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

# Maintenance costs for a particular production machine

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Maintenance costs for a particular production machine increase by \$1,000/year over the 5-yr life of the machine. The initial maintenance cost is \$3,000. Using an interest rate of 8% compounded annually, determine the present worth equivalent for the maintenance costs. P = \$3,000(P|A 8%,5) + \$1,000(P|G 8%,5) P = \$3,000(3.99271) + \$1,000(7.372.43) = \$19,350.56 or P = (\$3,000 + \$1,000(A|G 8%,5))(P|A 8%,5) P = (\$3,000 + \$1,000(1.846.47))(3.99271) = \$19,350.55 or

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition Example 2.29 Amanda Dearman made 5 annual deposits into a fund that paid 8% compound annual interest. Her first deposit was \$800; each successive deposit was \$100 less than the previous deposit. How much was in the fund immediately after the 5 th deposit? start

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Principles of Engineering Economic Analysis , 5th edition Example 2.29 Amanda Dearman made 5 annual deposits into a fund that paid 8% compound annual interest. Her first deposit was \$800; each successive deposit was \$100 less than the previous deposit. How much was in the fund immediately after the 5 th deposit? A = \$800 - \$100(A|G 8%,5) = \$800 - \$100(1.84647) = \$615.35 F = \$615.35(F|A 8%,5) = \$615.35(5.86660) = \$3,610.01
Principles of Engineering Economic Analysis , 5th edition Example 2.29 Amanda Dearman made 5 annual deposits into a fund that paid 8% compound annual interest. Her first deposit was \$800; each successive deposit was \$100 less than the previous deposit. How much was in the fund immediately after the 5 th deposit? A = \$800 - \$100(A|G 8%,5) = \$800 - \$100(1.84647) = \$615.35 F = \$615.35(F|A 8%,5) = \$615.35(5.86660) = \$3,6101.01 F =(FV(8%,5,,-NPV(8%,800,700,600,500,400)) F = \$3,610.03

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Principles of Engineering Economic Analysis , 5th edition
Principles of Engineering Economic Analysis , 5th edition P = G gradient series, present worth factor = G ( P | G i %, n ) A = G gradient-to-uniform series conversion factor = G ( A | G i %, n ) F = G gradient series, future worth factor = A ( F | G i %, n ) [ ] (1 + i ) n – (1 + ni ) i [(1 + i ) n – 1] [ ] 1 - (1 + ni )(1 + i ) - n i 2 [ ] (1 + i ) n – (1 + ni ) i 2

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Principles of Engineering Economic Analysis , 5th edition Geometric Series Formulas Geometric Series Formulas
Principles of Engineering Economic Analysis , 5th edition Geometric Series A t = A t-1 (1+j) t = 2,…,n or A t = A 1 (1+j) t-1 t = 1,…,n 0 1 2 3 n-1 n A 1 (1+j) 2 A 1 (1+j) n-2 A 1 (1+j) n-1 A 1 (1+j) A 1 Note: n-1 not n j is the interest rate being applied to the Annuity whereas i is the interest rate applied across time period.

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Principles of Engineering Economic Analysis , 5th edition Converting Geometric Series – I Converting a geometric series to a present worth j i j i i j A P n n ) 1 ( ) 1 ( 1 1 0 ) %, | )( %, | ( 1 1 j j i j i n i F P n j P F A P ) %, %, | ( ) 1 /( 1 1 1 n j i A P A P j i i nA P (2.42) (2.43) (2.44) (2.42)
Principles of Engineering Economic Analysis , 5th edition Converting Geometric Series – II Converting a geometric series to a future worth j i j i j i A F n n ) 1 ( ) 1 ( 1 0 ) %, | ( ) %, | ( 1 j j i j i n j P F n i P F A F F = nA 1 (1+ i ) n -1 i = j F = A 1 ( F|A 1 i %, j %, n ) Note: ( F|A 1 i %, j %, n ) = ( F|A 1 j %, i %, n ) (2.45) Notice the symmetry

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Principles of Engineering Economic Analysis , 5th edition Example 2.30 A firm is considering purchasing a new machine. It
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• Fall '17
• Mike Heny

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