engineering economics?

Hi All,

I wasn't sure which board to post this question but I'm assuming all disciplines will have engineering economics on their exam? Anyways I'm studying this topic right now to refresh my memory and I was stuck on the problem/solution below:

Two coatings are under consideration for a liquid chemical storage tank. A tank coating identified as Material A will have a first cost of $50,000 and a 10 year life, if repaired at the end of year five. Material B will cost $20,000 initially and $5,000 per year through its five year life. At an interest rate of 10% per year, the amount that could be spent for repairing Material A that would make the two breakeven is closest to:

Solution:

Equate present worth relations for 10 year service life. Let x = repair cost in year 5 for breakeven.

50,000 + x(P/F, 10%, 5) = 20,000 + 20,000(P/F, 10%, 5) + 5,000(P/A, 10%,10)
50,000 + x(0.6209) = 20,000 + 20,000(0.6209) + 5,000(6.1446)
0.6209x = 13,141
x = $21,164

On the right side of the equation, I don't know where 20,000(P/F, 10%, 5) came from? And for 5,000(P/A, 10%,10) , isn't n supposed to be 5? Also, instead of doing a present worth analysis, could this be solved with an annual worth equation instead?

Also generally speaking, how difficult can engineering economics be on the exam? I hope they are straight forward. Often times, it's the wording of the question that messes up your equation. The math itself is easy though lol.

engineer123 said:

Hi All,

I wasn't sure which board to post this question but I'm assuming all disciplines will have engineering economics on their exam? Anyways I'm studying this topic right now to refresh my memory and I was stuck on the problem/solution below:

Two coatings are under consideration for a liquid chemical storage tank. A tank coating identified as Material A will have a first cost of $50,000 and a 10 year life, if repaired at the end of year five. Material B will cost $20,000 initially and $5,000 per year through its five year life. At an interest rate of 10% per year, the amount that could be spent for repairing Material A that would make the two breakeven is closest to:

Solution:

Equate present worth relations for 10 year service life. Let x = repair cost in year 5 for breakeven.

50,000 + x(P/F, 10%, 5) = 20,000 + 20,000(P/F, 10%, 5) + 5,000(P/A, 10%,10)
50,000 + x(0.6209) = 20,000 + 20,000(0.6209) + 5,000(6.1446)
0.6209x = 13,141
x = $21,164

On the right side of the equation, I don't know where 20,000(P/F, 10%, 5) came from? And for 5,000(P/A, 10%,10) , isn't n supposed to be 5? Also, instead of doing a present worth analysis, could this be solved with an annual worth equation instead?

Also generally speaking, how difficult can engineering economics be on the exam? I hope they are straight forward. Often times, it's the wording of the question that messes up your equation. The math itself is easy though lol.

Click to expand...

$20,000 (P/F, 10%, 5) comes from assuming that you are looking at this as a 10 year solution whether using material A or B. for the Present worth of Material B you have to assume that over 10 years you will have to purchase 2 Material B's because its life is only 5 years. And 5000(P/A, 10%, 10) comes from the fact that you are comparing the two materials over 10 years, so purchasing 2 Material B's over 10 years still has $5000 maintenance cost per year so n = 10.

So for Present worth of B you have:

Pb = 20,000 (initial cost) + 20,000(P/F, 10%, 5) (Present worth of future cost to replace Material B at year 5) + 5000(P/A, 10%, 10) (you have to pay yearly maintenance for all 10 years)

Pb = $63,141

Pa = $50,000 + X (cost of repair at year 5)(P/F, 5, 10%)

X = ($63141-$50,000)/.6209 = X =$21,164

As for the exam, that is what they are trying to test you on, the ability to read the wording and apply the mathematics to it. If the problems were just math it wouldn't be the Principles and Practice of Engineering, it would be Principles and Practice of Math. They are testing your engineering knowledge and judgement on the process of solving problems.

@GeoDudeThanks I think I get it now! If I recall, when comparing different options - the life value n should be the same for both so its an equal comparison (in this case n = 10).

Would I be able to solve this using Annual Worth analysis? I wasn't sure why they chose Present Worth.

Thanks again.

Got another eng econ question if anyone can assist me

A new computer system will cost $25,000. Using the MACRS, the computer system has a useful life of 5 years. What is the estimated salvage value at the end of five years?

Heres my approach:

I looked at the MACRS table to find the recovery rate percent which is 11.52%, and calculated the depreciated value in year 5, D = (percent) x Cost = 0.1152 x 25,000 = 2880.

I then tried using the straight line equation and plugged in all my values to solve for the salvage value: D = (Cost - Salvage value) / useful life

….but after doing some research, apparently there is no salvage value for the MACRS method because the asset is always depreciated down to 0.

Zero was not given as a choice and the answer chosen was  $1,450. So I'm wondering if this is a mistake?