1. Shifted Uniform Series
A shifted uniform series starts at a time other than year 1
When using P/A or A/P, P is always one year ahead of first A
When using F/A or A/F, F is in same year as last A
P=?
A=10,000
i=10%
Example: The present worth of the cash flow shown below at i=10% is:
(a) $25, (b) $29, (c)$34, (d) $37,908
Solution: (1) Use P/A factor with n=5 to get P in year 1:
P1= 10,000(P/A,10%,5) =10,000(3.7908) = $37,908
(2) Use P/F factor with n=1 to move P back to year 0:
P0= P1(P/F,10%1) = 37,908(0.9091) = $34,462
Answer is (c)

2. Shifted Series & Random Single Amounts
These cash flows require the use of multiple factors
A=10,000
i=12%
6000
Example: The equivalent annual worth in yrs 1-7 for the
cash flow shown below at i=12% per year is:
(a) $ (b) $ (c) $10, (d) $11,780
Solution: The annual worth in yrs 1-7
can be found using either
the A/P or A/F factors
For A/P factor, find P in yr 0, then annualize with (A/P,12%,7)
For A/F factor, find F in yr 7, then annualize with (A/F,12%7)
Using A/F: A=[10,000(F/A,12%,6) + 6,000(F/P,12%,2)](A/F,12%,7)
=[10,000(8.1152) + 6,000(1.2544)]( )
=$8,789.80
Answer is (a)

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

4. Example : John Deere expects the cost of a
(a)$ (b)$ (c)$ (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:

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 $60, the present worth of the
cost thru year 10 at an interest rate of
12%/yr is closest to:
(a)$ (b)$ (c)$ (d)$397 1 2 3 10 4 5 60 65 70 95
Cash Flow Diagram
i=12%
The cash flow diagram is as follows:

6. 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 $60, the present worth of the
cost thru year 10 at an interest rate of
12%/yr is closest to:
(a)$ (b)$ (c)$ (d)$397 1 2 3 10 4 5
P=? 60 65 70 95 8
i=12%
The cash flow diagram is as follows:
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

7. 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 $60, the present worth of the
cost thru year 10 at an interest rate of
12%/yr is closest to:
(a)$ (b)$ (c)$ (d)$397 1 2 3 10 4 5
P=? 60 65 70 95 8
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

8. Example : John Deere expects the cost of a
(a)$ (b)$ (c)$ (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:
P=?
i=12% 1 2 3 4 5 10 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:
PA= 60(P/A,12%,2) = $101.41

9. 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 $60, the present worth of the
cost thru year 10 at an interest rate of
12%/yr is closest to:
(a)$ (b)$ (c)$ (d)$397 1 2 3 10 4 5
PT= ? 60 65 70 95
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:
PT = P0 + PA = $396.70
Answer is (d)

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

11. Negative Shifted Gradients
For arithmetic gradients, change sign in front of G term from + to -
For goemetric gradients, change signs in front of both g’s
All other procedures are the same as for positive gradients