# Example Equivalent Annual Worth with different lives: A mechanical engineer is considering two types

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? Type A First cost \$ Maintenance cost, \$/year Salvage value Life, years \$8,650.00 \$13,900.00 \$2,200.00 -\$1,900.00 \$0.00 \$3,000.00 5 7 1. Have available the Microsoft Excel software program. 2. Construct a table in the Microsoft Excel work sheet as shown below. The instructions are shown in the second table. A B C 7 N Rate = Year 0 – 1 5 0.1 Type A -8650 -2200 -2200 -2200 -2200 -2200 Type B -13900 -1900 -1900 -1900 2 -1900 10 – 4. 5 6 7 11 -1900 -1900 -1900 3000 12 13 14 EAC = 15 16 1. The construction of the Microsoft Excel sheets and the answered questions has a total grading value of ten (10) points. 2. The deadline for hand in this exercise is indicated at the end of the week of the fourth module. Questions (You only have to write in Excel one of the answers, the one you think is the correct answer to the question) 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