Lecture 6
ENGR 140
Arithmetic Gradient Factors (P/G and A/G)
Arithmetic gradient is used to model equivalent values for linearly increasing payments.
e.g. equipment maintenance costs may be assumed to increase a uniform amount each
year.
i=
P
1
2
3
4
5
0
G
The arithmetic gradient equations, A/G and P/G, are derived in your text. The equations
are:

n
1 
A=G  

n
 i   1  i   1
P
n
G  1  i   1
n 



n
i  i 1  i 
1  i  n 
Note: The payment at the end of the first period is the base amount. The gradient begins at the
end of the second period.
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Lecture 6
To model and solve problems involving arithmetically changing gradients, break the payment
into two components, the base amount and the arithmetically changing gradient. Model the base
amount as a uniform series, A. Model the arithmetically changing gradient as a gradient, G.
This technique is illustrated in the cash flow diagram below.
i=
P
1
2
3
4
0
5
A
G
To determine the present worth equivalent of the series, determine the present worth equivalent
of each component and sum them. The most direct solution method is to use the interest rate
tables. The present worth equivalent for the series is:
P = A(P/A, i, n) + G(P/G, i, n)
To determine the uniform series equivalent of the series, sum the base amount, A, and the
uniform series equivalent of the gradient (A/G). The uniform series equivalent for the series is:
Aeq = A + G(A/G, i, n)
Using Compound Interest Factor Tables to Solve Engineering Economic Problems
Notation
Name
Find/Given
Equation
(P/G, i, n)
Arithmetic gradient present worth factor
P/G
P = G(P/G, i, n)
(A/G, i, n)
Arithmetic gradient uniform series factor
A/G
A = G(A/G, i, n)
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Lecture 6
Example problem:
1. Maintenance cost for a new piece of manufacturing equipment is expected to be $5,000
in the first year and increase $250 per year for every year thereafter. If the cost of funds
is 10%, what is the present worth cost of maintenance for the next 5 years?
Given: A = $5,000; G = $250; i = 10%; n = 5; P = ?
P=?
i = 10%
1
2
3
4
0
5
A = $5,000
G = $250
Solution using compound interest tables:
P = A(P/A, i ,n) + G(P/G, i, n)
P = $5,000(P/A, 10%, 5) + $250(P/G, 10%, 5)
P = $5,000(3.7908) + $250(6.8618) = $20, 669
Solution using equations:
 1  i n  1 G  1  i  n  1
n 
P  A
 


n 
n
1  i  n 
 i 1  i   i  i 1  i 
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Lecture 6
 1  0.105  1  $250  1  0.10 5  1

5
P  $5,000 


= $20,669

5 
5
5 
0.10






0.10
1

0.10
0.10
1

0
.
10
1

0
.
10




Comment: The interest table method is easier by far!
2. A construction company plans to accelerate the payments on an equipment loan as
production increases. The initial payment is $10,000 per year and the plan is to increase
the payment, beginning in year 2, by an additional $1,000 each year through year 10.
Determine the equivalent annual payment if the loan interest rate is 12%.
Given: A = $10,000; G = $1,000; i = 12%; n = 10; Aeq = ?
i = 12%
1
2
3
4
5
6
7
8
9
10
A = $10,000
0
G = $1,000
Solution using compound interest tables
Aeq = A + G(A/G, i, n)
Aeq = $10,000 + $1,000(A/G, 12%, 10)
Aeq = $10,000 + $1,000(3.5847) = $13,585
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Lecture 6
3. Production from a gas well is expected to decline over the next five years until
production revenues equal operating costs, at which point the well will be shut down.
Net income for year 1 is projected to be $1.2 Million. Determine the gradient and the
present worth equivalent value of the five-year production from the well. The company
uses a 10% rate of return for its planning purposes.
i = 10%
P=?
$1.2 M
A
G=?
0
G=
1
 $1,200,000  $0 
1  5
2
3
=  $300,000
4
5
yr
P = A(P/A, i, n) – G(P/G, i, n)
P = $1,200,000(P/A, 10%, 5) – $300,000(P/G, 10%, 5)
P = $1,200,000(3.7908) – $300,000(6.8618) = $2.49 Million
Reading: Chapter 2, pp 65 – 71
Homework: Problems 2.27, 2.30, 2.33
Due Monday, October 3
Quiz 4, Monday October 3
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