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

# Bwhat single sum at t 10 is equivalent to the smp

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b)what single sum at t = 10 is equivalent to the SMP investment? c)what uniform annual series over the 10-year period is equivalent to the SMP investment?

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Principles of Engineering Economic Analysis , 5th edition Example 2.27 (Solution) P = -\$500,000 + \$92,500(P|A 10%,10) + \$50,000(P|F 10%,10) P = -\$500,000 + \$92,500(6.14457) + \$50,000(0.38554) P = \$87,649.73 P =PV(10%,10,-92500,-50000)-500000 P = \$87,649.62 F = -\$500,000(F|P 10%,10) + \$92,500(F|A 10%,10) + \$50,000 F = -\$500,000(2.59374) + \$92,500(15.93742) + \$50,000 F = \$227,341.40 F =FV(10%,10,-92500,500000)+50000 F = \$227,340.55
Principles of Engineering Economic Analysis , 5th edition Example 2.27 (Solution) A = -\$500,000(A|P 10%,10) + \$92,500 + \$50,000(A|F 10%,10) A = -\$500,000(0.16275) + \$92,500 + \$50,000(0.06275) A = \$14,262.50 A =PMT(10%,10,500000,-50000)+92500 A = \$14,264.57

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Principles of Engineering Economic Analysis , 5th edition P = A uniform series, present worth factor = A ( P | A i %, n ) =PV( i %, n ,- A ) A = P uniform series, capital recovery factor = P ( A | P i %, n ) =PMT( i %, n ,- P ) F = A uniform series, future worth factor = A ( F | A i %, n ) =FV( i %, n ,- A ) A = F uniform series, sinking fund factor = F ( A | F i %, n ) =PMT( i %, n ,,- F ) [ ] [ ] i (1 + i ) n (1 + i ) n – 1 (1 + i ) n – 1 i [ ] (1 + i ) n – 1 i (1 + i ) n [ ] (1 + i ) n – 1 i
Principles of Engineering Economic Analysis , 5th edition Remember Relationships among Remember Relationships among P, F, and A P, F, and A P occurs one period before the first A in a uniform series F occurs the same time as the last A in a uniform series Be careful in using the formulas

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Principles of Engineering Economic Analysis , 5th edition Gradient Series Gradient Series
Principles of Engineering Economic Analysis , 5th edition Gradient Series A t = 0 t = 1 = A t-1 +G t = 2,…,n = (t-1)G t = 1,…,n

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Principles of Engineering Economic Analysis , 5th edition Gradient Series A t = { 0 A t-1 +G t = 1 t = 2,…,n or A t = (t-1)G t = 1,…,n 0 1 2 3 n-1 n G 2G (n-2)G (n-1)G Note: n-1, not n P F
Principles of Engineering Economic Analysis , 5th edition Converting Gradient Series Converting gradient series to present worth P = G (2.35) P = G (2.36) P = G(P|G i%, n) (2.37) 2 ) 1 )( 1 ( 1 i i ni n i n i F P n n i A P ) %, | ( ) %, | (

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Principles of Engineering Economic Analysis , 5th edition Converting Gradient Series - II Converting gradient series to annual worth A = G A = G A = G(A|G i%, n) (2.38) 1 ) 1 ( 1 n i n i i n i F A n ) %, | ( 1
Principles of Engineering Economic Analysis , 5th edition Converting Gradient Series - III Converting gradient series to future worth F = G (2.39) F = G F = G(F|G i%, n) (not provided in the tables) 2 ) 1 ( ) 1 ( i ni i n i n n i A F ) %, | (

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Principles of Engineering Economic Analysis , 5th edition Example 2.28 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.
Principles of Engineering Economic Analysis , 5th edition Example 2.28 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.

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Principles of Engineering Economic Analysis , 5th edition Example 2.28
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• Fall '17
• Mike Heny

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