This document contains lecture notes on interest formulas including geometric series, uniform series, arithmetic gradient, geometric gradient, nominal and effective interest rates, and continuous compounding. It provides examples and explanations of formulas for present worth, future worth, and compound interest calculations for situations involving constant and increasing cash flows over time with single-rate and multiple compounding periods.

1. 5: More Interest Formulas
(continued…)
Dr. Mohsin Siddique
Assistant Professor
msiddique@sharjah.ac.ae
Ext: 29431
Date: 28/10/2014
Engineering Economics
University of Sharjah
Dept. of Civil and Env. Engg.

2. 2
Part I

3. Outcome of Today’s Lecture
3
After completing this lecture…
The students should be able to:
Understand geometric series compound interest formulas

4. More interest Formulas
4
Uniform Series
Arithmetic Gradient
Geometric Gradient
Nominal and Effective Interest
Continuous Compounding

5. Geometric Gradient Series
5
Instead of constant amount of increase, sometimes cash flows
increase by a uniform rate of increase g (constant percentage amount)
every subsequent period.
For example: If the maintenance costs of car are $100 for the first year
and they are increasing at a uniform rate, g, of 10% per year
100
110
121
133.1
146.41
1A
( )gAA += 112
( ) ( )2
123 11 gAgAA +=+=
10 2 3 4 5
( ) ( )4
145 11 gAgAA +=+=
( ) 1
1 1
−
+=
n
n gAA
...

6. Geometric Gradient Series
6
Let’s write a present worth value for
each period individually, and add them
up
Recall:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) nnnn
igAigA
igAigAiAP
−−+−−
−−−
++++++
++++++++=
1111
...11111
1
1
12
1
32
1
21
1
1
1
Eq. (1)
( )
( ) n
n
iF
iP
−
+=
+=
1P
1F
Multiply Eq. (1) by (1+g)/(1+i) to obtain
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) 1
1
1
1
43
1
32
1
21
1
11
1111
...11111111
−−−−
−−−−
++++++
+++++++++=++
nnnn
igAigA
igAigAigAigP
Eq. (2)

7. Geometric Gradient Series
7
Subtract Equation (2) from Equation (1) to yield
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )[ ]
( ) ( )
( ) 





−−+
++−
=
++−=−−+
++−=+−+
++−+=++−
−
−
−
−−−−
gi
ig
AP
igAgiP
igAAgPiP
igAiAigPP
nn
nn
nn
nn
11
111
11111
1111
11111
1
1
11
1
1
1
1
1
11
-
Eq. (3)
Eq. (4)Where gi ≠
Where is called geometric series
present worth factor and has notation
( ) ( )
( ) 





−−+
++−
−
gi
ig
nn
11
111
( )nigAP ,,,/
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) nnnn
igAigA
igAigAiAP
−−+−−
−−−
++++++
++++++++=
1111
...11111
1
1
12
1
32
1
21
1
1
1
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) 1
1
1
1
43
1
32
1
21
1
11
1111
...11111111
−−−−
−−−−
++++++
+++++++++=++
nnnn
igAigA
igAigAigAigP

8. Geometric Gradient Series
8
Example : Suppose you have a vehicle.The first year maintenance cost is
estimated to be $100.The rate of increase in each subsequent year is
10%.You want to know the present worth of the cost of the first five years
of maintenance, given i = 8%.
Solution:
1. Repeated Present-Worth (Step-by-Step) Approach:

9. Geometric Gradient Series
9
2. Using geometric series present worth factor formula

10. Multiple Compounding
10
The time standard for interest computations is OneYear.
Many banks compound interest multiple times during the year.
e.g.: 12% per year, compounded monthly (1% interest is paid
monthly)
e.g.: 8% per year, compounded semi-annually (4% interest is paid
twice a year or once every 6 months)
Compounding is not less important than interest.You have to know
all the info to make a good decision.

11. Multiple Compounding
11
You need to pay attention to the following terms:
Time Period –The period over which the interest is expressed (always
stated).
e.g.:“6% per year”
Compounding Period (sub-period) –The shortest time unit over which
interest is charged or earned.
e.g.: If interest is “6% per year compounded monthly”, compounding
period is one month
Compounding Frequency –The number of times (m) that compounding
occurs within time period.
Compounding semi-annually: m = 2; Compounding quarterly: m = 4
Compounding monthly: m = 12; Compounding weekly: m = 52;
compounding daily: m = 365

12. Nominal and Effective Interest Rate
12
Two types of interest are typically quoted:
1. Nominal interest rate, r, is an annual interest rate without considering
the effect of (sub-period) compounding.
2. Effective interest rate, i or ia , is the actual rate that applies for a stated
period of time which takes into account the effect of (sub-period)
compounding.
Sometimes one interest rate is quoted, sometimes another is
quoted. If you confuse the two you can make a bad decision.
Effective interest is the “real” interest rate over a period of
time; Nominal rate is just given for simplicity (per year)
All interest formulas use the effective interest rate

13. Nominal and Effective Interest Rate
13
Let r=nominal interest rate per interest period (usually one year)
i=effective interest rate per interest period (sub-period compounding)
m=Compounding frequency (No. of compounding sub period per time period)
r/m=interest rate per compounding sub-period
1=P
( )1
1 /11 mrF +=
( ) ( )21
12 /11/1 mrmrFF +=+=
1
0
2 3 4
( ) ( )41
34 /11/1 mrmrFF +=+=
...
interest period (one year)sub period
(quarter)
( )m
mrF /11 +=
( ) ( )11
1111 aiiF +=+=
( ) 1/1 −+=
m
a mri
( ) ( )m
a mri /11 +=+
Thus in general form we can
write
Moreover, we also know that
Thus

14. Nominal and Effective Interest Rate
14
Example:
Given an interest rate of 12% per year, compounded quarterly:
Nominal rate=r = 12%
Compounding frequency=m=4
Effective (Actual) rate =r/m= 12%/4 = 3% per quarter
Effective rate per year = [1+(0.12/4)]4-1= 0.1255=12.55%
Investing $1 at 3% per quarter is equivalent to investing $1 at 12.55%
annually

15. Nominal and Effective Interest Rate
15
Example: A bank pays 1.5% interest every three months.What are the
nominal and effective interest rates per year?
Solution:
Effective interest rate per three months=1.5%
Nominal interest rate per year = r = 4 x 1.5% = 6% a year
Effective interest rate per year= ia= (1 + r/m)m–1 = (1.015)4–1 =
=0.06136
=6.14% a year

16. Nominal and Effective Interest Rate
16
Example: $10K is borrowed for 2 years at an interest rate of 24% per
year compounded quarterly. If the same sum of money could be borrowed
for the same period at the same interest rate of 24% per year
compounded annually, how much could be saved in interest charges?
Solution
Interest=F-P
Interest charges for quarterly compounding:
10,000(1+24%/4)2x4-10,000 = $5938.48
Interest charges for annually compounding:
10,000(1+24%)2-10,000 = $5376.00
Savings: $5938.48 -$5376 = $562.48

17. Nominal and Effective Interest Rate
17
Example 4-15: A loan shark lends money on the following terms.“If I
give you $50 on Monday, then you give back $60 the following Monday.”
Solution
1.What is the nominal rate, r ?
The loan shark charges i= 20% a week:
60 = 50 (1+i) [Note we have solved 60 = 50(F/P,i,1) for i]
i= 0.2
We know m = 52, so r = 52 x i= 10.4, or 1,040% a year
2.What is the effective rate, ia?
ia= (1 + r/m)m–1 = (1+10.4/52)52–1 =13,104
This means about 1,310,400 % a year !!!!

18. Nominal and Effective Interest Rate
18
Example 4-16: You deposit $5,000 in a bank paying 8% nominal interest,
compounded quarterly.You want to withdraw the money in five equal
yearly sums, beginning Dec. 31 of the first year. How much should you
withdraw each year ?
08.0quarterlycompounded%8 == yearlyr
4
0
8 12 16
(year)
w w w w w
20 20 months
1 2 3 4 5 5 years
( ) ( ) yearlymri
m
a %24.814/08.011/1
4
=−+=−+=
$5,000
Effective annual interest:

19. Nominal and Effective Interest Rate
19
This diagram may be solved directly to determine the annual
withdrawal W with the capital recovery factor
The depositor should withdraw $1260 per year
4
0
8 12 16
(year)
w w w w w
20 20 months
1 2 3 4 5 5 years$5,000
( ) ( )
( )
1260$
11
1
%,,/ =





−+
+
== n
n
i
ii
PniPAPW

20. Continuous Compounding
20
Continuous compounding is when m is infinite
i.e. m ∞: r/m 0:
( )
( )mn
m
a
mrPF
mri
/1
1/1
+=
−+=
( )
( ) rnmn
rm
a
PemrPF
emri
=+=
−=−+=
/1
11/1

21. Continuous Compounding
21
Continuous compounding can sometimes be used to simplify
computations, and for theoretical purposes.
The previous equation illustrates that (er – 1) is a good
approximation of (1 + r/m)m -1 for large m (i.e., ∞).
This means there are continuous compounding versions of the
formulas we have seen earlier.
F = P ern is analogous to F = P (F/P,r,n): (F/P,r,n)∞= ern
P = F e-rn is analogous to P = F (P/F,r,n): (P/F,r,n)∞= e-rn

22. Continuous Compounding
22

23. 23
ThankYou
Feel Free to Contact
msiddique@sharjah.ac.ae
Tel. +971 6 5050943 (Ext. 2943)