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ISE 211
Engineering Economy
More Interest Formulas
(Chapter 4)
Uniform Series Compound Interest Formulas
 Many
times we will find situations where there are a uniform
series of receipts or disbursement.
 Examples: Automobile loan, house payments, etc.
 The series A is defined as: An end-of-period cash receipt or
disbursement in a uniform series, continuing for n periods – the
entire series equivalent to P or F at interest rate i.
Uniform Series Compound Interest Formulas (cont’d)
 The
uniform series is equivalent to a future worth value, F,
at interest rate i, and can be expressed as follows:
 In
functional notation:
 The term ___________________ is called the uniform series
compound amount factor.
 Or we can say that the term _________________ is called
the uniform series sinking fund factor.
Example 1
A man deposits $500 in a credit union at the
end of each year for five years. The credit
union pays 5% interest, compounded
annually. At the end of five years, immediately
following his fifth deposit, how much will he
have in his account?
Solution:
Example 2
Jim Hayes read that in western United States, a ten-
acre parcel of land could be purchased for $1000
cash. Jim decided to save a uniform amount at the
end of each month so that he would have the
required $1000 at the end of one year. The local
credit union pays 6% interest, compounded monthly.
How much would Jim have to deposit each month?
Solution:
How can we find P if we know A?
 The
factor __________________ is called the uniform
series capital recovery factor.
 The factor __________________ is called the uniform
series present worth factor.
Example 1
On January 1st a man deposits $5000 in a
credit union that pays 8% interest,
compounded annually. He wishes to withdraw
all the money in five equal end-of-year sums,
beginning December 31st of the first year. How
much should he withdraw each year?
Solution:
Example 2
An investor holds a time payment purchase contract
on some machine tools. The contract calls for the
payment of $140 at the end of each month for a five
year period. The first payment is due in one month.
He offers to sell you the contract for $6800 cash
today. If you otherwise can make 1% per month on
your money, would you accept or reject the
investor’s offer?
Solution:
Example 3
Suppose we decided to pay the $6,800 for
the time purchase contract in Example 2.
What monthly rate of return would we
obtain on our investment?
Solution:
Example 4
Using a 15% interest rate, compute the value of F
in the following cash flow:
Year
Solution:
Cashflow
1
+100
2
+100
3
+100
4
0
5
F
Example 5
Every month starting today, you save $50
for 12 months. If your account earns 3%
interest, compounded monthly, how much
will you have just after your last deposit?
Solution:
Example 6
Consider the following situation, and find
the value of P.
30
20
Solution:
20
0
i = 15%
P=?
Relationships Between Compound Interest Factors
 Single Payment:
 Compound amount factor = 1 / Present worth
factor
(F/P, i, n) = 1 / (P/F, i, n)
 Uniform Series:
 Capital recovery factor = 1 / Present worth factor
(A/P, i,n) = 1 / (P/A, i, n)
 Compound amount factor = 1 / Sinking fund factor
(F/A, i, n) = 1 / (A/F, i, n)
Miscellaneous Relationships

N
( P / A, i, n)   ( P / F , i, J )
J 1

N 1
( F / A, i, n)  1   ( F / P, i, J )
J 1

( A / P, i, n)  ( A / F , i, n)  1 i
Arithmetic Gradient
 In
an arithmetic series, a constant amount is added each period.
 Suppose the amount added = G.
 Examples: 2, 4, 6, 8, ….

(G=2).
: 225, 250, 275, 300, … (G=25)
 The cash flow diagram for such a situation is shown as:
Arithmetic Gradient (cont’d)
 In an arithmetic series, a constant amount is added each period.
 This
can be expressed as follows:
Arithmetic Gradient (cont’d)

The arithmetic gradient present
worth factor is given as follows:
 The arithmetic gradient uniform
series factor:
Example 1
A man purchased a new automobile. He wishes to set aside enough
money in a bank account to pay the maintenance on the car for the
first five years. It has been estimated that the maintenance cost of an
automobile is as follows:
Year
Maintenance Cost
1
$120
2
150
3
180
4
210
5
240
Assume the maintenance costs occur at the end of each year and that
the bank pays i=5%. How much should he deposit in the bank now?
Example 2
On a certain piece of machinery, it is estimated that the
maintenance expense will be as follows:
Year
Maintenance Cost
1
$100
2
200
3
300
4
400
What is the equivalent uniform annual maintenance
cost for the machinery if 6% interest is used?
Example 3
Consider the following cash flows
Year
Amount
1
$24,000
2
18,000
3
12,000
4
6,000
Assuming 10% interest, what is the equivalent
uniform annual series over 4 years?
Example 4
Compute the value of P in the diagram below, using
10% interest rate.
150
100
50
0
P
0
0
Geometric Gradient
 This illustrates the situation when the period-by-period change
is a uniform rate, g.
 Example: The maintenance costs for an automobile are $100
the first year and increasing at a uniform rate, g, of 10% per year.
Illustration:
Geometric Gradient (cont’d)
 From the previous table, we can calculate the maintenance in
any given year as follows:
$100(1+g)n-1
 Which can be written in general as follows:
An = A1 (1+g)n-1
Where g = geometric gradient -- uniform rate of cash flow
increase/decrease from period to period
A1 = value of cash flow at year 1
An = value of cash flow at any year n.
Geometric Gradient (cont’d)
 What if we are interested in obtaining the present worth of the
geometric series shown below?
P= ??
 This can be done using the following equation:
Geometric Gradient (cont’d)
 This
can be expressed as follows in functional
notation:
P = A1(P/A, g, i, n)
i≠g
 The term A1(P/A, g, i, n) is the geometric series
present worth factor where i  g.
 In the special case where i = g, the equation
becomes:
P = A1 n (1 + i)-1
Example 1
The first year maintenance for a new
automobile is estimated to be $100, and it
increases at a uniform rate of 10% per year.
What is the present worth of cost of the
first five years of maintenance in this
situation, using an 8% interest rate?
Solution:
Nominal and Effective Interest
Example: Consider a situation of a person depositing
$100 into a bank account that pays 5% interest
compounded semi-annually. How much would be in
the savings account at the end of one year?
Solution:
 What interest rate compounded annually, results in
this value of F = ______?
Solution:
Nominal and Effective Interest (cont’d)
Definitions:
r = nominal interest rate per interest period
(usually one year) – interest rate per interest period
without considering the compounding effect.
i = effective interest rate per interest period (e.g.,
year) – the interest rate taking into account the effecting
of compounding per interest period.
ia = effective interest rate per year (annum)
m= # of compounding sub-periods per time period
Nominal and Effective Interest (cont’d)

The effective interest rate per year is given as follows:
ia = (1 + r/m)m – 1
ia = (1 + i)m – 1
 Example: If a savings bank pays 1.5% interest
every three months, what are the nominal and
effective interest rates per year?
Nominal and Effective Interest (cont’d)
Example
You go to a check cashing store to get a $50 advance on your
pay check. They are willing to do so provided you write a check
for $60 dated for 1 week from now.
a) What nominal interest rate per year are you paying?
b) What effective interest rate per year are you paying?
c) How much would you have to write the check for if
your pay back period is 1 year instead of 1 week?
Solution:
What if compounding interval  cash flow period?
Example: A bank pays 8% nominal annual interest per
year, compounded quarterly. A person deposits $5000
now (at time 0). After the end of each of 5 years, she
wants to withdraw an equal amount of money. What is
this amount?
Solution # 1: