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
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