Homework SourseThe city engineer has prepared two plans for roads in the ci
The city engineer has prepared two plans for roads in the city park. Both plans meet anticipated requirements for the next 40 years. The city’s minimum attractive rate of return is 7%.
Plan A is a 3-stage development program: $300,000 is to be spent now followed by $250,000 at the end of 15 years and $300,000 at the end of 30 years. Annual maintenance will be $75,000 for the first 15 years, $125,000 for the next 15 years, and $250,000 for the final 10 years.
Plan B is a 2-stage development program: $450,000 is to be spent now followed by $50,000 at the end of 15 years. Annual maintenance will be $100,000 for the first 15 years and $125,000 for the subsequent years. At the end of 40 years, this plan has a salvage value of $150,000.
Use a conventional benefit-cost ratio analysis to determine which plan should be chosen.
Clear answer with formulation and steps needed.
Solution
Answer:
Part-A]
first calculate the toal cost needed for the project
Initial cost = $ 300,000
Maintenance cost = $ 75000 x15 Years = $ 1125000
= $ 125000 x15 Years = $ 1875000
= $ 250000 x10 Years = $ 2500000
Total maintenance cost = $ 1125000 + $ 1875000+ $ 2500000 = $ 5500000
Total cost = $ 5800000
Now calculate the total recovery cost
Recovery cost= $ 250000 x15 years = $3750000
= $ 300,000 x15 Years = $ 4500000
Total Recovery = $ 8250000
Total benefit = $ 8250000 - $ 5800000 = $ 2450000
B/C = $ 2450000 / $ 5800000
B/C = 0.42
Benefit cost ratio.
Part-B]
First calculate the total cost.
Initla cost = $ 450000
Maintence cost = $ 100000 x15 Years = $ 1500000
= $ 125000 x25 Years = $ 3125000
Total maintenance cost = $ 4625000
Total cost = $ 5075000
Recovery cost = $ 50000 *15 = $ 750000
= $ initial cost * ( 1 +i )n
= $ 450000 * ( 1+0.07 )40
= $ 6738506.02
Total Recovery cost = $ 750000 +$ 6738506.02 + $ 150000
= $ 7638506
Benefit = $ 7638506 - $ 5075000 = $ 2563506
B/C = $ 2563506 / $ 5075000
B/C = 0.50
so we should choose a Part-b] for B/C ratio 0.50.
