# CERM Practice Problems Ed. 12, Chapter 86, Problem 12 (ECON)

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### #1 Jayman_PE

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Posted 26 March 2012 - 04:41 AM

Ok - here's the deal. We got two options, so it's a EUAC we need to solve. "Option A" is an old bridge that can be strengthened at a cost of \$9,000. It also has a present salvage value of \$13,000. My question is this - when solving for the EUAC for "Option A" I need to subtract the present value (\$13,000, a negative cost) from the \$9,000 cost. Isn't that correct? If you see the problem Lindeburg adds the present cost to the present salvage value. I belive that is an error.

Let me know you what you all think.

thanks,
Jason

### #2 treyjay

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Posted 26 March 2012 - 03:14 PM

without having the full question, it is hard to reply. in these econ questions, one word can change the entire answer.

salvage value could be a cost if that is what it takes to demolish the bridge.

### #3 Jayman_PE

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Posted 02 April 2012 - 01:36 AM

Here is the full problem

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

Old Option
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

New Option
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?

thanks,
Jason

Edited by Jayman_10x, 02 April 2012 - 01:38 AM.

### #4 treyjay

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Posted 02 April 2012 - 01:02 PM

It looks like Lindeburg is considering the present salvage value as a present cost under the old option. This would make sense if the old option is selected because the salvage value would not be recovered and would have to be expensed some way since it was being treated as a "salvage" value. It would not be a complete analysis of the old option if the salvage value was not considered there. So, I think that Lindeburg is technically correct.

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.

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Posted 02 April 2012 - 05:01 PM

Jason .. I get the same answer you do. Your logic seems correct.

Edited by bradlelf, 02 April 2012 - 05:01 PM.

### #6 Jayman_PE

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Posted 02 April 2012 - 08:38 PM

Thanks Guys

Edited by Jayman_10x, 02 April 2012 - 08:38 PM.

### #7 miloc

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Posted 02 April 2012 - 08:51 PM

I agree as well but cannot understand how to compare both options from year 21 to 25. Any ideas?

### #8 bennyG19

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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.
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### #9 Jayman_PE

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Posted 05 April 2012 - 04:56 PM

Benny's right. The reason for finding an EUAC is the word "Equivalent." By reducing to an annual cost you are normalizing the variables, such as unequal years, varying intial costs, etc... Goswami's All-in-one book covers this topic very well. Short but sweet.

### #10 Dano_PE

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Posted 08 April 2012 - 02:08 PM

+1

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