Engineering Economy
Chapter 4: The Time Value of Money
Lecture 5
Sometimes cash flows change by a constant
amount each period.
We can model these situations as a uniform
gradient of cash flows. The table below
shows such a gradient.
End of Period
1
2
3
:
N
2
Cash Flows
0
G
2G
:
(N-1)G
Gradient Formulas
Arithmetic gradient series – Starts at base amount and
increases by constant gradient G in years 2 through n
Base
Amount
A
P = A(P/A,i%,n) + G(P/G,i%.n)
This total present worth of base and gradient
Linear Gradient Example
A1+n-1G
A1+n-2G
• Assume the following:
A1+2G
A1+G
0
1
2
3
n-1
N
This represents a positive, increasing arithmetic gradient
Arithmetic Gradients
•P/G factor formula for gradient only
•A/G factor for annual equivalent of gradient only
•
•P/G and A/G factors are in the tables at the rear of
text
Arithmetic Gradients
Example: Estimated annual revenue is $5,000 increasing
by $500 per year starting next year. Find P and A
equivalent over an 8-year period at i = 10%.
$8500
$8000
Gradient, G = $500
$7500
$7000
$6500
$6000
$5500
Base
$5000
A = $5000
0
1
2
3
4
5
6
7
8
Arithmetic Gradient Factors
• The “G” amount is the constant arithmetic change from one time period to the
next.
•The “G” amount may be positive or negative.
•The present worth point is always one time period to the left of the first cash flow
in the series or,
•Two periods to the left of the first gradient cash (G) flow.
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Present Worth Point…
$700
$600
$500
$400
$300
$200
$100
0
X
1
2
3
4
5
The Present Worth Point of the
Gradient
6
7
Present Worth: Linear Gradient
• The present worth of a linear gradient is the
present worth of the two components:
– 1. The Present Worth of the Gradient
Component and,
– 2. The Present Worth of the Base Annuity flow
– Requires 2 separate calculations.
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Present Worth: Gradient Component
• Three contiguous counties in Florida have agreed to pool
tax resources already designated for county-maintained
bridge refurbishment. At a recent meeting, the county
engineers estimated that a total of $500,000 will be
deposited at the end of next year into an account for the
repair of old and safety-questionable bridges throughout
the three-county area. Further, they estimate that the
deposits will increase by $100,000 per year for only 9 years
thereafter, then cease.
• Determine the equivalent (a) present worth and (b) annual
series amounts if county funds earn interest at a rate of 5%
per year.
10
11
Example
Determine the equivalent annual series amounts if county funds earn
interest at a rate of 5% per year.
Equations for P/G and A/G
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The annual equivalent of this series
of cash flows can be found by
considering an annuity portion of the
cash flows and a gradient portion.
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End of Year
1
2
3
Cash Flows ($)
2,000
3,000
4,000
4
5,000
End of Year
1
Annuity ($)
2,000
Gradient ($)
0
2
3
4
2,000
2,000
2,000
1,000
2,000
3,000
Uniform Gradient Series
Example: An engineer is planning for a 15 year retirement.
In order to
supplement his pension and offset the anticipated effects of inflation and
increased taxes, he intends to withdraw $5,000 at the end of the first year,
and to increase the withdrawal by $1,000 at the end of each successive
year. How much money must the engineer have in this account at the start
of his retirement, if the money earns 6% per year, compounded annually?
$19000
$18000
$8000
$7000
$6000
$5000
T=0
1
15
P
2
3
4
14
15
Want to Find: P
Given: A1, G, i, and n
Uniform Gradient Series
Example:
AT = (A1) + G(A/G,i,n)
A2 = G(A/G,i,n) = $1000 (A/G,6%,15) = $1000 (5.926) = $5926
AT = $5000 + $5926 = $10,926
P = AT (P/A,i,n) = $10,926 (P/A,6%,15) = $10,926 (9.7123) = $106,120
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17
Shifted Uniform Series
A shifted uniform series starts at a time other than period 1
The cash flow diagram below is an example of a shifted series
Series starts in period 2, not period 1
FA = ?
A = Given
0
1
2
3
4
Shifted series
usually
require the use of
multiple factors
5
PA = ?
Remember: When using P/A or A/P factor, PA is always one year ahead
of first A
When using F/A or A/F factor, FA is in same year as last A
Example Using P/A Factor: Shifted Uniform Series
The present worth of the cash flow shown below at i = 10% is:
(a) $25,304
(b) $29,562
(c) $34,462
P0 = ?
P1 = ?
0
1
0
(d) $37,908
i = 10%
2
3
1
4
2
5
3
Actual year
6
4
5
Series year
A = $10,000
Solution: (1) Use P/A factor with n = 5 (for 5 arrows) to get P1 in year 1
(2) Use P/F factor with n = 1 to move P1 back for P0 in year 0
P0 = P1(P/F,10%,1) = A(P/A,10%,5)(P/F,10%,1) = 10,000(3.7908)(0.9091) = $34,462
Answer is (c)
Example Using F/A Factor: Shifted Uniform Series
How much money would be available in year 10 if $8000 is deposited each year
in years 3 through 10 at an interest rate of 10% per year?
Cash flow diagram is:
FA = ?
i = 10%
Actual year
0
1
2
3
0
4
1
2
5
3
6
7
4
8
5
9
6
7
10
8
Series year
A = $8000
Solution:
Re-number diagram to determine n = 8 (number of arrows)
FA = 8000(F/A,10%,8)
= 8000(11.4359)
= $91,487
Shifted Series and Random Single Amounts
For cash flows that include uniform series and randomly placed single amounts:
Uniform series procedures are applied to the series amounts
Single amount formulas are applied to the one-time cash flows
The resulting values are then combined per the problem statement
The following slides illustrate the procedure
Example: Series and Random Single Amounts
Find the present worth in year 0 for the cash flows
shown using an interest rate of 10% per year.
PT = ?
0
i = 10%
1
2
3
4
5
6
7
9
8
A = $5000
PT = ?
0
10
$2000
i = 10%
1
2
3
0
4
1
5
2
6
3
7
4
A = $5000
Actual year
8
5
9
6
10
7
8
Series year
$2000
Solution:
First, re-number cash flow diagram to get n for uniform series: n = 8
Example: Series and Random Single Amounts
PA
PT = ?
i = 10%
0
1
2
0
3
4
1
5
2
6
3
7
4
A = $5000
8
5
9
6
Actual year
10
7
8
Series year
$2000
Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675
Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044
Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933
Now, add P0 and P2000 to get PT: PT = 22,044 + 933 = $22,977
Example Worked a Different Way
(Using F/A instead of P/A for uniform series)
The same re-numbered diagram from the previous slide is used
PT = ?
i = 10%
0
1
2
3
0
4
1
5
2
6
3
7
4
A = $5000
Solution:
FA = ?
8
5
9
6
10
7
8
$2000
Use F/A to get FA in actual year 10: FA = 5000(F/A,10%,8) = 5000(11.4359) = $57,180
Move FA back to year 0 using P/F: P0 = 57,180(P/F,10%,10) = 57,180(0.3855) = $22,043
Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933
Same as before
Now, add two P values to get PT: PT = 22,043 + 933 = $22,976
As shown, there are usually multiple ways to work equivalency problems
Example: Series and Random Amounts
Convert the cash flows shown below (black arrows) into
an equivalent annual worth A in years 1 through 8 (red arrows)
at i = 10% per year.
A=?
0
1
2
3
4
0
5
1
6
2
7
3
A = $3000
Approaches:
Solution:
i = 10%
8
4
5
$1000
1. Convert all cash flows into P in year 0 and use A/P with n = 8
2. Find F in year 8 and use A/F with n = 8
Solve for F: F = 3000(F/A,10%,5) + 1000(F/P,10%,1)
= 3000(6.1051) + 1000(1.1000)
= $19,415
Find A: A = 19,415(A/F,10%,8)
= 19,415(0.08744)
= $1698
Shifted Arithmetic Gradients
Shifted gradient begins at a time other than between periods 1 and 2
Present worth PG is located 2 periods before gradient starts
Must use multiple factors to find PT in actual year 0
To find equivalent A series, find PT at actual time 0 and apply (A/P,i,n)
Example: Shifted Arithmetic Gradient
John Deere expects the cost of a tractor part to increase by $5 per year beginning 4 years from
now. If the cost in years 1-3 is $60, determine the present worth in year 0 of the cost through
year 10 at an interest rate of 12% per year.
PT = ?
i = 12%
0
1
60
3
2
0
1
60
60
Actual years
4
2
65
10
5
3
70
95
G=5
Solution:
Gradient years
8
First find P2 for G = $5 and base amount ($60) in actual 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 and 2
PA = 60(P/A,12%,2) = $101.41
Finally, add P0 and PA to get PT in year 0
PT = P0 + PA = $396.70
Example: Shifted Arithmetic Gradient
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