# NCEES #102



## ndekens (Aug 5, 2008)

Okay so looking at the morning econ question #102 I dont get why they are saying that $50,000 * (1+(P/F,10%,2)??

Shouldn't we have a total period of n=4? also why 1+?? Can someone walk me through this im feeling stupid.


----------



## Flyer_PE (Aug 5, 2008)

The only way I can ever do these is by drawing the cash flow diagram.

They are investing 50,000 initially at n=0 with an additional 50k being invested at n=2. The objective is to determine the present value of both machines.

Present value of Machine 1=50k * (P/F,10%,0) where P/F,10%,0=1

Present value of Machine 2=50k * (P/F,10%,2)

Total hardware investment =(50k*1) + (50k*(P/F,10%,2))

The four-year period is applied to the maintenance and operating costs. Four years for the first machine and the final two years for both machines. They apply the maintenance costs for both machines for the entire four years and then subtract out the costs for the second machine from years 1 and 2.

Hope this helps.

Jim


----------



## ndekens (Aug 5, 2008)

Flyer_PE said:


> The only way I can ever do these is by drawing the cash flow diagram.
> They are investing 50,000 initially at n=0 with an additional 50k being invested at n=2. The objective is to determine the present value of both machines.
> 
> Present value of Machine 1=50k * (P/F,10%,0) where P/F,10%,0=1
> ...


That works....thanks!


----------



## ndekens (Aug 5, 2008)

Flyer_PE said:


> The only way I can ever do these is by drawing the cash flow diagram.
> They are investing 50,000 initially at n=0 with an additional 50k being invested at n=2. The objective is to determine the present value of both machines.
> 
> Present value of Machine 1=50k * (P/F,10%,0) where P/F,10%,0=1
> ...



Okay, so I got the first part however I did the next half of the problem a little different and dont understand why my answer is incorrect.

The book does it like this:

50K(1+(P/F,10%,2) + 20K(P/A,10%,4) - 10K(P/A,10%,2) = 137K

I did this:

50K(1+(P/F,10%,2) + 10K(P/A,10%,2) + 20K(P/A,10%,2) = 143K

So why am I wrong? I dont think my logic is incorrect. You pay 10K in maintenance for the first 2 years then the next 2 yrs you pay 20K in maintenance....what gives? At no time on the cash flow diagram are you paying 20K more then 2 years unless you go out beyond n=4 which this problem does not involve.


----------



## Flyer_PE (Aug 5, 2008)

I'm prone to the same mistake. I end up doing it the same way they do in their solution since it bases everything at n=0. You have to include the 10% interest on the yearly expenses also. For the present value calculation, you have to translate everything back to n=0. If you solve for each machine individually, you end up with a separate factor to translate the expenses for the second machine back to "n=0" dollars.


----------



## Chester95 (Aug 5, 2008)

My first post here so bear with me:

Your logic is correct, but to take the 20k using the P/A formula you need to bring that value back to t=0. Anytime you use the uniform series present worth, the value given will be at one time period before the uniform series begins, i.e. for this case the 20k begins at t=3 and goes to t=4, if you take 20000(P/A, 10,2), then you get 34,710 at t=2..you then have to take the 34,710 back to t=0. I hope this helps, by the way I have recently found this board and find it very beneficial. I am also planning on taking the PE this October.


----------

