PW @ 9% |
0 | 234000 | 234000 | 850000 | 850000 |
1 | 5000 | 4587.16 | 2000 | 1834.86 |
2 | 239000 | 201161.52 | 2000 | 1683.36 |
3 | 5000 | 3860.92 | 2000 | 1544.37 |
4 | 239000 | 169313.63 | 2000 | 1416.85 |
5 | 5000 | 3249.66 | 2000 | 1299.86 |
6 | 239000 | 142507.89 | 2000 | 1192.53 |
7 | 5000 | 2735.17 | 2000 | 1094.07 |
8 | 239000 | 119946.04 | 2000 | 1003.73 |
9 | 5000 | 2302.14 | 2000 | 920.86 |
10 | 5000 | 2112.05 | 2000 | 844.82 |
NPW | 885776.17 | 862835.32 | ||
PW Annuity | Factor | 6.4177 | 6.4177 | |
Annual Worth | 138021.7 | 134446.2 |
The annual worth can be calculated by dividing the NPW by the PW
annuity factor of 9% interest rate and 10 years.
Annual Worth = NPW / PW Annuity Factor
Option 1
885776.17 / 6.4177 = 138021.7
Option 2
862835.32 / 6.4177 = 134446.2
The annual cost of option 2 is lower so it should be selected.
b) The PW of the cash flow can be calculated by discounting the
cash flow by the given interest rate.
Option 2 ,
PW of year 0 = 850000
year 1 = ( 2000 / 1.09 ) = 1834.86
NPW of the option 1 = 885776.17
NPW of the option 2 = 862835.32
The NPW of the option 2 is lower so it should be
selected.