In this section you will learn to:
 Find the future value of a lumpsum.
 Find the present value of a lumpsum.
 Find the effective interest rate.
In the Section 2, we did problems involving simple interest. Simple interest is charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, the interest is paid (or charged) not only on the principal, but also on the past interest, and we say the interest is compounded.
Suppose we deposit $200 in an account that pays 8% interest. At the end of one year, we will have
$200+$200.08=$2001+.08=$216$200+$200.08=$2001+.08=$216 size 12{$"200"+$"200" left ( "." "08" right )=$"200" left (1+ "." "08" right )=$"216"} {}.
Now suppose we put this amount, $216, in the same account. After another year, we will have
$216+$216.08=$2161+.08=$233.28$216+$216.08=$2161+.08=$233.28 size 12{$"216"+$"216" left ( "." "08" right )=$"216" left (1+ "." "08" right )=$"233" "." "28"} {}.
So an initial deposit of $200 has accumulated to $233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of
$16.08=$1.28$16.08=$1.28 size 12{$"16" left ( "." "08" right )=$1 "." "28"} {}, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest.
Now suppose we leave this amount, $233.28, in the bank for another year, the final amount will be
$233.28+$233.28.08=$233.281+.08=$251.94$233.28+$233.28.08=$233.281+.08=$251.94 size 12{$"233" "." "28"+$"233" "." "28" left ( "." "08" right )=$"233" "." "28" left (1+ "." "08" right )=$"251" "." "94"} {}.
Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.
After one year, we had
$
200
1
+
.
08
=
$
216
$
200
1
+
.
08
=
$
216
size 12{$"200" left (1+ "." "08" right )=$"216"} {}
(16)After two years, we had
$
216
1
+
.
08
$
216
1
+
.
08
size 12{$"216" left (1+ "." "08" right )} {}
(17)But
$216=$2001+.08$216=$2001+.08 size 12{$"216"=$"200" left (1+ "." "08" right )} {}, therefore, the above expression becomes
$
200
1
+
.
08
1
+
.
08
=
$
233
.
28
$
200
1
+
.
08
1
+
.
08
=
$
233
.
28
size 12{$"200" left (1+ "." "08" right ) left (1+ "." "08" right )=$"233" "." "28"} {}
After three years, we get
$
200
1
+
.
08
1
+
.
08
1
+
.
08
$
200
1
+
.
08
1
+
.
08
1
+
.
08
size 12{$"200" left (1+ "." "08" right ) left (1+ "." "08" right ) left (1+ "." "08" right )} {}
(18)Which can be written as
$
200
1
+
.
08
3
=
$
251
.
94
$
200
1
+
.
08
3
=
$
251
.
94
size 12{$"200" left (1+ "." "08" right ) rSup { size 8{3} } =$"251" "." "94"} {}
(19)Suppose we are asked to find the total amount at the end of 5 years, we will get
$
200
1
+
.
08
5
=
$
293
.
87
$
200
1
+
.
08
5
=
$
293
.
87
size 12{$"200" left (1+ "." "08" right ) rSup { size 8{5} } =$"293" "." "87"} {}
(20)We summarize as follows:
The original amount
$200
=
$200
The amount after one year
$200
1
+
.
08
=
$
216
The amount after two years
$200
1
+
.
08
2
=
$
233
.
28
The amount after three years
$200
1
+
.
08
3
=
$
251
.
94
The amount after five years
$200
1
+
.
08
5
=
$
293
.
87
The amount after
t
years
$200
1
+
.
08
t
The original amount
$200
=
$200
The amount after one year
$200
1
+
.
08
=
$
216
The amount after two years
$200
1
+
.
08
2
=
$
233
.
28
The amount after three years
$200
1
+
.
08
3
=
$
251
.
94
The amount after five years
$200
1
+
.
08
5
=
$
293
.
87
The amount after
t
years
$200
1
+
.
08
t
size 12{ matrix {
"The original amount" {} # "$200"="$200" {} ##
"The amount after one year" {} # "$200" left (1+ "." "08" right )=$"216" {} ##
"The amount after thwo years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{2} } =$"233" "." "28" {} ##
"The amount after three years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{3} } =$"251" "." "94" {} ##
"The amount after five years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{5} } =$"293" "." "87" {} ##
"The amount after "t" years" {} # "$200" left (1+ "." "08" right ) rSup { size 8{t} } {}
} } {}
Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to onefourth of 8%, or 2%.
Now if we deposit $200 in the bank, after one quarter we will have
$2001+.084$2001+.084 size 12{$"200" left (1+ { { "." "08"} over {4} } right )} {} or $204.
After two quarters, we will have
$2001+.0842$2001+.0842 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{2} } } {} or $208.08.
After one year, we will have
$2001+.0844$2001+.0844 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4} } } {} or $216.49.
After three years, we will have
$2001+.08412$2001+.08412 size 12{$"200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"12"} } } {} or $253.65, etc.
The original amount
$200
=
$200
The amount after one quarter
$200
1
+
.
08
4
=
$
204
The amount after two quarters
$200
1
+
.
08
4
2
=
$
208
.
08
The amount after one year
$200
1
+
.
08
4
4
=
216
.
49
The amount after two years
$200
1
+
.
08
4
8
=
$
234
.
31
The amount after three years
$200
1
+
.
08
4
12
=
$
253
.
65
The amount after five years
$200
1
+
.
08
4
20
=
$
297
.
19
The amount after
t
years
$200
1
+
.
08
4
4t
The original amount
$200
=
$200
The amount after one quarter
$200
1
+
.
08
4
=
$
204
The amount after two quarters
$200
1
+
.
08
4
2
=
$
208
.
08
The amount after one year
$200
1
+
.
08
4
4
=
216
.
49
The amount after two years
$200
1
+
.
08
4
8
=
$
234
.
31
The amount after three years
$200
1
+
.
08
4
12
=
$
253
.
65
The amount after five years
$200
1
+
.
08
4
20
=
$
297
.
19
The amount after
t
years
$200
1
+
.
08
4
4t
size 12{ matrix {
"The original amount" {} # "$200"="$200" {} ##
"The amount after one quarter" {} # "$200" left (1+ { { "." "08"} over {4} } right )=$"204" {} ##
"The amount after two quarters" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{2} } =$"208" "." "08" {} ##
"The amount after one year" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4} } ="216" "." "49" {} ##
"The amount after two years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{8} } =$"234" "." "31" {} ##
"The amount after three years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"12"} } =$"253" "." "65" {} ##
"The amount after five years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{"20"} } =$"297" "." "19" {} ##
"The amount after "t" years" {} # "$200" left (1+ { { "." "08"} over {4} } right ) rSup { size 8{4t} } {}
} } {}
(21)Therefore, if we invest a lumpsum amount of
PP size 12{P} {} dollars at an interest rate
rr size 12{r} {}, compounded
nn size 12{n} {} times a year, then after
tt size 12{t} {} years the final amount is given by
A
=
P
1
+
r
n
nt
A
=
P
1
+
r
n
nt
size 12{A=P left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } } {}
If $3500 is invested at 9% compounded monthly, what will the future value be in four years?
Clearly an interest of .09/12 is paid every month for four years. This means that the interest is compounded 48 times over the fouryear period. We get
$35001+.091248=$5009.92$35001+.091248=$5009.92 size 12{$"3500" left (1+ { { "." "09"} over {"12"} } right ) rSup { size 8{"48"} } =$"5009" "." "92"} {}
(22)
How much should be invested in an account paying 9% compounded daily for it to accumulate to $5,000 in five years?
This time we know the future value, but we need to find the principal. Applying the formula
A=P1+rnntA=P1+rnnt size 12{A=P left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } } {}, we get
$5000=P1+.09365365×5$5000=P1.568225$3188.32=P$5000=P1+.09365365×5$5000=P1.568225$3188.32=P size 12{ matrix {
$"5000"=P left (1+ { { "." "09"} over {"365"} } right )"365" times 5 {} ##
$"5000"=P left (1 "." "568225" right ) {} ##
$"3188" "." "32"=P
} } {}
(23)For comparison purposes, the government requires the bank to state their interest rate in terms of effective yield or effective interest rate.
For example, if one bank advertises its rate as 7.2% compounded monthly, and another bank advertises its rate as 7.5%, how are we to find out which is better? Let us look at the next example.
If a bank pays 7.2% interest compounded monthly, what is the effective interest rate?
Suppose we deposit $100 in this bank and leave it for a year, we will get
$1001+.0721212=$107.44$1001+.0721212=$107.44 size 12{$"100" left (1+ { { "." "072"} over {"12"} } right ) rSup { size 8{"12"} } =$"107" "." "44"} {}
(24)Which means we earned an interest of
$107.44−$100=$7.44$107.44−$100=$7.44 size 12{$"107" "." "44"  $"100"=$7 "." "44"} {}
But this interest is for $100, therefore, the effective interest rate is 7.44%.
Interest can be compounded yearly, semiannually, quarterly, monthly, daily, hourly, minutely, and even every second. But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts. When interest is compounded "infinitely many times", we say that the interest is compounded continuously. Our next objective is to derive a formula to solve such problems, and at the same time put things in proper perspective.
Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be
$11+1=$2$11+1=$2 size 12{$1 left (1+1 right )=$2} {}.
If the interest is compounded semiannually, in one year we will have
$11+1/22=$2.25$11+1/22=$2.25 size 12{$1 left (1+1/2 right ) rSup { size 8{2} } =$2 "." "25"} {}
If the interest is compounded quarterly, in one year we will have
$11+1/44=$2.44$11+1/44=$2.44 size 12{$1 left (1+1/4 right ) rSup { size 8{4} } =$2 "." "44"} {}, etc.
We show the results as follows:
Table 1
Frequency of compounding 
Formula 
Total amount 
Annually 
$
1
1
+
1
$
1
1
+
1
size 12{$1 left (1+1 right )} {}

$2 
Semiannually 
$
1
1
+
1
/
2
2
$
1
1
+
1
/
2
2
size 12{$1 left (1+1/2 right ) rSup { size 8{2} } } {}

$2.25 
Quarterly 
$
1
1
+
1
/
4
4
$
1
1
+
1
/
4
4
size 12{$1 left (1+1/4 right ) rSup { size 8{4} } } {}

$2.44140625 
Monthly 
$
1
1
+
1
/
12
12
$
1
1
+
1
/
12
12
size 12{$1 left (1+1/"12" right ) rSup { size 8{"12"} } } {}

$2.61303529 
Daily 
$
1
1
+
1
/
365
365
$
1
1
+
1
/
365
365
size 12{$1 left (1+1/"365" right ) rSup { size 8{"365"} } } {}

$2.71456748 
Hourly 
$
1
1
+
1
/
8760
8760
$
1
1
+
1
/
8760
8760
size 12{$1 left (1+1/"8760" right ) rSup { size 8{"8760"} } } {}

$2.71812699 
Every second 
$
1
1
+
1
/
525600
525600
$
1
1
+
1
/
525600
525600
size 12{$1 left (1+1/"525600" right ) rSup { size 8{"525600"} } } {}

$2.71827922 
Continuously 
$
1
2
.
718281828
.
.
.
$
1
2
.
718281828
.
.
.
size 12{$1 left (2 "." "718281828" "." "." "." right )} {}

$2.718281828... 
We have noticed that the $1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name "ee size 12{e} {}" after the great mathematician Euler.
In mathematics, we say that as
nn size 12{n} {} becomes infinitely large the expression
1+1nn1+1nn size 12{ left (1+ { {1} over {n} } right ) rSup { size 8{n} } } {} equals
ee size 12{e} {}.
Therefore, it is natural that the number
ee size 12{e} {} play a part in continuous compounding. It can be shown that as
nn size 12{n} {} becomes infinitely large the expression
1+rnnt=ert1+rnnt=ert size 12{ left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } =e rSup { size 8{ ital "rt"} } } {}.
Therefore, it follows that if we invest
$P$P size 12{$P} {} at an interest rate
rr size 12{r} {} per year, compounded continuously, after
tt size 12{t} {} years the final amount will be given by
A=P⋅ertA=P⋅ert size 12{A=P cdot e rSup { size 8{ ital "rt"} } } {}.
If $3500 is invested at 9% compounded continuously, what will the future value be in four years?
Using the formula for the continuous compounding, we get
A=PertA=Pert size 12{A= ital "Pe" rSup { size 8{ ital "rt"} } } {}.
A
=
$
3500
e
.09
×
4
=
$
3500
e.36
=
$5016.65
A
=
$
3500
e
.09
×
4
=
$
3500
e.36
=
$5016.65
size 12{ matrix {
A=$"3500"e rSup { size 8{ "." "09" times 4} } {} ##
=$"3500"e rSup { size 8{ "." "36"} } {} ##
=$"5016" "." "65"
} } {}
(25)Next we learn a commonsense rule to be able to readily estimate answers to some finance as well as reallife problems. We consider the following problem.
If an amount is invested at 7% compounded continuously, what is the effective interest rate?
Once again, if we put $1 in the bank at that rate for one year, and subtract that $1 from the final amount, we will get the interest rate in decimals.
1e.07−11.0725−1.0725 or 7.251e.07−11.0725−1.0725 or 7.25 size 12{ matrix {
1e rSup { size 8{ "." "07"} }  1 {} ##
1 "." "0725"  1 {} ##
"." "0725"" or "7 "." "25"%
} } {}
(26)
If an amount is invested at 7%, estimate how long will it take to double.
Since we are estimating the answer, we really do not care how often the interest is compounded. Let us say the interest is compounded continuously. Then our problem becomes
Pe.07t=2PPe.07t=2P size 12{ ital "Pe" rSup { size 8{ "." "07"t} } =2P} {}
(27)We divide both sides by
PP size 12{P} {}
e.07t=2e.07t=2 size 12{e rSup { size 8{ "." "07"t} } =2} {}
(28)Now by substituting values by trial and error, we can estimate t to be about 10.
By doing a few similar calculations we can construct a table like the one below.
Table 2
Annual interest rate 
1% 
2% 
3% 
4% 
5% 
6% 
7% 
8% 
9% 
10% 
Number of years to double money 
70 
35 
23 
18 
14 
12 
10 
9 
8 
7 
The pattern in the table introduces us to the law of 70.
 Definition 1: The Law of 70:
The number of years required to double money = 70 ÷ interest rate
It is a good idea to familiarize yourself with the law of 70, as it can help you to estimate many problems mentally.
If the world population doubles every 35 years, what is the growth rate?
According to the law of 70,
35=70÷rr=235=70÷rr=2 size 12{ matrix {
"35"="70" div r {} ##
r=2
} } {}
(29) Therefore, the world population grows at a rate of 2%.
We summarize the concepts learned in this chapter in the following table:
If an amount
PP size 12{P} {} is invested for
tt size 12{t} {} years at an interest rate
rr size 12{r} {} per year, compounded
nn size 12{n} {} times a year, then the future value is given by
A
=
P
1
+
r
n
nt
A
=
P
1
+
r
n
nt
size 12{A=P left (1+ { {r} over {n} } right ) rSup { size 8{ ital "nt"} } } {}
(30) If a bank pays an interest rate
rr size 12{r} {} per year, compounded
nn size 12{n} {} times a year, then the effective interest rate is given by
r
e
=
1
1
+
r
n
n
−
1
r
e
=
1
1
+
r
n
n
−
1
size 12{r rSub { size 8{e} } =1 left (1+ { {r} over {n} } right ) rSup { size 8{n} }  1} {}
(31)
If an amount
PP size 12{P} {} is invested for
tt size 12{t} {} years at an interest rate
rr size 12{r} {} per year, compounded continuously, then the future value is given by
A
=
Pe
rt
A
=
Pe
rt
size 12{A= ital "Pe" rSup { size 8{ ital "rt"} } } {}
(32)
The law of 70 states that
The number of years to double money = 70 ÷ interest rate
"Reviewer's Comments: 'I recommend this book for undergraduates. The content is especially useful for those in finance, probability statistics, and linear programming. The course material is […]"