Example Problem A company is considering the purchase of a new piece of testing equipment that is expected to produce $8,000 additional income during the.

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Example Problem A company is considering the purchase of a new piece of testing equipment that is expected to produce $8,000 additional income during the first year of operation. This amount will decrease by $500 per year for each subsequent year of ownership. The equipment costs $20,000 and will have an estimated salvage value of $3,000 after 8 years of use. For a MARR of 15% compounded annually, determine the net present worth of this investment.

Problem 2 Project #1 costs $10,000 and has annual, end of the year revenues of $10,000 over its 5 year life. There is no salvage value. Project #2 costs $20,000 and has annual end of year revenues of $10,000 over its 10 year life. There is no salvage value. Conduct an economic analysis to select the preferred project using a MARR of 15% per year, compounded annually.

Calculating NPWs… NPW1 = $23,522 NPW2 = $30,188 Why is it wrong to select Project 2 based on this analysis?

(Net) Present Worth Analysis Two possible approaches when project lives are different: Common Multiple Period: Projects are assumed to be repeated until a common multiple point in time is established. Study Period: Select a study period for both projects and estimate cash flows to conform to the study period.

Problem 3 A firm is considering the purchase of one of two new machines. The data on each are as below: Machine A B Service Life 3 years 6 years Initial Cost $3,400 $6,500 Annual Net Operating Expense: $2,000 $1,800 Salvage Value $100 $500 Use a MARR of 12% compounded annually and the lowest common multiple assumption to determine the alternative to be selected.

Machine A B Service Life 3 years 6 years Initial Cost $3,400 $6,500 Annual Net Operating Expense: $2,000 $1,800 Salvage Value $100 $500 MARR is 12% compounded annually

Problem 4 Two alternatives are being considered regarding construction of a new high-voltage transmission line. Alternative I would build the transmission towers and the line at a capacity of 230 kVA, which is expected to be adequate for 15 years. After 15 years the 230 kVA lines would be removed and 560 kVA lines placed on the existing towers. Alternative II would build the transmission towers and the 560 kVA lines immediately. Given below are the pertinent data on the costs of these facilities. Expected Expected Item Present Cost Service Life Salvage Value Trans. Towers $15,000,000 55 years 0 after 30 yrs 230 kVA lines $8,000,000 15 years 10% of 1st cost 560 kVA lines $12,000,000 35 years 10% of 1st cost Salvage values for both transmission lines are 10% of first cost regardless of age at retirement. The cost of 560 kVA lines will inflate at the rate of 10% per year. The MARR is 15%. Use Present Worth analysis to determine which alternative is least expensive for a 35 year study period.