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

Must use multiple factors to find P in year 0

Arithmetic

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:

(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:

(a)$101 (b)$295 (c)$370 (d)$397

0 1 2 3 104 5

60 60 6065

70

95

The cash flow diagram is as follows:

Cash Flow Diagram

i=12%

First find P2 for the gradient ($5) and its base amount ($60) in year 2:

P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41

0 1 2 3 104 5

P2=?

60 60 6065

70

95

0 1 2 3 8

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:

(a)$101 (b)$295 (c)$370 (d)$397

The cash flow diagram is as follows:

i=12%

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

0 1 2 3 104 5

P0=?

60 60 6065

70

95

0 1 2 3 8

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:

(a)$101 (b)$295 (c)$370 (d)$397

The cash flow diagram is as follows:

i=12%

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

0 1 2 3 104 5

PA=?

60 60 6065

70

95

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:

(a)$101 (b)$295 (c)$370 (d)$397

The cash flow diagram is as follows:

i=12%

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:

PT = P0 + PAA = $396.70

Answer is (d)

0 1 2 3 104 5

PT= ?

60 60 6065

70

95

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:

(a)$101 (b)$295 (c)$370 (d)$397

The cash flow diagram is as follows:

i=12%

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

Geometric

Equation yields P for all cash 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)}