09/09/2018
Mathematics of Finance
Topic 1:
Simple Interest and Simple Discount
SIMPLE INTEREST AND SIMPLE
DISCOUNT
THE CONCEPT OF INTEREST
Money is invested or borrowed in thousands of transactions every day. When an
investment is cashed in or when borrowed money is repaid, there is a fee that is
collected or charged. This fee is called interest.
Interest can be viewed as the “rent” one pays for the
money.
use of somenone else’s
In any financial transaction, there are two parties involved: an investor, who is
lending money to someone, and a debtor, who is borrowing money from the
investor. The debtor must pay back the money originally borrowed, and also the
rent (or fee) charged for the use of the money, called interest.
The capital originally invested is called principal. The sum of the principal and
interest due is called the future or accumulated value. The rate of
interest is the ratio of the interest earned in one time unit on the principal.
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SIMPLE INTEREST
In these cases, we shall consider the interest is attracted only by the principal,
which remains unchanged during the period of investment. Therefore, under
simple interest, the interest earned (or paid) is computed on the principal
during the whole time.
We shall use the following notation:
P = principal, present value or the discounted value.
I = simple interest.
S = accumulated value or future value.
r = annual rate of simple interest.
t = time in years.
I=P·t·r
S=P+I=P·(1+r·t)
Short-term transactions often use simple interest.
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The unit of time will be one year. We shall convert any period
expressed in other units (days, weeks, months) into a fraction of a
year. When the time is given in months o days, then:
t = number of months / 12
t = number of days / 365 (exact interest)
t = number of days / 360 (ordinary interest)
EXAMPLE.- A loan of $15 000 is taken out. If the interest rate of
the loan is 7%, how much interest is due and what is the amount
repaid if the loan is due in 7 months?
I = P·r·t = 15000 · 0’07 · 7/12 = $612’50
S = P·(1+r·t) = 15000 · ( 1 + 0’07 · 7/12) = $15 612’50
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EXAMPLE.- Calculate the exact and ordinary simple interest on a 90days loan of $8000 at 8’5%
Exact interest ! I = 8000 · 0’085 · 90/365 = $167’67
Ordinary interest ! I = 8000 · 0’085 · 90/360 = $170’00
EXAMPLE.- A loan of $100 is to be repaid with $108 at the end of 8
months. What was the annual simple interest rate?
108 = 100 ( 1 + r · 8/12) ! r = 12%
EXAMPLE.- A deposit of $1500 is made into a fund for 140 days. The
fund earns simple interest at 5%. Then, the interest rate changes to
4’5% for 79 days. How much is in the fund at the end of the
investment?
S = 1500 + 1500·0’05·140/365 + 1500·0’045·79/365 = $1543’38
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EXERCISES TO WORK IN CLASS.
1) Determine the maturity value of a) a $2500 loan for 18 months at
12% simple interest, b) a $1200 loan for 120 days at 8’5% ordinary
simple interest, and c) a $10 000 loan for 64 days at 7% exact simple
interest.
2) At what rate of simple interest will a) $1000 accumulate to $1420 in
2’5 years, b) money double itself in 7 years, and c) $500 accumulate $10
interest in 2 months?
3) How many days will it take a) $1000 to accumulate to at least $1200
at 5’5% simple interest, b) $1600 to earn at least $30 of interest at 3’5%
simple interest, and c) $5000 to accumulate to at least $5100 at 9%
simple interest?
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When we calculate P from S, we call P the present value of S or the
discounted value of S. You can interpret the value P as the amount
needed to be invested today in order to have S in t years.
P = S (1+r·t)-1
EXAMPLE.- Three months after borrowing money, a person pays back
exactly $200. How much was borrowed if the $200 payment includes
the principal and simple interest at 9%?
P = 200 ( 1 + 0’09 · 3/12 )-1 ! P = $195’60
EXAMPLE.- Jennifer wishes to have $1000 in eight month’s time. If she
can earn 6% simple interest, how much does she need to invest? How
much interest does she earn?
P = 1000 ( 1 + 0’06 · 8/12)-1 = $961’54
I = 1000 – 961’54 = $38’46
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DISCOUNT
All financial decisions must take into account the basic idea that money
has time value. In order words, receiving $100 today is not the same
that receiving $100 one year ago.
In a financial transaction involving money due on different dates, every
sum of money should have an attached date, the date on which it falls
due. The mathematic of finance deals with dated values. This is one of
the most important facts in the mathematics of finance.
At a simple interest rate of 8%, $100 due in one year is considered to equivalent to $108
in two years since $100 would accumulated to $108 in one year. In the same way, $100
(1+0’08)-1 = $92’59 would be considered an equivalent sum today.
92’59
100
108
0
1
2
From a financial point of view, all of three values are the same, or are
equivalent.
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EXAMPLE.- A debt of $1000 is due at the end of 9 months. Using a
simple interest rate of 9%, determine an equivalent debt at the end of 4
months and at the end of 1 year.
At the end of 4 months ! Time goes back (simple interest):
P = 1000 (1+0’09·5/12)-1 = $963’86
At the end of 1 year ! Time moves forwards (simple discount):
S = 1000 (1+0’09·3/12) = $1 022’50
A sum of a set of dated values, due on different dates, has no meaning.
We must take into account the time value of money, which means we
have to replace all the dated values by equivalent dated values, due on
the same date.
Focal date is the particular date at which amounts of money payable at
different times can only be compared
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EXAMPLE.- A person owes $300 due in 3 months and $500 due in 8
months. The lender agrees to allow the person to pay off theses two
debts with a simple payment. What single payment a) now; b) in 6
months; c) in 1 year will liquidate these obligations if money is worth
8% simple interest?
EXAMPLE.- Debts of $500 due in 20 days ago and $400 due in 50 days
are to be settled by a payment of $600 now and a final payment 90 days
from now. Determine the value of the final payment at a simple interest
rate of 11% with a focal date at the present.
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EXAMPLE.- A person owes $300 due in 3 months and $500 due in 8
months. The lender agrees to allow the person to pay off theses two
debts with a simple payment. What single payment a) now; b) in 6
months; c) in 1 year will liquidate these obligations if money is worth
8% simple interest?
a) P = 300 (1+0’08·3/12)-1 + 500 (1+0’08·8/12)-1 = $768’80
b) P = 300 (1+0’08·3/12) + 500 (1+0’08·2/12)-1 = $799’42
c) P = 300 (1+0’08·9/12) + 500 (1+0’08·4/12) = $831’33
EXAMPLE.- Debts of $500 due in 20 days and $400 due in 50 days are to
be settled by a payment of $600 now and a final payment 90 days from
now. Determine the value of the final payment at a simple interest rate
of 11% with a focal date at the present.
500 (1+0’11·20/365)-1+400 (1+0’11·50/365)-1 = 600+X·(1+0’11·90/365)-1
503’01 + 394’06 = 600 + 0’9736 X ! X = $35’13
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EXAMPLE.- A person borrows $1000 at 6% simple interest. She is to
repay the debt with three equal payments, the first at the end of 3
months, the second at the end of 6 months, and the third at the end of
9 months. Determine the size of the payments. Put the focal date at the
present.
EXAMPLE.- A woman owes $3000 in 4 months with simple interest at
8% and another $4000 is due in 10 months with simple interest at 7%.
These two debts are to be replaced with a single payment due in 8
months. Determine the value of the single payment if money is worth
6’5% simple interest using the end of 8 months as the focal date.
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EXAMPLE.- A person borrows $1000 at 6% simple interest. She is to
repay the debt with three equal payments, the first at the end of 3
months, the second at the end of 6 months, and the third at the end of
9 months. Determine the size of the payments. Put the focal date at the
present.
1000 = X (1+0’06·3/12)-1 + X (1+0’06·6/12)-1 + X (1+0’06·9/12)-1
1000 = 0’98522X + 0’97087X + 0’95694X ! X = $343’28
EXAMPLE.- A woman owes $3000 in 4 months with simple interest at
8% and another $4000 is due in 10 months with simple interest at 7%.
These two debts are to be replaced with a single payment due in 8
months. Determine de value of the single payment if money is worth
6’5% simple interest using the end of 8 months as the focal date.
First Debt: 3000 (1+0’08·4/12) (1+0’065·4/12) = 3146’73
Second Debt: 4000 (1+0’07·10/12) (1+0’065·2/12)-1 = 4187’96
Single payment = 3146’73 + 4187’96 = $7334’69
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EXAMPLE.- A person borrows $1000 to be repaid with two equal
instalments, one in 6 months, the other at the end of 1 year. What will
be the size of these payments if the interest rate is 8% and the focal
date is at the present?
EXAMPLE.- Mrs. Adams borrows $2000 at 14%. He is to repay the debt
with four equal payments, one at the end of each 3-month period
during 1 year. Determine the size of the payments given a focal date at
the present time.
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EXAMPLE.- A person borrows $1000 to be repaid with two equal
instalments, one in 6 months, the other at the end of 1 year. What will
be the size of these payments if the interest rate is 8% and the focal
date at the present time?
1000 = X (1+0’08·6/12)-1 + X (1+0’08·1)-1
1000 = 0’961538X + 0’925926X ! X = $529’81
EXAMPLE.- Mrs. Adams borrows $2000 at 14%. He is to repay the debt
with four equal payments, one at the end of each 3-month period
during 1 year. Determine the size of the payments given a focal date at
the present time.
2000 = X(1+0’14·3/12)-1 +X(1+0’14·6/12)-1 +X(1+0’014·9/12)-1 +X(1+0’14·1)-1
= 0’966184X + 0’934579X + 0’904977X + 0’877193X = 3’682933X !
X = $543’05
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The calculation of present (or discounted) values over durations of less
than one year is sometimes based on simple discount rate (d).
In this situation, discount D is calculated on the final balance S and is
paid at the beginning of the term t (whereas in most financial
transactions, interest I is calculated on the beginning balance P and is
paid at the end of the term t).
Use years of 360 days.
The simple discount D on an amount S for t years at the discount rate d
is calculated by means of the formula: D = S·d·t
And the discounted value P of S is given by: P = S-D = S (1-d·t)
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EXAMPLE.- A person takes out a discounted loan with a face value of
$500 for 6 months from a lender who charges a 9’5% simple discount
rate.
a) What is the discount, and how much money does the borrower
receive?
b) What size loan should the borrower ask for if he wants to receive
$500 cash?
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EXAMPLE.- A person takes out a discounted loan with a face value
(future value) of $500 for 6 months from a lender who charges a 9’5%
simple discount rate.
a) What is the discount, and how much money does the borrower
receive?
b) What size loan should the borrower ask for if he wants to receive
$500 cash?
a) D = 500·0’095·6/12 = $23’75
P = 500 – 23’75 = 500·(1-0’095·6/12) = $476’25
b) 500 = S (1-0’095·6/12) ! S = $524’93
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EXAMPLE.- Calculate the discounted value of $1000 due in 1 year: a) at
a simple interest rate of 7%; b) at a simple discount rate of 7%.
EXAMPLE.- If $1200 is the present value of $1260 due at the end of 9
months, determine a) the annual simple interest rate, and b) the annual
simple discount rate.
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EXAMPLE.- Calculate the discounted value of $1000 due in 1 year: a) at
a simple interest rate of 7%; b) at a simple discount rate of 7%.
a) P = 1000 (1+0’07·1)-1 = $934’58
b) P = 1000 (1-0’07) = $930
Note the difference between the discounted value at a simple interest
rate and the discounted value at a simple discount rate. We can
conclude that a given simple discount rate results in a larger money
return to a lender than the same simple interest rate.
EXAMPLE.- If $1200 is the present value of $1260 due at the end of 9
months, determine a) the annual simple interest rate, and b) the annual
simple discount rate.
a) 1200 = 1260 (1+r·9/12)-1 ! r = 6’67% simple interest rate
b) 1200 = 1260 (1-d·9/12) ! d = 6’35% simple discount rate.
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SUMMARY
S : accumulate value of P at the end of t years at a simple
interest rate r is given by P (1+r·t).
P : discounted value of S due in t years at a simple interest
rate r is given by S (1+r·t)-1
P : discounted value of S due in t years at a simple discount
rate d is given by S (1-d·t).
Two rates are equivalent if they lead to the same
accumulated value of a fixed sum of money over the same
period of time.
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