slides session 3A TVOM Pt 2 class chrt 3.ppt

# Will have maintenance costs that increase 8 per year

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will have maintenance costs that increase 8% per year. An initial maintenance cost of \$1,000 is expected. Using a 10% interest rate, what present worth cost is equivalent to the cash flows for maintenance of the machine over its 15-year expected life?

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Principles of Engineering Economic Analysis , 5th edition Example 2.30 A firm is considering purchasing a new machine. It will have maintenance costs that increase 8% per year. An initial maintenance cost of \$1,000 is expected. Using a 10% interest rate, what present worth cost is equivalent to the cash flows for maintenance of the machine over its 15-year expected life? A 1 = \$1,000, i = 10%, j = 8%, n = 15, P = ? P = \$1,000(P|A 1 10%,8%,15) = \$1,000(12.03040) = \$12,030.40 P =1000*NPV(10%,1,1.08,1.08^2,1.08^3,1.08^4,1.08^5, 1.08^6,1.08^7,1.08^8,1.08^9,1.08^10,1.08^11,1.08^12, 1.08^13,1.08^14) = \$12,030.40
Principles of Engineering Economic Analysis , 5th edition Example 2.30 A firm is considering purchasing a new machine. It will have maintenance costs that increase 8% per year. An initial maintenance cost of \$1,000 is expected. Using a 10% interest rate, what present worth cost is equivalent to the cash flows for maintenance of the machine over its 15-year expected life? A 1 = \$1,000, i = 10%, j = 8%, n = 15, P = ? P = \$1,000(P|A 1 10%,8%,15) = \$1,000(12.03040) = \$12,030.40 P =1000*NPV(10%,1,1.08,1.08^2,1.08^3,1.08^4,1.08^5, 1.08^6,1.08^7,1.08^8,1.08^9,1.08^10,1.08^11,1.08^12, 1.08^13,1.08^14) P = \$12,030.40

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition Example 2.31 Mattie Bookhout deposits her annual bonus in a savings account that pays 8% compound annual interest. Her annual bonus is expected to increase by 10% each year. If her initial deposit is \$500, how much will be in her account immediately after her 10 th deposit? A 1 = \$500, i = 8%, j = 10%, n = 10, F = ? F = \$500(F|A 1 8%,10%,10) = \$500(21.74087) F = \$10,870.44 Excel)

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition Example 2.32 Julian Stewart invested \$100,000 in a limited partnership in a natural gas drilling project. His net revenue the 1 st year was \$25,000. Each year, thereafter, his revenue decreased 10%/yr. Based on a 12% TVOM, what is the present worth of his investment over a 20-year period? A 1 = \$25,000, i = 12%, j = -10%, n = 20, P = ? P = -\$100,000 + \$25,000(P|A 1 12%,-10%,20) P = -\$100,000 + \$25,000[1 – (0.90) 20 (1.12 )-20 ]/(0.12 + 0.10) P = \$12,204.15

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition P = A 1 geometric series, present worth factor i j P = nA 1 /(1 + i ) i = j P = A 1 ( P | A 1 i %, j %, n ) F = A 1 geometric series, future worth factor i j F = nA 1 (1 + i ) n -1 i = j F = A 1 ( F | A 1 i %, j%,n ) [ ] 1 - (1 + j ) n (1 + i ) - n i - j [ ] (1 + i ) n – (1 + j ) n i - j
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• Fall '17
• Mike Heny

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