# NCEES Morning Sample Exam #102 - ECON



## NVRSTOP (Mar 8, 2008)

NCEES Morning Sample Exam #102 - ECON

A small computing system costing $50,000 is installed initially. A second computer is installed 2 years later to handle additional production needs, also at a cost of $50,000. Each unit requires a maintenance cost of $10,000/year of operation.

The computers become obsolete very fast, hence zero salvage is expected by the owners.

Assume a 10% rate of return and a 10% interest rate.

For the first 4 years, the present worth of the investment and maintenance cost is most nearly?

Solution:

PW = 50,000[1+(P/F,10%,2)] + 20,000(P/A,10%,4) - 10,000(P/A,10%,2)

= 50,000[1+0.82645] + 20,000(3.1699) - 10,000(1.73554)

= 91,323 + 63,398 - 17,355

= 137,365

Could someone explain this solution to me. I think I understand the 50,000[1+(P/F,10%2)] - ROR and interest cancel each other out for the initial 50,000?

1. Where does the 20,000 come from?

2. Why are maintenance costs only for one machine for 2 years?

Thanks,

NVRSTOP


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## wilheldp_PE (Mar 8, 2008)

NVRSTOP said:


> Solution:
> PW = 50,000[1+(P/F,10%,2)] + 20,000(P/A,10%,4) - 10,000(P/A,10%,2)
> 
> = 50,000[1+0.82645] + 20,000(3.1699) - 10,000(1.73554)
> ...


Since you are trying to find the present value of the two systems, you have an investment of $50000 today, then another investment of $50000 2 years in the future. You have to discount that future amount by 10% over 2 years. That first term is actually 50000(P/F,10,0) + 50000(P/F,10,2). Since (P/F,10,0) = 1, you can simplify it to 50000[1+(P/F,10,2)].



> 1. Where does the 20,000 come from?2. Why are maintenance costs only for one machine for 2 years?
> 
> Thanks,
> 
> NVRSTOP


You have $20000/year maintenance costs for the two systems after year two which translates into 200000(P/A, 10, 4) - 10000(P/A, 10, 2). So you calculate how much it would be for $20000 over all 4 years, and then subtract the 10000 over the first 2 years that you don't incur since the 2nd system doesn't exist.

Make sense?


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## NVRSTOP (Mar 8, 2008)

wilheldp said:


> Since you are trying to find the present value of the two systems, you have an investment of $50000 today, then another investment of $50000 2 years in the future. You have to discount that future amount by 10% over 2 years. That first term is actually 50000(P/F,10,0) + 50000(P/F,10,2). Since (P/F,10,0) = 1, you can simplify it to 50000[1+(P/F,10,2)].
> 
> You have $20000/year maintenance costs for the two systems after year two which translates into 200000(P/A, 10, 4) - 10000(P/A, 10, 2). So you calculate how much it would be for $20000 over all 4 years, and then subtract the 10000 over the first 2 years that you don't incur since the 2nd system doesn't exist.
> 
> Make sense?



Yes, seems very clear now. Not sure why I couldn't see it before. Trying to make it much harder than it is I guess.

Thanks!


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## wilheldp_PE (Mar 8, 2008)

NVRSTOP said:


> Yes, seems very clear now. Not sure why I couldn't see it before. Trying to make it much harder than it is I guess.
> Thanks!


You're quite welcome.


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## IlPadrino (Mar 9, 2008)

NVRSTOP said:


> Yes, seems very clear now. Not sure why I couldn't see it before. Trying to make it much harder than it is I guess.
> Thanks!


I recommend you ALWAYS draw the cash flow diagram. Generally, conventional wisdom says not to write anything more than the minimum and you could certainly write the present value equations without a cash flow diagram... BUT... It is very difficult to get one of these problems wrong if you have the right cash flow diagram drawn. Take the extra thirty seconds to draw out the diagram. The rest becomes plug and chug with very little chance of doing something wrong.


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