Chapter 4 TVM Edited
Chapter 4 TVM Edited
Chapter 4 TVM Edited
MATHEMATICS OF FINANCE
1
n= Number of days
360
Maturity value (future value) represents the accumulated amount or value at the end of the time periods
given. Thus,
Future value (F) = Principal (P) + Interest (I)
Example
A credit union has issued a 3 year loan of Birr 5000. Simple interest is charged at a rate of 10% per year.
The principal plus interest is to be repaid at the end of the third year.
a. Compute the interest for the 3-year period.
b. What amount will be repaid at the end of the third year?
Solution
Given values in the problem
3 – Years loan = Principal = Birr 5000
Interest rate = i = 10% = 0.1
Number of years (n) = 3 years
a. I = p i n
I = 5000 x 0.1 x 3
I = Birr 1500
b. The amount to be repaid at the end of the third year is the maturity (future) value of the
specified money (Birr 5000). Accordingly, F = P + I
F = 5000 + 1500
F = Birr 6500
Or, using alternative approach,
F=P+I
Then, substitute I = P i n in the expression to obtain
F = P + Pin
F = P (1 + in)
Consequently, using this formula we can obtain
F = 5000 (1+ (0.1x3)
F = 5000 x 1.3
F = Birr 1500
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Example
A person “lends” Birr 10,000 to a corporation by purchasing a bond from the corporation. Simple interest
is computed quarterly (four times a year) at a rate of 2% per quarter, and a check for the interest is mailed
each quarter to all bondholders. The bonds expire at the end of 5 years, and the final check includes the
original principal plus interest earned during the last quarter. Compute the interest earned each quarter
and the total interest, which will be earned over the five-year life of the bonds.
Solution
Given values in the problem, P = Birr 10,000 i = 2% per quarter n = 5 years
Required: Interest per quarter and interest over the five-year periods
Interest per quarter (one quarter) = Pin
= 10000 x 0.2 x 1
= Birr 200
There, at each quarter the interest earned on Birr 10,000 is Birr 200.
Remark
The interest rate of 2% is given as at a quarterly rate. Hence, in computation of the interest we shall not
change it in to annual rate. As long as the interest rate is provided in the desired time interval, we shall not
make adjustment on the rate given as well as the period. Yet, if conversion from one time interval to
another is demanded, we have to be consistent. For example, in the above example the i = 2% is for a
quarter. The yearly rate will be 2x4 = 8%. By using this annual rate, we can compute the total interest as
follows.
I = Pin
= 10000 x0.08 x 5 = Birr 4000
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Solution for P, i, and n
In the computation of simple interest in some cases we will be required to find out the value of the
principal, interest rate and the time period. Such computation for P, i and n is simply made by driving the
formula for the unknown values from the formula we have used for simple interest.
Example
1. How long must one leave Birr 300 invested in order to learn Birr 28 interest at 3% per year?
2. At what rate will Birr 150 produce interest of Birr 20.25 in 4.5 years?
3. What principal is required to produce interest of Birr 38.50 in two year at 3.5 % per year?
Solution
1. The question involves determining the time period which is enough to earn an interest of Birr 28 on
Birr 300.
The given values in the problem are P = Birr 300 I = Birr 28 i = 3% and n = ?
I = Pin, now solve for n in this formula.
n= I = 28 = 28 = 4 years
Pi 300 x 0.30 9
2. Given values in the problem
P=Birr 150
i=Birr 20.25
n=4.5 years
The required value is the rate of interest.
I = pin, Solve for i
i=I = 20.25 = 20.25 = 0.03 or 3%
Pn 150x4.5 675
3. Given values in the problem
I = Birr 38.50
n =2 years
I = 3.5% per year
Required: Principal (P)
We can find out the value of P In the same manner with the above examples a follows.
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I = pin, solve for P
P=
P = Birr 550
Thus, Birr 550 is required to produce interest of Birr 38.5 in 2 years at 3.5% rate.
Solution
Given a) P=1460 b) P=1460
P = Birr 1460 n=72/365 n= 72/360
n = 72 days i= 0.1 i=0.1
i = 10% = 0.1 I=Pin =1460*72/365*0.1=28.8 I=Pin= 1460*72/365*0.1=29.2
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F=P+I but I = Pin
Thus, F = P + Pin
F = P (1+ in)
P=
If P is found by the above formula, we say that F has been discounted. The difference between F and P is
called the simple discount and is the same as the simple interest on P.
Example
1. 90 days after borrowing money a person repaid exactly Birr 870.19. How much money was
borrowed if the payment includes principal and arch nary simple interest at 9 ½ %?
2. What is the present value of Birr 645 due in 2 ½ years if the interest rate is 3%? What is the
simple discount?
Solution
1. Given values in the problem,
n in ordinary method = Number of days / 360
= 90 /360
n = 0.25
F = the amount repaid = Birr 870.19
i = 9 ½% = 9.5% = 0.095
Required:
The amount borrowed which is the same as simple present value, P.
F
1 + in
= 870.19 / (1+ (0.095 x 0.25))
P = 870.19 ÷ 1.024
P = Birr 849.795
2. Given values in the problem,
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F = Birr 645
n = 2.5 years
i = 3% = 0.03
P = F/ 1+ in
P = 645/ 1+ (0.03 x 2.5)
P = 645/1.075
P = Birr 600
A period, for this purpose, can be any unit of time. If interest is compounded annually, a year is the
appropriate compounding or conversion or interest period. If it is compounded monthly, a month is the
appropriate period. It is important to know that the number of compounding period/s within a year is/are
used in order to find the interest rate per compounding periods and it is denoted by i in the above formula.
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Consequently, when the interest rate is stated as annual interest rate and is compounded more than once a
year, the interest rate per compounding period is computed by the formula:
i = j / m, where j is annual quoted or nominal interest rate
m number of conversation periods per year or the
compounding periods per year
n = m x t, where t is the number of years
Example
Assume that we have deposited Birr 6000 at commercial Bank of Ethiopia which pays interest of 6% per
year compounded yearly. Assume that we want to determine the amount of money we will have on
deposit (our account) at the end of 2 years (the first and second year) if all interest is left in the savings
account.
Solution
Give values in the problem, P = Birr 6000, j = 6% = 0.06, t = 2 years
m = compounded annually = i.e. only once
n=mxt =1x2 = 2
i = j / m = 0.06 / 1 = 0.06
Then, the required value is the maturity or future value
F = P(1 + i )n
= 6000 (1 + 0.006)2
= Birr 6000 (1.06)2
= Birr 6741.6
Example
An individual accumulated Birr 30,000 ten years before his retirement in order to buy a house after he is
retried. If the person invests this money at 12% compounded monthly, how much will be the balance
immediately after his retirement?
Solution
Given values, P = Birr 30,000 , t = 10 years , i = 12% = 0.12
m = compounded monthly = 12
i = j / m = 0.12 / 12 = 0.01
n = m x t = 12 x 10 = 120 and what is required is the Future Value F.
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Then, F = P (1 + i) n
= 30,000 (1.01)120
= 30,000 (1.01)120
F = Birr 99,011.61
0 1 2 3 . . . n
Present Value (P) Future Value
P= n
= Fn (1+ i) n Fn = P(1+ i) n
(1+ i)
(Compound Amount)
Future value is obtained by compounding technique and the expression (1 + i) is called compound
factor. On the other hand, present value is obtained by discounting techniques and the expression (1 + i) -n
is referred to as the compound discount factor. The formula for present value of compound amount is
simply derived from compound amount formula by solving for P.
Examples
1. What is the present value of
a. Birr 5000 in 3 years at 12% compounded annually?
b. Birr 8000 in 10 years at 10% compounded quarterly?
2. Suppose that a person can invest money in a saving account at a rate of 6% per year compounded
quarterly. Assume that the person wishes to deposit a lump sum at the beginning of the year and have
that some grow to Birr 20,000 over the next 10 years. How much money should be deposited at the
beginning of the year?
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3. A young man has recently received an inheritance of birr 200,000. He wants to make a portion of his
inheritance and invest it for his late years. His goal is to accumulate Birr 300,000 in 15 years. How
much of the inheritance should be invested if the money will earn 8% per year compounded semi-
annually? How much interest will be earned over the 15 years?
Solution
1. (a) Given the values, Fn = F3 = Birr 5000, t = 3 years m = 1 (compounded annually)
n= txm =3x1=3
j = 12 % = 0.12
i = j / m = 0.12/1 = 0.12 and we are required to find Present Value P.
-n
Thus, P = Fn (1 + i)
= 5000 (1 + 0.12)-3
= 5000 (1.12-3)
= 5000 (0.7118)
P = Birr 3559
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i = j ÷ m = 0.08 ÷ 2 = 0.04
n = t x m = 15 x 2 = 30 interest periods/semi-annuals
P = how much of the inheritance should be invested now? P = Fn(1 + i)-n
I = Amount of interest?
= 300,000 (1+0.04)-30 = 300,000(1.04)-30 = 300,000(0.3083)
= Birr 92,490
The present value of Birr 300,000 after 15 years at 4% semi-annual interest rate is equal to Birr 92,490.
Therefore, from the total inheritances received Birr 92,490 needs to be deposited now.
Amount of compound interest = Future Value – Preset Value = 300,000 – 92,490
Amount of compound interest = Birr 207,510.
SECTION 3: ANNUITIES
Annuity refers to a sequence or series of equal periodic payments, deposits, withdrawals, or receipts made
at equal intervals for a specified number of periods. For instance, regular deposits to a saving account,
monthly expenditures for car rent, insurance, house rent expenses, and periodic payments to a person
from a retirement plan fund are some of particular examples of annuity.
Payments of any type are considered as annuities if all of the following conditions are present:
i. The periodic payments are equal in amount
ii. The time between payments is constant such as a year, half a year, a quarter of a year, a
month etc.
iii. The interest rate per period remains constant.
iv. The interest is compounded at the end of every time.
Annuities are classified according to the time the payment is made. Accordingly, we have two basic types
of annuities.
i. Ordinary annuity: is a series of equal periodic payment is made at the end of each interval or
period. In this case, the last payment does not earn interest.
ii. Annuity due: is a type of annuity for which a payment is made of the beginning of each
interval or period.
It is only for ordinary annuity that we have a formula to compute the present as well as future values. Yet,
for annuity due case, we may drive it from the ordinary annuity formula. To proceed, let us first consider
some important terminologies that we are going to use in dealing with annuities.
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i. Payment interval or period: it is the time between successive payments of an annuity.
ii. Term of annuity: it is the period or time interval between the beginning of the first payment
period and the end of the last one.
iii. Conversion or interest period: it is the interval between consecutive conversions of interest.
iv. Periodic payment/rent: it is the amount paid at the end or the beginning or each payment
period.
v. Simple annuity: is the one in which the payment period and the conversion periods coincides
each other.
Following the above basic overview about annuities, we shall progress to deal with practical business
problems, which relate with determining the maturity and present values of annuities with specific
application cases.
In attempting this problem, we should understand that the phrase at the end of every year implies an
ordinary annuity case. Likewise, we are required to find out the accumulated money immediately after the
last deposit which also indicate the type of annuity. Further, the term of the annuity is four years with
annual interest rate of 10%. Thus, we can show the pattern of deposits diagrammatically as follows.
Birr 1000
Birr 1100
Birr 1210
Birr 1331
Total Future Value = Birr 4641
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The first payment earns interest for the remaining 3 periods. Therefore, the compound amount of it at the
end of the term of annuity is given by,
F = P (1 + i) n = 1000 (1 + 0.1)3 = Birr 1331
In the same manner, the second payment earns interest for two periods (years). So,
F= 1000 (1+0.1)2 = 1210
The 3rd payment earns interest for only one period. So,
F=1000(1+0.1)1 = 1100
No interest for the fourth payment since it is made at the end of the term. Thus, its value is 1000 itself. In
total, the maturity value amounts to Birr 4641.
This approach of computing future value of ordinary annuity is complex and may be tiresome in case the
term is somewhat longer. Thus, in simple approach we can use the following formula for sum of ordinary
annuity (Future Value).
(1+ i)n - 1
i
(1 + 0.1)4 - 1
0.1
A person plans to deposit 1000 birr in a savings account at the end of this year and an equal sum at
the end of each following year. If interest is expected to be earned at the rate of 6% per year
compound semi-annually, to what sum will the deposit (investment) grow at the time of the fourth
deposit?
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2. A 12-year-old student wants to begin saving for college. She plans to deposit Birr 50 in a saving
account at the end of each quarter for the next 6 years. Interest is earned at a rate of 6% per year
compounded quarterly. What should be her account balance 6 years from now? How much interest
will she earn?
(1+ i) n -1
i
(1+ 0.03)4 -1
0.03
[
= 1000 (1.03)4 -1/ 0.03 ]
= 1000 x 4. 183627
F = Birr 4183.63
2. R = Birr 50
t = 6 years
m = quarterly = 4 times a year
n = t x m = 6 x 4 = 24 quarters
j = 6% = 0.06
i = j ÷ m = 0.06 ÷ 4 = 0.015
(1+ 0.015)24 -1
F 24 = 50
[
= 50 (1.01524 -1) ÷ 0.015 ]
F24 = Birr 1431.68
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= F24 -1200 = 1431.68 - 1200
= Birr 231.68
(1+ i) n -1
i
meet future obligations. Accordingly, we will be given the Future Amount, F, in n period and our interest
is to determine the periodic payment, R. Then we can drive the formula for R as follows.
i
(1 + i) n - 1
That is,
i (1+ i) n i
(1+ i) n -1 i (1+ i) n -1
i
Then, is the sinking found formula.
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In general, a sinking fund can be established for expanding business, buying a new building,
vehicles, settling mortgage payment, financing educational expenses etc.
A corporation wants to establish a sinking fund beginning at the end of this year. Annual deposits will
be made at the end of this year and for the following 9 years. If deposits earn interest at the rate of 8%
per year compounded annually, how much money must be deposited each year in order to have 12
million Birr at the time of the tenth deposit? How much interest will be earned?
2. Assume in the previous example that the corporation is going to make quarterly deposits and that
interest is earned at the rate of 8% per year compounded quarterly. How much money should be
deposited each quarter? How much less will the company have to deposit over the 10-year period as
compared with annual deposits and annual compounding.
3. A firm wishes to establish a sinking found for the purpose of expanding the production facilities at
one of its plants. The company needs to accumulate 500,000 birr over the next five years that
earn interest at 6% compounded semi-annually.
How much should the firm contribute to the found at the end of each semi-annual period in
order to achieve the goal?
Calculate the compound interest.
Prepare the fund accumulation schedule.
i
R =12,000,000
(1+ 0.08)10 - 1
0.08
R = 12,000,000
R = Birr 828,353.86
On the other hand, the amount of interest, I, is obtained by computing the difference between the maturity
value (Fn = 12,000,000) and the sum of all periodic payments made. Thus,
I = Fn - R (10)
= 12,000,000 – 823,353.86 (10)
= 12,000,000 – 8,283,538.6
= Birr 3,716,461.4
i
R =12,000,000
= 12,000,000 (0.016555747)
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R = 12,000,000
R = Birr 198,668.94
In further computation to determine the difference in amount of deposit by changing the length of
the conversion period, we see that as compared with the first case of annual conversion period, in
the quarterly conversation scheme, the corporation will deposit R (10) minus R (40) less over the
term of the annuity.
a. R=?
R = 500,000
= 500,000 (0.08723050506)
R = Birr 43,615.25
b. Compound Interest = Fn – (n x R)
= Fn - (R (10)
= 5000,000 - 43615.25(10)
= 500,000 - 436,152.5
Interest = Birr 63,847.5
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c. Fund Amortization Schedule
(1 + i) - n
i
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Example
A person recently won a state lottery. The term of the lottery is that the winner will receive annual
payments of birr 18,000 at the end of this year and each of the following 4 years. If the winner could
invest money today at the rate of 6% per year compounded annually, what is the present value of the five
payments?
Solution
R = Annual payments of Birr 18,000
Term of the annuity = t = this year and the following 4 years = 5 years
i = 6% = 0.06 (since the conversion period per year is annual)
n=5
Present value of payments = P = ?
- (1 + i) -n
P = 18,000 = 18,000
P = 18,000[4.212363785]
P = Birr 75,822.55
Example
A woman would like to borrow money from local microfinance institution which charges interest at 4%
compounded quarterly. If the woman is able to pay Birr 100 at the end of each quarter for one year,
a. How much should she receive from the institution at the time of borrowing?
b. How much interest will the woman be charged?
c. Prepare the debt repayment schedule (Amortization schedule).
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Solution
Interest charge rate = j = 4% = 0.04
Periodic payment by the woman = Birr 100
Term of the annuity (debt) = t = 1 year
Conversion period per year = quarterly = m = 4
Number of periods = n = t x m = 1 x 4 = 4 periods
Interest rate per conversion periods = j ÷ m = 0.04 ÷ 4 = 0.01
a. How much to receive now? That is, the present value of the annuities, p.
1 - (1+ i) -n
i
- (1.01)-4
= 100 (3.902)
Given the woman’s potential to pay Birr 100 at the end of each quarter for one year, she can
borrow Birr 390.2 at the beginning.
b. The debt repayment schedule is a table that shows a periodic status of payments that gradually make
the debt account balance zero. This table is also called amortization schedule. Now let us proceed
with preparing the schedule.
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Periodic Payment Ending Balance
(Debt + I – R)
Birr 100 Birr 294.102
Birr 100 Birr 197.043
Birr 100 Birr 99.013
Birr 100 Birr 0
As you observe in the above amortization schedule, in ordinary annuity periodic payment the last balance
becomes zero.
Mortgage Payments and Amortization
Another main area of application of annuities in to real world business situations in general and financial
management practices in particular is mortgage amortization or payment. Mortgage payment is an
arrangement where by regular payments are made in order to settle an initial sum of money borrowed
from any source of finance. Such payments are made until the outstanding debt gets down to zero. An
individual or a firm, for instance, may borrow a given sum of money from a bank to construct a building
or undertake something else. Then the borrower (debtor) may repay the loan by effecting (making) a
monthly payment to the lender (creditor) with the last payment settling the debt totally.
In mortgage payment, initial sum of money borrowed and regular payments made to settle the respective
debt relate to the idea of present value of an ordinary annuity. Along this line, the expression for mortgage
payment computation is derived from the present value of ordinary annuity formula. Our intention in this
case is to determine the periodic payments to be made in order to settle the debt over a specified time –
period.
1 - (1 + i) - n
i
Now, we progress to isolate R on one side. It involves solving for R in the above present value of
ordinary annuity formula. Hence, multiply both sides by the interest rate i to obtain:
P i = R [1 – (1 + i) –n]
i
1 - (1+ i) - n
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Further, we divide both sides by [1 – (1 + i) –n] and the result will be the mathematical expression or
formula for computing mortgage periodic payments as follows.
Emmanuel purchased a house for Birr 115,000. He made a 20% down payment with the remaining
balance amortized in 30 years mortgage at annual interest rate of 11% compounded monthly.
a. Find the monthly mortgage payment?
b. Compute the total interest.
2. Assume you borrowed Birr 11,500 from a bank to finance construction of a swimming pool and
agreed to repay the loan in 60 monthly equal installments. If the interest is 1.5% per month on the
unpaid balance,
How much is the monthly payment?
How much interest will be paid over the term of the loan?
i
1 - (1+ i) - n
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= 92,000 (0.009523233)
R = Birr 876.14
R = 92,000
b. Total Interest = (R x n) – P
= 876.14 x 360 – 92,000
= Birr 223,409.49
Over the 30 years period Emmanuel is going to pay a total interest of Birr 223,409.49, which is well
more than double of the initial amount of loan. Nonetheless, the high interest can be justified by the
fact that value of a real estate is usually tend to increase overtime. Therefore, by the end of the term
of the loan the value of the real estate (house) could be well higher than its purchase cost in addition
to owning a house to live in for the 30 years and more.
i
1 - (1+ i) - n
= 11,500 (0.025393)
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