Example Equivalent Annual Worth with different lives:

A mechanical engineer is considering two types of pressure sensors for a low pressure steam line. The costs are shown below. Which should be selected based on an annual worth comparison at an interest rate of 10% per year?

Construct a table in the Microsoft Excel work sheet as shown below. The instructions are shown in the second table.

Enter the information exactly as below. Write and answer the following questions using the Microsoft Excel spreadsheet.

1. When in cell C3 the interest rate is change from 10% to 5% the annual worth of Type A alternative Type A is $4,197.93.

a.True b. False

2. When in cell C3 the interest rate is change from 10% to 15% the annual worth of Type A alternative is $4,780.43

a. True b. False

3. When in cell C3 the interest rate is changed to 15% to 5% the present worth of this alternative is $4, 824.10.

a. True b. False

4. When in cell D1 the interest rate is changed to 10% to 5% the present worth of Type B alternative is -$3,933.74

a. True b. False

5. In this problem when you raise the rate of return (i), the higher the annual worth quantity?

a. True b. False

Type $8,650.00 $13,90000 First cost, $ Maintenance cost, year Salvage value Life, years $2,200.00 -$1,900.00 $0.00 $3,000.00