Question

Question 1

The HalX computer system cost $194,586. It will allow the company to save $89,837 per year. It will cost $22,588 per year to run. The system has a lifespan of 6 years. Using a MARR of 5% interest, what is the annual worth of this system?

Question 2 

Pacific Consolidated just bought a new building for $338,064. The utilities to run the building will be $8,981 per year. The company anticipates it will be able to collect $38,155 each year by renting out units in the building. The company knows it will need to do an overhaul of the heating system at the end of year 14, at a cost of $12,016. At the end of 23 years, Pacific believes it will be able to sell the building for $674,669. Using a MARR of 5%, what is the equivalent uniform annual worth of this building? 

Question 3 

The city of Haversham is going to construct a new bird watching park for the public. The initial cost of the project will be $802,088. The annual maintenance will be $25,259, with costs estimated to increase each year by 14%. 

A second phase of the project will be completed at the end of year 9 at a cost of $414,959. 

The city would like to know the equivalent uniform annual cost (EUAC) of this project using a MARR of 3% and a time frame of 23 years. 

Question 4

The city of Blimpson invested $155,509 in an account paying 5% interest, compounded annually. At the end of year 8, the city invested an additional $76,448. The city would like to use the money to fund a new scooter program for 8 years. If the city begins to make withdrawals at the end of year 15, how much can the city withdraw each year? (Assume the city withdraws the same amount each year.) 

Answer 1

1. AW of the system = -194586 * (A/P, 5%,6) + 89837 - 22588

= -194586 * 0.197017 + 89837 - 22588

= 28912.16

2.

AW of the building = -338064 *(A/P, 5%,23) - 8981 + 38155 -12016 * (P/F, 5%,14)*(A/P, 5%,23) + 674669 * (A/F, 5%,23)

= -338064 *0.074137 - 8981 + 38155 -12016 * 0.505068 *0.074137 + 674669 * 0.024137

= 19945.45

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