Please show work. Maintenance costs on a bridge are $3,000 every third year star
ID: 2612370 • Letter: P
Question
Please show work. Maintenance costs on a bridge are $3,000 every third year starting at the end of year 3. For analysis purposes, the bridge is assumed to have an infinite life. What is the Capitalized Equivalent (CE) cost of these infinite payments, assuming an annual interest rate of 9% compounded annually? Machine A costs $20,000, lasts 3 years and has a salvage value S of $3,000. Machine B costs $12,000, lasts 2 years and has a salvage value of $2,000. The machines can be purchased at the same price with the same salvage value in the future, and are needed for a 6 year project. Which Machine would you purchase and why? Provide justification using an Annualized Equivalent Cost analysis.Explanation / Answer
3) Capitalsed costs are used for those projects which can go on for an infinite period of time
The present value of a uniform amount is
P = A * [(1+i)n - 1] / [i * (1+i)n]
= (A / i) * [ (1+i)n/(1+i)n - (1/(1+i)n)]
= (A / i) * [1 - 1/(1+i)n)]
When n approaches infinity, the above equation becomes
P = A / i
where, P is the present value /Capitalised Equivalent Cost, A is the end of year payment ($3000) and i is the interest rate per year.
However, in the above case, maintenace costs of $3000 have to be paid at the end of every three years.
Therefore, in this case, i = (1 + annual interest rate)3-1
= ( 1 + 0.09)3-1
= 0.295 = 29.5%
Thus, Capitalised Equivalent Cost = 3000 / 0.295 = $10168.49
4) We will use the Annualised Equivalent Cost Analysis to determine which alternative to choose. The steps of the method are as follows:
1. Net Present Value of both the alternatives is calculated.
2. For each project, Annualised Equivakent Cost is calculated that is, the expected payment over the project's life, where the future value of the project would equal zero.
3. The alternative with a lower Annualised Equivalent Cost will be chosen as the present value will be negative.
We are assuming a discount rate of 9%
For Machine 1,
No of periods =3
Cost of Capital = 9%
PV = -$17683.44956
Annualised Equivalent Cost (calculated as PMT ) = $6985.93
For Machine 2
No of periods =2
Cost of Capital = 9%
PV = -$10316.64
Annualised Equivalent Cost (calculated as PMT ) = $5864.69
Since Annualised Equivalent Cost of Machine 2 is lower than Machine 1, I would purchase Machine 2.
Year Cash Flow Present Value of Cash Flow at 9% 0 -20000 -20000 1 0 0 2 0 0 3 3000 2316.55044 Present Value -$17683.44956