Question

An old covered wooden bridge can be reinforced at a cost of $12,000 (will last 18 years), or it can be replaced for $45,000 (will last 25 years). The present salvage value of the old bridge is $6,000. It is estimated that the reinforced bridge will last for 18 years, will have an annual cost of $500, and will have a salvage value of $6,000 at the end of 18 years. The salvage value of the new bridge after 25 years is $15,000. Maintenance for the new bridge would cost $100 annually. If the effective annual interest rate is 10%, which is the best alternative?

Answer 1

Equivalent uniform annual worth method will be used to solve the problem due to the unequal life of the alternatives.

R = 10%

When the bridge in reinforced,

n = 18 years

Present value of the total cost = initial value + reinforcement cost + present value annual cost – present value of the salvage value

Present value of the total cost = 6000 + 12000 + 500*(1-1/1.1^18)/.1 - 6000/1.1^18

Present value of the total cost = $21021.55

Let, equivalent uniform annual cost = E1

Then,

E1 = 21021.55/((1-1/1.1^18)/.1)

E1 = $2563.16                           

When New Bridge is implemented,

n = 25 years

Present value of the total cost = 45000 + 100*(1-1/1.1^25)/.1 - 15000/1.1^25

Present value of the total cost = $44523.26

Let, equivalent uniform annual cost = E2

Then,

E2 = 44523.26 /((1-1/1.1^25)/.1)

E2 = $4905.04

Since, the uniform annual cost for the reinforced bridge ($2563.16) is less than the uniform annual cost of new bridge ($4905.04), then the alternative of the reinforcement of the bridge should be pursued.