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Shifted gradients begin at a time other than between periods 1& 2

Shifted gradients begin at a time other than between periods 1& 2. Must use multiple factors to find P in year 0. Arithmetic. Shifted Gradients. (a)$101 (b)$295 (c)$370 (d)$397. Example : John Deere expects the cost of a

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Shifted gradients begin at a time other than between periods 1& 2

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  1. Shifted gradients begin at a time other than between periods 1& 2 Must use multiple factors to find P in year 0 Arithmetic Shifted Gradients

  2. (a)$101 (b)$295 (c)$370 (d)$397 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to:

  3. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: 1 10 3 0 2 4 5 60 60 60 65 70 (a)$101 (b)$295 (c)$370 (d)$397 95 Cash Flow Diagram i=12% The cash flow diagram is as follows:

  4. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: P2=? 1 10 3 0 2 4 5 0 1 2 3 8 60 60 60 65 70 (a)$101 (b)$295 (c)$370 (d)$397 95 P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 i=12% The cash flow diagram is as follows: First find P2 for the gradient ($5) and its base amount ($60) in year 2:

  5. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: P0=? 1 10 3 0 2 4 5 0 1 2 3 8 60 60 60 65 70 (a)$101 (b)$295 (c)$370 (d)$397 95 P0 = P2(P/F,12%,2) = $295.29 i=12% The cash flow diagram is as follows: First find P2 for the base amount ($60) & gradient ($5) in year 2: P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P2 back to year 0:

  6. (a)$101 (b)$295 (c)$370 (d)$397 PA= 60(P/A,12%,2) = $101.41 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: PA=? i=12% 1 10 3 0 2 4 5 60 60 60 65 70 95 The cash flow diagram is as follows: First find P2 for the base amount ($60) & gradient ($5) in year 2: P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P2 back to year 0: P0 = P2(P/F,12%,2) = $295.29 Next, find PA for the $60 amounts of years 1 & 2:

  7. PT= ? 1 10 3 0 2 4 5 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: 60 60 60 65 70 (a)$101 (b)$295 (c)$370 (d)$397 95 PT= P0 + PA = $396.70 i=12% The cash flow diagram is as follows: First find P2 for the base amount ($60) & gradient ($5) in year 2: P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P2 back to year 0: P0 = P2(P/F,12%,2) = $295.29 Next, find PA for the $60 amounts of years 1 & 2: PA= 60(P/A,12%,2) = $101.41 Finally, add P0 & PA to get PT in year 0: Answer is (d)

  8. Shifted gradients begin at a time other than between periods 1& 2 Equation yields P for allcash flow(i.e. base amt is included) For negative gradient, change signs in front of both g’s P=A{1-[(1+g)/(1+i)]n/(i-g)} Geometric Shifted Gradients

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