Engg. economics problem.

[winner9459]

It would be really appreciated if some one can help me solve this problem. I converted the values into NPV, taking into account the difference in main. cost after 15 years.

I arrived at at a total of $261,266.36 for the NPV and divided it by 25 to get the annual costs. But I don't think the last step is the one needed to be done here. Can someone throw some light on it.

Thanks.

 

[sac_engineer]

I'm getting $27,020 for part A. 100,000(A|P, 10%, 25) + (15000*15 + 20000*10)/25 - (25000/25)

For Part B, I'm intuitively guessing that the answer is 6.9% since money is being made from the investment, thus lowering the resulting interest rate below the original. As for the actual calculation, I'm not certain how it's derived.

 

[lady_j]

I took a stab at it and got answer "c" - $26,575

Here's what I did:

Machine 2, using i=10%, n = 25 years

Cost of machine - salvage value + annual maintenence costs years 1-15 (annuitized over 25 years) + annual maintenence costs years 16-25 (annuitized over 25 years)

100,000(A/P, 10%, 25) - 25,000(A/F, 10%, 25) + 15,000(P/A, 10%,15)(A/P,10%,25) + 20,000 (P/A, 10%, 10)(P/F,10%,15)(A/P, 10%, 25)

= 100,000 (0.1102) - 25,000(0.0102) + 15,000 (7.6061)(0.1102) +20,000(6.1446)(0.2394)(0.1102)

= 26,580

Perhaps I over complicated things, but I think that because those annual maintenence costs are variable, the only way I could think of to make them equivalent was to "bring" them all to present value and then calculate the annual payments from that point.

For part B - I cannot remember how to calculate this! Can anyone shed some light?

 

[winner9459]

Thank Sac for trying it out.

I think I got the answer. What I basically did was to convert everything back to the present values, then used (A/P, 10%, 25) to find the A, which is what is asked for in the first place. My Answer is $26,575, option C.

For the second part:

Step 1: Converted the salvage value into present value and subtracted it from the cost, then I have to total present cost.

Step 2: $25,000 is the revenue - ($18,000 other costs) = $7,000--->>A, which was used in P = A*((1+i)^n-1/(i*(1+i)^n)) and equated it to present cost value obtained in step 1.

Finally, I get some equation with i as the unknown, I substituted each given option to see which value satisfies the equation. I found 6.9% as the one that satisfies the equation.

Option A

 

[winner9459]

is there a simpler way to do part B. The above way is time consuming.

 

[ptatohed]

is there a simpler way to do part B. The above way is time consuming.

The way you went about it is smart. Plugging in the given answers is a great time saver over calculating i from scratch.

 

[civilized_naah]

is there a simpler way to do part B. The above way is time consuming.

Actually, if you take an approximate first step (by ignoring the salvage), it narrows down the number of alternatives you must 'plug and chug' Makes it a little better. See below.

If the rate of return is x, then present value = 0 for all cash flows:

-80,000 + 20,000(P/F,x,20) + (25,000 – 18,000)(P/A, x,20) = 0

– 80 + 20(P/F,x,20) + 7(P/A,x,20) = 0

Ignoring the middle term (i.e. ignoring salvage value), (P/A,x,20) = 80/7 = 11.4286

For x = 6%, P/A = 11.4699 and for x = 8%, P/A = 10.5940. Therefore, the x is slightly greater than 6%

However, if the salvage value is incorporated, the rate of return will be slightly higher. Try 6.9% (use 7% values from the table)

– 80 + 20(P/F,7%,20) + 7(P/A,7%,20) = – 80 + 20(.2584) + 7(10.5940) = -0.674 (practically zero). Therefore rate of return is 6.9%

 

[Peele1]

These problems are only an academic exercise and test prep. They aren't going to be used in real life. There are too many variables that may change a little bit, which affects the answer. My experience is that management would chose I because it has a lower initial cash outlay and the other variables are too close to really be a factor.

 

[BamaBino]

I took a stab at it and got answer "c" - $26,575Here's what I did:

Machine 2, using i=10%, n = 25 years

Cost of machine - salvage value + annual maintenence costs years 1-15 (annuitized over 25 years) + annual maintenence costs years 16-25 (annuitized over 25 years)

100,000(A/P, 10%, 25) - 25,000(A/F, 10%, 25) + 15,000(P/A, 10%,15)(A/P,10%,25) + 20,000 (P/A, 10%, 10)(P/F,10%,15)(A/P, 10%, 25)

= 100,000 (0.1102) - 25,000(0.0102) + 15,000 (7.6061)(0.1102) +20,000(6.1446)(0.2394)(0.1102)

= 26,580

Perhaps I over complicated things, but I think that because those annual maintenence costs are variable, the only way I could think of to make them equivalent was to "bring" them all to present value and then calculate the annual payments from that point.

That is the correct method and the correct numbers.

 

[BamaBino]

Step 1: Converted the salvage value into present value

Don't you already have to know what the Rate-of-Return is before you do that?

You could give it a guesstimate of 10% or 6% and continue to work it.