CERM Practice Problems Ed. 12, Chapter 86, Problem 12 (ECON)
Posted 26 March 2012 - 04:41 AM
Let me know you what you all think.
Posted 26 March 2012 - 03:14 PM
salvage value could be a cost if that is what it takes to demolish the bridge.
Posted 02 April 2012 - 01:36 AM
An old bridge can be strengthened at a cost of $9,000, or it can be replaced for $40,000. The present salvage value of the old bridge is $13,000. It is estimated that the reinforced bridge will last for 20 years, will have an annual cost of $500, and will have a salvage value of $10,000 at the end of 20 years. The estimated salvage value of the new bridge after 25 years is $15,000. Maintenance for the new bridge would cost $100 annually. The effective annual interest rate is 8%. What is the best alternative?
And here is how I understand it
P = 9,000
A = 500
F = 10,000
n = 20
EUACA= 500 + 9000(A/P,8%,20) -10,000(A/F,8%,20) = $1,200
P = 40,000 - 13,000 (13,000 = Old Bridge Present Salvage) = 27,000
A = 100
F = 15,000
n = 25
EUACB = 100 + 27,000(A/P,8$,25) - 15000(A/F,8%,25) = $2,400
Which clearly shows the Old Option is most favorable. And that's the ultimate answer, but Lindeburg get's EUACA = $2,500 by adding the 9,000 and 13,000, and that's where we disagree. The rest of his solution matches mine.
What do you think?
Edited by Jayman_10x, 02 April 2012 - 01:38 AM.
Posted 02 April 2012 - 01:02 PM
On the other hand, you treated the old option present salvage value as a deduction to the new option replacement cost (the value of the scrap), which is what would happen in the real world if the new option was selected.
I think this is a case where either way gets you the correct answer as long as nothing is double counted.
Posted 02 April 2012 - 05:01 PM
Edited by bradlelf, 02 April 2012 - 05:01 PM.
Posted 02 April 2012 - 08:38 PM
Edited by Jayman_10x, 02 April 2012 - 08:38 PM.
Posted 02 April 2012 - 08:51 PM
Posted 03 April 2012 - 11:26 AM
I agree as well but cannot understand how to compare both options from year 21 to 25. Any ideas?
I think that's why you find the annual cost for each.
Posted 05 April 2012 - 04:56 PM
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