Mathematics of
Finance and Investment
Prepared by
Prof. Dr
Omar Abd-Elgwad
Prof. Dr
Prof. Dr
Mohamed Nady Ezzat
Prof. Dr
Hosny Ahmed Elkholy
Eid Ahmed Abu-baker
Mathematics and Insurance department
Faculty of Commerce
Beni-suef university
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Introduction
Praise be to Allah, the Lord of the Worlds, and peace and blessings be upon
the Most Merciful of our Messengers Muhammad bin Abdullah, the faithful and
faithful Prophet, and upon his family and companions.
Mathematics is one of the oldest natural sciences to which mankind has
relied and which has been adopted by most other sciences as an input to the study,
analysis and measurement of many problems. Financial mathematics is one of the
branches of mathematics that contributes to the study and analysis of many
financial and administrative problems.
Capital is one of the main elements of production alongside land, labor and
organization (management), and each element of return or exchange or price for its
use or use. If the rent is the return of land, the reward or reward is the return of
labor, and profit is the return of regulation (management) Interest is the return of
capital, so the financial Mathematics is interested in the element of capital and its
return (interest) on the grounds that all economic and service projects, whether
public or private, whether individual or joint working to provide capital through
borrowing from individuals or entities specialized in the field Finance and
investment such as banks, insurance companies, or by issuing shares and bonds
and selling them on the stock market.
Financial Mathematics offers mathematical methods of financing and
investment through simple interest theory and compound interest theory. Simple
interest theory is used by social institutions and financial institutions such as social
security and social insurance, development and agricultural credit banks and
generally short-term loans. compound interest theory is used by Commercial
financial institutions such as banks and insurance companies.
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These mathematics methods are used by all interested in finance and
investment Mathematics and their applications in all aspects of life from
depositing, borrowing, buying, selling in installments or in installments.
This book deals with the two sectionsof financial Mathematics:
Section I: Simple Interest:
We consider the basic laws of the simple interest and its various
applications, the interest law and the amount (sum) of one principle and several
amounts of unequal invested or borrowed with one rate of interest , The law of the
present value and the discount (deduction) of one sum and several amounts of
unequal one rate, the discountd of commercial papers and the use of the application
of the laws of the sum and The present value of the settlement and replacement of
short-term debt. We also deal with sum laws and the present value of several equal
amounts paid at equal payments ( annuities) and periodic interest. Finally, we
address the methods of repayment or amortization of short-term loans.
Section II: Compound Interest:
We consider the basic laws of the compound interest and its various
applications, the interest law and the sum of one sum and several amounts invested
or borrowed at compound annual or non-annual compound interest rate, ie ,
interest is paid at less than one year (monthly, quarterly, semi-annual or one-third
annual). And then apply the laws of present value and discount for one or several
amounts. Then we address the application of the sum laws and the present value in
the settlement and replacement of long-term debt. Then we address the sum and
the present value of equal payments ( annuities). And finally the amortization of
long-term loans.
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Finally, we ask God to help us in presenting this author in an easy way that
serves our students in our institutes, colleges and universities, and our brothers, all
interested in finance and investment.
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Part one:
Simple Interest
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8
Chapter ( 1 )
Interset and Amount
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11
Chapter ( 1)
Interset and Amount
Basic Law for Interest:
Interest is defined as an increase in the capital resulting from its investment
for a certain period at a certain interest rate.
It is also defined as the return or financial compensation resulting from the
investment of funds or the borrowing of funds from others.
From the previous definition of interest it is clear to us that the amount of
interest due for any investment or borrowing depends on three basic elements:
A) The principal :
It is the amount borrowed or the amount deposited, the use of which
involves the performance of a financial compansation (interest) to which the debtor
(the borrower) owes the creditor (the owner of the capital).
B) The Rate of Interest :
The rate of interest - the return on investment of the capital unit at the end of
a one period of time. If the interest payable on the amount of 1000 L.E. at the end
of the year is 70 L.E., it can be said that the annual interest rate is 0.07. or 7%. It is
customary to use the year as a one period of time, as well as the use of 100 units of
money when determining the interest rate unless otherwise stated explicitly.
C) the period of time:
Means it the period after which the amount of the interest is payable. If the
creditor and the debtor agree to pay the interest of the invested capital once a year,
the time period or the unit of time is the year and therefore the rate used for interest
is an annual rate. If the interest is due every month, every quarter or every six
11
months, the period of time becomes one month or three months or six months,
respectively, and the rate is then reported at a rate of one period.
Elements of interest:
From the above it becomes clear that the amount of interest due from any
investment or borrowing depends on three elements:
1. Investor's principal ... symbolized by the symbol (a).
2. Period ... (duration or time of investment or borrowing) and symbolized
by the symbol (n).
3. Interest rate... ... denoted by (i).
It should be noted that the amount of interest due increases by increasing any
of the previous three elements with the stability of other racists, meaning that the
interest whether the interest of a loan or investment interest increases by increasing
the principal, time , rate of interest , and so on ... and vice versa .. That is, the
relationship is more positive increse incresing and decrease decreasing. It is
known that the interest of the amount of 1000 L.E. invested for a period of time at
a certain rate less than the interest of the amount of 2000 L.E. invested for the
same period and the same rate. As well as increase the interest of 1000 L.E.
invested for one year at a certain rate for the same interest rate invested at the same
rate and for half a year and the interest owed on the amount at the end of a certain
period at the rate of 9% necessarily exceeds if the user rate of 7%.
period (n)
interest rate (i %)
×
Present value (a)
12
Amount (S)
×
Interest (I)
Simple interest formula (I):
The principal of the investment is that the accrued interest is paid
periodically over the investment period and at the end of each agreed period
between the creditor (the owner of the capital) and the debtor (the borrower), such
as the period of one year, half year or quarter of the year. It may sometimes occur
that the debtor will not be able to pay the accrued interest at the end of each time
period or to agree with the creditor from the outset on the repayment of the
borrowed amount and all the accrued interest at the end of the loan term.
Therefore:
Interest = principal × rate × time
I=a× i× n
Principle amount (a), interest rate (i), period (n). Therefore, one of these
elements can be found if the three elements are known as follows:
1- Find (a)
I
a = ‫ــــــــــــــــــ‬
i× n
2- Find (i)
I
i = ‫ــــــــــــــــــ‬
a× n
3- Find ( n)
I
n = ‫ــــــــــــــــــ‬
a× i
Conclusion amount formula (S):
Amount = principle + interest
S=a+I
Substitute by (I) in the previous equation.
S=a+(a × i× n)
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S = a (1 + i × n )
In order to use the previous laws of interest and amount , in any law, (n) , (i)
must be of the same time units.
In the sense that:
(i) Yearly
,
(n) years
(i) Half-monthly
(i) Monthly
,
.
(n) in half years
(n) in months
And so on:
In the sense that when applying the interest and amount law in any of the
previous
equations, the period of the rate (unit of interest rate time) must
correspond to the time units on which the period is calculated. If the interest rate is
annual, the period must be calculated in a years. Calculation of rate of interest is a
month , The period must be calculated in months. If the period is in years, either
we convert the rate from a monthly rate to an annual rate or we convert the period
from years to months.
Meaning that the interest rate used corresponds to the periods of time. If the
time period is one year, the interest rate must be annual and in cases where the rate
is stated for a period of time less than one year, the annual rate must be obtained by
multiplying the rate by Period in the number of periods in which a full year is
included.
For example:
- If the interest rate is 5% semiannual.
The annual rate = 5% × 2 = 10%.
- If the quarterly interest rate = 3%
The annual rate = 3% x 4 = 12.
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The name of the rate varies according to its time unit, if:
- Time unit of time is years, the rate of interest is called annual rate.
- Time unit of time is 6 months, the rate of interest is called semi-annual rate.
- Time unit of time is 3 months, the rate of interest is called quarterly rate.
- The unit of time is 4 months, the rate is called the rate of one-third annual.
- The unit time of time is One month, the rate of interest is called monthly rate.
Problems of period:
1. If the period of investment or borrowing in years, in this case there is no
problem since (n) years.
2. If the period of investment or borrowing is months, it must be converted to
years.by divided 12 . ( n = m /12 )
3. If the period of investment or borrowing is days, it must be converted to
years.divided by (360 , 365 , 366 ).
( n = d /360 ) ,
( n = d /365 )
, ( n = d /366 )
4. If the period is in days and it is not specified whether the year is simple or leap,
the year is considered simple. (number of days = 365)
5. If the period of the investment or borrowing part there of falls in a simple year
and another part falls in a leap year.
{n = (m1 /365) + (m2 /366 )}
6. If the period of the investment or borrow is not explicitly given but gives the
date of depsit (or the date of borrowing) and the date of withdrawal (or the date of
payment), the period shall be calculated as follows:
Period = Number of days remaining in the month of deposit (or month of
borrowing) + Number of days of full months + Number of investment days
(borrowing) in the month of withdrawal (or month of repayment).
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7. The date of deposit (date of borrowing) and the date of withdrawal (date of
payment) may occur from the beginning, middle or end of the month, in which
case the period between the two dates is calculated in months, For example:
- If the date of deposit is the first of January and the date of withdrawal is the
last month of August, the investment period in this case is eight months as
follows: (January, February, March, April, May, June, July, August).
From 1/1 / ..... to 31/8
the period = 8 months
- If the borrowing date is 15/3/2020 and the repayment date is 15/8/2020 , the
borrowing period in this case is 5 months.
8. The financial year may be a simple year or leap year. The leap year is the year in
which February is 29 days and the number of days is 366 days. While the simple
year is the year in which February is 28 days and the number of days is 365 days,
determine the kind of the leap year , the year number is divided by 4 and is
outside the integer without fractions. However, the century year (100 multiples)
must divide the year number by 400 and be outside the integer without fractions
the leap year, while the simple years is outside the integer with fractions of 0.25 or
0.5 or 0.75.
The years of 1996 , 2000 , 2004 , 2008
,
2012 , 20016 , 2020 is a
leap year, wheale the years 2010 , 2011 , 2013 , 2014 , 2015 , 2017 , 2018 ,
2019 is a simple year
Example: Ahmed borrowed L.E. 20000 from the Bank of Alexandria, which
calculates a simple interest rate of 12% per annum and agreed with the bank to
repay it after 10 years.
Required: Calculate the interest payable on it and calculate the total amount paid
by the end of the period.
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Sulotion:
a = 20000
, i = 12% annually
,
n = 10 years
I=a× i × n
I = 20000 × 0.12 × 10 = 24000
S=a+ I
S = 20000 + 24000 = 44000
S = a (1+ i × n)
Or:
S = 20000( 1+ 0.12 × 10) = 44000
Example: A person borrowed 5000 L.E. from a bank at an interest rate of 12% per
annum for 10 months.
Required: Calculate the amount payable.
Sulotion:
a = 5000
,
i = 12% annually , n = 10 months = 10/12 year
I=a× i × n
I = a × i × ( m /12)
I = 5000 × 0.12 × (10/12) = 500
S= a+I
S = 5000 + 500 = 5500
Or :
S = a {1+ i × ( m /12 )}
S = 5000 { 1+ 0.12 × (10 /12 ) } = 5500
Example: A person invested 10000 L.E. at a simple interest rate of 12% per
annum for 9 years, 9 months and 10 days.
Required: Calculate of the amount at the end of the period.
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Sulotion:
a = 10000
,
i = 12% annually
months + 10 days = 9 + (9/12) + (10/360)
= 9 + 0.75 + 0.028 = 9.778
years
I=a× i × n
I = 10000 × 0.12 × 9.778 = 11733.6
S= a+I
S = 10000 + 11733.6 = 21733.6
Or:
S = a (1+ i × n )
S = 5000( 1+ 0.12 × 9.778 ) = 21733.6
Or
Interest for years
I1 = 10000 × 0.12 × 9= 10800
Interest for months
I2 = 10000 × 0.12 × (9 /12) = 900
Interest for days
I3 = 10000 × 0.12 × (10 /360) = 33.33
Total interest
I = I1 + I2 + I3
I = 10800 + 900 + 33.33 = 11733.33
S= a+I
S = 10000 + 11733.6 = 21733.6
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n = 9 years + 10
How to calculate the period between two dates:
To calculate the period between two dates, the following considerations must be
taken into account:
1. Ignore the day of depoists ( or borrowing) or the day of withdrawal (or payment)
and often neglects the day of depoists.
2. Note that the number of days of the month varies, some months have 30 days
and others have 31 days. This is unlike the month of February, which is 28 days in
the simple year, which is 29 days in the leap year, as follows:
January
February
March
April
May
June
31days
28 or 29 day
31 days
30 days
31 days
30 days
July
August
September
October
November
December
31 days
31 days
30 days
31 days
30 days
31 days
According to the previous two considerations, the period of the days is determined
between two dates.
Period = Number of days remaining in the month of deposit (or borrowing) +
Number of days of full months between the month of deposit (borrowing) and the
month of withdrawal (repayment). + Number of days in the month of withdrawal
(payment)
The following identify each of these elements:
1. The number of days remaining in the month of deposit (borrowing):
This number is determined by subtracting the date of the deposit day from
the number of days of the deposit month, so we have neglected the day of deposit
and therefore when calculating the number of days in the month of withdrawal (or
payment) do not neglect the day of withdrawal or payment.
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For example, if the deposit date is March 17, the number of days remaining
of the deposit month = the number of days in March (31) - the day of deposit (17)
= 14 days.
For example, if the day of borrowing is 25 June, The remaining days of the
borrowing month = 30 - 25 = 5 days.
2. Number of days of full months between the month of deposit and the month of
withdrawal:
This number is calculated by adding the number of days of those months.
3. Number of days in the month of withdrawal (or payment):
This number equals the date of the withdrawal day of the month of withdrawal, so
we have taken within the period of the draw day. For example, if the amount is
withdrawn on March 24, the number of days of the withdrawal month = 24 days.
To determine the period between two dates, we use the following formula:
period = Number of days remaining in the month of deposit (month of borrowing)
+ Number of days of full months + Number of days in the month of withdrawal
(month of payment)
Example: Calculate the period in the following cases:
1- The period from 25/3/2017 to 16/8/2017
2- The period from 11/1/2017 to 22/4/2017
3- The period from 22/9/2017 to 13/3/2018
4- The period from 12/2/2018 to 18/5/2018
5- The period from 1/6/2018 to 1/11/2018
6- The period from 15/3/2018 to 15/8/2018
7- The period from 1/7/2018 to 31/12/2018
8- The period from 5/3/2018 to 5/8/2018
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Solution:
1- The period from 25/3/2017 to 16/8/2017
Period = Mar
Apr May June July Aug
= (31- 25) + 30 + 31 + 30 + 31 + 16 = 144 days
2- The period from 11/1/2017 to 22/4/2017
Period = Jan
Feb
March Apr
= (31- 11) + 28 + 31 + 22 = 101 days
3- The period from 22/9/2017 to 13/3/2018
Period = Sep
Oct Nov Dec Jan Feb March
= (30- 22) + 31 + 30 + 31 + 31 + 28 + 13 = 173days
4- The period from 12/2/2018 to 18/5/2018
Period = Feb March
Apr
May
= (28- 12) + 31 + 30 + 18 = 96 days
5- The period from 1/6/2018 to 1/11/2018
Period = 5 Months
6- The period from 15/3/2018 to 15/8/2018
Period = 5 Months
7- The period from 1/7/2018 to 31/12/2018
Period = 6 Months
8- The period from 5/3/2018 to 5/8/2018
Period = 5 Months
In cases 5, 6, 7 and 8, it is noted that the periods have been calculated in months. If
the date of deposit (borrowing) agreed with the date of withdrawal or repayment,
the period shall be calculated in months.
Note: If the deposits date corresponds to the withdrawal date, so that the same day
of deposit is the same on the day of the withdrawal or the day of deposit and
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withdrawal from the first or the last or the middle of the month, the period shall be
calculated in months.
The following examples illustrate how to use previous relationships:
Example: On 13/5/2019, a person borrowed a sum of L.E. 30000 from a bank
which calculate a simple interest at the rate of 10% per annum and agreed with the
bank to pay the amount on 13/12/2019.
Required: Calculate the interest and the amount payable.
Solution:
Calculation of the period between two dates from 13/5 to 13/12, since there
is an agreement between the date of borrowing and the date of payment, and
therefore the period is calculated in months.
The period from 13/5/2019 to 13/12/2019
Period = 7 Months
a = 30000
, i = 10% annually
,
n = 7 months
I=a× i × n
I = 30000 × 0.10 × (7/12) = 1750
S=a+I
S = 30000 + 1750 = 31750
Or:
S = a (1+ i × n)
S = 30000 { 1+ 0.12 × (7 /12)} = 31750
Example: a person deposits L.E.5000 in a bank. If the total amount he has at the
end of a specified period is L.E. 5500 . If the bank calculates a simple interest at
the rate of 12% per annum.
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Required: calculate the period during which the amount was invested (calculate
the period of investment) .
Solution:
a = 5000
, i = 12% annually
,
S = 5500
I=S- a
I = 5500 – 5000 = 500
I=a× i × n
500 = 5000 × 0.12 × n
I
n = ‫ــــــــــــــــــ‬
a× i
500
n = ‫ = ــــــــــــــــــــــــ‬0.833
5000 × 0.12
converted into the period to months
= 0.833 × 12 = 10 months
Or:
S = a (1+ i × n )
5500 = 5000( 1+ 0.12 × n )
5500 / 5000 = ( 1+ 0.12 × n )
1.1 = ( 1+ 0.12 × n )
1.1 – 1 = 0.12 × n
0.1 = 0.12 × n
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year
0.1
n = ‫ = ــــــــــــــــــ‬0.833
0.12
converted into the period to months
year
= 0.833 × 12 = 10 months
Example: Find a simple interest and amount of L.E 15000 for 5 years,with a
quarterly interest rate of 3%.(Using two methods).
Solution:
a = 15000
, i = 3% quarterly
,
n = 5 years
Thus, there must be a correlation between the period of time and the period
of rate
Period = Number of years x Number of periods per year
Time = 5 × 4 = 20 periods
I=a× i × n
I = 15000 × 0.03 × 20 = 9000
S= a+I
S = 15000 + 9000 = 24000
Anther Solution:
a = 15000
, i = 3% quarterly
,
n = 5 years
Thus, there must be a correlation between the period of time and the period
of rate
The rate of interest = rate × Number of periods per year
i = 3% × 4 = 12 %
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I = 15000 × 0.12 × 5 = 9000
S = a+ I
S = 15000 + 9000 = 24000
Example: A person invested L.E. 50000 in a bank that calculates a simple interest
rate of 12%
per annum if the date of deposit 13/10/2018 and the date of
withdrawal 3/3/2020.
required: Find the interest and the total sum at the end of the period.
Solution:
a=50000
(S)
i = 12 %
×
13\10\2018
2019
- Period in the year 2018 .
Period = Oct
Nov Dec
= (31- 13) + 30 + 31 = 79 days
- Period in the year 2019 .
Period = 365
- Period in the year 2020 .
Period = Jau
Feb Mar
= 31 + 29 + 3 = 63 days
Total period = 79 + 365 + 63 = 507 days
I=a× i × n
I = 50000 × 0.12 × (507\360) = 8450
25
×
3\3\2020
S= a+I
S = 50000 + 8450 = 58450
S = a (1+ i × n )
Or:
S = 50000 { 1+ 0.12 × (507\360 )} = 58450
How to determine the date of deposit (borrowing) or the date of withdrawal
(payment):
1. Determine the date of withdrawal (payment):
If you give the date of deposit and the period of the investment and ask you
to determine the withdrawal date (payment ), to calculate the withdrawal date, the
investment period will be calculated in the different months starting from the
month of deposit and the following months with the subtraction of the period of
each month of the investment period to determine the remaining period until the
last of each month of investment. We reach less than the number of days of the
next month and the month of withdrawal is the following month and the date of
withdrawal is the same as the number of days calculated.
Example: If the depoists date is February 17, 2019 and the withdrawal is after a
lapse of 125 days, find the date of withdrawal.
Solution:
The year of 2019 is simple year , thus February = 28 days
the month
February
Number of investment
days per month
28 - 17 = 11 days
The remaining investment period
until the end of the month
125 - 11 = 114
March
31
114 - 31 = 83
April
30
83 - 30= 53
May
31
53 - 31 = 22
June
22
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The remaining period of investment in the last month of May is 22 days.
Therefore, the withdrawal month is the following month , the month of June and
the date of withdrawal is the same as the number of days remaining (calculated
from the month of withdrawal).
The withdrawal date is 22 June 2019.
2. Determination of the date of deposit (borrowing):
If you give the date of withdrawal and the period of the investment
(borrowing) and ask you to determine the date of deposit (borrowing ). To
calculate the date of deposit, the period of investment shall be calculated in the
different months starting from the month of withdrawal and the preceding months (
in reverse) The period of the investment expiring until we reach a period less than
the number of days of the previous month then the previous month is the date of
deposit (borrowing) and that period is the investment period calculated from this
month and thus the date of deposit (borrowing) is the number of days of the month
of deposit (borrowing) Of which the investment period calculated per month.
Example: If the withdrawal date is August 13, 2019 and the investment period is
110 days, what is the date of deposit.
Solution:
Months of
investment
reversed
August
Number of investment
days per month
13
The period of the investment
expiring until the first of each
month
110 - 13 = 97
July
31
97 - 31 = 66
June
30
66 - 30 = 36
May
31
36 - 31 = 5
April
5
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As the number of investment days until the first of May is 5 days, which is
less than the number of days of the previous month is the month of April, so the
month of deposit is the month of April.
Date of deposit
= Number of days of deposit (April) - the period calculated
= 30 - 5 =25
The depoist date is 25 April 2019.
Example: Someone deposited a sum in a bank on March 19, 2019 and withdrew it
after 175 days, calculate the date of withdrawal of this amount.
Solution :
Months of
investment
reversed
March
Number of investment
days per month
31 - 19 = 12
The period of the investment
expiring until the first of each
month
175 - 12 = 163
April
30
163 - 30 = 133
May
31
133 - 31 = 102
June
30
102 - 30 = 72
July
31
72 - 31 = 41
August
31
41 - 31 = 10
September
10
Where the remaining period at the end of August is 10 days and less than the
number of days of the following month is September, and the month of withdrawal
is the month of September and the date of withdrawal is on 10 September.
Example: On 25/7/2019, a person withdraws a sum of money that he deposited in
a bank. If he knows that the bank has interest on the basis of an investment period
of 127 days, calculate the date of deposit .
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Solution:
Inverse
months
Number of investment
days per month
July
25
The period of the investment
expiring until the first of each
month
127 - 25 = 102
June
30
102 - 30 = 72
May
31
72 - 31 = 41
April
30
41 - 30 = 11
March
11
The number of days of investment in the first month of April is 11 days,
which is less than the number of days of the previous month (March), so the month
of deposit is the month of March.
Date of deposit
= Number of days of deposit month - Investment period in the first month of April
= 31 – 11 = 20
The date of depoist is March 20.
Example: Ahmed borrowed L.E. 100000 from NCB bank on 15/2/2019 and he
paid this loan from a certain date. The interest on this loan amounted to L.E.
4166.6. If you know that the bank calculates the simple interest rate of 15% per
annum.
Calculate:
- the loan period .
- Loan repayment date.
Solution: It does not specify the type of interest used and therefore is considered a
commercial interest.
a = 100000
, i = 15% annually
,
n = ??
I = 4166.6
29
I= a × i × n
4266.6 = 100000 × 0.15 × n
I
n = ‫ــــــــــــــــــ‬
a× i
4266.6
n = ‫ــــــــــــــــــــــــــــــــــ‬
10000 × 0.15
converted into the period to months
= 0.2777
year
= 0.2777 × 360 = 100 days
That is, the loan period = 100 days
- Determination of loan repayment date:
the month
February
Number of
investment days per
month
28 - 15 = 13
The remaining investment
period until the last of each
month
100 - 13 = 87
March
31
87 - 31 = 56
April
30
56 - 30 = 26
May
26
As the remaining period at the end of April is 26 days, which is less than the
number of days of the following month is the month of May.
* Month of payment is May, payment date is 26/5/2019
Example: A person deposits a sum of L.E. 180000 in a bank on a given date for
investing in a simple interest at the rate of 15% per annum. If you know that the
total (sum) of him on 15/11/2019 is L.E.190500. Calculate the date of deposit.
31
Solution:
a = 180000
,
i = 15% annually
,
S = 190500
I=S- a
I = 190500 – 180000 = 10500
I=a× i × n
10500 = 180000 × 0.15 × n
I
n = ‫ــــــــــــــــــ‬
a× i
10500
n = ‫ــــــــــــــــــــــــــــــــــ‬
180000 × 0.15
= 0.3888
year
converted into the period to days
= 0.38888 × 360 = 140 dayss
Or:
S = a (1+ i × n )
190500 = 180000( 1+ 0.15 × n )
(190500 / 180000) = ( 1+ 0.15 × n )
1.05833 = ( 1+ 0.15 × n )
1.05833 – 1 = 0.15 × n
0.05833 = 0.15 × n
0.05833
n = ‫ــــــــــــــــــــــــــــــــــ‬
0.15
31
= 0.38888 year
converted into the period to months
= 0.38888 × 360 = 140 days
Specify the date of the deposit:
Reverse
months
Nov
Number of
investment days
per month
15
The period of the investment
expiring until the first of each
month
140 - 15 = 125
October
31
125 - 31 = 94
September
30
94 - 30 = 64
August
31
64 - 31 = 33
July
31
33 – 31 = 2
June
2
The number of investment days on the first of July is 2 days (two days), which is
less than the number of days in the previous month (June). It reflects the difference
between the date of deposit and the number of days of the month.
* Date of deposit
= Number of days of deposit month - Investment period from the first of July
= 30 - 2 = 28
Thus, the month of deposit is in June.
The deposit date is 28/6/2019.
Computing the Exact interest and Ordinary interest:
- Exact interest equal:
- if the year is simple:
n = (d /365)
Ie = a × i × (d /365)
And Se = a + Ie
32
Se = a (1+ i × (d /365))
- if the year is leap:
n = (d /366)
Ie = a × i × (d /366)
And S = a + Ie
Se = a (1+ i × (d /366))
- Ordinary interest equal:
n = (d /360)
Io = a × i × (d /360)
And So = a + Io
So = a (1+ i × (d /360))
If there is a part of the period, it falls in a simple year (d1) and another part falls
into a leap year (d2).
- Exact interest equal:
n = (d1 /365) + (d2 /366)
Ie = I = a × i × {(d1 /365) + (d2 /366)}
- ordinary interest equal:
n = {(d1 + d2) /360}
Io = a × i × {(d1 + d2) /360}
We notes the following:
1.Ordinary interest (Io) is always greater than the Exact interest (Ie).
2. If the type of interest is not specified in the exercise, it is always the Ordinary
interest (Io).
3. If not specified in the exercise type of year (simple or leap) is a simple year (the
number of days 365 days).
33
4. If part of the period occurs in a simple year and the other part of the period in a
leap year, the number of days in a simple year should be divided by 365 days and
the number of days in the leap year to 366 days. If we assume that the number of
days in a simple year is d1, and that the number of days in the leap year is d 2, then
the period when calculating the Exact interest equal:
n = {(d1 /365) + (d2 /366)}. While the period when calculate ordinary interest
equal: n = {(d1 + d2) /360}
5. We have stated that Ordinary interest is always greater than the Exact interest,
and therefore the use of Ordinary interest is in the interest of the creditor, so it has
traditionally been used in commercial financial transactions.
Example: Calculate the Exact interest and Ordinary interest of L.E. 10000 at a
simple interest rate of 10% per annum if the period of investment for this amount
from 10/1/2019 to 14/6/2019. Then calculate the amount with the Exact interest
and Ordinary interest.
Solution:
a = 10000
, i = 10% annually
,
n = 155 days
The period from 10/1/2019 to 14/6/2019
Period = Jan
Feb March Apr May June
= (31- 10) + 28 + 31 + 30 + 31 + 14 = 155 days
- Exact interest (Ie) :
Ie = a × i × (d /365)
Ie = 10000 × 0.10 × (155 /365) = 424.66
The amount with the Exact interest:
S e = a + Ie
Se = 10000 + 424.66 = 10424.66
Or:
Se = a ( 1 + i × n)
34
Se = 10000 { 1 + 0.10 × (155 /365)} =10424.66
- Ordinary interest ( Io) :
Io = a × i × (d /360)
Io = 10000 × 0.10 × (155 /360) = 430.56
The amount with the Ordinary interest:
So = a + Io
So = 10000 + 430,66 = 10430,56
So = a ( 1 + i × n)
Or:
So = 10000 { 1 + 0.10 × (155 /360)} =10430,56
The Exact interest is less than Ordinary interest.
Example: Abu-Liath borrowed an amount of L.E. 50,000 on 5/2/2019 from Union
International Bank, which calculates the simple interest rate at 12% per annum.
Calculate the Exact interest and Ordinary interest payable on 15/8/2019.
Solution:
The period from 5/2/2019 to 15/8/2019
Period = Feb March Apr May June July Aug
= (28- 10) + 31 + 30 + 31 + 30 + 31 + 15 = 191 days
a = 50000
, i = 12% annually
- Exact interest (Ie) :
,
n = 191 days
Ie = a × i × (d /365)
Ie = 50000 × 0.12 × (191 /365) = 3139.73
- Ordinary interest ( Io) :
Io = a × i × (d /360)
Io = 50000 × 0.12 × (191 /360) = 3183,33
35
Example: Ahmed borrowed an amount from the National Bank on 5/1/2019 at a
simple interest rate of 12% per annum. On 31/5/2019 he found that the Ordinary
interest owed to him amounted to L.E. 7300.
Required:
1. Calculate the loan principal.
2. Exact interest.
Solution:
The period from 5/1/2019 to 31/5/2019
Period =
Jan Feb March Apr May
= (31- 5) + 28 + 31 + 30 + 31 = 146 days
a = ???
, S = 7300 , i = 12% annually ,
n = 146 days
Io = a × i × (d /360)
- Ordinary interest ( Io) :
7300 = a × 0.12 × (146 /360) = 3183.33
a = 7300 / (0.12 × (146 /360) = 150000
- Exact interest (Ie) :
Ie = a × i × (d /365)
Ie = 150000 × 0.12 × (146 /365) = 7200
Example: A person borrowed L.E. 200000 from Arab Bank on 24/10/2019. If you
know that the bank calculates the simple interest at the rate of 12% per annum.
find the total amount due on that person on 20/4/2020, using ordinary interest as
well as using the exact interest .
Solution:
Calculate period (n), noting that part of the period is in a simple year (2019) and
another is in leap year (2020).
36
a = 200000
, i = 12% annually
,
n =
days
The period from 24/10/2019 to 20/4/2020
Period = Oct
Nov Dec Jan Feb March Apr
= (31- 24) + 30 + 31 + 31 + 29 + 31 + 20 = 179 days
It is noted in this example that part of the period occurs in a simple year (2019)
and the other part of the period in a leap year (2020), the number of days in a
simple year ( 68) should be divided by 365 days and the number of days in the leap
year (111) should be divided by 366 days.
then the period when calculating the Exact interest equal:
n = {(68 /365) + (111/366)}.
While the period when calculating ordinary interest equal:
n = {(68 + 111) /360}
First: Finding the total due to the debtor using the Exact interest (Ie):
- Exact interest (Ie) :
-
n = {(d1 /365) + (d2 /366)}
Ie = a × i × {(d1 /365) + (d2 /366)}
Ie = 200000 × 0.12 × {(68 /365) + (111/366)} = 11736
Thus, the total amount with the Exact interest at the end of the period (Se):
Se = a + Ie
Se = 200000 + 11736 = 211736
Or:
Se = a ( 1 + i × n)
Se = 200000 { 1 + 0.12 × {(68 /365) + (111/366)} } =211736
Second: finding the total receivable on the debtor using commercial interest (I o).
- Ordinary interest ( Io) :
Io = a × i × (d1 +d2 /360)
Io = 10000 × 0.12 × {(68 + 111) /360} = 11933.33
37
Thus, the total amount with the Ordinary interest at the end of the period (So).
So = a + Io
So = 200000 + 11933,33 = 211933,33
Or:
So = a ( 1 + i × n)
So = 200000 { 1 + 0.12 × (179 /360)} =211933,33
Thus, the total receivable at the end of the period (S).
The relationship between Ordinary interest (Io) and Exact interest (Ie):
A) - Relationship ratio between Ordinary interest and Exact interest:
1. If the year is simple:
Io / Ie = 73 / 72
it is possible to calculate the value of one of the two interests in the other.
Io = (73 / 72) Ie
Thus , Io is 1 /72 more than Ie
Io = Ie + (1 / 72) Ie
and
Ie = (72 /73) Io
so , Ie is 1 /73 less than Io
Ie = Io - (1 / 73) Io
2. If the year is a leap:
Io / Ie = 61 / 60
then it is possible to calculate the value of one of the two interest in the other.
Io = (61 /60) Ie
Thus , Io is 1/60 more than Ie
Io = Ie + (1 / 60) Ie
and
Ie = (60 /61) Io
so , Ie is 1 /61 less than Io
Ie = Io - (1 / 61) Io
38
B) - The relationship differenace between Ordinary interest and Exact interest:
1. If the year is simple:
Io - Ie = (73/72) Ie - Ie
Io - Ie = (1 /72) Ie
Io - Ie = Io - (72 /73) Io
Io - Ie = (1 /73) Io
2. If the year is a leap:
Io - Ie = (61/60) Ie - Ie
Io - Ie = (1 /60) Ie
Io - Ie = Io - (60 /61) Io
Io - Ie = (1 /61) Io
Example: Find the Exact interest and the Ordinary interest if the difference
between the two interest L.E. 1000 for the same amount , period and rate.
The solution:
1. If the year is simple:
Io - Ie = (1 /72) Ie
1000 = (1 /72) Ie
Ie = 72 × 1000 = 72000
Io - Ie = (1 /73) Io
1000 = (1 /73) Io
Io = 73 × 1000 = 73000
Note: After finding the Exact interest = L.E.72.000, Ordinary interest can be found.
Io = (Ie) + Difference between the two interests
Also, after finding a Ordinary interest of L.E. 73,000, the Exact interest can be
found.
Ie = (Io) - Difference between the two interests
39
Where the Ordinary interest is greater than the Exact interest.
2. If the leap year:
Io – Ie = (1 /60) Ie
1000 = (1 /60) Ie
Ie = 1000 × 60 = 60000
Io – Ie = (1 /61) Io
1000 = (1 /61) Io
Io = 1000 × 61 = 61000
Note: You can find Ordinary interest after finding the Exact interest or you can
find the Exact interest after finding the Ordinary interest:
Io = (Ie) + Difference between the two interests
Ie = (Io) - Difference between the two interests
Example: A person deposited a sum of L.E. 20,000 at Banque Misr on February 9,
2019 to invest at a simple interest rate. On June 9 of the same year, he found that
his Ordinary interest amounted to L.E. 854, calculate the Exact interest and then.
Find the public investment rate.
Solution:
2019 is a simple year and so when calculating the Exact interest.
Ie = (72 /73) Io
Ie = (72 /73) 854 = 842.3
Calculate the period :
The period from 9/2/2019 to 9/6/2019
Period = Feb Mar Apr May June
= (28 - 9) + 31 + 30 + 31 + 9 = 120 days
Or:
The period from 9/2/2019 to 9/6/2019
41
Period = 4 months
Calculate the rate :
- By using Ordinary interest ( Io) :
Io = a × i × (d /360)
854 = 20000 × i × (120 /360)
i = 854 / {20000 × (120 /360} = 12.8%
or :
- By using Exact interest ( Ie) :
Ie = a × i × (d /365)
842.3 = 20000 × i × (120 /365)
i = 842.3 / {20000 × (120 /365} = 12.8%
Example: If the difference between the ordinary interest and the Exact interest of
the amount L.E. 150000 was invested in a simple interest for 235 days in 2019 is
L.E. 150 .
Find the following:
1. the ordinary and the Exact interest
2. Rate of investment used.
Solution:
2019 is a simple year and so when calculating the Exact interest.
Io - Ie = (1 /72) Ie
150 = (1 /72) Ie
Ie = 72 × 150 =10800
Io - Ie = (1 /73) Io
150 = (1 /73) Io
41
Io = 73 × 150 = 10950
Calculate the rate :
- By using Ordinary interest ( Io) :
Io = a × i × (d /360)
10950 = 150000 × i × (235 /360)
i = 10950 / {150000 × (235 /360} = 11,2%
or :
- By using Exact interest ( Ie) :
Ie = a × i × (d /365)
10800 = 150000 × i × (235 /365)
i = 10800 / {150000 × (235 /365} = 11.2%
Example: If the difference between the ordinary and the Exact interest of the
amount of L.E. 150 and the simple interest rate used 12% per annum if this
amount invested from 24/6/2019 until 15/12/2019.
Required: Find the amount invested , period , principle.
Solution:
2019 is a simple year and so when calculating the Exact interest and ordinary
interest:
Io - Ie = (1 /72) Ie
150 = (1 /72) Ie
Ie = 72 × 150 =10800
Io - Ie = (1 /73) Io
150 = (1 /73) Io
42
Io = 73 × 150 = 10950
Calculate the period :
The period from 24/6/2019 until 15/12/2019.
Period = June July Aug Sep Oct Nov Dec
= (30 - 24) + 31 + 31 + 30 + 31 + 30 + 15 = 174 days
Io = 10950
, i = 12% annually
,
n = 174 days
Calculate the principle :
- By using Ordinary interest ( Io) :
Io = a × i × (d /360)
10950 = a × 0.12 × (174 /360)
a = 10950 / {0.12 × (174 /360} = 188793.1
or :
- By using Exact interest ( Ie) :
Ie = a × i × (d /365)
10800 = a × 0.12 × (174 /365)
a = 10800 / {0.12 × (174 /365} = 188793.1
Example: On 13/5/2019 Zaki Labib borrowed L.E. 30000 from a bank that
calculates simple interest at the rate of 15% per annum and agreed with the bank to
repay the loan on 18/8/2019.
Required: Calculate the interest and the amount with the ordinary interest once and
with the exact interest.
Solution:
a = 30000
, i = 15% annually
,
n = 97 days
The period from 13/5/2019 to 18/8/2019
43
Period = May June July Aug
= (31- 13) + 30 + 31 + 18 = 97 days
- Exact interest (Ie) :
Ie = a × i × (d /365)
Ie = 30000 × 0.15 × (97 /365) = 797.26
The amount with the Exact interest:
Se = a + Ie
Se = 30000 + 797.26 = 30797.26
Se = a ( 1 + i × n)
Or:
Se = 30000 { 1 + 0.15 × (97 /365)} = 30797.26
- Ordinary interest ( Io) :
Io = a × i × (d /360)
Io = 30000 × 0.15 × (97/360) = 808.33
The amount with the Ordinary interest:
So = a + Io
So = 30000 + 808.33 = 30808.33
So = a ( 1 + i × n)
Or:
So = 30000 { 1 + 0.15 × (97 /360)} =30808.33
Calculate interest and amount for several unequal amounts :
Financial transactions are not limited to the deposit or borrowing of one
amount but can be deposited or borrowed several amounts and each amount has its
own period, whether this period months or days, the interest can be found on these
amounts.
Under this method, if interest rate (i) is held constant for all amounts, the total
interest on the amounts can be calculated by the following steps:
44
1- Total interest :
 I = I 1 + I2 + I3 + I4
I = (a1 × i × n1) + (a2 × i × n2) + (a3 × i × n3) + (a4 × i × n4)
I = i { (a1 × n1) + (a2 × n2) + (a3 × n3) + (a4 × n4)}
 I = i% ×  (ai × ni)
 {(ai)× (ni)}
 I = i% × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12 or 360 or 365 or 366
2- Total amount
S = S1 + S2 + S3
+ S4 + S5
S = (a1 + I1 ) + (a2 + I2 ) + (a3 + I3 )
S = {a1 + (a1 × i × n1)} + {a2 + (a2 × i × n2)} + {a3 + (a3 × i × n3)} + {a4 +
(a4 × i × n4)} + {a5 + (a5 × i × n5)}
S = {a1 + a2 + a3 + a4 + a5 } + i { (a1 × n1) + (a2 × n2) + (a3 × n3) + (a4 ×
n4)} + { (a5 × n5)}
S =  ( ai ) + { i ×  (ai × ni)}
 {(ai)× (ni)}
S =  ( ai ) + i % × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12 or 360 or 365 or 366
Example: Abu Ali invested the following amounts in 2019 using a simple interest
rate of 12% per annum.
10000 L.E. for 50 days.
20000 L.E. for 100 days.
45
40000 L.E. for 200 days.
25000 L.E. for 300 days.
Calculate the total interest on these amounts as well as the total receivable in both
cases:
1. Using Ordinary interest .
2. Using the Exact interest .
Solution:
- First: Total interest Using Ordinary interest (Io):
I = I 1 + I2 + I3
I = (a1 × i × n1) + (a2 × i × n2) + (a3 × i × n3) + (a4 × i × n4)
Io = {10000 × 0.12 × (50\360)} + {20000 × 0.12 × (100\360)} + {40000 × 0.12
×(200\360)} + {25000 × 0.12 × (300\360)}
I = 166.67 + 666.67 + 2666.67 + 2500 = 6000
- Total amount Using Ordinary interest (Io):
S = S1 + S2 + S3
S = (a1 + I1 ) + (a2 + I2 ) + (a3 + I3 )
So= a1 ( 1 + i × n1) + a2 ( 1 + i × n2) + a3 ( 1 + i × n3)
So= 10000{1 + 0.12 × (50\360)} +20000 {1 + 0.12 × (100\360)} + 40000{1 +
0.12 × (200\360)} + 25000{1 + 0.12 × ( 300\360)}
So= 10166.67 + 20666.67 + 42666.67 + 27500 = 106000
- Second: Total interest Using Exact interest:
- Exact interest (Ie) : 2019 is a simple year
I = I 1 + I2 + I3
I = (a1 × i × n1) + (a2 × i × n2) + (a3 × i × n3) + (a4 × i × n4)
46
Ie = {10000 × 0.12 × (50\365)} + {20000 × 0.12 × (100\365)} + {40000 × 0.12
×(200\365)} + {25000 × 0.12 × (300\365)}
Ie = 5917.8
- Total amount Using Exact interest:
S = S1 + S2 + S3
S = (a1 + I1 ) + (a2 + I2 ) + (a3 + I3 )
Se= a1 ( 1 + i × n1) + a2 ( 1 + i × n2) + a3 ( 1 + i × n3)
Se= 10000{1 + 0.12 × (50\365)} +20000 {1 + 0.12 × (100\365)} + 40000{1 +
0.12 × (200\365)} + 25000{1 + 0.12 × ( 300\365)}
Se= 105917.8
Anther Soluation:
Calculate the sum of ( The amounts × Number of days):
The amounts
(ai)
Number of days
(ni)
The amounts × Number of
10000
50
500000
20000
100
1000000
40000
200
8000000
25000
300
7500000
95000

18000000
days {(ai)× (ni)}
- First: Total interest Using Ordinary interest (Io):
Total interest = i% ×
 ( The amounts × Number of days)
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
47
 {(ai)× (ni)}
 Io = i% × ‫ــــــــــــــــــــــــــــــــــ‬
360
18000000
 Io = 0.12 × ‫ = ــــــــــــــــــــــــــــــــــــــــــــ‬6000
360
The total amount with the Ordinary interest:
Total due to him at the end of the period =Total amounts + Total Ordinary interest:
S o =  a +  Io
So = 95000 + 6000 = 106000
- Second: Total interest Using Exact interest:
- Exact interest (Ie) :
Total interest = i% ×
 ( The amounts × Number of days)
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
365
 {(ai)× (ni)}
 Ie
= i% × ‫ــــــــــــــــــــــــــــــــــ‬
365
 Ie = 0.12 ×
18000000
‫ = ــــــــــــــــــــــــــــــــــــــــــــ‬5917.8
365
The total amount with the Ordinary interest:
Total due to him at end of period = Total amount + Total
S e =  a +  Ie
Se = 95000 + 5917.8 = 105917.8
48
Exact interest:
Example: A person borrowed money from Al-Orouba Bank, which calculates
simple interest at the rate of 12% per annum.
40000 L.E. for 4 months.
60000 L.E. for 5 months.
50000 L.E. for 6 months.
Find the accured interest on that person and total amount.
Solution:
- First: Total interest
I = I 1 + I2 + I3
I = (a1 × i × n1) + (a2 × i × n2) + (a3 × i × n3) + (a4 × i × n4)
Io = {40000 × 0.12 × (4\12)} + {60000 × 0.12 × (5\12)} + {50000 × 0.12
×(6\12)}
I = 1600 + 3000 + 3000 = 7600
- Total amount Using Ordinary interest (Io):
S = S1 + S2 + S3
S = (a1 + I1 ) + (a2 + I2 ) + (a3 + I3 ) = ai ( 1+ i × ni)
S = a1 ( 1 + i × n1) + a2 ( 1 + i × n2) + a3 ( 1 + i × n3)
S = 40000{1 + 0.12 × (2\12)} +60000 {1 + 0.12 × (5\12)} + 50000{1 + 0.12
×(6\12)}
S = 41600 + 63000 + 53000 = 157600
49
Anther Solution:
Where the periods of months and therefore:
The amounts
(ai)
Number of
months (ni)
The amounts × Number of
40000
4
160000
60000
5
300000
50000
6
300000
001111

760000
months {(ai)× (ni)}
 ( The amounts × Number of Months)
Total interest = i% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
 {(ai)× (ni)}
 I = i% × ‫ــــــــــــــــــــــــــــــــــ‬
12
760000
 I = 0.12 × ‫ = ـــــــــــــــــــــــــــــــــــــ‬7600
12
The total amount :
Total due to him at the end of the period =Total amounts + Total interest:
S=  a +  I
S = 150000 + 7600 = 157600
Or
 {(ai)× (ni)}
S =  ( ai ) + i × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
Example: Hadi owes the following amounts:
5000 L.E. due in (or payment on) 1/3/2018.
6000 L.E. due in (or payment on) 1/5/2018.
51
7000 L.E. due in (or payment on) 1/6/2018.
If Hadi is unable to pay its due dates, all of them are required to be paid on
1/12/2018 at a simple interest rate of 12% per annum.
Solution:
5000
6000
7000
S
12%
1\3\2018
1\5
1\6
1\12
On the maturity date of the new debt 1\12\2018
Thus, the first debt period of the debt is 9 months.
Thus, the second debt period is 7 months.
Thus, the period of the third debt is 6 months.
Using the method, you can calculate interest to be paid:
The amounts
(ai)
Number of
months (ni)
The amounts × Number of
5000
9
45000
6000
7
63000
7000
6
63000
18000

171000
months {(ai)× (ni)}
 ( The amounts × Number of Months)
Total interest = i% ×
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
 {(ai)× (ni)}
 I = i% × ‫ــــــــــــــــــــــــــــــــــ‬
12
51
 I = 0.12 ×
171000
‫ = ــــــــــــــــــــــــــــــــــــــــــــ‬1710
12
- The total amount :
Total due to him at the end of the period =Total amounts + Total interest:
S=  a +  I
S = 18000 + 1710 = 19710
Or
 {(ai)× (ni)}
S =  ( ai ) + i% × ‫ـــــــــــــــــــــــــــــــــــــــــــــ‬
12
Example: A trader borrowed the following amount from a bank:
30000 L.E. on 25/3/2020.
40000 L.E. on 18/5/2020.
80000 L.E. on 29/6/2020.
All of which were to be repaid on December 15, 2018 at an interest rate of 8% per
annum.
Required: Calculate of the amount to pay for the Exact
interest as well as
Ordinary interest.
Solution:
- Calculate the period :
Period = Mar Apr May June July Aug Sep Oct Nov Dec
= 6 + 30 + 31 + 30 +31 + 31+ 30 + 31 + 30 + 15 = 265
=
=
13 + 30 +31 + 31+ 30 + 31 + 30 + 15 = 211
1 +31 + 31+ 30 + 31 + 30 + 15 = 169
52
The amounts
(ai)
Number of days
(ni)
The amounts × Number of
30000
265
7950000
40000
211
8440000
80000
169
13520000
150000

29910000
days {(ai)× (ni)}
- First: Total interest Using Ordinary interest:
Total interest = i% ×
 ( The amounts × Number of days)
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
 {(ai)× (ni)}
 Io
= i% × ‫ــــــــــــــــــــــــــــــــــ‬
360
29910000
 Io = 0.08 × ‫ = ــــــــــــــــــــــــــــــــــــــــــــ‬6646.67
360
The total amount with the Ordinary interest:
Total due to him at the end of the period =Total amounts + Total Ordinary interest:
S o =  a +  Io
So = 150000 + 6646.67 = 156646.67
- Second: Total interest Using Exact interest:
Exact interest (Ie) : 2020 is a leap year
 ( The amounts × Number of days)
Total interest = i% ×
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
366
53
 {(ai)× (ni)}
= i% × ‫ــــــــــــــــــــــــــــــــــ‬
366
29910000
 Ie = 0.08 × ‫ = ــــــــــــــــــــــــــــــــــــــــــــ‬6537.7
366
The total amount with the Exact interest:
 Ie
Total due to him at end of period = Total amount + Total
Exact interest:
S o =  a +  Ie
Se = 150000 + 6537.7= 15637.7
Thus, It is noted that the total ordinary interest is greater than the total of the exact
interest .
54
Exercise
1) Find the simple interest in each of the following cases, assuming that the year
contains 360 days when the time expressed in days:
1
2
3
4
Principal
(L.E.)
15000
3000
5000
2000
Interest Rate
(%)
8
7.5
9
6.8
5
9000
9.2
6
4000
8.4
No
Time
4 years
6 years
185 days
18 months
3 years, 3
months
75 days
2) Determine the number of days using the exact and the approximate methods in
each of the following cases:
From
To
1. January 13. 2020
May 16. 2020
2. April 6.2020
July 27.2020
3. October 8.2020
March 21, 2021
4. December 24.2020
April 9.2021
5. November 28.2020
May 13.2021
6. February 14.2019
June 18.2019
3) Determine the times as a fraction of a year for calculating ordinary interest and
exact interest in each of the following cases:
No
1
2
3
4
5
6
From
January 3.2020
April 8.2021
October, 3.2020
December 4.2020
November 8.2020
February 4.2019
To
May 6.2020
July 7.2021
March 2.2021
April 5.2021
May 3.2021
June 8.2019
4) Find the ordinary interest using the formula method and the exact interest by
the relationship between the two interests in each of the following cases:
55
No
1
2
3
4
5
6
Principal (L.E.)
5000
7000
5000
1000
8000
4000
Interest Rate (%)
8
7.5
9
6.8
9.2
8.4
Time
140 days
175 days
150 days
173 days
125 days
173 days
5) For each of the following (I0 - Ie) find Io then Ie:
1.) 30 L.E.
3. ) 40 L.E.
2.) 50 L.E.
4. ) 80 L.E.
6) For each of the following (I0 - Ie ) find Io then Ie:
1.) 16 L.E.
3. 4 L.E.
2.) 15 L.E.
4.) 8 L.E.
7) For each of the following (I0 + Ie) find Ie then Io:
1.) 605 L.E.
3.) 968 L.E.
2.) 726 L.E.
4.) 1331 L.E.
8) For each of the following (I0 + Ie) find Io then Ie:
1.) 209 L.E.
3.) 1015 L.E.
2.) 870 L.E.
4.) 725 L.E.
9) In each of the following problems, find the unknown symbols:
No
p
i
n
I
S
1
10000
10%
6 months
?
?
2
15000
?
4 months
?
1560
3
2000
8%
? years
400
?
4
30000
?
150 days
?
3045
5
50000
6%
? months
?
200
10)
a)
b)
c)
Solve the following problems :
Find the simple interest and the amount on 40000 L.E. at 6% for 90 days?
Find the simple interest and the amount on 50000 L.E. at 7.5% for 2.5 years?
A woman borrowed a loan of 20000 L.E. from a bank at 12% for 9 months.
How much must she repay at the end of this period?
d) If a loan of 50000 L.E. is borrowed at 10%, what is the amount due at the
end of 2.75 years?
56
e) If a loan of 20000 L.E. is borrowed at 11% for 9 months, what is the amount
due at the end of the period?
f) A man borrowed 10000 L.E. for 80 days at 10%. How much he must repay?
g) On February 22, 2012, a man borrowed 24000 L.E. at 8.5% for 160 days.
What is the amount must he repay? And when?
h) At what interest rate will 20000 L.E. yield 212.3 L.E. in 140 days? And what
is the amount?
i) A man borrowed 10000 L.E and paid 10712.5 L.E. after 9 months. What
was the interest rate charged for this debt?
j) How many years are needed for 5010 L.E. to yield 1200 L.E simple interest
at 8% rate of interest?
k) How many days will be required for 18000 L.E. to yield 202.5 L.E. interest
at 9%?
l) How many months are necessary for 12000 L.E. to yield 640 L.E. interest at
8%?
57
58
Chapter ( 2 )
Present value and discount
95
06
Chapter ( 2)
Present value and discount
Introduction:
In the previous chapter, we discussed how to calculate the sum of an amount
at the end of a given investment or borrowing period (n), by indicating the value of
the principle (a) and the rate of interest (i). In some cases, however, the value is
the sum of the amount due at the end of a certain period, which is usually called
the matuitry value (or face value) of the loan in the case of borrowing or the sum
of the amount invested in the case of investment or deposit. It is required to know
the value of the loan on a date preceding the date of entitlement, , This value is
called the present value and is usually less than matuitry value of the amount
called the discount, since the creditor usually gives the debtor a discount in case of
repayment of the loan before the date of entitlement.
present value (a)
Discount× (D)
period (n)
Rate of interest (i%)
Rate of discount (d%)
Amount (S)
×
We have explained above that the owner of the capital is entitled to the
operation of his money so-called interest whether in the form of investment interest
or loan interest, and that this interest is due to the exploitation of capital, if it was a
limited number of days, so that the capital does not remainidle one day, A person
for the capital of another person obliges the debtor to pay the interest of such
money at an agreed interest rate and for the duration of the debtor's possession of
06
the money. If a person borrows L.E. 1,000 from another person and agrees that the
debtor will pay the amount at the end of one year from now,with the rate of interest
12% Annually, this means that the debtor is obliged to pay the amount of L.E.
1120 at the end of the year, which is For the principal amount plus interest for the
year at a rate of 12%.
If a person is a creditor of another person in the amount of L.E. 1000 , which
is worth paying after one year, this means that the amount is not equal to L.E.
1000 except on the due date, ie after one year. If the debtor wants to pay his debt
now( befor the due date) he will the obtains discount, This is called the discount
(deduction) of the acceleration of the payment and the reward for repaying its debt
before the payment date. Therefore, what is being paid now is the present value of
the debt rather than the original debt and the difference between them is the
discount.
Basic concepts:
Suppose if a person borrowed from a bank an amount today and this amount is
payable at the end of a certain period, the value of the amount on the due date or
payment called maturity (or fase) value, and the amount on the date of borrowing
is called the present value.
If we assume that the debtor wanted to repay the debt before the maturity date, it
will pay a value less than the maturity value by the amount of deduction it is
entitled to for the period from the date of repayment of the debt until the date of
entitlement.
From the above, the following concepts can be reached:
- Maturity value or ( face value) : Is the value of the debt on the maturity
date and is the sum of the amount, symbolized by the symbol (S).
- Present value: Is the value of the debt at a date prior to the date of
entitlement (maturity date), symbolized by the symbol (a).
06
-
Simple present value: Is the difference between the maturity value and the
simple discount. This difference, if invested by simple interest for the
period (duration) of the debt and at the rate used, is at the end of the period
equal to the maturity value and is symbolized by the symbol (ae).
- Bank present value or commercial present value (proceeds) : Is the
difference between the maturity value and the bank discount. This
difference, if invested by simple interest rate (or discount rate) for the period
(duration) of the debt and the rate used, is at the end of the period equal to
the maturity value and is symbolized by the symbol (a0).
-
Discount : The amount that to be wavied by the creditor to the debtor for
the repayment of the debt before the due date or the interest due to the debtor
as a result of the repayment of the debt on any date prior to maturity. This is
the difference between the maturity value and the present value and is
symbolized by the symbol (X).
- Simple discount: Which is the difference between the maturity value and
the simple present value or the interest of the simple present value of the
due to the debtor as a result of repayment of the debt on a date prior to
maturity date and is symbolized by the symbol (Xe).
- Bank discount: Is the difference between the maturity value and the bank
(commercial) present value or the interest of the bank present value of the
due to the debtor as a result of repayment of the debt on a date prior to
maturity and is symbolized by the symbol (X0).
- Discount Date:Is the date on which the debt is discounted, ie its present value is
determined before the due date.
- Duration of the discount (discount period):Is the period between the date of
the discount and the due date.
Equation of calculating present value and discount:
06
In general,
Maturity Value = Present Value + Discount
S=a+X
Such as:
Present Value = Maturity Value - Discount
a=S–X
Discount = Maturity Value - Present Value
X= S–a
The Present value and discount laws can be expressed as follows:
1- Simple present value and simple discount:
-
Simple present value:
Simple Present Value = Maturity Value - Simple Discount
ae = S - Xe
or
S
ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
- Simple discount:
Simple Discount = Maturity Value - Simple Present Value
Xe = S – ae
Or
Xe = ae × i × n
Or
Xe =
S× i× n
‫ـــــــــــــــــــــــــ‬
1+ i× n
2- Commerial present value or bank present value( proceeds) and bank
discount:
- Commerial present value or bank present value( proceeds):
06
Bank Present Value = Maturity Value – Bank Discount
ao = S – Xo
ao = S ( 1 – i × n)
or
- Bank discount:
Bank Discount = Maturity Value - Bank Present Value
Xo = S – ao
Or
Xo = S × i × n
From the above we note that:
- Bank discount (X0) is always greater than the Simple discount (Xe) because the
Bank discount is calculated on the basis of the maturity value ,while, the Simple
discount is calculated on the basis of the present value, and where the maturity
value is greater than the present value:
Xe < X0.
- The Commerial present value ( proceeds) (ao) is always less than the Simple
present value (ae) where:
ae = S – Xe
ao = S – Xo
Since:
Xe < X0
is:
a o < ae
- If the desired present value type and the type of discount required are not
specified in the exercise, the bank present value and the bank discount is required.
- We can calculate the simple discount if we know the bank discount by using the
equation:
X0
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
09
- We can calculate the bank discount if we know the simple discount by using the
equation:
Xo = Xe ( 1 + i × n)
Example: A debt whose maturity value is L.E. 20000 is payable after 18 months
with a simple interest rate of (equal discount rate ) 12% per annum.
Required:
1- Simple present value and simple discount.
2- Commerial present value and bank discount.
Soluation:
S = 20000
,
n = 18 months = 18 \ 12= 1.5
,
i = 12 %
1- Simple present value and simple discount:
-
Simple present value:
Simple Present Value = Maturity Value - Simple Discount
ae
ae = S - Xe
S
= ‫ـــــــــــــــــــــــــ‬
1+i× n
20000
ae = ‫ـــــــــــــــــــــــــ‬
( 1+ 0.12 × 1.5)
20000
ae = ‫ = ـــــــــــــــــــــــــ‬16949.15
1.18
- Simple discount:
Simple Discount = Maturity Value - Simple Present Value
Xe = S – ae
Xe = 20000 – 16949.15 = 3050.85
Or:
Xe = ae × i × n
00
Xe = 16949.15 × 0.12 × 1.5 = 3050.85
Note: We can calculate the simple discount , if we know the meturity value by
using the following formula:
S× i× n
‫ـــــــــــــــــــــــــ‬
1+ i× n
Xe =
Xe =
(20000 × 0.12 × 1.5)
‫ـــــــــــــــــــــــــــــــــــــــــــــــــ‬
( 1+ 0.12 × 1.5)
Xe =
(3600)
‫ = ــــــــــــــــــــــــــــ‬3050.85
(1.18)
Note: We can calculate the simple peresnt value , if we know the simple discount
by using the following formula:
Simple Present Value = Maturity Value - Simple Discount
ae = S – Xe
ae = 20000 – 3050.85 = 16949.15
2- Commerial present value or bank present value (or proceeds):
ao = S ( 1 – i × n)
ae = 20000 ( 1 – 0.12 × 1.5) = 16400
- Bank discount:
Bank Discount = Maturity Value - Bank Present Value
Xo = S – ao
Xo = 20000 – 16400 = 5600
Or:
Xo = S × i × n
Xo = 20000 × 0.12 × 1.8 = 3600
Note: We can calculate the simple discount , if we know the simple discount by
using the following formula:
Xo = Xe ( 1 + i × n)
06
Xo = 3050.85 ( 1 + 0.12 × 1.5) = 5600
Note: We can calculate the peresnt value , if we know the bank discount by using
the following formula:
Bank Present Value = Maturity Value – Bank Discount
ao = 20000 – 3600 = 16400
Note: We can calculate the simple discount , if we know the bank discount by
using the following formula:
X0
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
3600
Xe = ‫ = ـــــــــــــــــــــــــ‬3050.85
( 1+ 0.12 × 1.5)
Example: A person who owes a loan that is repayable at the end of 6 months. The
simple present value of this loan is calculated on the basis of a discount rate of 9%
per annum, which is L.E. 60000.
Required: Calculate the simple discount and matuirty value of the loan.
Soluation:
ae = 60000 ,
n = 6 months = 6 \ 12= 0.5
,
i=9%
- Simple discount:
Xe = ae × i × n
Xe = 60000 × 0.09 × 0.5 = 2700
-
Maturity Value = Present Value + Discount
S = ae + Xe
S = 60000 + 2700 = 62700
06
Or :
S= p ( 1+ i × n)
S= 60000 ( 1+ 0.09 × 0.5) = 62700
Example: A maturity value of L.E. 60000, payable at the end of 6 months, with a
bank present value of L.E. 528,000.
Required: Calculate the simple present value and the simple discount.
Soluation:
ae = 60000 ,
n = 6 months = 6 \ 12= 0.5
,
ao = 52800
- Calculate the bank discount
Bank discount = maturity value - bank present value.
Xo = S – ao
Xo = 60000 – 52800 = 7200
- Calculate the rate of interest:
Xo = S × i × n
7200 = 60000 × i × 0.5
7200
i = ‫ = ــــــــــــــــــــــــــــــــــــــ‬24%
( 60000 × 0.5)
- Simple present value:
S
ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
60000
Xe = ‫ = ــــــــــــــــــــــــــــــــــــــ‬53571.43
( 1+ 0.24 × 0.5)
05
- Simple discount:
Simple Discount = Maturity Value - Simple Present Value
Xe = S – ae
Xe = 60000 – 53571.43 = 6428.57
Or : you can calculate the simple discount using the formula:
Xe = ae × i × n
Xe = 53571.43 × 0.24 × 0.5 = 6428.57
Or Note: We can calculate the discount , if we know the meturity value by using
the following formula:
S× i× n
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
(60000 × 0.24 × 0.5)
Xe = ‫ = ــــــــــــــــــــــ ـــــــــــــــــــــــــ‬6428.57
( 1+ 0.24 × 0.5)=
Or We can calculate the simple discount if we know the bank discount by using
the equation:
X0
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
2700
Xe = ‫ = ـــــــــــــــــــــــــ‬6428.57
( 1+ 0.24 × 0.5)
Example: If the bank discount of a debt is due (repayable) at the end of 18
months at a simple interest rate of 8% per annum is L.E. 1080 .
66
Required: Calculate the simple discount for this debt as well as the matuirty value,
simple present value, and commercial present value.
Soluation:
Xe = 1080 ,
-
n = 18 months = 18 \ 12= 1.5
,
i=8%
calculate the simple discount if we know the bank discount by using the
equation:
X0
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
1080
Xe = ‫ = ـــــــــــــــــــــــــ‬964.29
( 1+ 0.08 × 1.5)
- calculate the matuirty value , if we know the bank discount by using the
following formula:
Xo = S × i × n
1080 = S × 0.08 × 1.5
1080
S = ‫ = ـــــــــــــــــــــــــ‬9000
(0.08 × 1.5)
- calculate the peresnt value , if we know the bank discount by using the
following formula:
Bank Present Value = Maturity Value – Bank Discount
ao = S – Xo
ao = 9000 – 1080 = 7920
- Simple present value:
S
ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
66
9000
Xe = ‫ = ـــــــــــــــــــــــــ‬8035.71
( 1+ 0.08 × 1.5)
or : Simple Present Value = Maturity Value - Simple Discount
ae = S – Xe
ae = 9000 – 694.29 = 8035.71
Relationship between the bank discount (X0 ) and the simple discount (Xe ):
1. The ratio between the bank discount simple discount:
Xo
S× i× n
‫ـــــــــــــــــــــــــــــــــ = ــــــــــــــ‬
Xe

ae × i × n
Xo
S
‫ـــــــــــــــ = ــــــــــ‬
Xe
S
 Xo = ‫× ــــــــــــــــ‬
ae
Xe
ae
ae
 Xe = ‫× ــــــــــــــــ‬
Xo
S
Previous relationships are used to find one of the discount if the other
opponent knows the face value and the simple present value.
S× i× n
 Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n

Xo
S× i× n
‫ــــــــــــــــــــــــــــــــــــــــــــــــ = ــــــــــــــ‬
Xe
(S × i × n)\ 1 + i × n
Xo
 ‫ = ــــــــــــــ‬1 + i × n
Xe
66
Previous relationships are used to find one of the discount if the opponent
knows the other and the interest rate and period.
2. The difference between the bank discount (Xo) and the simple discount (Xe):
 Xo
S × i2 × n2
- Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
Or:
Xo = S × i × n
Xo × i × n
 Xo - Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
or
S× i× n
Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
 Xo - Xe = Xe × i × n
Or
S
 ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
 Xo - Xe = ae × i2 × n2
Use previous relationships if the exercise gives the difference between the
bank discount and the simple discount, the interest rate, the period, and ask for the
face value, the bank discount, the simple discount, and the simple present value.
Example: a debtor owes L.E. 9900 due in (payment after) a certain period,
calculated the simple present value of this debt was L.E. 9000.
Required: Find bank present value and bank discount.
66
The solution:
S = 9900
,
a0 = 9000
- Simple discount:
Simple Discount = Maturity Value - Simple Present Value
Xe = S – ae
Xe = 9900 – 9000 = 900
Xo
S

‫ـــــــــــــــ = ــــــــــ‬
Xe

ae
Xo
9900
‫ـــــــــــــــ = ــــــــــ‬
900
9000
9900
 Xo = ‫× ــــــــــــــــ‬
900 = 990
9000
- Bank present value:
Bank Present Value = Maturity Value – Bank Discount
ao = S – Xo
ao = 9900 – 990 = 8910
Example: A debt whose face value is L.E. 36630 , with a bank present value of
L.E. 36297 , what is the simple present value of this debt.
The solution:
S = 36630
,
ao = 36297
- Bank discount (Xo):
Xo = S – ao
Xo = 36630 – 36297 = 333
66
in which :
Xe = S – ae
Xe = 36630 – ae

333
‫= ــــــــــ ــــــــــ‬
36630
‫ـــــــــــــــ‬
(36630 - ae )
 333 ae =
ae
36630 ( 36630 - ae )
ae = 1341756900 \ 36960 = 36300
Example: Caculate the difference between the bank discount and the simple
discount for a debt that is repayable after 15 months, it is found that it is L.E.
1085.21 with a simple interest rate of 12% per annum.Find the face value.
The solution:
Xo - Xe = 1085.21 , n = 15 months = 15\12 = 1.25 , i = 12%
 Xo
S × i2 × n2
- Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
S = 55466.29
Example: A person owes the bank a sum of L.E. 40000 payable at the end of the
year. If you know that the ratio of the bank discount to the simple discount is
equal to 1.04 and that the interest rate is equal to the discount rate: caculate
- Simple present value and bank present value.
- The simple discount and bank discount.
- The interest rate (discount rate).
69
The solution:
S = 40000 , n = 1 year
, Xo \ Xe = 1.04
- Calculate the simple present value:
Xo
S
‫ـــــــــــــــ = ــــــــــ‬

Xe
ae
S
1.04 = ‫ـــــــــــــــ‬
ae
40000
1.04 = ‫ـــــــــــــــ‬
ae
ae = 40000 \ 1.04 = 38461.54
- Calculate the simple discount:
Xe = S – ae
Xe = 40000 – 38461.54 = 1538.46
- Calculate the bank discount:
Xo
1.04 = ‫ـــــــــــــــ‬
Xe
Xo
1.04 = ‫ـــــــــــــــ‬
1538.46
 Xo = 1538.46 ×
- Calculate the bank present value:
Xo = S – ao
60
1.04 = 1600
Xo = 40000 – 1600 = 38400
-
Calculate the rate of interest :
Xo = S × i × n
Xo = 40000 × i × 1
i = 1600 \ ( 40000 × 1) = 4%
or
X e = ae × i × n
1538.46 = 38461.54 × i× 1 = 4%
Example: Caculate the difference between the bank discount and the simple
discount for a debt that is repayable one year from now, it is found that it is L.E.
15.38 , if the simple interest rate of 4% per annum.
Required:
- The mayurity value, the simple present value, the bank present value.
- Bank discount, simple discount.
The solution:
Xo - Xe = 15.38
, i = 4% per year
, n = one year
- Calculate the face value:
 Xo
S × i2 × n2
- Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
 S = 10000
- Calculate the simple present value:
S
ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
66
10000
Xe = ‫ = ـــــــــــــــــــــــــ‬9615.38
( 1+ 0.04 × 1)
- Calculate the simple discount:
Xe = S – ae
Xe = 10000 – 9615.38 = 384.62
Or
X e = ae × i × n
Xe = 9615.38 × 0.04 × 1 = 384.62
Or
S× i× n
 Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
- Calculate the bank present value:
ao = S ( 1 – i × n)
ao = 1000 ( 1 – 0.04 × 1) = 9600
or
ao = S – Xo
ao = 10000 – 400 = 9600
- Calculate the bank discount:
Xo = S × i × n
Xo = 10000 × 0.04 × 1 = 400
Example, on January 1, 2019, Abu Jihad owed L.E. 20000 payable after one year
and on 1 March of that year he wanted to repay the debt. If the ratio between the
bank discount and the simple discount is 1.15, then the discount rate is equal to
the interest rate.
66
Required:
- Simple present value and bank present value.
- The simple discount and bank discount.
- The interest rate (discount rate).
The solution:
S = 20000 , n = 10\12 year
, Xo \ Xe = 1.15
on 1 \ 3 \ 2019 calculate the present value of the debt
S = 20000
1\1\2019
Present value
1\3\2019
i= %
- Calculate the rate of interest.
Xo
 ‫ = ــــــــــــــ‬1 + i × n
Xe
1.15 = 1 + i × (10\12)
1.15 – 1 = i × (10\12)
0.15 = i × (10\12)
i = 18%
- Calculate the simple present value:
S
ae = ‫ـــــــــــــــــــــــــ‬
1+i× n
20000
Xe = ‫ = ـــــــــــــــــــــــــ‬17391.3
1+ 0.18 ×(10\12
65
- Calculate the simple discount:
Xe = S – ae
Xe = 20000 – 17391.3 = 2608.7
Or
X e = ae × i × n
Xe = 17391.3 × 0.18 ×(10\12) = 2608.7
Or
S× i× n
 Xe = ‫ـــــــــــــــــــــــــ‬
1+i× n
- Calculate the bank present value:
ao = S ( 1 – i × n)
ao = 20000 { 1 – 0.18×(10\12) } = 17000
or
ao = S – Xo
ao = 20000 – 3000 = 17000
- Calculate the bank discount:
Xo = S × i × n
Xo = 20000 × 0.18 × (10\12) = 3000
Calculate present value and discount for several unequal amounts :
We used the same method to find the amount and interest for several unequal
amounts, whether these amounts in years, months or days, and since the discount is
not very different from interest, to find the present value of several amounts we
first get the total discount, and then subtract from The total matuirty value ( or face
value).
Present values can be found by subtracting the previous total discount obtained
from the sum of maturity values, as the following:
66
- Total discount : Di = Si × d × ni
 D = D1 + D2 + D3
+ D4
 Di = (S1 × d × n1) + (S2 × d × n2) + (S3 × i × n3) + (S4 × d × n4) + (S5 ×
d × n5)
 D = (S1 × d × n1) + (S2 × d × n2) + (S3 × d × n3) + (S4 × d × n4) + (S5 × d
× n5)
 D = d% { (S1 × n1) + (S2 × n2) + (S3 × n3) + (S4 × n4)}
 D = d%  (Si × ni)
 {(Si)× (ni)}
 D = d % × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12 or 360 or 365 or 366
- Total present values = total maturity values - Total discount
ai = Si – Di
ai = Si – (Si × d × ni)
ai = Si ( 1 – d × ni)
a i = a 1 + a 2 + a 3 + a 4 + a5
 ai = S1 ( 1 – d × n1) + S2 ( 1 – d × n2) + S3 ( 1 – d × n3) + S4 ( 1 – d × n4)
+ S5 ( 1 – d × n5)
 ai = {S1 + S2 + S3 + S4 + S5 } – d% { (S1 × n1) + (S2 × n2) + (S3 × n3) +
(S4 × n4)} + { (S5 × n5)}
 ai =  ( Si ) – d%  (Si × ni)
66
 {(Si)× (ni)}
 ai =  ( Si ) – d % × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12 or 360 or 365 or 366

 ai =  ( Si ) –  D
The following examples illustrate how to apply this method:
Example: Abu Lahab owes the following amounts:
6000 L.E. payment (due in) after 8 months.
12000 L.E. payment (due in) after 10 months.
15000 L.E. payment (due in) after 15 months.
Calculate what he should pay now to repay this debt if the interest rate used is 7%
per annum.
The solution:
What must be paid now to pay off these debts (present value):
 ai = S1 ( 1 – d × n1) + S2 ( 1 – d × n2) + S3 ( 1 – d × n
 ai = 6000 { 1 – 0.07× ( 8\12)} + 12000 { 1 – 0.07× ( 10\12)} + 15000 { 1 –
0.07× ( 15\12)}
 ai = 5720 + 11300 + 13687.5 = 30707.5
Anather solution:
- Calculate the total discount.
 D = D1 + D2 + D3
 Di = (S1 × d × n1) + (S2 × d × n2) + (S3 × i × n3)
66
 Di = 6000 × 0.07 × (8\12) + 12000 × 0.07 × (10\12) + 15000 × 0.07 ×
(15\12) = 2292.5
- Calculate the total present value.
Total present values = total maturity values - Total discount
 ai =  Si –  Di
 ai = = 33000 – 2292.5 = 30707.5
Anather solution:
Total present values = total maturity values - Total discount
The amounts
(Si)
Number of
months (ni)
The amounts × Number of
6000
12000
15000
33000
8
10
15
48000
120000
225000
393000
months (Si × ni)

- Calculate the total discount.
 ( The amounts × Number of Months)
Total discount = i% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
12
393000
Total discount = 0.07 × ‫ = ـــــــــــــــــــــــــــــــــــ‬2292.5
12
- Calculate the total present value.
66
Total present values = total maturity values - Total discount
Total present values = 33000 – 2292.5 = 30707.5
Example: Muhannad Mahab is debted to the following amounts:
5000 L.E. payment (due in) on 23/12/2018
6000 L.E. payment (due in) on 15/10/2018
7000 L.E. payment (due in) on 13/9/2018
He wanted to repay all his debts on 20/5/2018 at a simple interest rate of 9% per
annum.
Required: Find the amount paid on 20/5/2008 to repay this debt.
The solution:
Period = May June July Aug Sep Oct Nov Dec
= 11 + 30 + 31 + 31+ 30 + 31 + 30 + 23 = 217
= 11 + 30 + 31 + 31+ 30 + 15 + -- + -- = 184
= 11 + 30 + 31+ 31 + 13 + -- + -- + -- = 116
The amounts
(Si)
Number of days
(ni)
The amounts × Number of
5000
217
1085000
6000
148
8880000
7000
116
812000
18000

2785000
days (Si × ni)
Thus,
First : Calculate the total bank present value (proceeds ) and total bank discount:
- Calculate the total bank discount.
66
 ( The amounts × Number of days)
Total discount = i% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
360
2785000
Total discount = 0.09 × ‫ = ـــــــــــــــــــــــــــــــــــ‬696.25
360
- Calculate the total bank present value (proceeds ).
Total present values = total maturity values - Total discount
Total present values = 18000 – 696.25= 17303.75
secand: Calculate the total simple present value and total simple discount
- Calculate the total simple discount.
- 2018 is a simple year
 ( The amounts × Number of days)
Total discount = i% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
365
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
365
Total discount = 0.09 ×
2785000
‫ = ـــــــــــــــــــــــــــــــــــ‬684.84
365
69
- Calculate the total simple present value.
Total present values = total maturity values - Total discount
Total present values = 18000 – 684.84= 17315.16
Example: A person who owes the following amounts:
10000 L.E. payment after 40 days
15000 L.E. payment after 60 days
20000 L.E. payment after 80 days
25000 L.E. payment after 90 days
If the person wants to pay all his debts today and the discount rate in the market
was 12% per annum.
Calculate the total discount and the sum of the present values paid for these
amounts.
The solution:
The amounts
(Si)
Number of days
(ni)
The amounts × Number of days
10000
40
400000
15000
60
600000
20000
80
1600000
25000
90
2250000
70000

5150000
(Si × ni)
Calculate the total bank present value (proceeds ) and total bank discount:
- Calculate the total bank discount.
60
 ( The amounts × Number of days)
Total discount = d% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
360
Total discount = 0.12 ×
5150000
‫ = ـــــــــــــــــــــــــــــــــــ‬1716.67
360
- Calculate the total bank present value (proceeds ).
The amount payable now is the sum of the present values:
Total present values = total maturity values - Total discount
Total present values = 70000 – 1716.67= 68283.33
Example: Abu Muhannad borrowed funds from the Arab Bank:
10000 L.E. payment (due in) after 4 months.
20000 L.E. payment (due in) after 8 months
30000 L.E. payment (due in) after 10 months.
If Abu Muhannad wants to pay all his debts today.
Calculate the amount of bank discount he gets if the discount rate is 12% per
annum and then calculate the amount he is paying now for these debts.
66
The solution:
The amounts
(Si)
Number of
months (ni)
The amounts × Number of
10000
4
40000
20000
8
160000
30000
10
300000
60000

500000
months (Si × ni)
The amount payable now is the sum of the present values of the previous amounts.
- Calculate the total discount.
 ( The amounts × Number of Months)
Total discount = d% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
12
500000
Total discount = 0.12 × ‫ = ـــــــــــــــــــــــــــــــــــ‬5000
12
- Calculate the total present value.
Total present values = total maturity values - Total discount
Total present values = 60000 – 5000 = 55000
Example: On 1 January 2019, Abu Mehab was owed the following amounts:
5000 L.E. payment (due in) after 3 months.
10000 L.E. payment (due in) after 5 months.
66
6000 L.E. payment (due in) after 6 months.
On the first of February of the same year, Abu Mahab wanted to pay all his debts
once.
Calculate the value he will pay if the interest rate is 15% per annum.
The solution:
on 1 \ 2 \ 2019 calculate the present value of the debit
1\1\2019
5000
10000
3
5
1\2\2019
2
Present value
6000
6
4
5
i = 15%
The amounts
(Si)
Number of
months (ni)
The amounts × Number of
5000
2
10000
10000
4
40000
6000
5
30000
21000

80000
months (Si × ni)
The amount payable on 1 \ 2 \ 2019 is the sum of the present values of the
previous amounts.
- Calculate the total discount.
65
 ( The amounts × Number of Months)
Total discount = d% × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
12
Total discount = 0.15 ×
80000
‫ = ـــــــــــــــــــــــــــــــــــ‬1000
12
- Calculate the total present value.
Total present values = total maturity values - Total discount
Total present values = 21000 – 1000 = 20000
56
Exercise:
1- In each of the following problems, find the Principal:
No
1
2
3
4
5
p
?
?
?
?
?
i
10%
8%
6%
12%
9%
n
?
4 months
6 months
120 days
5 years
I
5000
4000
?
?
9000
S
10500
?
41200
31200
?
2- Solve the following problems:
a)
Find the principal that yield 100 L.E. in 6 months at 10%?
b) What principal will yield 288 L.E. in two years at 8% simple interest?
c)
What principal will amount 2080 L.E. in eight months at 6% simple interest?
d) What principal will accumulate 2550 L.E. in 80 days at 9% simple interest?
e)
How much money must you invest at 7% for 2.5 years in order to receive 4700
L.E. at end of the period?
f)
What is the present value of 2000 L.E. due in three years if the money is worth
9%? And how much in the simple interest?
g) What is the present value of 2500 L.E. due at the end of 8 months if the money
is worth 8%? And what is the simple discount?
h) A debt of 4000 L.E. is due in two years, if the debt is settled now and the
simple rate is 6%, what are the present value and the simple discount?
i)
A debt of 5000 L.E. is due nine months from now, what are the present vale
and the simple discount if the debt is settled now at 8% simple interest?
j)
Mr. Ayman purchased a TV and made a down payment of 1500 L.E., he
agreed to pay 500 L.E. after two months and 400 L.E. after four months. If the
rate of simple interest is 10%, what is the cash price of the TV?
56
k) Ms. Mona plans to purchase LED. Set. She is offered the option of paying
1000 L.E. dowr and 1000 L.E. in 5 months, or of paying 1200 L.E. and 600
L.E. in 4 months. If the rate of wimple interest in 12%, which option would be
a better offer for her?
l)
A man borrowed 2000 L.E. on April 4, 2013, and agreed to repay the loan plus
19% simple interest in 6 months. He wishes to pay the loan on May 4, 2013,
by discounting it at 9% simple interest rate. How much should he pay?
m) In the above problem, if the loan discounted at 11% simple interest rate, how
much should the debtor pay?
56
Chapter (3)
Discount of commercial papers
82
83
Chapter three
Discount of commercial papers
Definition of Commercial Papers:
Commercial paper represents instruments of a certain form and condition,
and in practice it is customary to use them as a means of dealing with futures. The
bill of exchange and promissory notes are the most important types of commercial
paper used in the process. Any type of commercial paper used is a document
proving the existence of the value worth a specific date in the future.
If a person borrowed a sum of money from another person or financial
institution and undertook to repay the debt on a certain date, in such a case a
commercial paper such as the cash or promissory note is issued at the value of the
loan and the date of payment of that value - the maturity date of the promissory
note - Sometimes the creditor wishes to obtain the value of the bill or the
promissory note before the specified maturity date. Therefore, the banks have
provided a banking service to the commercial market, which is the deduction of
these commercial papers, meaning that the creditor receives the present value of
the commercial paper today - the date of the discount - rather than waiting. Until it
dissolvesd.
It is noted that the Bank gives the creditor a value less than the maturity
(face) value of the commercial paper by the amount of discounts (deductible
expenses of commercial paper).
From the above, it is clear that commercial securities are instruments that
have certain terms and conditions on which credit is based. These are bills of
exchange and promissory notes, which represent either an order from the creditor
to the debtor to pay an amount to another person on a known date or a pledge by
the debtor to pay a certain amount on a given date.
84
Banks and other financial institutions in the financial and commercial
markets accept these securities from their owners for an appropriate deduction
from their face value. This process is called discounting, cutting or selling of
commercial paper.
The purchasers of commercial papers try to increase the value of the
discount in all possible ways, such as using the commercial discount instead of the
simple discount, adding one or more days to the discount period and usually asking
for collection of the face value of the debtor and commission or transfer expenses,
Discount again.
The set of discount values and transfer charges is called the total discount
expense and this sum is deducted from the face value to obtain the net worth of the
paperholder.
Thus, when a person discounts a commercial paper or several commercial
paper at a bank, the bank usually imposes its conditions to complete the discount
process which does not exceed the following conditions:
- Bank Discount :
In the case of a single commercial paper:
D = S × d% × n
In the case of several commercial papers:
 ( The amounts × Number of Months or days)
Total discount = d% × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
or 360
- Collection commission:
Calculated based on a percentage or proportion per thousand of the Face value.
Collection commission = Face value × Commission rate
85
- Collection expenses:
Calculated on the basis of a percentage or proportion per thousand of the Face
value
Collection Expenses = Face Value × Expense Ratio
The Bank may set a minimum collection expense and compare it to the calculated
ratio and choose the larger value.
The discount rate is calculated on the basis of total discount, which requires the
calculation of the total annual discount rate, which includes both the bank
discount, the collection expenses and the collection commission. Naturally, the
overall discount rate increases with increase these expenses.
Steps commercial paper discount:
From the above, the steps of discounting commercial papers are as follows:
1. calculate the discount period.
2. Caculate bank discount.
* If the commercial paper one:
D = S × d% × n
* If there are more than commercial paper:
 ( The amounts × Number of Months or days)
Total discount = d% × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
or 360
 {(Si)× (ni)}
 D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
12 or 360
3. Calculation of collection commission:
Commission = Face Value x ratio
4. Calculation of collection expenses:
Collection Charges = Face Value × Ratio
86
5. Overall (Total) discounts:
= Bank discount + commission + collection expenses
6. Net receivable to the client ( Net present value)
7. Total discount rate:
- In the case of one sheet:
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
face value × period without time limit
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ S× n } without time limit
- In the case of several commercial papers:
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
 { (The amounts × Number of days or months) /360 or 12} without
time limit
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
 {(Si)× (ni)} without time limit
Example: A trader's discount on 3/10/2018 with a bill of face value of L.E. 10000
payable on 31/12/2018. If the bank's conditions for deducting the commercial
paper are:
- Bank discount rate of 12% per annum.
- Collection expenses 0.25% with a minimum of L.E. 30 .
- Collection commission 0.5%.
- The Bank adds a one-day repayment period.
87
Required:
1. Net receivable to the customer.
2. The total discount rate achieved by the bank.
Solution:
1. Period of the discount:
Period = Oct Nov Dec limt
n = (31 - 3) + 30 + 31 + 1 = 90 days
2. Commercial or bank Discount:
D= S × d × n
D = 10000 × 0.12 × (90 /360) = 300
3. Collection expenses:
= S × ratio
= 10000 × 0.0025 = 25
Where the minimum L.E. 30 , which is greater than the expenses calculated L.E.
25 and therefore take the minimum (the largest value).
4. Collection commission:
= S × ratio
= 10000 × 0.005 = 50
5. Total discounts:
= Bank discount + Commission + Collection expenses
= 300 + 50 + 30 = 380
6. Net receivable to the customer:
= Face value - total discounts
= 10000-380 = 9620
7. Total discount rate:
Total discounts = face value × Total discount rate × Period without time limit.
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
face value × period without time limit
88
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ S× n } without time limit
380
Total discount rate = ‫ = ـــــــــــــــــــــــــــــــــــــــــــــــــــــ‬15.4%
{ 10000 × 89) /360 }
Example: A merchant's discount on a bill dated 3/10/2018 is due on 11/2/2019 and
receives a net value of L.E. 28612.5. If the bank's conditions for deducting
commercial paper are:
- Bank discount rate of 12% per annum.
- Collection commission of 0.1%.
- Collection expenses of 0.25%.
- The bank add 4 day payment period.
Required: Calculate the face value of the bill.
Solution:
1. Period of discount (n):
Period = Oct Nov Dec Jan Fab limt
n = (31 - 3) + 30 + 31 + 31 +11 + 4 = 135 days
2. Bank Discount:
=S×d×n
D = S × 0.12 × (135 /360) = 0.045 S
3. Collection commission:
= S × ratio
= S × 0.0001 = 0.001 S
4. Collection expenses:
= S × ratio
= S × 0.0025 = 0.0025 S
5. Total discounts:
=Bank (Commercial) discount + Commission + Collection expenses
= 0.045 S + 0.001 S + 0.0025 S = 0.0485 S
Net receivable to the customer = face value - total discounts
011
28612.5 = S - 0.0485 S
28612.5 = 0.9515 S
S = (2861.5 / 0.9515) = 30070.94
For example: a dealer's discount on 10/10/2018 is payable on 20/1/2019 with a
face value of L.E. 50000. The total discount rate is 16% per annum. If you know
that the bank calculates a collection commission at a rate of 0.03% and a collection
expenses (fee) of 0.05% The bank has add payment time 3 days.
Required: Calculate the discount rate used by the bank.
Solution:
1. Calculate the Period of the discount:
Period = Oct Nov Dec Jan limt
n = (31 - 10) + 30 + 31 + 20 + 3 = 105 days
2. Calculate Commercial Discount:
X = 5000 × d × (105 /360) = 14583.33 d
3. Collection expenses:
= S × ratio
= 50000 × 0.0005 = 25
4. Collection commission:
= S × ratio
= 50000 × 0.003 = 15
5. Total discounts:
= Bank discount + commission + collection expenses
= 14583.33 d + 15 + 25 = 14583.33 d + 44
Total discounts
6. Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
face value × period without time limit
 D
d% = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ S× n } without time limit
010
(14583.33 d + 44)
0.16 = ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ 50000 × 102) /360 }
(14583.33 d + 44)
0.16 = ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ 14166.67 }
Conducting interactive beatings:
14583.33 d + 44 = 14166.67 × 0.16
14583.33 d + 44 = 2266.67
14583.33 d = 2266.67 - 44
14583.33 d = 2222.67
d = (2222.67 /14583.33) = 15.24 %
Example: Mennat Allah Electrical Appliances Company deducted a Commercial
paper (bill) face value of L.E. 200000 of payment on 12/3/2020 at Bank of
Alexandria. The company received the net due to it and its capacity is L.E.
180,000. If the bank's conditions in the discount process are as follows:
- Bank discount rate of 15%.
- Collection commission of 0.2%
- Collection expenses 0.05%.
Required: Calculate the date of discount of the bank.
Solution:
whereas:
Net due to the company = face value - total discounts
Total Discount = face Value - Net due to the company
= 200000 - 180000 = 20000
The total discounts are Bank discounts, commission and collection expenses.
- Bank discount:
=S×d×n
011
= 200000 × 0.15 × (D /360) = 833.33 D
Where the duration is unknown.
- Commission:
= S × ratio
= 200000 × 0.002 = 400
- Collection expenses:
= S × ratio
= 200000 × 0.0005 = 100
Thus, the total discounts:
= Bank discount + commission + collection expenses
= 833.33 D + 400 + 100
= 833.33 D + 500
20000 = 500 + 833.33 D
19500 = 833.33 D
D = (19500 /833.33) = 234 days
Thus, the period between the date of deduction of the bill and the maturity date is
234 days.The date of the bill is calculated as follows:
Months
The number of discount
Elapsed time until the
reversed
days each month
first of each month
March
12
234 – 12 =222
February
29
222 – 29 = 193
January
31
193 – 31 = 162
Dec
31
162 – 31 = 131
Nov
30
131 – 30 = 101
October
31
101 – 31 = 70
September
30
70 – 30 = 40
August
31
40 – 31 = 9
July
9
012
The bill's discount date is July:
= 31 – 9 = 22 July
Thus, the date of deduction of the bill is 22/7/2019.
Example: On 1/2/2019, the following dealer's securities were traded at the
National Bank of Egypt:
- A bill of face value of L.E. 18000 payable to a customer in Giza Governorate on
17/3/2019.
- A bill of face value of L.E. 15,000 payable to a customer in Fayoum Governorate
on 15/6/2019.
- A bill of face value of L.E. 25,000 payable to a customer in Beni Suef
Governorate on 25/8/2019.
If the conditions of the bank in discounting the commercial papers are:
- the Bank discount rate of 8%.
- - Collection commission of 0.1% .
- Collection fees (expenses) on the second and third papers at a rate of 0.1%, so
that the collection expenses of the paper is not less than L.E. 20 .
- The bank add 2 day payment period.
Required: Net receivable to the customer.
Solution:
1. Calculate the duration of the discount:
Period = Fab Mar Apr
May June July Aug limt
n1 = (28 - 1) + 17 -
-
-
-
- + 2 = 46 days
n2 = (28 - 1) + 31 + 30 + 31 + 15 - - + 2 = 136 days
n3 = (28 - 1) + 31 + 30 + 31 +30 + 31 + 25 + 2 = 207 days
2. Calculate the bank discount .
Di = (S1 × d × n1) + (S2 × d × n2) + (S3 × i × n3)
013
Di = 18000 × 0.08 × (46\360) + 15000 × 0.08 × (136\360) + 25000 × 0.08×
(207\360) = 1787.33
Or:
 ( The amounts × Number of Months or days)
Bank discount = d% × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
The amounts
The amounts ×
Number of days
18000
46
Number of days
828000
15000
136
2040000
25000
207
5175000
58000
8043000
 {(Si)× (ni)}
D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
360
8043000
D = 0.08× ‫ = ــــــــــــــــــــــــــــــــــ‬1787.33
360
3. Commission:
= Total face values × commission rate
= total S × ratio
= 58000 × 0.001 = 58
4. Collection expenses on the second and third papers:
- Secand paper = S × ratio
= 15000 × 0.0001 = 15
- Third paper = S × ratio
014
= 25000 × 0.0001 = 25
The collection fees for the second paper are L.E. 15 , which is less than the
minimum and therefore we take the minimum L.E. 20 , while the collection
expenses of the third paper L.E 25 is greater than the minimum and therefore take
the calculated expenses.
Thus, collection expenses = 20 + 25 = 45
5. Total discounts:
= Bank discount + Commission + Collection expenses
= 1787.33 + 58 + 45 = 1885.33
Net due to trader:
= Total face values - total deductions
= 58000 - 1885.33 = 56114.67
Example:
I want the dealer to deduct the following commercial papers on
15/3/2018:
- A bill of face value of L.E. 10000 drawn to a merchant in Alexandria and
payable on 20/6/2018.
- A bill of face value of L.E. 12000 drawn to a merchant in Mansoura and payable
on 10/8/2018.
- A bill of face value of L.E. 15000 drawn to a dealer in Cairo and payable on
25/9/2018.
- A bill of face value of L.E. 18000 drawn by a merchant in Tanta and payable on
15/10/2018.
If the Bank of Egypt calculates a discount rate of 15% per annum and takes a
commission rate of 0.3% and expenses of collection of L.E. 5 per paper, and adds
the bank one-day payment of the withdrawal for the paper withdrawal of customers
015
in Cairo, while adding a two-day payment deadline for the paper withdrawal of
customers outside Cairo.
While Bank of Cairo calculates a bank discount of 15.8% per annum and does not
calculate commission and does not calculate collection costs, while adding a oneday payment period to the customers in Cairo as well as two days to customers
outside Cairo.
Required: Compare theconditions of the two banks to know the best for the trader
and calculate the total discount rate .
The solution:
1. Computation of discount periods:
Period = Mar Apr May June July Aug Sep Oct limt
n1 = 16 + 30 + 31 + 20
-
- + 2 = 99 days
n2 = 16 + 30 + 31 + 30 + 31 + 10 - - + 2 = 150 days
n3 = 16 + 30 + 31 + 30 +31 + 31 + 25 + 2 = 195 days
n3 = 16 + 30 + 31 + 30 +31 + 31 + 30 + 15 + 2 = 206 days
2. Calculate the bank discount .
Di = (S1 × d × n1) + (S2 × d × n2) + (S3 × i × n3) + (S4 × i × n4)
Di = 10000 × 0.15× (99\360) + 12000 × 0.15 × (150\360) + 15000 × 0.15 ×
(195\360) + 18000 × 0.15 × (206 \360) = 3926.25
Or:
 ( The amounts × Number of Months or days)
Bank discount = d% × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
016
First: For Bank Misr:
The amounts
The amounts ×
Number of days
10000
99
Number of days
990000
12000
150
1800000
15000
195
2925000
18000
206
3708000
55000
9423000
 {(Si)× (ni)}
D = d% × ‫ــــــــــــــــــــــــــــــــــ‬
360
9423000
D = 0.15 × ‫ = ــــــــــــــــــــــــــــــــــ‬3926.25
360
- Commission = Total face values × commission rate
= total S × ratio
= 55000 × 0.003 = 165
- Collection expenses: = 4 × 5 = 20
- Total discounts: = 3926.25 + 165 + 20 = 4111.25
- Total discount rate (gross):
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
 { The amounts × Number of days or months) /360 or 12} without
time limit
017
‫× ‪The amounts‬‬
‫‪Number of days‬‬
‫‪The amounts‬‬
‫‪Number of days‬‬
‫‪970000‬‬
‫‪97‬‬
‫‪10000‬‬
‫‪1776000‬‬
‫‪148‬‬
‫‪12000‬‬
‫‪2910000‬‬
‫‪194‬‬
‫‪15000‬‬
‫‪3672000‬‬
‫‪204‬‬
‫‪18000‬‬
‫‪55000‬‬
‫‪9328000‬‬
‫‪ D‬‬
‫= ‪d%‬‬
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
‫‪ {(Si)× (ni)} without time limit‬‬
‫‪4111.25‬‬
‫‪ = 15.58%‬ـــــــــــــــــــــــــــــــــــــ‬
‫=‬
‫‪d%‬‬
‫‪9328000‬‬
‫‪Second: For Bank of Caire:‬‬
‫‪- bank discount.‬‬
‫)‪Di = (S1 × d × n1) + (S2 × d × n2) + (S3 × i × n3) + (S4 × i × n4‬‬
‫‪Di = 10000 × 0.158× (99\360) + 12000 × 0.158 × (150\360) + 15000 × 0.158‬‬
‫‪× (195\360) + 18000 × 0.158 × (206 \360) = 4135.65‬‬
‫‪Or:‬‬
‫)‪ ( The amounts × Number of Months or days‬‬
‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ × ‪Bank discount = d%‬‬
‫‪360‬‬
‫})‪ {(Si)× (ni‬‬
‫ــــــــــــــــــــــــــــــــــ × ‪D = d%‬‬
‫‪360‬‬
‫‪018‬‬
9423000
D = 0.158 × ‫ = ــــــــــــــــــــــــــــــــــ‬4135.65
360
- Total discount rate (gross):
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
 { The amounts × Number of days or months) /360 or 12} without
time limit
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
 {(Si)× (ni)} without time limit
4135.65
d%
=
‫ = ـــــــــــــــــــــــــــــــــــــ‬15.98%
9328000
Since Bank of Cairo does not charge a commission and does not calculate
collection expenses, the total discount is equal to the bank discount = 4135.65.
Since the total discounts at Bank of Misr are less than the total discounts at Bank
of Cairo, the conditions of Bank of Misr for the deduction of commercial paper are
better than those of Bank of Cairo.
As the total discount rate at Bank of Misr is lower than the total discount
rate at Bank of Cairo and thus the conditions of Bank of Misr are better.
Example: On 5/1/2019, a discount merchant wanted a bill of face value of L.E.
70000 payable on 4/4/2019. The trader had the conditions of Bank of Cairo
Amman and the conditions of the Arab Bank as follows:
001
Conditions
of Bank
of
Caire Conditions of Arab Bank
Amman
-Commercial discount 12% per
-Commercial discount 12.5% per
annum
annum
-Commission 0.1%
-Commissions 0.25%
-Collection expenses 0.2%
-Collection expenses 0.15%
-Two-day payment period
Required: Any conditions of the two banks better?
The solution:
First: For the conditions of Cairo Amman Bank:
1. calculate the discount period.
Period = Jan Fab Mar Apr
n = (31 - 5) + 28 + 31 + 4 = 89 days
2. Bank Discount:
D=S×d×n
D = 70000 × 0.12 × (89 /360) = 2076.67
3. Collection commission:
= S × ratio
= 70000 × 0.001 = 70
4. Collection expenses:
= S × ratio
= 70000 × 0.002 = 140
5. Total discounts:
= bank discount + commission + collection expenses
= 2067.67 + 70 + 140 = 2286.67
6. Net receivable to the customer:
= face value - total discounts
= 70000 – 2286.67 = 67713.33
7. Total discount rate (gross):
Total discounts = face value × Total discount rate × Period without time limit.
000
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
face value × period without time limit
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ S× n } without time limit
2286.67
Total discount rate = ‫ = ـــــــــــــــــــــــــــــــــــــــــــــــــــــ‬13.2%
{ 70000 × 89) /360 }
Second: For the conditions of the Arab Bank:
1. calculate the discount period.
Period = Jan Fab
Mar Apr limt
n = (31 - 5) + 28 + 31 + 4 + 2 = 91 days
2. Bank Discount:
D=S×d×n
D = 70000 × 0.125 × (91 /360) = 2211.81
3. Collection commission:
= S × ratio
= 70000 × 0.0025= 175
4. Collection expenses:
= S × ratio
= 70000 × 0.0015 = 105
5. Total discounts:
= bank discount + commission + collection expenses
= 2211.81 + 175 + 105 = 2491.81
6. Net receivable to the customer:
= face value - total discounts
= 70000 + 2491.81 = 67508.19
7. Total discount rate (gross):
Total discount rate = total discounts / face value × period without time limit
001
Total discounts
Total discount rate = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
face value × period without time limit
d%
=
 D
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
{ S× n } without time limit
2491.81
Total discount rate = ‫ = ـــــــــــــــــــــــــــــــــــــــــــــــــــــ‬14.4%
{ 70000 × 89) /360 }
As the total discounts at Arab Bank are greater than the total discounts in the
Bank of Cario Amman. Accordingly, the net present value received from Arab
Bank is lower than that received from Bank of Cairo Amman. As the total
discount rate in the Arab Bank is greater than the total discount rate at Bank of
Cairo Amman.
The conditions of Bank of Cairo Amman are better than those of Arab
Bank.
002
Chapter (4)
Settlement and the Replacement
of short-term debt
115
116
Chapter four
Settlement and the Replacement
of short-term debt
Introduction:
Debt settlement is a direct application of the basic interest rate laws that have been
studied in the previous chapters. For the amount and present value, usually when
borrowing or depositing capital, specific maturity dates are agreed upon to match
the conditions of both debtors and creditors and what they expect. the future.
However, unforeseen changes may occur to debtors that prevent them from making
payments on the basis of previously agreed payment methods and dates. (Creditor
and debtor) to modify the debt or settle it by new dates and dates commensurate
with the circumstances of the new debtors.
Debt settlement methods:
1) Repayment of one debt is due before (previous) to the maturity dates of old
debts.
2) Repayment of one debt is due after (subsequent) to the maturity dates of the old
debts.
3) Repayment of one debt is due during the maturity dates of old debts.
4) Repayment in several debts (installments) is due during the maturity dates of the
old debts.
5) Repayment of old debts with one face value The total face value of the debts
(average maturity date) is adjusted.
117
1) Repayment of old debts (original debts or pre-settlement debts) with one debt
payable prior to maturity of old debt:
In this case, the debt is settled on the basis of replacing the original (old) debts
group with one debt that is due before the maturity date of the original debt, and
then it is preferable to use the present value equation at settlement as follows:
The present value of the old debts on the maturity date of the new debt = the
value of the new debt at the date of maturity.
In this case, it is noted that the debtor wishes to make payment before the
agreed date and previously agreed, and therefore it receives a deduction for the
amounts paid before the due date as in the following form:
S1
now
n1
S2
n2
S3
n3
x
It is noted from the previous figure that the debtor wishes to pay before maturity
dates.
Example: If Abdel Fattah owes the following amounts:
20000 L.E. due in 15/8/2019.
40000 L.E. due in 1/10/2019.
60000 L.E. due in 23/11/2019.
If Abdelfattah wants to repay this debt with a debt that is repayable on May 5,
2019, what is the value of the new debt if the interest rate used is 6% per annum.
Solution:
On the maturity date of the new debt (settlement date):
118
20000
5\5
40000
60000
7
5
3
15\8
1\10
23\11
x
The value of the new debt = the present value of the old debts on the maturity
date of the new debt.
Therefore, the period of each old debt must be calculated on the basis of the period
between the settlement date (the maturity date of the new debt) and the date of
payment and the maturity date of the new debt as follows:
Period = May June July Aug Sep Oct nov
n1 = 26 + 30 + 31 + 15 -
-
-
n2 = 26 + 30 + 31 + 31 + 30 + 1
= 102 days
- = 149 days
n3 = 26 + 30 + 31 + 31 + 30 + 31 + 23 = 202 days
Total present values = total maturity values - Total discount
-
Calculate Total discount : Di = Si × d × ni
 D = D1 + D2 + D3
+ D4
 Di = (S1 × i × n1) + (S2 × i × n2) + (S3 × i × n3)
 D = {20000 × 0.06 × (102\360)} + {40000 × 0.06 × (149\360)} + {60000
× 0.06 × (202\360)} = 3353.333
the present value of old debt = Total face values - Total discounts
ai =  ( Si ) –  (Di)
= 120000 - 3353.333 = 116646.667
119
Another solution:
The previous example can be solved by finding the present value of each amount
separately.
The value of new debt = The present value of the first debt + the present value of
the second + the present value of the third debt.
ai = Si ( 1 – d × ni)
ai = a1 + a2 + a3
 ai = S1 ( 1 – i × n1) + S2 ( 1 – i × n2) + S3 ( 1 – I × n3)
 ai = 20000{ 1 – 0.06 × (102\360)} + 40000 { 1 – 0.06 × (149\360)} +
60000 { 1 – 0.06 × (202\360)} = 116646.667
Anather soluation
The present value is calculated by using the method:
The amounts × Number of days
The amounts
(ai)
Number of
days (ni)
20000
102
{(ai) × (ni)}
2040000
40000
149
5860000
60000
202
12120000
120000

20120000
Thus,
the present value of old debt = Total face values - Total discounts
 {(Si)× (ni)}
 ai =  ( Si ) – i % × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
20120000
 ai = 120000 – 0.06 × ‫ـــــــــــــــــــــــــــــــــــــــ‬
360
121
 ai = 120000 - 3353.333 = 116646.667
Thus, the face value of the new debt is L.E. 116646.667
Example: On 1/3/2019 Aya Beauty Products Company debted the following
amounts to a bank:
10000 L.E. due in (payable after) 7 months.
25000 L.E. due in (payable after) 10 months.
60000 L.E. due in (payable after) one year.
On May 1, 2019, the Company wanted to repay these debts on the basis of a simple
interest rate of 9%.
Calculate the face value of the new debt.
Solution:
On the maturity date of the new debt (settlement date):
10000
1\3
1\5
7
25000
10
5
60000
12
8
10
x
The value of the new debt = the present value of the old debts at the new maturity
date.
At the settlement date 1/5/2019 which is the maturity date of the new debt and
therefore:
the value of the new debt (face value) = the present value of the old debts.
- the present value of the old debts:
121
The amounts × Number of
The amounts
(ai)
Number of
months (ni)
10000
5
months {(ai) × (ni)}
50000
25000
8
200000
60000
10
600000
95000

850000
The present value of the old debts = total face values - Total discounts
 ai = {S1 + S2 + S3 } – d% { (S1 × n1) + (S2 × n2) + (S3 × n3) }
 {(Si)× (ni)}
 ai =  ( Si ) – d % × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12
850000
 ai = 95000 – 0.09 × ‫ـــــــــــــــــــــــــــــــــــــــ‬
 ai = 95000 - 6375 = 88625
12
Thus, the face value of the new debt is L.E. 88625.
Another solution: The previous example can be solved by finding the present
value of the old debts as follows:
Present value of the old debts = Present value of the first debt + Present value of
the second debt + Present value of the third debt.
= 10000 { 1- 0.09 × (5\ 12)} + 25000 { 1- 0.09 × (8\ 12)}+ 60000 { 1- 0.09 × (10\
12)}
= 10000 × 0.9625 + 25000 × 0.94 + 60000 × 0.925
= 9625 + 23500 + 55500 = 88625
Thus, the face value of the new debt is L.E. 88625.
122
2) Repayment of old debts (debts before settlement) with one debt due at a later
date to maturity dates of old debts:
In this case, the original (pre-settlement ) debts are replaced by one debt that is due
on after or subsequent to the original debts maturities. Therefore, when the value
of the new debt is found, it is preferable to use the amount (sum) laws.
This can be illustrated by the following figure:
S1
S2
S3
n
n1
n2
n3
S
Therefore, at the maturity date of the new debt (settlement date):
The face value of the new debt = the total of the old debts at the maturity date of
the new debt.
Example: Arzak Textile Company is a debted with the following amounts:
10000 L.E. due in 1/4/2020.
20000 L.E. due in 15/5/2020.
30000 L.E. due in 15/6/2020.
If the company wants to pay this debt with a single debt that is repayable on
10/9/2020 using a simple interest rate of 6% per annum.
Solution:
It is considered that the settlement date is the maturity date of the new debt and
thus:
123
On the maturity date of the new debt (settlement date):
10000
20000
30000
10\9
1\4
15\5
15\6
x
The face value of the new debt = the total of the old debts at the maturity date of
the new debt.
Period = Apr May June July Aug Sep Oct nov
n1 = 29 31 + 30 + 31 + 31 + 10
= 162 days
n2 = - + 16 + 30 + 31 + 31 + 10
= 118 days
n3 = - + - + 15 + 31 + 31 + 10
= 87 days
Thus, The value of amount debt can be calculated as follows:
The previous example can be solved by calculating the sum of each sum separately
as follows:
S = S1 + S2 + S3
S = (a1 + I1 ) + (a2 + I2 ) + (a3 + I3 ) = ai ( 1+ i × ni)
S = a1 ( 1 + i × n1) + a2 ( 1 + i × n2) + a3 ( 1 + i × n3)
S = 10000{1 + 0.06 × (162\360)} + 20000 {1 + 0.06 × (118\360)} + 30000{1
+ 0.06 × (87\360)}
S = 10270 + 20393.33 + 30435 = 61098.33
Thus, the value of the new debt is: L.E. 61098.333 .
124
Another solution:
The amounts × Number of
The amounts
(ai)
Number of days
(ni)
10000
162
days {(ai) × (ni)}
1620000
20000
118
3260000
30000
87
2610000
60000

6590000
Thus,
Total old debts = Total Amounts + Total Interest.
 {(ai)× (ni)}
S =  ( ai ) + i × ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
12 or 360 or 365 or 366
6590000
S = 60000 + 0.06 × ‫ـــــــــــــــــــــــــــــــــــــ‬
360
= 60000 + 1098.333 = 61098.333
Thus, the value of the new debt is: L.E. 61098.333 .
Example: On 3/4/2020 Hridi for building materials debted with the following
amounts:
70000 L.E. due in 30/7/2020.
50000 L.E. due in 29/8/2020.
30000 L.E. due in 27/11/2020.
On May 31, 2020 the company paid the amount of L.E. 60000 in cash and the rest
was released by an instrument of credit (bond) payable on 27/12/2020 at an interest
rate of 6% per annum.
Required: Calculate the face value of the bond.
125
Solution:
This example can be resolved by using the present value as follows:
10000
3\4
20000
30000
31\5
27\12
30\7
29\8
27\11
x
60000 cash
On the settlement date (31/5/2020):
Present value of old debts = cash value + Present value of new debt.
The calculation of the period between the settlement date and the due date of the
old debts is as follows:
Period = June July Aug Sep Oct Nov Dec
n1 = 30 + 30 + -- + - - + - - + -- + --
= 60 days
n2 = 30 + 31 + 29 + -- + -- + -- + --
= 90 days
n3 = 30 + 31 + 31 + 30 + 31 + 30 + 27 = 180 days
Period of the new debt:
n = 30 + 31 + 31 + 30 + 31 + 30 + 27 = 210 days
Thus,
The present value of the old debts at settlement date is calculated as follows:
The amounts × Number of days
The amounts
(ai)
Number of days
(ni)
70000
60
{(ai) × (ni)}
4200000
50000
90
4500000
30000
180
5400000
150000

14100000
126
Present value of old debts = total face values - total deductions
 {(Si)× (ni)}
 ai =  ( Si ) – i % × ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
360
14100000
 ai = 150000 – 0.06 × ‫ـــــــــــــــــــــــــــــــــــــــ‬
360
 ai = 150000 – 2350 = 147650
present value of old debts = Cash amount + Present value of new debt :
147650 = 60000 + x { 1- 0.06 × (210\ 360)}
147650 = 60000 + 0.965 x
147650 - 60000 = 0.965 x
87650 = 0.965x
X = 87650 \ 0.965 = 90829
- the face value of the bond = L.E. 90 829
3) Repayment of the old debts with one debt is due during the maturity dates of
old debts:
In this case, the old debts (original debts) is settled and repaid in one debt that is
due over the maturity dates of the old debts, meaning that there are a number of old
debts that fall after the date of maturity of the new debt; and that there are a
number of old debts that fall befor the date of maturity of the new debt; The
settlement is made on the basis that the pre-maturity debt is added to the interest
(deferred or delayed debt), while the debt that falls after the maturity date
(accelerated or outstanding debt) is granted a discount, whereas the debt is at the
same maturity date The new value of the name of be as it is:
127
On the maturity date of the new debt (settlement date):
S1
S2
S3
S4
x
At the maturity date of the new debt, the amount for the first and second debt is
calculated and the present value of the third and fourth debt is calculated.
Example: On 1/1/2020 Noor Lighting Company has the following amounts:
20000 L.E. due in after 4 months.
30000 L.E. due in after 5 months.
50000 L.E. due in after 8 months.
If the company wants to replace these debts with one debt that is repayable on
1/7/2020.
Find: The value of this debt if the simple interest rate is 9% per annum.
Solution:
On the maturity date of the new debt (settlement date):
20000
30000
1\1\2020
50000
1\7
1\5
1\6
1\9
x
On the maturity date of the new debt,
The face value of the new debt = the amount of the first debt + the amount of the
second debt + the present value of the third debt.
The face value of the new debt
128
= 20000 { 1+ 0.09 × (2\ 12)} + 30000 { 1+ 0.09 × (1\ 12)}+ 50000 { 1- 0.09 × (2\
12)}
= 20000 × 1.015 + 30000 × 1.0075 + 50000 × 0.985
= 20300 + 30225 + 49250 = 99775
Thus, the face value of the new debt = 99775 L.E.
another solution:
As the settlement date is now: 1/1/2020
Thus, the present value of the old debts = the present value of the new debt
- the present value of the old debt:
The amounts × Number of
The amounts
(ai)
Number of
months (ni)
20000
4
months {(ai) × (ni)}
80000
30000
5
150000
50000
8
400000
100000

630000
Thus, the present value of the old debt:
Present value of old debts = total face values - total discount
= 100000 – { 0.09 × (630000\12)} = 4725
= 100000 - 4725 = 95275
- Present value of new debt:
= x (1-0.054) = 0.955x
Thus, at the focal date :
Present t value of old debts = Present value of new debt
95275 = 0.955x
X = 95275 \ 0.955 = 99764.4
129
4) Repayment of old debts with several new debts due to be repaid during the
maturity dates of old debts:
We have noted from the methods of settlement and replacement of previous debts
that we have added interest when the debtor defaults on the payment and give it
discount when accelerating the repayment of the principal amounts. However, if
several new debts are settled and dates may fall between the old and some later
dates of the old debts, The date of the agreement on the basis that the settlement
date and the calculation of the value of the old debts as well as the value of the new
debts at the settlement date, the debtor may also wish to modify and settle its old
debts with several new debts or may wish to repay part of the old debt now in cash
(or settlement date) A bond, a bill or both is payable on different dates, so that the
present value equation of the old debts (now at the settlement date) will be used
with the present value of the new debts on the same date.
Therefore:
At the settlement date (date of the agreement):
Present value of old debts = Present value of new debts
Example: Aya Car Trading Company has the following amounts:
40000 L.E. due in after 4 months.
80000 L.E. due in after 8 months
50000 L.E. due in after 9 months.
If the company wants to replace and settle these debts with two equal amounts of
face value, the first is due after 6 months and the second after one year. If the
interest rate used is 9% annually, find the face value of both debts.
131
Solution:
40000
now
80000
50000
6
12
4
8
9
x
x
On settlement date (now):
Present value of old debts = Present value of new debts
First: Calculate the present value of the old debts :
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
40000
4
{(ai) × (ni)}
160000
80000
8
640000
50000
9
450000
170000

1250000
Thus, the present value of the old debt:
Present value of old debts = total face values - total discount
= 170000 – { 0.09 × (1250000\12)} = 9375
= 170000 - 9375 = 160625
Second: Present value of new debts:
The amounts
(ai)
Number of
months (ni)
x
x
2x
6
12
The amounts × Number of months
{(ai) × (ni)}
6x
12x
18x

Thus, the present value of the new debts:
Present value of old debts = total face values - total discount
131
= 2 x – { 0.09 × (18 x \12)}
= 2x - 0.135x = 1.865x
Therefore:
Present value of old debts = Present value of new debts
160625 = 1.865x
X = 160625 \ 1.865 = 86126
Accordingly, the face value of each new debt is equal to L.E. 86126.
Example: A person who owes the following amounts:
50000 L.E. due in after 9 months.
100000 L.E. due in after 10 months.
200000 L.E. due in after 11 months.
If the person agrees to repay his debts :
- Payment of L.E. 35000 in cash.
- The balance he agrees to repay his debts with two debts, first bill is twice the
face value of the second bill and is payable after 6 months, 15 months respectively,
if the simple interest rate used is 12% per annum. Calculate the face value of each
bill.
Solution:
Therefore, at the settlement date (now):
50000
100000
9
10
6
60000 cash
200000
11
15
2x
x
On the settlement date (now):
Present value of old debts = Present value of new debts
- First: the present value of the old debts:
132
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
50000
9
{(ai) × (ni)}
450000
100000
10
1000000
200000
11
2200000
350000

3650000
Thus, The present value of the old debts
= Total Amounts - Total Discount
= 350000 – { 0.12 × (3650000\12)}
= 350000 - 36500 = 313500
- Second: the present value of the new debts:
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
2x
6
{(ai) × (ni)}
12x
x
15
15x
3x

27x
Thus, the present value of the new debts:
Present value of old debts = total face values - total discount
= 3 x – { 0.12 × (27 x \12)}
= 3x - 0.27x = 2.73x
Therefore:
Present value of old debts - Cash amount = Present value of new debts
313500 - 35000 = 2.73x
2.73x = 278500
X = 278500\ 2.73 = 102014.652
133
Therefore:
Face value of first bill (2x):
= 2 × 102014.652 = 204029.3
Face value of the second bill (x):
= 102014.652
Example: A person who owes the following amounts:
40000 L.E. due in after 120 days.
50000 L.E. due in after 150 days.
60000 L.E. due in after 180 days.
If the person agrees to repay his debts with two debts, the face value of the first
debt is half the face value of the second debt and is payable 100 days later, 130
days respectively. If the interest rate used is 9% per annum, calculate the face
value of each debt is determined.
Solution:
Assume that the face value of the first debt is = x
Assume that the face value of the second debt is = 2x
Therefore, at the settlement date (now):
40000
now
120
100
x
50000
150
130
2x
On the settlement date (now):
Present value of old debts = Present value of new debts.
134
60000
180
- First: the present value of the old debts:
The amounts × Number of days
The amounts
(ai)
Number of days
(ni)
40000
120
{(ai) × (ni)}
4800000
50000
150
7500000
60000
180
10800000
150000

23100000
Thus, The present value of the old debts:
= Total Amounts - Total Discount
= 150000 – { 0.09 × (23100000\360)}
= 150000 - 5775 = 144225
Second: the current value of the new debts:
The amounts × Number of days
The amounts
(ai)
Number of days
(ni)
x
100
{(ai) × (ni)}
100x
2x
130
260x
3x

360x
The present value of the new debt:
Present value of new debts = total face values - total discount
= 3 x – { 0.09 × (360 x \360)}
= 3x - 0.09x = 2.91x
Therefore:
Present value of old debts = Present value of new debts
144225 = 2.91x
X = 144225\ 2.91 = 49561.86
135
Therefore:
The face value of the first debt (x) = 49561.86 L.E.
The face value of the second debt (2x) = 49561.86 × 2 = 99123.72 L.E.
Example: A merchant owes the following amounts:
60000 L.E. due in after 6 months.
90000 L.E. due in after 10 months.
40000 L.E. due in after 13 months.
If the trader wants to settle his debts with three face notes of equal face value and
mature after 8 months, 9 months, one year, respectively. If the interest rate used is
9% per annum, calculate the face value of each bond.
Solution:
Therefore, at the settlement date (now):
60000
90000
6
10
now
8
x
9
x
40000
13
12
x
On the settlement date (now):
Present value of old debts = Present value of new debts
- First: the present value of the old debts:
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
60000
6
{(ai) × (ni)}
360000
90000
10
900000
40000
13
520000
190000

1780000
136
The present value of the old debts =
= Total Amounts - Total Discount
= 190000 – { 0.09 × (1780000\12)}
= 190000 - 1335 = 176650
Second: the present value of the new debts:
The amounts × Number of
The amounts
(ai)
Number of
months (ni)
x
8
months {(ai) × (ni)}
8x
x
9
9x
x
12
12x
3x

29x
The present value of the new debts:
Present value of new debts= total face values - total discount
= 3 x – { 0.09 × (29 x \12)}
= 3x - 0.2175x = 2.7825x
Therefore:
Present value of old debts = Present value of new debts
176650 = 2.7825x
X = 17665\ 2.7825 = 63486.1
Thus, the face value of each of the three bonds is L.E. 63486.1
Example: On 1/1/2020 Al-Manal Trading Company had debt the following
amounts:
13000 L.E. payment in 1/4/2020.
15000 L.E. payment in 1/6/2020.
30000 L.E. payment in 1/9/2020.
On 1/3/2020 the company wanted to repay its debts as follows:
- Payment of 8000 L.E. in cash.
137
- with three face bonds ,The third face value of the second bond is three times the
face value of the first bond and the ratio between the face value of the third bond
and the face value of the second bond as a 3: 1 ratio. The three bonds are maturing
after 3 months, 4 months and 5 months respectively, User interest is 10% per
annum. Calculate the face value of each bond.
Solution:
At the settlement date (date of the agreement is 1/3/2020):
13000
1\1
1\3
15000
1\4
1\6
3
8000
cash
30000
x
3x
1\9
4
9x
On the settlement date 1\3\2020
Present value of old debts = Present value of new debts
- First: the present value of the old debts:
The amounts × Number of
The amounts
(ai)
Number of
months (ni)
13000
1
months {(ai) × (ni)}
13000
15000
3
45000
30000
6
180000
58000

238000
Thus, the present value of the old debts:
= Total Amounts - Total Discount
= 58000 – { 0.10 × (238000\12)}
= 58000 - 1983.33 = 56016.67
- Second: the present value of the new debts:
The value of the new debt is calculated as follows:
The face value of the first debt = x
138
5
The face value for the second debt = 3x
The face value for the third debt = 9x
Calculate the present value of the new debts:
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
x
3
{(ai) × (ni)}
3x
3x
4
12x
9x
5
45x
13x

60x
The present value of the new debts:
= Total Amounts - Total Discount
= 13x – { 0.10 × (60x\12)}
= 13x - 0.5x = 12.5x
Thus,
Present value of old debts = Present value of new debts
56016.67 = 8000 + 2.5 x
56016.67- 8000 = 2.5 x
48016.67 = 2.5 x
X = 48016.67 \ 2. 5 = 3841.33
Therefore:
The face value of the first debt (x) = 3841.33
The face value for the second debt (3x) = 115233.99
The face value for the third debt (9x) = 34571.29
Example: On 1/1/2018 Al-Manal Trading Company had the following amounts:
3000 L.E. payment after 6 months.
5000 L.E. payment after 7 months.
6000 L.E. payment after 9 months.
139
On 1/2/2018 the Company settled these debts as follows:
- Payment of L.E. 2580 in cash.
- The balance three face bonds , of the face value of the first bond to the face
value of the second bond as the ratio of 2: 3 and the ratio of the face value of the
second bond to the face value of the third bond as the ratio of 2: 4 and the first due
in after two months and the second due in after 3 months and the third due in after
4 months, if the interest rate is 9% per annum. Calculate the face value of each
bond.
Solution:
To settle this debt, the following general formula is used to settle the settlement at
the settlement date of 1/2/2018.
At the settlement date (date of the agreement is 1/2/2018):
1\1
5000
6
7
1\2
2
2580
3000
cash
4x
6x
3
4
5
6
12x
On the settlement date 1\3\2020
Present value of old debts = Present value of new debts
- First: the present value of the old debts:
The amounts ×
The amounts
(ai)
Number of months
(ni)
3000
5
Number of months
{(ai) × (ni)}
15000
5000
6
30000
6000
8
48000
14000

93000
141
6000
9
8
Thus, the present value of the old debts:
= Total Amounts - Total Discount
= 14000 – { 0.909 × (69000\12)}
= 14000 - 697.5 = 13302.5
- Second: the present value of the new debts:
present value of new debts (after adjustment)
From the previous representation of the ratios between the par value of the three
bonds it can be said that the ratios take the following values:
It assumes that the face value of the first bond = 4x
The face value of the second bond = 6x
The face value of the third bond = 12x
Total face values of the three bonds = 22x
Calculate the present value of the new debts:
The amounts × Number of months
The amounts
(ai)
Number of
months (ni)
4x
2
{(ai) × (ni)}
8x
6x
3
18x
12x
4
48x
22x

74x
The present value of the new debt:
= Total Amounts - Total Discount
= 22x – { 0. 09 × (74x\12)}
= 22x - 0.555x = 21.445x
Thus,
Present value of old debts = cash value + Present value of new debts
13302.5= 2580 + 21.445 x
13302.5 - 8000 = 21.445 x
141
10722.5= 21.445 x
X = 10722.5\ 21.445 = 500
Therefore:
The face value of the first year (4x) = 2000
The face value for the second year (6x) = 3000
The face value for the third year (12x) = 6000
5) Repayment of old debts with one debt whose face value equals the total face
value of the old debts (average maturity date):
In this case, the old debts are settled by one debt whose face value equals the sum
of the face values of the old debts. The date on which the debt is due is called the
average maturity date. Where the maturity date of this new debt is maturing,
meaning that it falls between the maturity date of the first debt and the maturity
date of the last debt. The main advantage of this method is the interest of delaying
some of the original outstanding debts with the deductions granted to some original
debts to be expedited. The maturity date of the new debt (maturity date of the
average).
Example: On January 1, 2019, Ahmed Sporting Goods Company had the
following amounts:
20000 L.E. due in 1/3/2019
30000 L.E. due in 30/6/2019
40000 L.E. due in 31/8/2019
If the company wishes to repay all its debt with a face value equal to the total face
value of the old debt.
calculate the maturity date of the new debt.
142
Solution:
20000
1\1\2019
30000
2
6
40000
??
8
1\3
30\6
90000
31\8
Face value of new debt = total face value of old debts
= 20000 + 30000 + 40000 = 90000
- First: the present value of the old debts:
The amounts × Number of
The amounts
(ai)
Number of
months (ni)
20000
2
months {(ai) × (ni)}
40000
30000
6
180000
40000
8
320000
90000

540000
Therefore:
{ The amounts × Number of months) /12}
Average maturity date = ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
total face value of old debts
{ 540000) /12}
Average maturity date = ‫ = ــــــــــــــــــــــــــــــــــــــــــــــ‬6 months
90000
the average maturity date is 1/7/2009.
143
Exercise
1- In each of the following problems, find value of the new debts.
No
No. of
Due
Money
No. of
Due
Values
old debts
dates
value
new
dates
debts
1
1
2000 L.E 4 Ms
12%
1
6 Ms
2
1
4000 L.E 5 Ms
10%
1
8 Ms
3
2
1500 L.E 9 Ms,
9%
2
4 Ms,
4000 L.E 8 Ms
Equals
6 Ms
4
2
2600 L.E
1
2
8 Ms,
8%
5000 L.E Year,
Equals
10 Ms
6 Ms
5
3
6000 L.E 4 Ms
3
3 Ms
7.5%
400 L.E
6 Ms
Equals
7 Ms
800 L.E
8 Ms
10 Ms
2- Statement problems :
a)
A debt of 4000 L.E. is due in 6 months. If the interest rate in 10%, what is the
value of the debt if it is paid 4 months hence.
b) A man owes :
2000 L.E. due in 8 months.
1500 L.E. due in 6 months.
1000 L.E. due in 9 months.
He agreed with his creditor to settle the old debts by 2 equal new obligations,
one due in 7 months and the other due in 10 months. Find the value of each
obligation if the interest rate is 8% and the focal date is 7 months hence.
c)
Find the size of each obligation in the above Problem if the settlement date is
10 months hence.
d) As obligation of 8000 L.E. due in 4 months is to be paid by 4 equal payments
due 6, 8, 9, 10, and the is worth 9%. Find the value of each payment using 8
months hence as a focal date.
144
e)
Find the size of each payment in the above Problem if the focal date is 4
months hence.
f)
Answer the question in the problem (d) if the focal date is 10 months hence.
g) Answer the question in the problem (d) if the focal date is 12 months hence.
145
Chapter ( 5)
Annuities or Equal payments
(Amount and present value)
747
748
Chapter ( 5)
Annuities or Equal payments
(Amount and present value)
Introduction:
Annuities are a set of regular and consquance amounts payable regularly and
at equal intervals, each of which is called the principal of the annuities.
The principal of the annuity may be taken as an installment (payment) of the
loan or a rent payment or interest and wages. If these amounts are equal, the
annuity is called equal annuities. If the payment amounts are not equal, the annuity
is called variable annuities. In simple interest annuities are often equal, If the
period between the two dates is a full year, if the period is six months the payment
is semi-annual the annuity is annually, if the period is 3 months is called a
quarterly annuity, The period was one month The annuity was named monthly.
The annuity may be certain (specific) and the annuity may be (uncertain) and
the interest in the simple interest is limited to certain annuities only.
The annuities are divided into:
749
Annuities
temporary Annuities
Straight
Annuity
deferred
Annuity
due immediate due
premanent Annuities
Straight
Annuity
deferred
Annuity
immediate due immediate due
immediate
Temporary annuity: The value of the annuity will be paid for spesific or fixed time
(n numbers of years) that could be 5 , 10 or more.
premanent annuity: The value of the annuity will be paid for infinate.
Straight annuity: The value of the annuity will be paid starting from that year.
annuities that start paying now meaning that the first annuity is being paid now.
Deferred annuity: The value of the annuity will be paid after (m) number of years.
annuities that start after a certain period of time, meaning that the first annuity will
begin after the expiry of the period following the date of the contract (the deferred
period).
Immediate annuity: The value of the annuity will be paid at the end of each year.
Due annuity: The value of the annuity will be paid at the beginning of each year.
In this chapter we will examine how to derive the basic laws for the
calculation of the amount (sum) and the present value of equal annuities using
simple interest laws, as follows:
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First: Amount (sum) of an annuities (equal payments):
The equation of annuities can be deducted on the basis of the calculation of
the sum of each of the amounts of the amounts separately and in the usual way and
then collect these amounts to get the total annuities required.
- If the annuities is ordinary (paid at the end of each period) , then:
a
1
a
1
a
a
a
1
i= %
,
n = Number
Sn
Note that the first annuity (installment) invested for 11 months, the second
annuity invested for 10 months, the third annuity invested for 9 months, and so on
... until the last annuity as it did not stay any time in the bank and thus be as it.
- If the annuity is due , (paid at the beginning of each period), then:
a
1
a
1
a
a
1
i= %
,
n = Number
Sn
Note that the first annuity (installment) invested for 12 months, the second
annuity invested for 11 months, the third annuity invested for 10 months, and so on
.... until the last annuity , where it invested for one month.
Thus: Total annuities = total amount + total intrests
Sn = ( a × n ) + { n ( I1 + I2 ) }
2
Sn = ( a × n ) + { a × i ×
1×
12
n( L1 + L2 ) }
2
757
whereas:
Sn = Sum of annuities:
a = value of annuity.
i = Investment rate (interest).
n = number of annuity.
L1 = Period of investment the first annuity.
Ln = Period of investment the last annuity.
From the above we note the following:
1. The sum of the annuities shall be equal to the sum of the total sum of this
annuity at the expiry date of the annuity period.
Total annuities = total amount + total interest until the expiry date of annuity
2. In case of ordinary annuities:
- The period of investment of the first annuity annuity L1 is equal to the annuity
period of annuity minus one period of time.
- The period of investment last amount is L2 is equal to zero.
3. In case of due annuities:
- The period of investment of the first annuity annuity L 1 is equal to the annuity
period of all annuities.
- The period of the last investment annuity is L2 equal to one time period.
4. The sum of due annuities is greater than the sum of ordinary annuities.
5. If there is a payment delay period, it is added to L1 , L2 .
Example: Find the sum of monthly annuity ordinary L.E. 100 invested for one
year at an interest rate of 9% per annum.
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Solution:
Where the annuity is ordinary ,therefore pays at the end of every month:
100
1
100
1
100
100
100
1
i = 9%
1
12×
Sn = ( a × n ) + { a × i ×
,
n = 12
n
(2L1 + L2 ) }
1
12×
Sn = ( 100 × 12) + {100 × 0.09 ×
12
( 11 + 0) }
2
= 1200 + 49.5 = 1249.5
Example: Find the sum of monthly annuity due L.E. 100 invested for one year at
an interest rate of 9% per annum.
Solution:
Where the annuity is due ,therefore pays at the begining of every month:
100
100
1
1
1
i = 9%
Sn = ( a × n ) + { a × i ×
100
1
12×
Sn = ( 100 × 12) + {100 × 0.09 ×
,
n = 12
n
(2L1 + L2 ) }
1
×
12
= 1200 + 58.5 = 1258.5
751
12
( 12 + 1) }
2
100
100
Sn
Example: Ahmad will deposit L.E. 300 per month for two years in the economy
bank, which gives interest rate 10% per annum. Required: Calculate the sum of
what is made up of Ahmed in the bank if:
a) If the the deposit at the beginning every month ( due annuity).
b) If the the deposit at the end every month ( ordinary annuity).
Solution:
a) Where the annuity is due ,therefore pays at the begining of every month:
300
300
1
1
i = 10%
300
Sn
1
,
n = 24
1
12×
Sn = ( a × n ) + { a × i ×
300
n
2( L1 + L2 ) }
Sn = ( 300 × 24) + {300 × 0.09 ×
1
12×
24
2( 24 + 1) }
= 7200 + 750 = 7950
b) Where the annuity is ordinary ,therefore pays at the end of every month:
300
300
1
i = 10%
Sn = ( a × n ) + { a × i ×
300
1
300
1
,
n = 24
1
12×
(nL1 + L2 ) }
2
754
300
300
Sn = ( 300 × 24) + {100 × 0.09 ×
1
12×
24( 23 + 0) }
2
= 7200 + 690 = 7890
From the two previous examples, we note that the due annuity clause is greater
than the ordinary annuity.
Example: Aya Cosmetics Company will deposit a sum of L.E. 1000 for the first
and middle of each month of 2020 at Bank Cairo, which gives an interest rate of
12% per annum. Calculate the total composition of the company at the end of the
year.
Solution:
The above example can be resolved in more than one way:
1000 1000 1000
i = 12%
,
1000
n = 24
The example can be resolved as an equal semi-monthly annuity due.
Sn = ( a × n ) + { a × i ×
1
12×
Sn = ( 1000 × 24) + {1000 × 0.12 ×
n
(2L1 + L2 ) }
1
12×
24
(212 + 0.5) }
= 24000 + 1500 = 25500
Another solution: The example can be resolved on the basis that it is two
annuities:
= sum monthly payment due + sum semi-monthly payment due:
Sn = ( a × n ) + { a × i ×
1
12×
n
2( L1 + L2 ) }
755
1
12×
+ (a× n)+ {a× i ×
(nL1 + L2 ) }
2
1
12×
Sn = ( 1000 × 12) + {1000 × 0.12 ×
1
12×
+ ( 1000 × 12) + {1000 × 0.12 ×
12
( 12 + 1) }
2
12
( 11.5 + 0.5) }
2
= {1200 + 780} + {1200 + 720} = 25500
Example: Al-Manal Company deposited L.E. 1000 in the first month of each
month in 2020 and was withdrawing L.E. 500 in the middle of each month of the
same year. If the interest rate on deposits and withdrawals is 10% per annum,
Calculate the
accumulated
balance of the company at the end of the year
(31/12/2020).
Solution:
1000
500
1000
500
500
i = 10 %
1000
500
,
n = 12
Made up the balance of the company reach trade at the end of
the year:
Balance = Total deposits - total withdrawals
First: Total Deposits:
Sn = ( a × n ) + { a × i ×
1
12 ×
Sn = ( 1000 × 12) + {1000 × 0.10 ×
n
(2L1 + L2 ) }
1×
12
756
12
( 12 + 1) }
2
= 12000 + 650 = 12650
Second: Drawings:
Sn = ( a × n ) + { a × i ×
1
12 ×
n( L + L ) }
2
2 1
1
Sn = ( 500 × 12) + {500 × 0.10 × 12 ×
= 6000 + 300 = 6300
12
2( 11.5 + 0.5) }
Third: The accumulated balance:
Balance = Total deposits - total withdrawals
= 12650 - 6300 = 6350
Example: Mennat Allah for import and export deposit L.E. 3000 at the end of
every 3 months of 2019 and then increased the deposit duable at the end of every 3
months of 2020, found the total of the company in the bank at the end of 2020 if
the rate of interest used 9% per year.
Solution:
This example can be resolved in more than one way as follows:
3000 3000
3000
6000
2019
6000
2020
6000
i = 9%
Example can be resolved on the basis of:
The total value of the company at the end of 2020 = total annuities in 2019 at the
end of 2020 + total annuities in 2020 at the end of the year.
- Total deposits made during 2019 at the end of 2020:
Sn = ( a × n ) + { a × i ×
1
12 ×
n( L + L ) }
2
2 1
757
Sn = ( 300 × 4) + {300 × 0.09 ×
1
12×
4( 21 + 12) }
2
= 12000 + 14850 = 13485
- Total deposits made during 2020 at the end of the year:
Sn = ( a × n ) + { a × i ×
1
12×
Sn = ( 600 × 4) + {600 × 0.09 ×
n( L + L ) }
2
2 1
1
12×
4
2( 9 + 0) }
= 24000 + 810 = 24810
Thus, the total of the company consists = total deposits in 2019 at the end of the
period + total deposits in 2020
Total = 13485 + 24810 = 38295
Another solution: The previous example can be solved considering that the deposit
deposited in 2019, which is L.E. 3000 also continue the same value during 2020,
and therefore the value of the payment deposited in 2010 becomes only L.E. 3000
instead ofL.E. 6000 Thus,
The balance formed at the end of the year 2020 Equals:
= Total annuity of L.E. 3000 during 2019, 2020 + total annuity of L.E. 3000
during 2020.
- The sum of the first annuity of the amount of L.E. 3000 and the number of
amounts 8 annuities:
Sn = ( a × n ) + { a × i ×
1
12 ×
Sn = ( 300 × 8) + {300 × 0.09 ×
n( L + L ) }
2
2 1
1×
12
= 24000 + 1890 = 25890
758
8( 21 + 0) }
2
- The sum of the second annuity of the amount of L.E. 3000 and the number of
amounts 4 annuities:
Sn = ( a × n ) + { a × i ×
1 ×
12
Sn = ( 300 × 4) + {300 × 0.09 ×
n( L1 + L2 ) }
2
1
12×
4( 9 + 0) }
2
= 12000 + 405 = 12405
- Total balance formed = First annuity + Second annuity
= 25890 + 12405 = 38295
A third solution: The previous example can be solved by finding the total of the
2019 annuities at the end of 2019, then finding the accrued interest on the total
amounts for one year and finding the total annuities in 2020 as follows:
2019 total at the end of 2019:
Sn = ( a × n ) + { a × i ×
1
12×
Sn = ( 300 × 4) + {300 × 0.09 ×
n( L + L ) }
2
2 1
1
12×
4( 9 + 0) }
2
= 12000 + 405 = 12405
and then find the total interest amounts for one year:
S= a(1+i× n)
S = ( 3000 × 4 ) { 1 + 0.09 × 1 } = 1080
The total annuities for 2019 at the end of 2020 is:
= 12405+ 1080 = 13485
Find the 2020 batch as follows:
Sn = ( a × n ) + { a × i ×
1
12×
n( L + L ) }
2
2 1
759
1
12×
Sn = ( 600 × 4) + {600 × 0.09 ×
4
2( 9 + 0) }
= 24000 + 810 = 24810
Total = 13485 + 24810 = 38295
Example: Ahmed deposits the amount in the National Bank for the first every 3
months for 3 years, Ahmed withdrew the total of what he has and find it equal to
L.E. 1317 , if the bank calculates simple interest rate of 6% per annum. Calculate
the amount of the quarterly annuity (the amount deposited).
Solution:
a
a
a
Sn =1317
i=6%
,
n = 24
Where the annuities period is 3 years (36 months) , every 3 months. so that:
The number of annuties = 36 \ 3 = 12
2019 total at the end of 2019:
Sn = ( a × n ) + { a × i ×
1
12×
Sn = ( a × 12) + {a × 0.06 ×
n( L + L ) }
2
2 1
4( 36 + 3 ) }
2
1
12×
1317 = 12 a + 1.17 a
1317 = 13.17 a
Thus, (a) the amount of the annuity:
a = 1317 \ 13.17 = 100
761
Example: a regular, semi-annual annuity immediate of L.E. 2000 and a period of
3 years, calculated the sum ( or amount) as a simple interest of L.E. 13,200,
which is the annual interest rate used.
Solution:
2000
20000
2000
Sn =13200
i=?%
,
n=6
Where the annuity every semi-annual annuity , Where the annuities period is 3
years (36 months)
The number of annuties = 36 \ 6 = 6
Amount = Total Amount of annuity + Total Interest
Sn = ( a × n ) + { a × i ×
1
12×
Sn = ( 2000 × 6) + {2000 × i ×
n
2( L1 + L2 ) }
1
12×
6( 30 + 0) }
2
13200 = 12000 + 15000i
13200 - 12000 = 15000i
1200 = 15000i
i = 1200 \ 15000 = 0.08
i=8%
Example: Menna Electrical Equipment Company deposits the amount of the first
month of each month of 2020, and duable of this amount in the middle of each
month of the same year, if the total of the company at the end of the year L.E.
15140, the bank uses a simple interest rate of 10% annually. Find the amount
767
Menna deposited at the first of each month and the amount you deposit in midmonth.
Solution:
a
2a
a
2a
Sn =15140
i = 10%
The total of the company at the end of the period is:
The sum of the first annuity made by (a) + the sum of the second annuity which
has been made (2a)
sum monthly annity due + sum semi-monthly annuity due:
1
Sn = ( a × n ) + { a × i × 12 ×
n
2( L1 + L2 ) }
1×
12
n
2( L1 + L2 ) }
+ ( 2a × n ) + { 2a × i ×
Sn = ( a × 12) + {a × 0.10 ×
1
12 ×
12
2( 12 + 1) }
1
12
+ ( 2a × 12) + {2a × 0.10 × 12 ×
( 11.5 + 0.5) }
2
= {12 a + 0.65 a } + {24 a + 1.2 a } = 37.85 a
15140 = 37.85 a
a = 15140 \ 37.85 = 400
Thus, it compensates for the total annuities:
Thus, the amount to be deposited first every month is L.E. 400 and the amount
deposited in the middle of each month is L.E. 800.
761
Secand:Present value of an annuities (equal payments):
The present value is the sum of the present value of the first annuity, the present
value of the second annuity, the present value of the third annuity, and so on …. to
the present value of the last annuity.
In our study of the present value and discount of one or several unequal sums
(Chapter 2) we found that there are two ways to calculate the present value of a
sum due at a later date, namely the simple present value (Ae) and the bank present
value (Ao), and this is what we will calculate in value present of annuities.
The bank present value ( proceed) (Ao):
The bank present value is usually used in the business life. The bank discount
should first be determined in the light of face value and we subtract this discount
from the face value. We obtain the bank present value, as follows:
- If the annuity is due :
a
a
a
a
An
i%
,
n years
- If the annuity is immediate:
a
a
a
An
i%
,
n years
- The bank present value of the payments is (Ao).
- The maturity value or face value is (Sn).
761
a
- Interest rate or discount rate is (i) or (d).
- The discount period is (n).
whereas :
Total annuities = total amounts + total intrests
Bank present value of annuities= Total amount - Total discounts
where it invested for one month.
Thus:Bank present value = maturity value – bank discount value
Ao = S - Xo
Ao = ( a × n ) - {
n
2 ( D1 + D2) }
1
Ao = ( a × n ) - { a × i × 12 ×
n( M1 + M2 ) }
2
whereas:
Ao = Bank present value of annuities:
a = value of annuity.
i or d = Interest rate or discount rate.
n = number of annuity.
M1 = Period of discount the first annuity.
Mn = Period of discount the last annuity
Notes that:
1. If the present value is calculated at the beginning of payment annuities:
- If the annuity is due , then:
M1 = equal to zero,
Mn = the entire period minus one period.
- If the annuity is ordinary or immadiate , then:
764
M1 = equal to one time of period,
Mn = the entire period ( total period )
2. If the present value is calculated at a time prior to the beginning of payment
annuities:
- This period is added to each M1, Mn
The simple present value (Ae):
By studying the concept of the present value of an amount that is shown to be the
value that, if invested at a agreed rate of interest or discount, will be equal to the
maturity value at the maturity date.
- The simple present value of the payments is (Ae).
- The face value or sentence is (Sn).
- Interest rate or discount is (I) or (d).
- The discount period is (n).
The present value (Ae).:
Accordingly, the:
Therefore, when calculating the simple present value of the annuities, we first
calculate the total annuities, whether due annuities or immediate annuities, and
then find the simple present value of this sum or amount by dividing by (1 + rate
of interest × period ).
Sn
Ae = ‫ـــــــــــــــــــــــــ‬
(1 + i × n )
(a× n)+ {a× i ×
1×
12
n
2( L1 + L2 ) }
Ae = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
(1 + i × n )
765
Example: Calculate the simple present value and bank present value (proceeds) of
an ordinary semi-annual payment of L.E. 3000 and its period of 3 years, if the
discount rate equals the interest rate equal to 12% per annum.
Solution:
3000 3000
3000
i = 12%
3000
,
3000
n = 3 years
- The number of annuties = 36 \ 6 = 6
- The bank present value ( proceed) (Ao):
Where the payment is ordinary,
M1 = equal to one time of period,
M1 = 6
Mn = the entire period ( total period ),
Mn = 36
Thus: Total payments = total payments - total discounts
Ao = ( a × n ) - { a × i ×
1
12 ×
Ao = ( 3000 × 6 ) - { 3000 × 0.12 ×
n
2( M1 + M2 ) }
1
12×
6( 6 + 36) }
2
= 18000 - 3780 = 14220
- The simple present value (Ae):
Where the annuity is ordinary or immdiate, whereas:
L1 = the period of the first investment amount.
L1 = 30
L2 = the period of the last investment amount.
L2 = 0
766
3000
(a× n)+ {a× i ×
1×
12
n( L + L ) }
2
2 1
Ae = ‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
(1 + i × n )
( 3000 × 6 ) + { 3000 × 0.12 ×
1
12×
6( 30 + 0) }
2
Ae = ‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
(1 + 0.12 × 3 )
Ae = 18000 + 2700
1.36
Ae = 20700 = 15220.29
1.36
It is noted that the bank present value( proceeds value) of the annuities is
lower than the simple present value of this annuities.
Example: an ordinary payment of L.E. 4000 payable at the end of every 4 months
for period of 4 years. If the discount rate or interest rate equals 9% per annum,
find the bank present value (proceeds value). if:
A) Before paying of the first amounts over a period of 4 months.
B) before paying the first amounts over a period of 10 months.
Solution:
A) Before paying the first installment for a period of 4 months (ordinary
payment):
Thus, the bank present value( proceeds value). (Ao):
Solution:
4000 4000
4000
4000
767
4000
4000
i = 9%
Ao = ( a × n ) - { a × i ×
,
1
12 ×
Ao = ( 4000 × 12 ) - { 4000 × 0.09 ×
n = 4 years
n
2( M1 + M2 ) }
1
12 ×
12
2( 4 + 48) }
= 48000 - 9360 = 38640
B - before paying the first amounts over a period of 10 months:
The example can be solved as an ordinary deferred annuity of 6 months, so in this
case the bank present value of the ordinary annuity (as in the previous case) is
found (ie before the first annuity is due in four months) and then subtracting the
amount of annuities (total payments only) For the remaining period of 6 months as
follows:
The present value of ordinary annuity and deferred:
= Present value of the ordinary payment - Deduction of payment amounts for 6
months.
- Present value of the ordinary annuity has already been found as in the previous
case:
A0 = 38640
- Discount for the amount of payments for 6 months:
= (a × n ) × i × n
= (4000 × 12 ) × 0.09 × (6 \ 12) = 2160
Thus, the present value of the annuity before maturity:
= 38640 - 2160 = 36480
768
another solution:
The previous example can be resolved by adding the previous period to both M1
and Mn
Ao = ( a × n ) - { a × i ×
n
2( M1 + M2 ) }
1×
12
whereas:
M1= the period of the first annuity discount on the previous date.
M2 = the period of the last annuity discount on the previous date.
Ao = ( 4000 × 12 ) - { 4000 × 0.09 ×
1
12 ×
12( 10 + 54) }
2
= 48000 - 11520 = 36480
Example: Ahmad bought a car from the Aya exhibition for L.E. 23120 cash and
paid for it in equal monthly annuities for one year. He agree to pay the first
annuity at the time of the contract and received the car. If the exhibition uses a
discount rate or interest rate of 8% per annum.
Caculate the value of the monthly annuity?
Solution:
- Cash price 23120 represents the present value of the annuities.
a
a
a
a
A0 = 23120
i=8%
- The due annuity value (a) is required and thus:
Ao = ( a × n ) - { a × i ×
1
12 ×
23120 = ( a × 12 ) - { a × 0.08 ×
n
2( M1 + M2 ) }
1×
12( 0 + 11) }
12
2
769
23120 = 12 a
- 0.44 a
23120 = 11.56 a
a = 23120 \ 11.56 = 2000
Thus, the monthly annuity paid by Ahmed is L.E. 2000.
771
Chapter (6)
Repayment and Amortization
Of short-term loan
171
171
Chapter ( 6)
Repayment and Amortization
Of short-term loan
Methods of repayment of short-term loans:
The repayment or amortization of loans is the repayment of the loans with their
interest once or in equal and unequal installments. Consequently, there are many
ways of consuming and repaying loans:
- The first case: repayment of the loan at the end of the period:
This case includes the following ways:
1. Repayment of the loan and interest at the end of the borrowing period
2. Repayment of the loan at the end of the period , while (with) the interest is paid
in whole or in part in advance.
3. Repayment of the loan at the end of the borrowing period, while interest is
repaid in the form of regular payments (annuities).
- The second case: repayment of the loan in equal installments:
This case includes the following ways:
1. Repayment of the loan in equal premuims ( or installments) of the principal and
interest together.
2 - Repayment of the loan in equal premuims ( or installments) of the principal
only with the payment of interest in advance.
3 - repayment of the loan in equal premuims ( or installments)
with the payment of interest on the balance.
171
of the principal
- The third case: The repayment of the loan in premuims ( or installments) are
not equal with the payment of interest on the balance:
Methods of repayment and amortization of loans are as follows:
- The first case: repayment of the loan at the end of the period:
In this case, the loan is repaid at the end of the period of the borrowing, while the
interest is repaid with the loan at the end of the borrowing period or is paid in
whole or in part at the beginning of borrowing or paid in the form of regular
payments (annuities).
The first method: repayment of the loan and interest at the end of the borrowing
period:
Under this method, the creditor and the debtor agree to repay the loan and its
interest in full at the end of the period of the loan
(amount of the loan). This is
what was dealt with in the first chapter including a sum of simple interest as
follow:
Amount of the loan = loan + Interest
S=a+I
S = a + (a × i × n)
S = a (1 + i × n)
Example: Ahmed borrowed L.E. 5000 from the Bank Masr, which calculates
simple interest at the rate of 12% per annum. If Ahmed wants to repay the loan and
interest after 3 years, calculate the sum paid by Ahmed to the bank at the end of the
period.
171
Solution:
S=a+I
I= a×i×n
I = 5000 × 0.12 × 3 = 1800
S=a+I
S = 5000 + 1800 = 6800
Another solution: The previous example can be resolved using the relationship:
S = a (1 + i × n)
S = 5000 (1 + 0.12 × 3) = 6800
Example: On 1/1/2019, Baari borrowed L.E. 6000 from the Bank of Industry and
Trade, he agreed to repay it and interest on 31/10/2019, if the bank calculates
interest at the rate of 12% per annum.
Find the interest and the amount.
Solution:
n = 10 months
,
a = 6000
,
i = 0.12
I= a×i×n
I = 6000 × 0.12 × (10\12) = 600
The the amount is:
S=a+I
S = 6000 + 600 = 6600
The amount can also be calculated as follows:
S = a (1 + i × n)
S = 6000 {1 + 0.12 × (10\12)} = 6600
171
The secand method: Repayment of the loan principal at the end of the period,
while the interest is paid in whole or in part in advance:
Under this method, the creditor and the debtor agree that the creditor will
calculate the interest payable on the loan and all or part of the interest on the loan,
and the debtor will receive the rest of the loan (the loan minus the interest or the
part agreed to discount).
In this way, the creditor has an opportunity to reinvest interest again, thus
increasing its investment rate as well as increasing the real interest rate on the
debtor. This may prompt the debtor to request an increase in the loan amount to
offset the interest Benefit from the amount received after deduction of interest.
Based on this method, the creditor calculates the interest first as follows:
- Interest I:
I=S×i×n
- Interest is deducted from the loan amount (or part of the interest under the
agreement)
Net receivable for the debtor = loan - interest
a=S-I
- We then calculate the real interest rate of the debtor
I
i = ‫ــــــــــــــــــ‬
a× n
Example: Fahmi Fahim borrowed an amount of L.E. 8000 from Banque du Caire,
which calculates a simple interest rate of 12% per annum, repaying the loan after
two years. If the agreement between Fahmi and the bank provides for repayment of
the principal at the end of the period and discount the interest advanced .
171
Required:
- Calculate the amount received by my understanding when borrowing.
- The real interest rate that has come to my understanding.
Solution:
1. We calculate interest:
I= a×i×n
I = 8000 × 0.12 × 2 = 1920
2. The Bank shall deduct the interest in advance when borrowing, and the net
amount received by the borrower shall represent the loan minus the interest,
provided that the debtor pays the loan in full.
Net amount receivable by the debtor (Fahmi) when borrowing:
a=S-I
= 8000 - 1920 = 6080
3. The debtor will pay the full amount of L.E. 8000 at the end of the period. This
means that the interest paid is L.E. 1920 for the loan amount of L.E. 6080.
Accordingly, the actual interest of the debtor is higher than the nominal interest
rate.
I=a×i×n
I
i = ‫ــــــــــــــــــ‬
a× n
1920
i = ‫ = ــــــــــــــــــــــــ‬15.8%
8000 × 2
177
We note that the interest rate of the debtor is higher than the rate at which the
bank deals.
Example: Bilal borrowed L.E. 5000 from a bank for 3 years at a simple interest
rate of 9% per annum, provided that two thirds of the interest is deducted in
advance and the balance is paid with the principle of loan at the end of the period.
Required:
1 - Calculation of the amount received by Bilal from the bank.
2- The amount that Bilal commits to repay at the end of the borrowing period.
3. If the bank invests its money at a simple interest rate of 10% per annum, find the
overall investment rate achieved by the bank from this process.
Solution:
1- Calculation of the interest due on the debtor (Bilal):
I= a×i×n
I = 5000 × 0.09 × 3 = 1350
2. The Bank shall deduct two thirds of the interest in advance.
What the bank deducts:
= 2\3 (I)
= 2\3 (1350) = 900
3 - Net receivable by the debtor of the loan:
= principle of loan - what was deducted from interest
= 5000 - 900 = 4100
4. The amount to be paid by the debtor at the end of the borrowing period:
= Loan principal + interest portion remaining
171
= 5000 + 450 = 4550
5. The bank invests its funds at the rate of 10% per annum and therefore the bank
invests the interest that was deducted from the debtor interest interest investment
which the bank deducted in advance.
I= a×i×n
I = 900 × 0.10 × 3 = 270
Thus, the Bank has achieved three benefits:
- Interest deducted in advance = 900
- Interest on investment interest deducted in advance = 270
- The rest of the interest received by the bank when paying the loan = 450
- Total interest earned by the Bank = 1620
Accordingly, the Bank's interest rate:
I
i = ‫ــــــــــــــــــ‬
a× n
1620
i = ‫ = ــــــــــــــــــــــــ‬13.2%
4100 × 3
The third method: Repayment of the loan at the end of borrowing while interest
is paid in the form of regular payments (annuities):
Under this method, the creditor and the debtor agree to repay the principal of the
loan at the end of the term of the loan while the interest on the loan is repaid
periodically at the end of each time period (annuities) and agreed upon between the
creditor and the debtor of the debtor may not pay some periodic interest on time
and deferred some to the date of repayment of the principal loan or to any date
171
subsequent to the date of the loan, in which case the creditor would impose delay
interest on the debtor in the case of partial or full repayment of periodic interest at
a rate higher than the interest rate used mainly in calculate of the periodic interest.
Since the creditor is investing the interest received from the debtor either on the
date of receipt or after a certain period.
Accordingly, the:
1. Calculation of the number of periodic interest payable by the debtor:
The number of periodic interest = period of the loan \ interest payment period
The number of periodic interest = N \ n
2. Calculation of periodic interest value:
= Loan x interest rate x interest payment period
In = a × i × n
3. Calculation of the number of delaying interest:
= Number of periodic interest - Number of interest paid
4. Calculation of the total delaying interest in:
S n = ( In × n ) + { In × i ×
1
12 ×
n( L + L ) }
2
2 1
5. If there is a delay period for the loan, the loan delay interest is calculated as
follows:
Loan delay interest = Loan principal x Delay rate x Delay period
I=a×i×n
111
6. Calculation of the total of the debtor's obligation at the end of the borrowing
period as follows:
The debtor's obligation = the principal of the loan + the sum of the delaying
interest + the interest of the loan delay.
7. If the bank invests the interest paid, the:
Total interest paid = Total interest paid + Total interest on investment
S n = ( In × n ) + { In × i ×
1
12 ×
n( M + M ) }
2
2 1
8. Total interest earned by the Bank is as follows:
= Total delaying interest + Total interest paid + Loan delay interest
9. General Investment Rate:
= Total interest earned by the bank / loan amount × Total borrowing period
Example: Amr borrowed L.E. 50000 from one of the banks to pay him after 5
years, to pay the interest periodically at the end of every 3 months at a simple rate
of 12% per annum, if Amr paid the first eight periodic interests and agreed with
the bank to pay the rest of the remaining periodic interests with the principle loan
at the end of the period at an interest rate of 15% per annum.
Required: Calculate what is required by Amr at the end of the borrowing period.
Solution:
1500 1500
3
1500
i = 15%
, Loan =50000
3
i = 12%
3
1- The number of periodic interest .
The number of periodic interest = N \ n
111
3
3
3
The number of periodic interest = ( 5 × 12) \ 3 = 20
2- Calculating the value of periodic interest.
= Loan x interest rate x interest payment period
In = a × i × n
In = 50000 × 0.12 × (3\12) = 1500
3 - Calculation of the number of delaying interest:
= Number of periodic interest - Number of interest paid
= 20 - 8 = 12
4. Calculation of the total delaying interest in:
= ( In × n ) + { In × i ×
1
12 ×
n( L + L ) }
2
2 1
= ( 1500 × 12 ) + { 1500 × 0,15 ×
1
12×
12
( 33 + 0 ) }
2
= 18000 + 3712.5 = 21712.5
whereas :
L1: Period for the first delayed interest.
L2: Period for the Late Delayed Interest.
5. Obligations of the debtor at the end of the period of the loan:
= original loan + total delaying interest
= 50,000 + 21712.5 = 71712.5
Example: Abu Amro borrowed L.E. 100,000 from the Bank of Credit. Abu Amr
agreed with the bank to repay the loan after two years. The interest to pay the
interest periodically at the end of every 2 months at an annual rate of 9%. If the
111
debtor (Abu Amro) pays the first four periodic interest at time, And to ask the bank
to deferred the repayment of the remaining periodic interest and the principle of
loan beyond the end of the loan term by six months, if the bank calcualate interest
delay at the rate of 12% per annum.
Calculate: What the obligations of the debtor at the end of the deferret period.
Solution:
1500 1500
2
1500
i = 15%
,
Loan =100000
2
i = 12%
2
2
2
2
6manths
1- The number of periodic interest .
The number of periodic interest = N \ n
The number of periodic interest = ( 2 × 12) \ 2 = 12
2- Calculating the value of periodic interest.
= Loan x interest rate x interest payment period
In = a × i × n
In = 100000 × 0.09 × (2\12) = 1500
3 - Calculation of the number of delaying interest:
= Number of periodic interest - Number of interest paid
= 12 - 4 = 8
4. Calculation of the total delaying interest in:
= ( In × n ) + { In × i ×
1 ×
12
= ( 1500 × 8 ) + { 1500 × 0,12 ×
n( L + L ) }
2
2 1
1
12×
111
8
(220 + 6 ) }
= 12000 + 1560 = 13560
whereas :
L1: Period for the first delayed interest.
L1 = 14
L2: Period for the Late Delayed Interest.
L2 = 0
1-
It is noted that the delay period (6 months) is added to both M1 , M2
5. Interest delay of the principle loan:
I =a×i×n
I = 100000 × 0.12 × (6\12) = 6000
6. Obligations of the debtor at the end of the period of the loan:
= original loan + total delaying interest + Interest delay of the principle loan
= 100,000 + 13560 + 6000 = 119560
Example: Assuming that the bank in the previous example invests the periodic
interest paid on time as soon as they receive a 10% interest rate.
Calculate :the overall rate of investment achieved by the bank.
Solution:
In order to find the overall rate of investment achieved by the bank from this
process, the previous steps of the solution are completed so as to calculate the total
interest paid and then calculate the total interest achieved by the bank from this
process and finally calculate the overall rate of investment as follows:
7. If the bank invests the interest paid, the:
Total interest paid = Total interest paid + Total interest on investment
= ( In × n ) + { In × i ×
1
12 ×
n
2( M1 + M2 ) }
111
1
12×
= ( 1500 × 4 ) + { 1500 × 0.10 ×
4( 28 + 22 ) }
2
= 6000 + 1250 = 7250
8. Total interest earned by the Bank from this process:
= Total delaying interest 13560 + Total interest paid 7250 + Loan delay interest
6000 = 26810
9. The overall investment rate achieved by the bank:
= Total interest earned by the bank / (loan × the entire period)
= 26810 \ ( 100000 × 2.5) = 10.72 %
In other words, the rate of public investment achieved by the Bank is 10.724%
per annum.
Example: Karim borrowed L.E. 50,000 from Al-Nahda Bank and agreed with the
bank to pay him two years and half,
and paying the interest on the loan
periodically every 3 months with an interest rate of 9% per annum. If the debtor
(Karim ) pays the first four periodic interest at time, And to ask the bank to
deferred the repayment of the remaining periodic interest and the principle of loan
beyond the end of the loan period by six months, to pay interest delay at the rate of
12% per annum. If the bank invests interest paid after repayment of a month at an
interest rate of 10% per annum.
Calculated: the overall rate of investment achieved by the bank.
Solution:
1125 1125
3
1125
i = 15%
, Loan = 50000
3
i = 9%
3
3
111
3 3
6manths
1- The number of periodic interest .
The number of periodic interest = N \ n
The number of periodic interest = ( 2.5 × 12) \ 3 = 10
2- Calculating the value of periodic interest.
= Loan x interest rate x interest payment period
In = a × i × n
In = 50000 × 0.09 × (3\12) = 1125
3 - Calculation of the number of delaying interest:
= Number of periodic interest - Number of interest paid
= 10 - 4 = 6
4. Calculation of the total delaying interest in:
= ( In × n ) + { In × i × 12 ×
1
n( L + L ) }
2
2 1
= ( 1125 × 6 ) + { 1125 × 0,12 ×
1×
12
6( 21 + 6 ) }
2
= 6750 + 911.25 = 7661.25
whereas :
L1: Period for the first delayed interest.
L1 = 15
L2: Period for the Late Delayed Interest.
L2 = 0
5- It is noted that the delay period (6 months) is added to both M1 , M2
6. Interest delay of the principle loan:
I =a×i×n
111
I = 50000 × 0.12 × (6\12) = 3000
7. Obligations of the debtor at the end of the period of the loan:
= original loan + total delaying interest + Interest delay of the principle loan
= 50,000 + 7661.25 + 3000 = 60661.25
follows:
8. If the bank invests the interest paid, after repayment of a month at an interest
rate of 10% per annum.
Total interest paid = Total interest paid + Total interest on investment.
= ( In × n ) + { In × i ×
1
12 ×
n(M +M )}
1
2
2
1
12×
= ( 1125 × 4 ) + { 1125 × 0.10 ×
4(32 + 22 ) }
2
= 4500 + 1031.25 = 5531.25
9. Total interest earned by the Bank from this process:
= Total delaying interest (7661.25) + Total interest paid on investment (5531.25)
+ Loan delay interest (3000) = 16192.5
9. The overall investment rate achieved by the bank:
= Total interest earned by the bank / (loan × the entire period)
= 16192. 5 \ ( 50000 × 3) = 10.795%
The second case: repayment of the loan in equal installments ( payments):
In this case, the loan is repayable in equal installments (payments), The
installments may include part of the principle and interest together or part of the
principle only, while interest is paid in advance when borrowing.
117
The first method: Repayment of the loan in equal installments of principal and
interest together:
In this way, the creditor and the debtor agree that will repay the loan principal and
its interest in the form of equal payments (regular) called (premuim) so that the
installment includes part of the loan principal and part of the interest.
Amount of the Loan = Amount of the Premiums
Whereas:
Loan amount = principal + interest
S=a+I
S = a ( 1+ i × n )
Or:
Total premiums: the sum of payments:
1
= ( p × n ) + { p × i × 12 ×
n (L +L )}
2
2 1
Example: Abu Karim bought a car price L.E. 75000 and agreed with the seller to
pay the price in installments equal to the principle and interest together every 3
months for a period of 3 years and at an interest rate of 12% per annum.
Calculate of the equal payments.
Solution:
p
i = 12 %
p
,
p
p
p
n = 12
Amount of the Loan = Amount of the Premiums
111
p
n
2 ( L1 + L2 )}
1
12×
a ( 1+ i × n ) = ( p × n ) + { p × i ×
75000( 1+ 0.12 × 3 ) =
1
12 ×
( p × 12 ) + { p × 0.12 ×
12( 33 + 0)}
2
Amount of the Loan = Amount of the Premiums
102000 = 12 p + 1.98 p
102000 = 13.98 p
P = 102000\ 13.98 = 7296.14
Example: Ahmed bought a car price L.E. 100000 and paid the advanced price of
L.E. 15000 in cash. Ahmed agreed with the seller to pay the rest of the price on 15
equal installments of the principle and interest together that the installment is paid
at the end of every two months, and at an interest rate of 12% per annum. Calculate
the value of the equal premuim.
Solution:
The price of the car = 100000
Cash advance = 15000
The rest of the purchase price of the car (loan)
= the price of the car - the amount of cash
= 100000 - 15000 = 85000
p
i = 12 %
p
,
p
p
p
n = 15
Amount of the Loan = Amount of the Premiums
111
p
a ( 1+ i × n ) = ( p × n ) + { p × i ×
1
12×
n
2( L1 + L2 )}
1×
12
15( 28 + 0)}
2
85000( 1+ 0.12 × 2.5 )
= ( p × 15 ) + { p × 0.12 ×
Amount of the Loan = Amount of the Premiums
110500 = 15 p + 2.1 p
110500 = 16.1 p
P = 110500 \ 16.1= 6462
The secand method: Repayment of the loan in equal installments from the
principle only with interest payment in advance:
In this way, the interest is calculated and deducted from the loan amount
when borrowing (the contract). The remaining debtor receives the obligation to
repay the principal of the loan in equal installments (from the principle only). In
this case, the bank invests the interest that was deducted from the loan value when
borrowing, Investment of installments paid by the debtor.
Example: Mennat-allah company for importing cars borrowed L.E. 200000 from
Bank Misr which calculates a simple interest rate of 10% per annum. If the
company agrees with the bank to pay the principal of the loan in the form of equal
installments every 3 months and for two years, If the bank invests its funds at the
same rate of borrowing (10%) annually.
Required: Find the overall investment rate achieved by the bank from this process.
Solution:
1- Calculating the value of interest.
= Loan x interest rate x period
I=a×i×n
111
I = 200000 × 0.10 × 2 = 40000
2- As interest is deducted from the loan when borrowing,
the company received = loan - interest
= 200000 - 40000 = 160000
While the company will repay the principal of the loan in equal installments every
three months:
3- The number of premuim .
The number of premium = N \ n
The number of periodic interest = ( 2 × 12) \ 3 = 8
4- Value of the equal amount of the principle loan:
The value of premium = Loan \ number of premuim
The value of premium = 200000 \ 8 = 25000
But the bank invests its money at the time of receipt of 10% per annum:
- Interest on the investment of interest deducted from the loan
I=a×i×n
I = 40000 × 0.10 × 2 =8000
- Interest paid loan installments:
1
n
= { p × i × 12 ×
2 ( L1 + L2)}
1
8( 21 + 0)} = 17500
= { 25000 × 0.12 × 12 ×
2
Thus, the Bank has achieved the following interest:
=
Interest
deducted
when
borrowing
(40000)
+
Interest
of
Interest
Deductible(8000) + Benefits of investment installments paid (17500) = 65500
111
Thus, the overall investment rate achieved by the Bank:
I=a×i×n
General Investment Rate:
= Total interest earned by the bank / loan amount × Total borrowing period
General Investment Rate:
= 65500 / (160000 × 2) = 20.5%
Example: Letha borrowed L.E. 60000 from the Arab Bank, which calculates a
simple interest at the rate of 9% per annum, provided that the interest is deducted
from the loan amount at the time of the contract and the debtor receives the
remainder. The loan contract stipulates repayment of the principal of the loan in
equal installments at the end of every two months, for two years .
Required: Calculate the general interest rate achieved by the bank if it invests its
funds at the same rate of the loan.
Solution:
1. Calculate the interest on the loan.
= Loan x interest rate x period
I=a×i×n
I = 60000 × 0.09 × 2 = 10800
2- As interest is deducted from the loan when borrowing,
the company received = loan - interest
= 60000 - 10800 = 49200
3. The principal of the loan shall be repaid in equal installments every two months
and for a period of two years.
111
The number of premium = N \ n
The number of periodic interest = ( 2 × 12) \ 2 = 12
4- Value of the equal amount of the principle loan:
The value of premium = Loan \ number of premuim
The value of premium = 260000 \ 12 = 5000
5- the bank invests its money at the time of receipt of 9 % per annum:
- Interest on the investment of interest deducted from the loan
I=a×i×n
I = 10800 × 0.09 × 2 = 1944
- Interest paid loan installments:
= {p× i×
1 ×
12
= { 5000 × 0.09 ×
n ( L1 + L2)}
2
1 ×
12
12( 22 + 0)} = 4950
2
Thus, the Bank has achieved the following interest:
=
Interest
deducted
when
borrowing
(10800)
+
Interest
of
Interest
Deductible(1944) + Benefits of investment installments paid (4950) = 617694
Thus, the overall investment rate achieved by the Bank:
I=a×i×n
General Investment Rate:
= Total interest earned by the bank / loan amount × Total borrowing period
General Investment Rate:
111
= 17694 / (49200 × 2) = 18 %
The third method: Repayment of the loan in equal installments from the
principle only with interest on the balance:
Under this method, the principal of the loan will be repaid in equal periodic
installments (equale annuities) over the period of the loan with interest on the
remaining balance. Therefore, the remaining balance of the loan on any date equals
the sum of the remaining installments or equal to the principal of the loan loan less
the equal installments paid as follows:
Loan balance from any date = total remaining installments at that date
or
Loan balance at any date = Loan - Total premiums paid up to that date
In calculating the interest on the remaining balance of the loan, the calculated
interest decreases steadily in form constaint, in order to reduce the remaining
balance of the loan by the amount of the equal amount paid. The decrease in
interest takes the form of a numerical sequence. The sum of the numerical
sequence (total interests) can be calculated as follows:
the total interest = number of interest \ 2 (first interest + last interest).
The solution is to:
1 - equal installment = loan / number of installments
P=L\n
2. Interest = Balance First year x Rate x Interest payment period
I=a×i×n
3. Outstanding installment = equal installment + interest
111
4. Balance of the year = Balance of the year - equal amount
5. The first balance of any year = the balance of the last previous year
This is repeated for each year (period of time) and then the table of consumption of
the loan as follows:
Loan amortization table
Years (number The balance at
of premiums)
Equal
Interest
The
The
the beginning
installment
(i)
premium
balance
of each year
( payment)
paid (p”)
at the
(a)
(p)
end of
each year
1
2
3
.
.
n
Example: Amr borrowed L.E. 5000 from Banque du Caire Amman, which
calculates interest at the rate of 10% per annum. Amr agreed with the bank to
repay the loan in equal installments of the principle and the number of five annual
installments with interest on the remaThe secand method: Repayment of the loan
in equal installments from the principle only with interest payment in advance:
In this way, the interest is calculated and deducted from the loan amount
when borrowing (the contract). The remaining debtor receives the obligation to
111
repay the principal of the loan in equal installments (from the principle only). In
this case, the bank invests the interest that was deducted from the loan value when
borrowing, Investment of installments paid by the debtor.
Example: Mennat-allah company for importing cars borrowed L.E. 200000 from
Bank Misr which calculates a simple interest rate of 10% per annum. If the
company agrees with the bank to pay the principal of the loan in the form of equal
installments every 3 months and for two years, If the bank invests its funds at the
same rate of borrowing (10%) annually.
Required: Find the overall investment rate achieved by the bank from this process.
Solution:
1- Calculating the value of interest.
= Loan x interest rate x period
I=a×i×n
I = 200000 × 0.10 × 2 = 40000
2- As interest is deducted from the loan when borrowing,
the company received = loan - interest
= 200000 - 40000 = 160000
While the company will repay the principal of the loan in equal installments every
three months:
3- The number of premuim .
The number of premium = N \ n
The number of periodic interest = ( 2 × 12) \ 3 = 8
4- Value of the equal amount of the principle loan:
111
The value of premium = Loan \ number of premuim
The value of premium = 200000 \ 8 = 25000
But the bank invests its money at the time of receipt of 10% per annum:
- Interest on the investment of interest deducted from the loan
I=a×i×n
I = 40000 × 0.10 × 2 =8000
- Interest paid loan installments:
1
n
2 ( L1 + L2)}
= { p × i × 12 ×
1
= { 25000 × 0.12 × 12 ×
8 ( 21 + 0)} = 17500
2
Thus, the Bank has achieved the following interest:
=
Interest
deducted
when
borrowing
(40000)
+
Interest
of
Interest
Deductible(8000) + Benefits of investment installments paid (17500) = 65500
Thus, the overall investment rate achieved by the Bank:
I=a×i×n
General Investment Rate:
= Total interest earned by the bank / loan amount × Total borrowing period
General Investment Rate:
= 65500 / (160000 × 2) = 20.5%
Example: Letha borrowed L.E. 60000 from the Arab Bank, which calculates a
simple interest at the rate of 9% per annum, provided that the interest is deducted
from the loan amount at the time of the contract and the debtor receives the
117
remainder. The loan contract stipulates repayment of the principal of the loan in
equal installments at the end of every two months, for two years .
Required: Calculate the general interest rate achieved by the bank if it invests its
funds at the same rate of the loan.
Solution:
1. Calculate the interest on the loan.
= Loan x interest rate x period
I=a×i×n
I = 60000 × 0.09 × 2 = 10800
2- As interest is deducted from the loan when borrowing,
the company received = loan - interest
= 60000 - 10800 = 49200
3. The principal of the loan shall be repaid in equal installments every two months
and for a period of two years.
The number of premium = N \ n
The number of periodic interest = ( 2 × 12) \ 2 = 12
4- Value of the equal amount of the principle loan:
The value of premium = Loan \ number of premuim
The value of premium = 260000 \ 12 = 5000
5- the bank invests its money at the time of receipt of 9 % per annum:
- Interest on the investment of interest deducted from the loan
I=a×i×n
111
I = 10800 × 0.09 × 2 = 1944
- Interest paid loan installments:
1
n ( L1 + L2)}
2
= { p × i × 12 ×
= { 5000 × 0.09 ×
1 ×
12
12( 22 + 0)} = 4950
2
Thus, the Bank has achieved the following interest:
=
Interest
deducted
when
borrowing
(10800)
+
Interest
of
Interest
Deductible(1944) + Benefits of investment installments paid (4950) = 617694
Thus, the overall investment rate achieved by the Bank:
I=a×i×n
General Investment Rate:
= Total interest earned by the bank / loan amount × Total borrowing period
General Investment Rate:
= 17694 / (49200 × 2) = 18 %
The third method: Repayment of the loan in equal installments from the
principle only with interest on the balance:
Under this method, the principal of the loan will be repaid in equal periodic
installments (equale annuities) over the period of the loan with interest on the
remaining balance. Therefore, the remaining balance of the loan on any date equals
the sum of the remaining installments or equal to the principal of the loan loan less
the equal installments paid as follows:
111
Loan balance from any date = total remaining installments at that date
or
Loan balance at any date = Loan - Total premiums paid up to that date
In calculating the interest on the remaining balance of the loan, the calculated
interest decreases steadily in form constaint, in order to reduce the remaining
balance of the loan by the amount of the equal amount paid. The decrease in
interest takes the form of a numerical sequence. The sum of the numerical
sequence (total interests) can be calculated as follows:
the total interest = number of interest \ 2 (first interest + last interest).
The solution is to:
1 - equal installment = loan / number of installments
P=L\n
2. Interest = Balance First year x Rate x Interest payment period
I=a×i×n
3. Outstanding installment = equal installment + interest
4. Balance of the year = Balance of the year - equal amount
5. The first balance of any year = the balance of the last previous year
This is repeated for each year (period of time) and then the table of consumption of
the loan as follows:
111
Loan amortization table
Years (number The balance at
of premiums)
Equal
Interest
The
The
the beginning
installment
(i)
premium
balance
of each year
( payment)
paid (p”)
at the
(a)
(p)
end of
each year
1
2
3
.
.
n
Example: Amr borrowed L.E. 5000 from Banque du Caire Amman, which
calculates interest at the rate of 10% per annum. Amr agreed with the bank to
repay the loan in equal installments of the principle and the number of five annual
annual installments with interest on the remaining balance.
Required:
1. The premium equal installments.
2 - the loan amortization table.
3 - Calculate the total interests.
Solution:
Equal installment (payment) (p)= Loan / Number of Installments
P=L\n
111
P = 5000 \ 5 = 1000
First year:
- Interest for the first year (I1)
= Balance at the first year (loan) × Rate x 1
I1 = a × i × n
I1 =5000 × 0.10 × 1 = 500
- Premium paid for the first year = equal installment + first year interest :
P”1 = p1 + I
P”1 = 1000 + 500 = 1500
- Balance at the end of first year = balance at the beginning of First year Equivalent premium
= 5000-1000 = 4000
Second Year:
- Second Year Interest (I2)
= Balance at the secand year × Rate x 1
I1 =4000 × 0.10 × 1 = 450
- Premium paid for second year p2:
P”1 = p1 + I
P”1 = 1000 + 400 = 1400
- Balance at the end of the second year = Balance at the beginning of second year
– Equivalent:
= 4000-1000 = 3000
111
This is done for all remaining years (third year, fourth year, fifth year). This is
illustrated by the depriction of the loan amortization table.
Loan amortization table
Years
The
Equal
Interest
The
The
(number of
balance
installment (
(i)
premium
balance
premiums)
at the
payment)
paid (p”)
at the
beginning
(p)
end of
of each
each year
year (a)
1
5000
1000
500
1500
4000
2
4000
1000
400
1400
3000
3
3000
1000
300
1300
2000
4
2000
1000
200
1200
1000
5
1000
1000
100
1100
-
- Total interest = interest for the first year + interest for the second year + .... +
interest for the fifth year.
= 500 + 400 + 300 + 200 + 100 = 1500
Or the sum of the benefits can be found by representing the number sequence of
the first (I1), only the last (I5), and its number (n):
= n \ 2 ( I1 + I5)
= 5 \ 2 ( 500 + 100) = 1500
For example: Karim borrowed L.E. 200,000 from Banque Misr to set up a poultry
breeding project. Karim agreed with the bank to repay the loan in equal annual
111
installments of only 8 installments with interest on the remaining balance at an
interest rate of 8% per annum.
Required:
1 - depiction of the loan amortization table.
2 - Calculation of the total interest borne by Karim.
Solution:
Equal installment (payment) (p)= Loan / Number of Installments
P=L\n
P = 200000 \ 8 = 25000
First year:
- Interest for the first year (I1)
= Balance at the first year (loan) × Rate x 1
I1 = a × i × n
I1 =200000 × 0.08 × 1 = 16000
- Premium paid for the first year = equal installment + first year interest :
P”1 = p1 + I
P”1 = 25000 + 16000 = 41000
- Balance at the end of first year = balance at the beginning of First year Equivalent premium
= 200000 - 25000 = 175000
Thus for the rest of the years as follows:
111
Loan amortization table
Years
The balance at
Equal
Interest
The
The balance at
(number of
the beginning
installment (
(i)
premium
the end of
premiums)
of each year
payment) (p)
paid (p”)
each year
(a)
1
200000
25000
16000
41000
175000
2
175000
25000
14000
39000
150000
3
150000
25000
12000
37000
125000
4
125000
25000
10000
35000
100000
5
100000
25000
8000
33000
75000
6
75000
25000
6000
31000
50000
7
50000
25000
4000
29000
25000
8
25000
25000
2000
27000
-
Total interests borne by Karim:
= I1 + I2 + ………………..+ I8 =
= 16000 + 14000 + 12000 + ……………+ 2000 = 272000
Or
Total interest = n \ 2 ( I1 + I5)
= 8 \ 2 ( 16000 + 2000) = 27200
111
Part Two
207
208
Introduction
Mathematics of finance offers mathematical tools
for financing and investment Operations through
Simple Interest Theory and compound interest Theory.
Where Simple Interest Theory is used by non –profit
financial
Institutions,
such
as
social
security,
Development and Agricultural Credit Bank, and in
general for Short–Term Loans. While compound
interest is used in other commercial Financial
Institutions such as Commercial Banks and Insurance
Companies.
The
Authors
deal
with
these
Tools
in
Mathematics of Finance and its various applications.
Including the sum of One and several amounts, Present
Value and discount for One and Several Amounts,
Settlement of Debts, Sum and Present Value of
Annuities, Periodic Benefits, Methods of Settlement of
Loans and Short –Term Discount of Commercial
Papers.
It is worth mentioning that these tools are of
interest to all interested in finance and investment
209
practices and their applications in various aspects of
life. Such as Depositing, Borrowing, and Buying or
Selling on Time. Especially after the spread of
installment systems in commercial transactions, banks
and other financial institutions at fixed interest rates
has been sanctioned by some Muslim Scholars,
including Dr.Mohamed Sayed Tantawi the Grand
sheikh of Al- Azhar and the former Mufti of the Arab
Republic of Egypt (May Allah have mercy him), as
mentioned in Jordanian (El Rai.) newspaper issued on
Monday December 16,1998 in response to a question:
“ where does the Grand Sheikh of Al – Azhar invest
his money:” he said: there is no difference between an
Islamic bank or a Non Islamic bank “ and said there no
anything to prevent me from depositing it in Banque
Misr, National Bank, Alexandria Bank or elsewhere,
because I know that these Banks are an Egyptian
Banks and do not invest money except in the case of
God Almighty, and if it did otherwise, it is responsible
to God and I am not responsible. When he had been
asked Why he does not invest his money in a project,
210
rather than deposing it in a commercial bank, he
replied “If I run an investment project I think it will fail
because it needs experience and this experience is not
available to me that I am a scholar not an investor or a
businessman.
The newspaper also mentioned that he demanded
at a seminar held at the faculty of commerce –Cairo
University not to follow the claims of some extremists
for not dealing with Banks. Since the bank benefits are
not irrigation as that defined by Islam and the work in
banks leads with honor and trust.
The Authors refer to many models and examples
for the various applications on simple and compound
interest as well as a series of exercises not solved at the
end of each chapter to allow the student the
opportunity to practice and training.
Finally, we ask God that we have succeeded in
providing financial mathematics in a convenient
manner that serves our students in our Egypt ion and
Arab Universities and all those interested in finance
and investment.
211
212
The Compound Interest
213
214
Chapter (1 )
Compound Interest
At Annual Rate.
215
216
Chapter (1 )
Compound Interest
At Annual Rate.
It has already been shown in the study of simple
interest that Interest is the return of capital whether
it is invested or borrowed. Also the interest is usually
paid periodically to the Owner of the capital, such as
the returns of other factors of production not to
accumulate on the capital and lose its original nature
and turn to the same nature of capital.
The sum of money invested or borrowed in a
transaction is called The principal.
Therefore, the principal in simple interest does
not change during periods of investment and remains
fixed until the end of period is paid to the owner of the
capital plus the total interest which does not change
from time to time due to the stability of the capital.
While the investment of money in compound
interest over a given period means that the interest
payable on that amount at a certain rate is calculated at
217
the end of each period of investment, then add to the
principal in order to get a new asset for the next period.
The Compound amount at the end of each period
shall be the principal invested at the start of period plus
the interest of that period. This sum represents the
amount invested in the following period.
The Difference between Simple and Compound
Interest
Based on the above the similarities and
differences between simple and compound interest can
be summarized as follow:
1-
Both represent a return of capital.
2-
The simple interest is equal to the compound
interest at the end of the first period for the same
principal and the same rate.
3-
The difference between them is that the simple
interest remains constant from period to period
as a result of the stability of the capital,
While compound interest increases from time to
time as a result of increasing capital by adding interest
to it at the end of each period.
218
EIGHTH WONDER OF THE WORLD
When Baron Rothschild, one of the World’s
Wealthiest Bankers, was asked if he could name the
seven Wonders of the World, he said: “ No, but I can
tell you what The Eighth Wonder is. This wonder
should be utilized by all of us to accomplish what we
want, It is “ Compound Interest “
Table (1) shows the impressive growth that results
When Compound interest is applied Compared to
Simple Interest:
Growth of $ 100 at Compound & Simple interest
Time Simple Interest
in
6%
8%
years
10
160
180
20
220
260
30
280
340
40
340
420
50
400
500
We can see
Compound Interest
10%
6%
8%
10%
200
$ 179,o8
$215,89 $259,37
300
320,71
466,10
672,75
400
574,35
1.006,27 1.744,94
500
1.028,57 2.172,45 4.525,93
600
1.842,02 4.690,16 11.739,09
from Table (1) the amount in
Compound interest for 8% is not halfway between the
corresponding values at 6% and 10%, On the other
219
hand, with Simple interest the values for 8% will be
halfway between 6% and 10% values.
The Concept is basically simple, as the
following example shows, Compound interest is simple
interest applied over and over to a Sum that is
increased by the simple interest each Time it is earned.
An investment of $ 1000 at 8% Simple interest
earns $ 80 per year. In 3 years the simple interest
would amount to $ 240.
However, if the interest as it is earned is added
to the principal and this new principal draws interest,
the investment increases more rapidly than it would
with simple interest.
Interest paid on an increasing principal in this
way is known as “ Compound Interest “. The
following example shows the increase in investment of
$ 1000 if the interest rate is 8% compounded annually:
Original Principal
Interest for first year at 8%
Principal at the start of second year
Interest for second year at 8%
Principal at the start of the third year
Interest for third year at 8%
220
1000
80
1080
86, 4
1166,4
93, 312
Amount at the end of 3 years
1259, 712
Thus the compound interest earned on the
original investment is $ 259,712 as compared to $ 240
at simple interest in the same length of time. The
difference of $ 19,712 is interest earned on interest.
The Total $ 259,712 is called The Compound
amount.
Elements of Compound Interest (I):
The value of compound interest is determined as
simple interest with three elements:
1-Principal:
Which means the amount in the first term. since
the compound interest is affected other than the simple
interest, by increasing the value of the asset over a
period of time. The value of asset as an element of
determining compound interest is not a fixed element,
but a variable in itself as it changes with the compound
interest.
Therefore, the original amount at the start of
initial period is indicated by the symbol (a), the amount
at the start of the second period is indicated by (a1), and
221
at the start of the third by (a2) ……at the start of the
last period by (a n-1).
2-Time (Interest Period):
It means the number of periods at the end of
which the principal of amount is paid with its interest,
it symbolized by (n).
3-Rate of compound Interest:
It means the return of the unit money at the end
of each period, it symbolized by (r).
Equation of Compound Interest:
To conclude the compound interest equation
assume that we have an original principal its value (a)
invested by a compound interest rate (r) for (n) periods.
Using the concept of compound interest we find
that the asset is increased at the end of each period by
the value of interest due to that period as follow:
Interest payable on the principal (a) at the end of
first period (I1) since:
I1 = a × r × 1 = a × r.
The compound amount of (a) at the end 0f first period
= a + I = a + a × r = a (1 + r) = a1
222
(Which represents the asset invested at the
start of the second period)
Interest payable on the amount (a1) at the end of
second period (I2) since I2 = a1 × r = a (1 + r) × r.
The compound amount of (a1) at the end 0f
second period
= a1 + I2 = a1 + a1 × r = a1 (1 + r) = a (1 + r) (1 + r) =
a (1 + r)2 = a2.
(Which represents the asset invested at the
start of the third period).
Interest payable on the amount (a2) at the end of
third period (I3) since
I3 = a2 × r = a (1 + r)2 × r.
The compound amount of (a2) at the end 0f third
period
= a2 + I3 = a2 + a2 × r = a2 (1 + r) = a (1+ r)2 (1 + r)
= a (1 + r)3 .
And so we can conclude The compound amount
at the end of fourth period = a (1 + r)4.
… and at the end of fifth period = a (1 + r)5.
Conclusion:
223
The Compound amount of (a) invested by a
compound interest rate
(r) for (n) periods = S ……… since
S = a (1 + r)n . …………………(1)
Since:
a = The principal.
r = The rate per conversion period.
n = The number of conversion periods.
S = The amount at Compound interest.
Note that:
“ This Law is true for all Correct and fraction
values of r ; n “.
The factor (1 + r)n is called the accumul-ation
factor, or amount of 1.
Or The compound amount of unit money
invested by a compound interest rate (r) for
(n) periods.
Accordingly, the Compound Interest will be:
The Compound amount minus the principal:
I = S – a = a (1 + r)n _ a.
I = a [ (1 + r)n _ 1 ] ………………… (2) .
224
Example (1):
Compare simple interest to compound interest
for $ 1000 at 10% annually in the end of 5 years.
Solution:
Simple interest:
I=a×r×n
= 1000 × 0.10 × 5 = $ 500.
Compound interest:
I = a [ (1 + r)n _ 1 ]
I = 1000 [ (1 + 0.10)5 _ 1 ]
= 1000 [ 1.61051 - 1 ] = $ 610.51 .
Methods of calculating accumulation factor (1 + r):
1-
Theoretical or Traditional methods
2-
The Practical method (by using financial Tables)
3-
By using Calculators.
Compound
Amount
Traditional methods:
1-Ordinary Multiplication:
225
by
Theoretical
or
We can calculate the value of the “Amount of 1”
(1 + r)n from multiple it in itself number of (n) times …
So that
(1 + r)n = (1+ r) (1 + r) (1 + r)……….n times.
Note that this method can only be used if
(n) a correct and small number.
Example (2):
Find the compound amount of $ 1000 at 8%
annually for 3 years.
Solution:
S = a (1 + r)n
= 1000 (1 + 0.08)3
= 1000 (1.08)(1.08)(1.08).
= 1000 (1.259712) = $ 1259.712.
2-Binomial theory:
The general formula of Binomial theory as
follow:
(X + Y)n = n cr Xr y n-r .
(X + Y)n = n c0 X0 Y n +
Yn-2 +…… +
n
cn-1 Xn-1 Y +
n
226
n
c1 X1 Yn-1 +
cn Xn Y0.
n
c2 X0
We can find the value of factor, (1 + r)n by using
Bin0mial Theory … Since
(1 + r)n = 1 + n × r + n (n-1) / (2×1) × r2 + n (n-1) (n2) / (3×2×1) × r3 + ………
Note:
This formula is correct for all values of
(r) if (n) is a positive and correct number, but if (n) is
incorrect number, this formula will be incorrect, only if
(r) less than (1), in compound interest we need to
Calculate
(1 + r)n …since (r) practically less than
correct 1 .
So that, this formula is correct for all values of
(r) either correct or fraction numbers.
Example (3):
Find the compound amount in example (1) by
using Bin0mial Theory
Solution:
S = 1000 (1 + 0.08)3
= 1000 [ 1 + 3 × 0.08 + 3 × 2 / 2 (0.08)2
+3 × 2 ×1 /6 (0.08)3 ] .
S = 1000 × (1.259712) = $ 1259.712
227
Logarithmic Method:
Logarithms can be used when compound interest
Tables are not available or when the value of n ; r is
not found in the table
a = 10000,
r = 0.o63, n = 5 + 3/12 = 5.25
Example (4):
Find the compound amount of $ 10000 at 6.3%
annually for 5 years and 3 months.
Solution:
S = a (1 + r)n
S = 10000 (1 + 0.063)5.25
We take the Logarithm of both sides:
Log S = Log 10000 + 5.25 × Log (1.063)
= 4 + 5.25 × .02653326 = 4 +. 1393 =
4.1393.
Looking up the antilog of 4.1393:
We find that S = $ 13781.611.
The Practical method to find the different
applications
of
compound
(by using financial Tables):
228
interest
There are Tables of Finance were prepared for
this purpose to all practical values of r, n .
Note that the values in these tables to a unit of money
for
r (from
¼%
to 12%)
and
for
n
(from 1 to 50).
0.015
0.03
0.045
0.075
0.105
0.0125
0.0275
0.0425
0.07
0.1
0.0075
0. 005
0.0025
0.0225
0.02
0.0175
0.375
0.035
0.0325
0.06
0.055
0.05
0.09
0.085
0.08
0.12
0.115
0.11
Here are a few examples explain how these
tables are used to find the various applications of
compound interest which include 5 tables:
Table (1):
This table shows the Amount of 1 invested by a
compound interest = (1 + r)n
Example (5):
Find the compound amount of $ 1000 at 6%
annually for 30 years.
Solution:
We search in column (1) in front of 30 periods
under the rate 6%, will find the value 5.743491.
229
So
that
the
compound
amount
=
1000
(5.743491) = $ 5743.491 .
Appendix of Table (1):
This table shows The Amount of 1 for periods
from one Month to eleven Months and for one day.
Example (6):
Find the compound amount of $ 3000 at 9%
annually for 8 months.
Solution:
By searching in
appendix table
(1) for the
compound amount of unit money (1 + r)n
in front of
9% under 8 months (8/12) Will find it = 1.059134 .
So that the compound amount will be:
S = 3000 (1.059134) = $ 3177.4
Table (2):
This table shows the Present Value of a unit
money invested by compound interest, Which symbol
Vn = 1/ (1 + r)n = (1+ r)-n
Example (7):
Find the Present Value of $ 19348.422 due after
20 years if money is worth 7% annually.
230
Solution:
We search in table (2) under 7% in front of 20
periods, we find V20 7% = 0.25419
a = 19348.422 × (0.25419) = $ 5000
Present value (a) = S × vnr% = S ÷ (1 + r)n
a = S × (1 + r)-n ………….………..(3)
Table (3):
This table shows the Amount of Ordinary
Annuity of a unit money invested by compound
interest
n
[ (1 + r)n - 1 ] ÷ r .
Example (8):
A person deposits $ 500 yearly (at the end of all
year) in an account paying 8.5% compound annually
in Misr Bank.
What amount has accumulated at the end of 10
years?
Solution:
a = 500, r = 8.5%, n = 10
Solution:
231
S1
8.5%
= 14.835098 …... (from table 3)
Amount of Annuity = 500 (14.835098) = $
7417.6.
Table (4):
This table shows the present value of an
Ordinary Annuity for a unit money invested with
compound interest:
an
r%
= [ 1 - (1 + r)-n ] ÷ r
= [ 1 - Vn ] ÷ r.
Example (9):
Find the present value of an ordinary annuity of
$ 500 for 10 years at 8.5% annually.
Solution:
a 1
8.5%
= 6.561348…(from table 4).
The present value = 500 (6.561348) = $ 3280.67.
Table (5):
This table shows the equal settlement (from
principal & interests)
Of a debt its value a unit money invested by
compound interest at (r) rate and (n) periods.
= 1/ a n
1 [(1- Vn) ÷ r ]
232
= [ r / (1 - Vn ] .
Example (10):
Someone borrowed $ 10000, and agreed with the
creditor to repay the Loan by equal settlement from
principal and interests at the end of all year for 15
years at 10% annually, Find the equal yearly Premium?
Solution:
a = 10000,
r = 10%, n = 15
The Premium = 10000 (0.131474) = $ 1314.74.
(from table 5).
The Practical method to find the compound interest
(by using financial Tables):
Using financial Tables to find compound amount
if (r) or (n) or both are not found in tables:
If (r) is found and (n) is not found:
Periods (n) is not found in the following cases:
a-
Periods (n) is correct number but over than
table scope:
In this case, (n) will divided to correct numbers
each of them take a symbol such as x, y, z … since a
233
value of each less than table scope (50 periods) …
since
S = a (1 + r)n = a (1 + r)x (1 + r)y (1 + r)z
Since…. n = X + Y + Z
Example (11):
If Mothamd Omar deposit 5000 in Misr Bank
at 9% annually for 70 years, Find the compound
amount.
Solution:
S
= a (1 + r)n
= 5000 (1 + 0.09)70
= 5000 (1 + 0.09)35(1 + 0.09)35
= 5000 (20.413968)(20.413968)= 5000 (416.73)
= $ 2083650.
b-
Periods (n) is no correct number either or less
or over than a
In this case, will divided (n) to a correct number
and a fraction.
Example (12):
234
If Wessam Omar deposit $ 3000 in Misr Bank
at 7.25% annually for 80 years and 8 months, Find the
compound amount.
Solution:
S
= a (1 + r)n
a
= 3000, r = 0.0725, n = 80 + 8/12
= 3000 (1 + 0.0725)80+8/12
= 3000 (1 + 0.0725)40(1 + 0.0725)40 (1 +
0.0725) 8/12
(1 + 0.0725)40 = 16.43963 ……(from table 1)
(1 + 0.0725)8/12 = 1.047767....(from appendix table 1).
S = 3000 × 16.43963 × 16.43963 × 1.047767
= 3000 × (283.171) = $ 849513.04 ..
C-
If (r) and (n) are not found in Tables of
finance
Example (13):
If Amr deposited $ 10000 in a bank at 7.3%
annually for 60 years, Find the compound amount.
Solution:
S = 10000 (1 + 0.73)60
235
We can divided the solution to three stages:
1-
Will find the compound amount for 60 years at
7%..(table 1)
2-
Will find the compound amount for 60 years at
7.5%.(table 1
3-
Will find the compound amount for 60 years at
7.3%..(by Proportion & proportionality).
(1 + 0.07)60 = (1 + 0.07)30 × (1 + 0.07)30
= 7.612255 × 7.612255 = 57.946427 …(1)
(1 + 0.075)60 = (1 + 0.075)30 × (1 + 0.075)30
= 8.754955 × 8.754955 = 76.64924 … (2)
By subtracting (1) from (2) = 18.702813, this
difference in the compound amount equal Difference
in the rate its value 0.5%
So that:
The difference in the compound amount which
equal difference in the rate its value 0.3 = 18.702813 ×
0.3 /0.5 = 11.221688.
S = 10000 (1 + 0.73)60
= 10000 (57.946427 + 11.221688)
= 10000 × 69.168115 = $ 691681.15
236
By using Calculator:
We can find the factor (1 + r)n at any value of (r)
and for any number of (n) by using Calculator, which
consider the easies and fastest method from all
previous methods, that any student has one of it. from
using exponent keys
(Xy) or (YX) in other calculators’ as the following
example:
Example (14):
Find the value of (1.075)25.25?
Solution:
1-
Write the basis (1.075)
2-
Clicking the conversions key (which is found in
Top at any calculator on left side, and is written
under it,Shift / [
Shift
3-
], or Inv / [
,
Inv
].
,
2ndf
Clicking the key which is written under it (Xy)/
[
] or (YX)/ [
]
4-
Write the exponent (25.25).
5-
Clicking the key which is written 0n it [=]
237
6-
We find the value = 6.2096014.
Note: If (Xy)or (YX) is written directly upon the key
such as [ Xy ] or.[ YX ], we do not need to click the key
of conversion ”.
Example (15):
Find the compound amount and the compound
interest if $ 10,000 is invested for 10 years at 7%.
Solution:
S = a (1 + r)n
a = 10,000, r = 0.07, n = 10
S
= 10,000 (1 + 0.07)10
S
= 10,000 (1.967151) = $ 19671.51.
I
= 19671 – 10,000 = $ 9671.51.
S0 the compound amount = $ 19671.51 and the
compound interest = $ 9671.51.
Example (16):
The day a boy was born, his father invested $
2000 at 10% compounded annually. Find the value of
the fund on the boy’s 18 th birthday.
Solution:
a= 2000, r = 0.10, n = 18
238
S = a (1 + r)n
S = 2000 (1 + 0.1)18 = 2000 (5.559917) = $ 11119.83.
Example (17):
A town increased in population at 2% a year
from 2000 to 2010.
If the population was 180,000 in 2010, what is
the estimated population to the nearest hundred for
2020, assuming that the rate of growth remains the
same?
Solution:
S = a (1 + r)n
a
= 180,000, r = 0.02, n = 10
S
= 180,000 (1 + 0.02)10 = 180,000 (1.218994) =
219,400.
The estimated population for 2020 = 219,900.
239
Exercises
1-
If Islam deposed $ 5000 in a bank at 10%
annually, compound interest, find the compound
amount for 5 years, by using:
Traditional methods:
a- Ordinary Multiplication.
b- Logarithmic method.
c- Binomial theory.
2-
If Hosam borrowed $ 3000 from a bank, for 4
years and 3 months at 8% annually, find the
compound amount by using:
a- One of the Traditional methods ($ 4160.756).
d- Tables of Finance.
3-
If Amr deposited $ 10,000 in a bank to invest it
with compound interest at 8.2% annually, Find
the compound amount for:
a- 4 years.
b- 5 years and 2 months.
c- 3 years, 6 months and 20 days.
4-
The day a boy was born, his father invested $
2000 at 7.5% compounded annually. Find the
240
value of the fund on the boy’s 21 birthday.($
9132.88)
5-
On a girl’s 8th birthday, her parents place $ 1000
in her name in an investment paying 8%
compounded annually. How much will she have
to her credit on her 18 st compounded annually
birthday?
6-
If Dr. Hosni deposited $10000 for his son who is
old now 5 years, 7 months and 6 days, in a bank
at 7.2% annually. Find the compound amount
when his son will be has 22 years.
7-
If Mohamed Omar invested now $ 8000 in Cairo
Bank, and after 5 years withdraw the compound
amount and deposited it in National Bank, find
the compound amount in last bank after 3 years,
9 months and 18 days
if you knew: ($
14615.322)
- Cairo Bank used rate 6.4% annually.
- National Bank used 8% annually.
8-
A town increased in population at 2% a year
from 2000 to 2010. If the population was 2
241
Millions in 2010, what is the estimated
population to the nearest hundred for 2020,
assuming that the rate of growth remains the
same?
242
Chapter (2 )
Compound Interest
At Non Annual Rate
243
244
Chapter (2 )
Compound Interest
At Non Annual Rate
In the preceding chapter the compound interest
was computed and add to principal every year. In many
business
transactions,
the
interest
is
computed
annually, semiannually, quarterly, monthly. daily, or at
other time interval.
The
time
between
successive
interest
computations is called the conversion, or interest,
period. This basic unit is used in all compound interest
problems. The important rate is the interest rate per
conversion period.
In that cases we can conclude the formula of
compound amount as follow:
a:
The principal invested or borrowed.
r:
Interest rate per conversion period.
n:
Total number of conversion periods.
S = (1 + r)n
245
Which the same formula of compound amount at
annual rate, but at Interest rate per conversion period,
and number of conversion periods.
Example (1):
Find the compound amount for $ 10000 at 1%
monthly for 5 years.
Solution:
A = 10000, r = 1% monthly, n = 5 × 12 = 60 periods.
Conversion Periods per year, since n = time in years ×
number of periods in a year.
S = 10000 (1 + 0.01)60
By searching in table (1) in front of 60 periods
under rate 1%
Will find (1 + 0.01)60 = 1.816697 ……….so
S = 10000 (1.816697) = $ 18166.97
I = 18166.97 – 10000 = $ 8166.97.
Example (2):
Find the compound amount for $ 10000 at 12%
yearly for 5 years.
Solution:
A = 10000, r = 12% yearly, n = 5 years
246
S = 10000 (1 + 0.12)5
= 10000 (1.762342) = $
17623.42
Note: (1 + 0.12)5 ≠ (1 + 0.01)60, ……
(1 + 0.01)60 > (1 + 0.12)5
Since, in (1 + 0.12)5 the interest will add to
principal once in a year, but in (1 + 0.01)60 the interest
will add 12 times in a year.
Example (3):
Find the compound interest of $ 20000 at 4%
semiannually for 10 years.
Solution:
a = 20000,
r = 0.04 semiannually, n = 10 × 2 = 20
periods. S = (1 + 0.04)20.
By searching in table (1) in front of 20, under r =
0.04 we find (1 + 0.04)20 = 2.191123.
Compound amount (S) = 20000 (2.191123) = $
43822.46.
Compound interest (I) = 43822.46 – 20000 = $
23822.46.
Example (4):
247
Calculate the compound interest in previous
example time of investment will be 10 years and 3
months.?
Solution:
a = 20000, r = 0.04, n = (10 + 3/12) × 2 = 20 + 6/12
periods.
S
= 20000 (1 + 0.04)20 + 6/12
= 20000 (1 + 0.04)20 (1 + 0.04)6/12
= 20000 (2.191123)(1.019804) =$ 44690.32.
I = 44690.32 – 20000 = $ 24690.32.
Example (5):
Calculate the compound interest for $ 10000
invested for 10 years and 3 months at 2.2% quarterly?
Solution:
a = 10000, r = 0.022, n = (10 + 3/12) × 4 = 41
periods.
S = 10000 (1 + 0.022)41
We can find the value of (1 + 0.022)41 by
proportion and proportionality as follow:
(1 + 0.025)41 = 2.7521904 …………………(1)
(1 + 0.02)41 = 2.2522005 ………………….(2)
248
= 0.49999 ……………..by subtracting
(2) from (1)
0.49999 = Deference in compound amount
Equivalent Deference in the rate its value 0. 5%
So: The Deference in compound amount Equivalent
Deference in the rate its value 0. 2% = 0.49999 ×
0.2/0.5 = 0.1999996
(1 + 0.022)41 = 2.2522005 + 0.1999996 = 2.4522001
S = (1 + 0.022)41 = 10000 (2.4522001) = $ 24522.
I = 24522 – 10000 = $ 14522 .
Note: Sometimes the interest instead of add more
than once throw a year, it can be add once for more
than one year.
In this case, the time will be converted to periods
more than a year to consist with the rate.
Example (6):
Hesham borrowed $ 5000 from Misr Bank for
10 years, and agreed with bank to calculate the interest
at 12% every two years, find the compound amount.
249
Solution:
a = 5000, r = 12% every two years, n = 10 ÷ 2 = 5
periods.
S
= 5000 (1 + 0.12)5
= 5000 (1.762342) = $ 8811.71.
Effective rate Of Interest & Nominal rate Of
Interest:
Effective rate Of Interest: which its period equivalent
the period of addition the interest to the principal .
In this case, the Rate still as it, only the Time
must convert to periods equivalent to the period of the
rate.
Nominal rate Of Interest: which its period deference
from the period of addition the interest to the principal.
In this case, the Rate must convert to Effective
rate and the Time must convert to periods equivalent to
the period of the rate.
Relation
between
Effective
annual
rate
and
Nominal annual rate:
We assume that: Nominal annual rate takes symbol
(rm), Effective annual rate (r), since (m) = number
250
times of addition the interest per year. If the nominal
annual rate is 12% converted annually, the effective
rate will also be 12%. But if the nominal rate is 12%
converted semiannually, the amount of $ 1 at the end
of one year will be (1.06)2 = $ 1.1236. This calculation
is simply the accumulation factor for a rate per period
of 6% and two periods. The interest on $ 1 for 1 year is
then $ 1.1236 – 1.0000 = $ 0.1236.This result is
equivalent to an annual rate of 12.36%. Thus 12.36%
converted annually would result in the amount of
interest as 12% converted semiannually. The computations used to get the effective rate of 12.36% in this
case can be summarized as follow:
1.1236 = (1.06)2, r = 1.1234 - 1 = 12.36.%
To obtain an equation to find the effective rate in
general, we assume that the effective rate produces the
same amount (S) from the given principal (a) in 1 year
as the compound interest. We then have,
The rate of one period = rm/m
The interest of first period = rm/m
251
The amount of $ 1 at the end of first period will be (1+
rm/m).
The interest of second period = (1+ rm/m) × rm/m .
The amount of $ 1 at the end of second period will be:
(1+ rm/m) + (1+ rm/m) × rm/m = (1+ rm/m)[ 1 +rm/m] =
[ 1 +rm/m]2
Thus we can find the amount of $ 1 at the end of
(m) periods of first year = [ 1 +rm/m]m = The amount of
$ 1 at the end of one year at effective rate (r), so that:
(1+ r) = (1 + rm/m)m
r = (1 + rm/m)m – 1 …… (3)
Example (7):
Find the Effective annual rate equivalent to the
Nominal annual rate 8% if the interest is compounded:
1- Annually.
2- Semiannually.
3- Quarterly.
4- Monthly.?
Solution:
r = (1 + rm/m)m - 1 ………………….(3)
r = (1 + 0.08/1)1 - 1 = 0.08 = 8%
r = (1 + 0.08/2)2 - 1 = (1 + 0.04)2 = 1.0816 – 1 =
0.0816 = 8.16%
252
r = (1 + 0.08/4)4 – 1 = (1 + 0.02)4 = 1.082432-1 =
0.082432 = 8.243%
r = (1 + 0.08/12)12 -1 = (1 + 0.00667)12 = 1.082999 – 1
= 0.082999 = 8.2999%.
Compound Amount by Nominal Rate Of Interest:
Example (8):
A principal of $ 20000 is deposited at 12%
annually for 10 years. What will be the compound
amount and the compound interest if the interest is
compounded
annually,
semiannually,
quarterly,
monthly.?
Solution:
The required factors and results are summarized
in the following Table. As the frequency of conversion
is increased, interest is added to principal more often,
so the depositor has a larger amount credited:
1-
Rate = 0.12
annually, the interest
is
compounded annually…Effective rate, n = 10
years.
2-
Rate = 0.12
annually, the interest
compounded semiannually,
253
is
Nominal rate, n = 10 × 2 = 20 periods, r = 0.12
÷ 2 = 0.06
3-
Rate = 0.12 annually, the interest is compounded,
quarterly,
Nominal rate, n = 10 × 4 = 40 periods, r = 0.12 ÷ 4 =
0.03
4-
Rate = 0.12
annually, the interest
is
compounded monthly,
Nominal rate, n = 10 × 12 =120 periods, r = 0.12 ÷ 12
=0.01
“ Note: Effective rate = Nominal rate ÷ Number of
periods in a year.“
Frequency
Of Compounding
Rate
Per
Period
(r)
Annually
Semiannually
Quarterly
Monthly
0.12
0.06
0.03
0.01
Numbe Amount
r Of
Of 1
Conve (1 + r)n
rsion
Period
s (n)
10
3.105848
20
3.207135
40
3.262038
120
3.300387
Compoun
d
Amount
S
Compou
nd
Interest
I
62116.96
64142.71
65240.76
66007.74
42116.96
44142.71
45240.76
46007.74
Example (9):
A depositor planned to leave $ 10000 in a
savings and loan association paying 10% annually for a
254
period of 5 years. At the end of 2.5 years the depositor
had to withdraw $5000. What amount will be in the
account at the end of the original 5-year period if the
interest is compounded semiannually.
Solution:
First we find the amount in the account at the
end of 2.5 years, a= 10000, Effective rate r = 0.10 ÷ 2
= 0.05, n = 2.5 × 2 = 5 periods,
S = 10000 (1 + 0.05)5 = 10000 (1.276282) = 12762.82
After Withdrawal of $ 5000 the depositor has a balance
of $ 7762.82.
During the remainder of the 5 years this amount will
grow to,
S = 7762.82 (1 + 0.05)5 = 7762.82 × (1.276282) =$
9907.54.
Alternate Solution:
S = 10000 (1 + 0.05)10 - 5000 (1 + 0.05)5
= 10000 × 1.628895 - 50000 × 1.276282.
= 16288.95 - 6381.41 = $ 9907.54
Note:
The compound interest Law is often called the
Law of organic growth. This Law can be applied to
anything that is changing at a constant rate.
Example (10):
During the period 2000 – 2010, the population of
a city increased at a rate of about 3%.If the population
255
in 2000 was 2000,000, what is the predicted population
in 2020.
Solution:
a = 2000,000, r = 0.03, n = 10
in formula s = a (1 + r)n …………. we have
S = 2000,000 (1 + 0.03)10 = 2000,000 × 1.343916 =
2,687,832.
Finding the variables of compound interest’s law:
S = a (1 + r)n
1-Find the principal (a):
a = s / (1+ r)n
a = S (1 + r)-n
………. (4)
Example (11):
Find the principal of the compound amount $
59177.55 at 7.5% annually for 15 years.
Solution:
S = 59177.55, r = 0.075 annually, n = 15.
a = S (1 + r)-n
a = 59177.55 (1 + 0.075)-15
a = 59177.55 (0.337966) = $ 20000.
2- Find the Rate of Interest (r):
S = a (1 + r)n
256
(1 + r)n = S / a
(1 + r) = [s/a ](1/n)
r = [s/a ](1/n) - 1 ………….(5)
Example (11):
If Amr deposed $ 20000 in Cairo bank and at
the end of 5 years, the compound amount was
$
29604.9, Find the rate of interest, if the interest is
compounded semiannually?
Solution:
a = 20000, S = 29604.9, r =??, m = 2, n = 5 × 2 = 10
periods.
r = [s/a ](1/n) - 1
r = [29604.9/20000 ](1/10) - 1
r = (1.480245)0.1 - 1 = 1.04 – 1 =.04 = 4%.
Example (12):
For a sum of money to double itself in 5 years,
what must be the rate of interest converted annually?
Solution:
Assuming that, a = 1, s = 2, n = 15, r =??
r = [s/a ](1/n) - 1
r = [2/1 ](1/15) - 1 = [2 ](1/15) - 1 = 1.04729 – 1 =
0.04729
r = 4.73%
Example (13):
257
If $ 10,000 accumulates to $ 15,000 in 5 years,
Find the nominal rate converted semiannually.
Solution:
a = 10,000, S = 15,000, n = 5 × 2 = 10 periods, r =??
r = [s/a ](1/n) - 1
r = [15,000/10,000 ](1/10) - 1 = [1.5 ](0.1) - 1 = 1.04138
– 1 = 0.04138
The nominal rate = 0.04138 × 2 = 0.08276 = 8.276.%
Example (14):
A firm organized in 1995 had sales of $ 300,000
that year. in 2015 the sales were $ 120,000 . What was
the annual rate of increase, assuming that it was
approximately constant from year to year?
Solution:
a = 300,000, S = 120,000, n = 20, r =??
r = [s/a ](1/n) - 1
r = [120,000 ÷ 30,000 ](1/20) - 1 = [4](0.05) - 1 =
1.07177 – 1 = 0.07177
The annual rate of increase = 0.07177 = 7.18.%
3Find the Time (n):
S = a (1+ r)n
(1 + r)n = S/a
We take the Logarithm of both sides:
n Log (1 + r) = Log (S) - Log (a)
n = [ Log (S) - Log (a) ] / Log (1 + r) ……….. (6)
Example (15):
Wessam Omar deposited $ 40000 in a bank at
7% annually, and after a Time she found the compound
amount $ 61896.8, Find the Time of investment?
Solution:
a = 40000, r = 0.07, S = 61896.8, n =??
258
n = [ Log (61896.8) - Log (40000) ] / Log (1 + 0.07).
n = [ 4.7916682 - 4.60206 ] / (0.02938378)
n = [ 0.1896082 ] / (0.02938378) = 6.45282 years.
Since 0.45282 is a fraction of year, must convert
to months and days, 0.45282 × 12 = 5.433822 months.
0.433822 × 30 = 13 days.
The Time is: 6 years, 5 months, 13 days.
Alternate Solution:
We can find the Time by using Tables of
Finance (Table 1):
S = a (1 + r)n
61896.8 = 40000 (1 + 0.07)n
(1 + 0.07)n = 61896.8 / 40000 = 1.54742.
By searching in table (1) for amount of 1 under the
rate 7%
We will find the value 1.54742 located between 6, 7
years (since the rate is annually).
so the time is 6 years + fraction
from the table (1) we will find
(1 + 0.07)7 = 1.605781 ……….. (1)
(1 + 0.07)n = 1.54742 …………….(2)
(1 + 0.07)6 = 1.500730 …………… (3).
Subtract (3) from (1) = 0.105051 this difference
Equivalent a year
Subtract (3) from (2) = 0.04669 this difference
Equivalent a Fraction Δ.
Δ = 0.04669 /0.105051 = 0.444451 fraction of year,
The time (n) = (6 + 0.444451) years.
0.444451 × 12 = 5.3333412 months,
0.3333412 × 30 = 10 days.
The Time is: 6 years, 5 months, 10 days.
259
Example (16):
How long will it take $ 5000 to 7500 at 8%
compounded semiannually?
Solution:
a= 5000, S = 7500, r = 0.08/2 = 0.04, n =???
Substituting in the compound interest formula, we have
5000 (1.04)n = 7500
(1.04)n = 7500/5000 = 1.5
We now look under 4% for an accumulation
factor of 1.5.
We do not find this exact value. If n is 10, the
factor is 1.480244, and if n is 11, the factor is 1.539454
from the table (1) we will find
(1 + 0.04)10 = 1.4802443 ………....(1)
(1 + 0.04)n = 1.5000000 …………….(2)
(1 + 0.04)11 = 1.539454 …………..…(3).
Subtract (1) from (3) = 0.05921 this difference
Equivalent a year
Subtract (1) from (2) = 0.0197557 this difference
Equivalent
a Fraction Δ
Δ = 0.0197557 /0.05921 = 0.333655 fraction of year,
The time (n) = (10 + 0.333655) periods semiannually
.
n = 10.333655 ÷ 2 = 5.166828 years
0.166828 × 12 = 2 months,
The Time is: 5 years and 2 months.
Alternate Solution:
a = 5000, r = 0.04, S = 7500, n =??
n = [ Log (S) --- Log (a) ] ÷ Log (1 + r)
n = [ Log (7500) - Log (5000) ] ÷Log (1 + 0.04).
260
n = [ 3.875061 - 3.69897 ] ÷ (0.0170333)
n = [ 0.176091 ] ÷ (0.0170333) = 10.338 periods.
n = 10.338 ÷ 2 = 5.169 years
0.169 × 12 = 2 months,
The Time is: 5 years and 2 months.
261
Exercises
1-
Find the compound amount if $ 10000 is
invested for 10 years at:
(a) 1% compounded monthly.
(b) 3% compounded quarterly
(c) 5% compounded semiannually.(d) 12% all two
years.
2-
What is the compound amount of $ 5000
invested at 7% compounded semiannually for 40
years.($ 78,378.69)
3-
How much must be invested today at 7%
converted
semiannually to amount to $
1,000,000 in 25 years?.($ 179,053.37)
4-
What is the effective rate of interest equivalent
to 9% converted:
(a)
semiannually? (b) quarterly? (c) monthly?
5-
Which gives the better annual return on an
investment, 6.5% converted annually or 6%
converted quarterly?
262
6-
If $ 10000 amount to $ 20000 in 10 years with
interest compounded quarterly, what is the rate if
interest.? (1.748%)
7-
For a sum of money to double itself in 10 years,
what must be the rate of interest converted
annually?
8-
If the population of a city increases from
400,000 to 500,000 at
Approximately a constant rate during a decade,
what is the annual rate of increase?
9-
A firm organized in 1990 had sales of $100,ooo
that year. In 2010 the sales were $ 350,000.
What was the annual rate of increase, assuming
that it was approxim-ately constant from year to
year?
10-
Values of disposable personal income for
selected years are:
Year
Personal income (millions)
2005
2284.5
2010
3836.2
2015
5666.4
263
Find the annual compound rates of growth for
2005 – 2010, 2010 – 2015, 2005 – 2015.
11-
How Long will it take $ 5000 to amount to $
7500 at 7% compounded semiannually.
12-
Eayd has 4 years old is left $ 15000. The money
is to be invested until it amounts to $ 25000. If
the money is invested at 7.5% annually,
How old will the child be when the $ 25000 is
received?
264
Chapter ( 3)
Present Value
At Compound Interest
265
266
Chapter ( 3)
Present Value
At Compound Interest
The present Value is defined as the principle that will
amount to the given sum at the specified future date. In
business transactions there are many times when it is
necessary to determine the Present Value.
The Compound discount is the difference between
the future amount and its present value. To find the
present value of a future amount, we solve the
compound interest formula for (a) by dividing both
sides by
(1 + r) n, so that:
a = S ÷ (1 + r) n, or:
a = S × (1 + r)- n
We use the symbol Vn to indicate for the
discount factor (1 + r)- n, Or The present value of 1
so: a = S × Vnr% ……………..…….. (7)
Where:
a
= the principal or present value
267
S
= the amount due in the future
r
= the rate per period
n
= the number of periods.
since: Vnr% = 1÷ (1 + r) n = (1 + r)- n
,:
V r%= 1÷ (1 + r) = (1 + r)-1
Numerical values of the discount factor Vr% for
common interest rates are given in the “ Present worth
of 1 ˮ column in Table (2)
We will follow the common practice of using a
negative exponent that a sum due in the future is to be
discount. we can use a calculator to solve the present
value by applying formula (7) either by division or by
using a reciprocal key (1/x) to calculate (1 + r)-n .
The “ Present worth of 1 ˮ column gives the
present or current value of a one dollar or one pound to
be paid in the future, taking into account compound
interest. For example, $1 due in 5 years at 7%
compound annually, has a present - worth - of- 1 factor
of 0.712986. In other words, $ 0.71 will grow to $1.00
in 5 years if it earns 7% compound annually. The
present value of any sum due in the future can be
obtained by multiplying the future amount by the
appropriate present –worth–of– factor from Table (2).
268
The concept of present value is one of the most
useful tools in economic analysis since it enables the
analyst to take sums of money due in the future and
determine how much they are worth now.
Example (1):
Find the present value of $10000 due in 5 years
if money is worth 7.5% compounded annually.
Solution:
a = S × Vnr%
Substituting: S = 10000, r = 0.075, n = 5 and using the
present –worth-of-1 factor from Table (2), we have:
a = 10000 × V57.5%
a = 10000 (0.696559) = $ 6965.59
This means that if $ 6965.59 had been put at
interest for 5 years at 7.5% compounded annually, the
amount would be $ 10000.
We can compute V57.5% = (1 + 0.075)-5, directly
by using calculator.
Example (2):
Find the present value of $20000 due in 4 years
if money is worth 12% annually compounded monthly.
269
Solution:
a = S × Vnr%
Substituting: S = 20000, m = 12, r = 0.12/12= 0.01, n
= 4×12= 48 periods and using the present –worth-of-1
factor from Table (2), we have:
a
= 20000 × V481%
a
= 20000 (1 + 0.01)-48 = 20000 (0.6202604) = $
12405.2.
Note: Using a calculator with exponential capability
allows for the use of any interest rate and in any time
(periods).
Example (3):
How much must be invested in an account
paying 7.2% annually compounded monthly in order to
accumulate to $25000 in 5 years?
Solution:
The rate, r = 7.2%/12 = 0.6%, (since m = 12) is not
found in Table (2), By using a calculator: S = 25000, r
= 0.006, n = 5 × 12 = 60
We have: a = 25000 (1 + 0.006)-60 = 25000
(0.698426) = $17460.68.
270
Example (4):
Hossam can buy a piece of property for $ 10,000 cash
or for $ 5000 now and $ 6500 in 3 years. If Hossam
has
money
earning
7%
annually
converted
semiannually, which is better purchase plan and by
how much now?
Solution:
We get the present value of $ 6500 due in 3
years at 7% annually converted semiannually:
a = S × Vnr%
S = 6500, m = 2, r = 3.5%, n = 3 × 2 = 6 periods a
= 6500 × V63.5%
a = 6500 (1 + 0.035)-6 = 6500 (0.8135006) =
$ 5287.75
Adding this amount to the $ 5000 down payment
makes the present value of the time payment plan $
10287.75. By paying $ 10000 cash is better, since the
buyer saves $ 287.75 now.
Example (5):
ESLAM can purchase a piece of property for $
21000 cash now or for $ 24000 in 2 years. Which is
271
better plan for ESLAM if money is worth 8%
annually, compounded quarterly? Find the cash
(present value) equivalent of the savings made by
adopting the better plan.
Solution:
We can compare alternative purchase plans by
bringing all payments to the same point in time and
seeing which plan is better. It is usually best to make
the comparison on a present value basis. get the present
value of $ 24000 due in 2 years at 8% annually,
compounded quarterly:
a = S × Vnr%
S = 24000, m = 4, r = 8/4 = 2%, n =2 × 4 = 8 periods
a = 24000 × V82%
a = 24000 (1 + 0.02)-8 = 24000 (0.8534904) = $
20483.77.
Since the present value of the $ 24000 due in 2
years is less than the cash payment, it is better to pay
later. The cash equivalent of the savings is
21000 – 20483.77 = $ 516.23.
272
Note: A different rate of interest could lead to a
different decision.
If the buyer's money was earning only 6%
compounded quarterly, the present value of $ 24000
would be:
24000 (1 + 0.015)-8 = 24000 (0.887711) = $
a =
21305,07.
In this case it is better to pay cash and save
21305.07 – 21000 =$ 305.07
on a cash basis.
Example (6):
A note with a maturity value of $10,000 is due in
3 years and 8 months.
What
is
its
present
value
at
6%
compound
semiannually?
Solution:
a = S × Vnr%
S = 10000, m = 2,
r = 6/2 =3%, n = (3 + 8/12) × 2
= 7.3333 periods.
a = 10000 (1 + 0.03)-7.3333 = 10,000 (0.8051203)
= $8051.2
273
Example (7):
Find The present value of $ 12,000 due in 18 months at
9% annually compounded quarterly.
Solution:
S = 12,000, m = 4, r = 0.09/4= 0.0225, n = (18/12) × 4
= 6 periods
a = 12,000 (1 + 0.0225)-6 = $ 10,500.29
Example (8):
Wessam Omar owns a note for $ 25,000 due in
5 years. What should a buyer wishing money to earn
12% annually converted monthly pay for the note?
What is the compound discount?
Solution:
The buyer should pay the present value of
$25000 due in 5 years at 12% converted monthly,
a = S × Vnr%
S = 25,000, m = 12,
r = 0.01, n = 5 × 12 = 60
periods,
a = 25,000 × V601%
a = 25,000 (1 + 0.01)-60 = 25,000 (0.5504496) = $
13761.24
274
The compound discount = 25,000 - 13761.24
= $ 11238.76
Example (9):
On April 5, 2018, Mr Amr buys some equipment
from El Arabi Company. He sings a note promising to
pay $ 3000 with interest at 14% compounded
semiannually in 18 months. On July 5, 2018, El Arabi
Company sells the note to a finance company that
charges an interest rate of 16% converted quarterly for
discounting. How much does El Arabi Company get
for the note?
Solution:
The compound amount of $ 3000 at 14%
compounded semiannually in 18 months:
a = 3000, r =0.14/2 = 0.07%, m = 2, n = (18/12) × 2 =
3 periods
S = 3000 (1.07)3 =$ 3675.129.
The present value of $ 3675.129, On July 5,
2018, at interest rate of 16% converted quarterly:
m = 4, r = 0.16 /4 = 0.04, n = (15/12)× 4 = 5 periods
a = S × Vnr%
275
S = 3675.129
a = 3675.129 × V54%
a = 3675.129 (1 + 0.04)-5= $ 3020.688.
El Arabi Company gets for the note $ 3020.688.
Example (10):
Mr Mohamed Omar gets a note that calls for
repayment of $ 5000 in 5 years with compound interest
at 10% converted quarterly. He sells the note
immediately to a finance company which charges an
interest rate of 9% converted semiannually for
discounting. How much does Mr Mohamed Omar get
for the note?
Solution:
The compound amount of $ 5000 at 10%
converted quarterly in 5 years:
a = 5000, m = 4, r =0.10/4 = 0.025, n = (5)× 4= 20
periods
S = 5000 (1.025)20 = $ 8193.082.
The present value of $ 8193.082, at interest rate
of 9% converted semiannually:
m = 2, r = 0.09 /2 = 0.045, n = (5)× 2 = 10 periods
276
a = S × Vnr%
…..
S = 8,193.082
a = 8,193.082 × V104.5%
a = 8,193.082 (1 + 0.045)-10 = $ 5,275.752 .
Mr Mohamed Omar gets for the note $ 5,275.752
Example (11):
A note of $ 100,000 is due in 5 years with
interest at 8% annually. At the end of 3 years it is
discounted at 9% annually. What are the proceeds at
the time of discounting?
Solution:
The compound amount of $ 100,000 is due in 5
years with interest at 8% annually:
a = 100,000,, r =0.08, n = 5 periods
S = 100000 (1.08)5 = $ 146,932.81 .
The present value of $ 146,932.81, at interest rate of 9
annually. At the end of 3 years:
r = 0.09, n = 2 periods
…..
S = 146,932.81
a = S × Vnr%
a =146,932.81 × V29%
a = 146,932.81 (1 + 0.09)-2 = $ 123670.41.
277
Example (12):
On October 15, 2016, Miss Reham borrows $
4000 from a bank. she gives the bank a note promising
to repay the money in 5 years with interest at 11%
annually. On March 15, 2018, if the bank sells the
note to another buyer to charges a rate of 13%
compounded semiannually for discount purposes. How
much does the bank get for the note?
Solution:
The compound amount of $ 4000 is due in 5
years with interest at
11% annually:
a = 4000,, r =0.11, n = 5 periods
S = 4000 (1.11)5 = $ 6740.23
The present value of $ 6740.23, at rate of 13%
compounded semiannually for discount On March 15,
2018 (period of discount will be 3.5 years):
r = 0.065, m = 2 periods, n = (3.5)× 2 = 7
a = S × Vnr%
a =6740.23 × V76.5%
a = 6740.23 (1 + 0.065)-7 = $ 4337.38.
278
The Practical method to find the present
value (by using financial Tables):
Using financial Tables to find present value if (r)
or (n) or both are not found in tables:
If (r) is found and (n) is not found:
Periods (n) is not found in the following cases:
1-
Periods (n) is correct number but over than table
scope:
In this case, (n) will divided to correct numbers
each of them take a symbol such as x, y, z …since a
value of each
less than table scope (50 periods)
…since
a = S × Vnr%
a = s (1 + r)-n = s (1 + r)-x (1 + r)-y (1 + r)-z
ince…. n
X +Y +Z
Example (13):
How much must invested today at 8%
compounded semiannually to amount to $1,000,000 in
50 years.
Solution:
a = S × Vnr%
…. ,
a = s (1 + r)-n
279
m = 2, r = 0.08/2 = 0.04, n = 50 × 2 = 100 periods,
From table (2):
= 1,000,000 (1 + 0.04)-100
a
= 1,000,000 (1 + 0.04)-50(1 + 0.04) -50
= 1,000,000 (0.140713)(0.140713)
= 1,000,000 (0.140713)2 = $ 19800.15
2-
Periods (n) is no correct number either or less
or over than table:
In this case, will divided (n) to a correct number
and a fraction.
Example (14):
Suppose the interest in Example (13) was
compounded monthly at 9% annually for 20 years and
3 months, Find the present value.
Solution:
m = 12, r = 0.09/12 = 0.0075, n = (20 + 3/12)× 12 =
243 periods
a = S × Vnr%
….,
a = s (1 + r)-n
a = 1,000,000 V243 0.75%
= 1,000,000 (1 + 0.0075)-243
By using calculator directly:
280
a = 1,000,000 (0.162724) = $ 162,724.
From financial Tables:
a
= 1,000,000 (1.0075)-50 × (1.0075)
-50
×
(1.0075) -50 × (1.0075) -50 × (1.0075)-43
= 1,000,000 [ (1.0075)-50 ]4 × (1.0075)-43
= 1,000,000 [ 0.688252 ]4 × (0.725208)
= 1,000,000 (0.224383) (0.725208)
= $ 162,724.35
3-
If (r) and (n) are not found in Tables of finance
Example (15):
Suppose the interest in Example (14) was
compounded at 7.3% annually for 60 years, Find the
present value?
Solution:
r
= 0.073,
n = 60 periods
a
= S × Vnr%
a
= 1,000,000 V60 7.3%
a
= 1,000,000 (1 + 0.073)-60
….,
a = s (1 + r)-n
We can divided the solution to three stages:
1-
Will find the present value for 60 years at
7%..(table 2)
281
2-
Will
find the present value for 60 years at
7.5%.(table 2
3-
Will find the present value for 60 years at
7.3%..(by Proportion & proportionality).
(1 + 0.07)-60 = (1 + 0.07)-30 × (1 + 0.07)-30
= 0.131367 × 0.131367 = 0.017257 … (1)
(1 + 0.075)-60 = (1 + 0.075)-30 × (1 + 0.075)-30
= 0.114221 × 0.114221 = 0.013046 … (2)
By subtracting (2) from (1) = 0.004211, this
difference in the present value equal the Difference in
the rate its value 0.5%
So that: the difference in the present value which
equal the difference in the rate its value 0.3 = 0.004211
× 0.3 /0.5
= 0.0025266.
a
= 1,000,000 (1 + 0.073)-60
= 1,000,000 (0.017257 - 0.0025266)
= 1,000,000 × 0.0147304 = $ 14730.4
By using calculator directly:
a = 1,000,000 (1 + 0.073)-60
= 1,000,000 (0.0145887) = $ 14,588.7.
282
Not that:
(1 + 0.07)-60 > (1 + 0.073)-60 > (1 + 0.075)-60
(1 + 0.07) 60 < (1 + 0.073)60 < (1 + 0.075)60
Example (16):
Find the present value of $ 20,000 due in 10
years, one month and 6 days at 8.2% annually
compounded semiannually.
Solution:
m =2, r = 0.082 / 2 = 0.041, n = (10 + 1/12 + 6/360) ×
.2 = 20.2 periods
a = S × Vnr%
….,
a = s (1 + r)-n
a = 20,000 V20.2 4.1%
a = 20,000 (1 + 0.041)-20.2
We also can divided the solution to three stages:
1-
Will find the present value for 20 periods at
4.%.(table 2)
2-
Will find the present value for 20 periods at
4.5%.(table 2)
3-
Will find the present value for 20 years at 4.1%.
(by Proportion & proportionality).
283
(1 + 0.04)-20 = 0.456387 ……..……….. (1)
(1 + 0.045)-20 = 0.414643 ……………… (2)
By subtracting (2) from (1) = 0.041744, this
difference in the present value equal the Difference in
the rate its value 0.5%
So that: the difference in the present value which
equal the difference in the rate its value 0.1 = Δ =
0.041744 × 0.1 /0.5
= 0.0083488
(1+ 0.041)-20
= (1 + 0.04)-20 - Δ
(1+0.041)-20 = 0.456387 - 0.0083488 = 0.448038 (3)
We can also find (1 + 0.041)-21 as follow:
(1 + 0.04)-21 = 0.4388336 ……………… (4)
(1 + 0.045)-21 = 0.3967874 ……..….….. (5)
By subtracting (5) from (4) = 0.041744, this
difference in the present value equal the Difference in
the rate its value 0.5%
So that: the difference in the present value which
equal the difference in the rate its value 0.1 = Δ =
0.0420462 × 0.1 /0.5 = 0.00840924
(1+ 0.041)-21 = (1 + 0.04)-21 - Δ
284
(1+
0.041)-21
=
0.
0.4388336
-
0.00840924=
0.4304244………..(6)
By subtracting (6) from (3) = 0.0176136, this
difference in the present value equal the Difference in
the time its value one period
So that: the difference in the present value which
equal the difference in the time its value 0.2 = Δ =
0.0176136 × 0.2 =0.00352272.
(1+ 0.041)-20.2 = (1 + 0.041)-20 - Δ
= 0.448038 - 0.00352272 = 0.4445153.
The present value will be:
a
= 20,000 (1 + 0.041)-20.2
= 20,000 (0.4445153) = $ 8890.31
By using calculator directly:
a
= 20,000 (1 + 0.041)-20.2
= 20,000 (0.4441148) = $ 8882.3 .
The Practical method to find the Rate and the Time
(by using financial Tables):
1- The Time:
Example (17):
285
If the present value of $ 9000 at 4.5% annually
compounded semiannually was $ 3690, Find the Time.
Solution:
m = 2, r = 4.5/2 = 2.25%, S = 9000, a = 3690.
a = S × Vnr%
3690 = 9000 Vn 2.25%
Vn2.25% = 3690 ÷ 9000 = 0.41 ……………(1)
By searching about the value (0.41) in table (2)
of the present value
under the rate 2.25% will be
between 40, 41 periods
So: The number of the periods will be 40 + fraction:V402.25% = 0.410646 ………………. (2)
V412.25% = 0.401610………………..(3)
By subtracting (3) from (2) = 0.009036, this
difference in the present value equal the Difference in
the time its value one period
By subtracting (1) from (2) = 0.000646, this
difference in the present value equal the Difference in
the time its value Δ
Δ = 0.000646 ÷ 0.009036 = 0.071492
n = (40 + 0.071492) /2 = 20.035746 years.
286
To return the fraction into months and days:0.035746 × 12 = 0.42895 months
0.42895 × 30 = 12.868 = 13 days
So: n = 20 years and 13 days.
By calculator: _
3690 = 9000 Vn 2.25%
Vn 2.25% = 3690 ÷ 9000 = 0.41
(1 + 0.0225)-n = 0.41
-n log (1 + 0.0225) = log 0.41
-n = log 0.41 ÷ log (1.0225) = - 40.0707289
periods
n = 40.0707289 ÷ 2 = 20.0353645 years
0.0353645 × 12 = 0.424374 months
0.424374 × 30 = 12.73 = 13 days.
n = 20 years and 13 days.
2- The Rate:
Example (18):
If the present value of $ 5000 dui in 8 years was
$ 1830.2, find the rate if the interest compounded
quarterly.
287
Solution:
m = 4, n = 8 (4) = 32 periods., S = 5000, a = 1830.2
a = S Vn r%
1830.2 = 5000 V32 r%
V32 r% = 1830.2 ÷ 5000 = 0.36604 …… (1)
By searching about the value (0.36604) in table
(2) of the present value under the Time 32 will be
between 3%, 3.5% rates
So: The Rate will be 3% + fraction:-
V323%
=
0.388337 ……………………….(2)
V323.5% = 0.3325897…………….. (3)
By subtracting (3) from (2) = 0.0557473, this
difference in the present value equal the Difference in
the rate its value 0.5%
By subtracting (1) from (2) = 0.022297, this
difference in the present value equal the Difference in
the rate its value Δ
Δ = 0.022297 (0.5) ÷ 0.0557473 = 0.199999 = 0.2
So: r = 3 + 0.2 = 3.2% quarterly
By calculator:1830.2 = 5000 V32 r%
288
V32 r% = 1830.2 ÷ 5000 = 0.36604
(1 + r)-32 = 0.36604
1 + r = (0.36604)(-1÷32) = 1.0319
r = 0.0319 = 3.19%
289
Exercises
1-
Find the present value of $ 20,000 due in 3 years
if money is worth 7% compounded annually.
2-
Find the present value of $10,000 due in 4 years
if money is worth 12% annually compounded
monthly.($ 2419.99)
3-
How much must be invested in an account
paying 7.4% annually compounded mon-thly in
order to accumulate to $50000 in 5 years?
4-
Hossam can buy a piece of property for $ 20000
cash or for $ 10000 now and $ 12000 in 4 years.
If Hossam has money earning 8% annually
converted
semiann-ually,
which
is
better
purchase plan and by how much now?
5-
ESLAM can purchase a piece of property for $
40000 cash now or for $ 45000 in 4 years.
Which is better plan for ESLAM if money is
worth 8% annually, compounded quarterly? Find
the cash (present value) equivalent of the savings
made by adopting the better plan.
290
6-
A note with a maturity value of $20000 is due in
4 years and 4 months. What is its present value
at 9% compound semiannually? ($ 13657)
7-
Find The present value of $ 15000 due in 15
months
at
10%
annually
compounded
quarterly.($ 13257.81)
8-
Wessam Omar owns a note for $ 50000 due in 5
years. What should a buyer wishing money to
earn 12% annually converted monthly pay for
the note? What is the compound discount? ($
2752.25,
$ 2247.75).
9-
On Jun 5, 2018, Mr Alli-Eldeen buys some
equipment from El Arabi Company. He sings a
note promising to pay $ 25000 with interest at
12% compounded semian-nually in 18 months.
On September 5, 2018, El Arabi Company sells
the note to a finance company that charges an
interest rate of 16% converted quarterly for
discounting. How much does El Arabi Company
get for the note? ($ 24473.2).
291
10-
Mr Mohamed Omar gets a note that calls for
repayment of $ 25000 in 5 years with compound
interest at 12% converted quarterly. He sells the
note immediately to a finance company which
charges an interest rate of 10% converted
semiannu-ally for discounting. How much does
Mr Mohamed Omar get for the note? ($
27719.89).
11-
A note of $ 200,000 is due in 5 years with
interest at 9% annually. At the end of
2 years IT is discounted at 10% annually. What
are the proceeds at the time of discounting
12-
On November 15, 2016, Miss Reham borrows $
10000 from a bank. she gives the bank a note
promising to repay the money in 5 years with
interest at 12% annually. On April 15, 2018, if
the bank sells the note to another buyer to
charges a rate of 14% compounded semiannually
for discount purposes. How much does the bank
get for the note.?
292
13-
A note of $ 75,000 is due in 6 years with interest
at 9%.at the end of 3 years the note is discounted
at 10%. What are the proceeds at the time of
discounting?. ($ 94502.26).
14-
On September 10, 2016, Miss Reham borrows $
8000 from a bank. She gives the bank a note
promising to repay the money in 5 years with
interest at 12%. On March 10, 2018, If the bank
sells the note to a another buyer who charges a
rate of 14% compound semiannually for
discount purposes. How much does the bank get
for the note? ($ 8779.98).
293
294
Chapter ( 4)
Settlement Of Debts
At Compound Interest
295
296
Chapter ( 4)
Settlement Of Debts
At Compound Interest
Debt Settlement is a direct application of Amount
and Present value Laws at compound interest for a unit
money. Where it is assumed that the debtor will borrow
certain amounts to agree with the creditor or creditors
on certain payment methods and certain maturity dates
for such debts. So that these dates coincide with his
future circumstances, However, there may be changes
in future income that prevent the debtor from repaying
its debts on due dates. Which prompting the parties, the
debtor and the creditor to adjust the ways and dates of
maturity of these debts (Settlement of debts) in new
ways and dates. Which called (Debt Scheduling). In
business transactions it is often necessary to exchange
one set of obligations for another set of different
amounts due at different times. To do this, it is
necessary to bring all the obligations to a common date
called a focal date. Then we set up an equation of
297
value in which all the original obligation at the focal
date equal all the new obligations at the focal date.
This procedure is based on the fact that we can find the
value of any sum of money at any time by
accumulating it at compound interest if we take it into
the future, or discounting it if we bring it back in time.
Debt Settlement Methods:
1-
Repayment of old debts with one previous debt
to maturity dates of old debts (One – Former
Debt).
2-
Repayment of old debts with one debt due
during the maturity dates of old debts.
3-
Repayment of old debts with several debts due
during the maturity dates of old debts.
4-
Repayment of old debts with one debt whose
nominal value equals the total nominal value of
the old debts.
First: Repayment of old debts with one previous debt
to maturity dates of old debts (One – Former
Debt):
298
Example (1):
If Ali al –Din owes the following amounts:
$ 20000 due in 3 years
$ 40000 due in 5 years
$ 50000 due in 6 years
If he agreed with the creditor to pay these debts
in a debt due in 2 years. Find the nominal value of the
new debt at compound interest 8% annually.
Solution:
.r = 8% annually
_______ S_____20000___ __40000____50000
(2 years) 2
3
5
6
Assuming that the nominal value of the new debt is
(S) and a focal date is now:
so the equation of settlement will be:
The present value of new debt = the present value of
the old debts
S V28% = 20000 V38% + 40000 V58% + 50000 V68%
S (1+ 0.08)-2 = 20000 (1+ 0.08)-3 + 40000 (1+ 0.08)-5 +
50000 (1+ 0.08)-6
S
(0.857339)
=
20000
(0.793832)
(0.680583) + 50000 (0.630170) = 74608,44
299
+
40000
S = 74608,44 ÷ 0.857339 = $ 87023,3.
Another Solution:
_______ S_____20000___ __40000____50000
(2 years) 2
3
5
6
A focal date is the maturity date of the new debt:
S = 20000 V8% + 40000 V38% + 50000 V48%
= 20000 (1.08)-1 + 40000 (1.08)-3 + 50000 (1.08)-4
= 20000 (0.925926) + 40000 (0.793832) + 50000
(0.7350298)
= $ 87023.3 ……..the same result.
(Note That: The Difference in the settlement date
does not result a difference in outcome).
Second: Repayment of old debts with one debt due
during the maturity dates of old debts:
In this case the old debts are replaced by a debt
due in a previous date of some debts and followed date
for others. Then we can select any date and considered
it as a date of settlement.
Example (2):
If Hossam was owed the following amounts:
$ 40000 due in 4 years
$ 50000 due in 5 years
300
$ 80000 due in 8 years
If he agreed with the creditor to pay these debts
in a debt due in 6 years. Find the nominal value of the
new debt at compound interest 5% semiannually.
Solution:
r = 5% semiannually, m = 2
_______ 40000_ _50000__ S _ 80000
(4 years)
4
5
6
8
Assuming that the nominal value of the new debt
is (S)
A focal date is the maturity date of the new
debt (After 6 years):
Then The first debt is invested for 2 years (for 4
periods), the second debt is invested for one year (2
periods), but the third debt is discounted for 2 years (4
periods). So:
S = 40000 (1 + 0.05)4 + 50000 (1 + 0.05)2 + 80000
V4%5
= 40000 (1.215506)
+ 50000 (1.1025)
(0.8227025) = $ 169561,44.
301
+ 80000
Another Solution: a focal date is now:
r = 5% semiannually, m = 2
_______ 40000__50000___ S _ 80000
(4 years) 4
5
6
8
So the equation of settlement will be:
The present value of new debt = the present
value of the old debts.
S V125% = 40000 V85% + 50000 V105% + 80000 V165%
S (1.05)-12 = 40000 (1.05)-8 + 50000 (1.08)-10 + 80000
(1.08)-16
S (0.5568374) = 40000 (0.6768394) + 50000
(0.6139133) + 80000 (0.4581115)
= 94418,163
S = 94418,17 ÷ 0.556837
= $ 169561,46 …..
a same result
A Third solution:
A focal date is the date of first debt:
r = 5% semiannually, m = 2
_______ 40000__50000__ S _ 80000
(4 years) 4
5
6
8
So the equation of settlement will be:
302
The present value of new debt = The present
value of the old debts.
S V45% = 40000 + 50000 V25% + 80000 V85%
S (1.05)-4 = 40000 + 50000 (1.08)-2 + 80000 (1.08)-8
S (0.8227025) = 40000 + 50000 (0.9070295) + 80000
(0.6768394) = 139498,62
S = 139498,62 ÷ 0.8227025 = $ 169561,44. ….. a
same result.
THIRD-- Repayment of old debts with several debts
due during the maturity dates of old debts:
In case of multiple new debts, settlement is
preferred on the date of the agreement between the
creditor and the debtor on the settlement.
So the equation of settlement will be:
The present value of new debts on a focal date =
the present value of the old debts on the same date.
Example (3):
On January 1, 2017, Esslam was owed the
following amounts:
$ 20000 due in 3 years
$ 40000 due in 5 years
303
$ 60000 due in 7 years
On January 1, 2019 he agreed with the creditor
to pay these debts
a-
$ 54996,3 Cash payment.
b-
Pay the remaining with three Notes, the nominal
value of the first is twice the nominal value of
the second, the ratio of the nominal value of the
second to the third as percentage 2: 3, and this
notes due in 2, 3, 4 years respectively, at
compound interest 8% annually. Find the
nominal value of the three notes.
Solution:
r =% 8 annually
_______
20000_ _40000___60000
1/1/2017 1/1/2019 3
5
7
A focal date is 1/1/2019:
The present value of old debts = 20000 V8% + 40000
V38% + 60000 V58% .
=
20000
(1.08)-1
+
40000
(1.08)-3
+
60000
(1.08)-5
=
20000
(0.9259259)
+ 40000
60000 (0.6805832) = 91106,798
304
(0.7938322)
+
The present value of new debts = 91106,798 - 54996,3
= 36110,498. …………… (1)
Assuming that the nominal value of the three notes as
follow:
First
Second
Third
2
:
1
Δ
2
: 3
4
:
2
:
3
If the nominal value of the first note = 4 S
The nominal value of the second note = 2 S
The nominal value of the third note = 3 S
4S 2S 3S
1/1/2019 2
3
4
The present value of new debts = 4S (V2)8% + 2S
(V3)8% + 3S(V4)8%
= 4S (1.08)-2 + 2S (1.08)-3 + 3S (1.08)-4
=
4S
(0.857339)
+
2S
(0.793832)
(0.73503) = 7,22211 (S)……….. (2)
From (1) and (2) We conclude that:
7,22211 (S) = 36110,498
S
= 36110,498 ÷ 7,22211 = 5000
305
+
3S
The nominal value of the first note = 4 (5000) = $
20000.
The nominal value of the second note = 2 (5000) =$
10000
The nominal value of the third note = 3 (5000)
=$
15000
Example (4):
If Wessam was owed the following debts;
$ 10000 due in 4 years
$ 30000 due in 6 years
$ 40000 due in 8 years
and she agreed with the creditor to pay this debts as
follow:
a)
$10000 cash payment
b)
Pay the remaining with two equal debts, the
normal value of each $20551,1, the first debt due
in 2 years, find the maturity date of the second
debt at compound interest 10% annually.
Solution:
In this case it preferred using the following
equation for settlement:
306
The present value of old debts = the present
value of new debts now
The old debts:
10000 30000 40000
4
6
8
The present value of old debts =
10000(V4)10% + 30000(V6)10% + 40000(V8)10%
= 10000(1.1)-4 + 30000 (1.1)-6 + 40000 (1.1)-8
= 10000(0.683013) + 30000 (0.564474) +
40000 (0.466507) = 42424.63
The remaining after cash payment = 42424.63 -10000
= 32424.63
So: The present value of new debts = 32424.63
The new debts:
20551,1
20551,1
2
?
32424.63 = 20551,1(V2)10% + 20551,1(Vn)10%
= 20551(1.1)-2 + 20551,1(Vn)10%
= 20551(0.826446) + 20551,1(Vn)10%
= 16984.3 + 20551,1(Vn)10%
32424.63 - 16984.3 = 20551,1(Vn)10%
15440.33 = 20551,1(Vn)10%
307
(Vn)10% = 15440.33 ÷ 20551,1 = 0.7513148
By searching in table (2) under 10% about the value
0.7513178,
We find it front of 3 period
So …. n = 3, the rate = 10% annually
then the second debt due in 3 years.
By using calculator and logarithmic:
Vn10% = 0.7513148
n log V10% = log 0.7513148
n log (1.1)-1 = log 0.7513148
n log (0.9090909) = log 0.7513148
n = log 0.7513148 ÷ log (0.9090909) = 3, then n = 3
years
Example (5):
If Reham was owed the following debts;
$ 40000 due in 4 years
$ 50000 due in 5 years
$ 60000 due in 6 years
and she agreed with the creditor to pay this debts as
follow:
a)
$22495 cash payment
308
b)
Pay the remaining with two equal debts in
normal value, due in 2, 3 years respectively at
compound interest 12% annually converted
quarterly. Find the normal value of the new
debts.
Solution:
m = 4, r = 0.12 /4 = 0.03
The old debts:
40000 50000 60000
16
20
24
The present value of old debts =
40000 (V16)3% + 50000 (V20)3% + 60000 (V24)3%
= 40000 (1.03)-16 + 50000 (1.03)-20 + 60000
(1.03)-24
=
40000
(0.623167)
+
50000
(0.553676)
60000 (0.491934) = 82126,52.
The remaining after cash payment =
82126,52 – 22495 = 59631,52
So The present value of new debts = 59631,52
The new debts:
59631,52
S
S
8
12
8
12
= S (V) 3% + S (V) 3%
309
+
= S [ (1.03)-8 + (1.03)-12 ]
= S [ 0.789409 + 0.70138 ]
59631,52
S
= S (1.4907889)
= 59631,52 ÷ 1.4907889 = $ 40000.
Example (6):
On January 1, 2018 Hesham was owed three
equal debts in normal value due in 3, 4,7 years
respectively, and he demanded from creditor to delay
the repayment of the three debts for a period of two
years after their due date, then the creditor offers the
two following alternatives:
a) $ 13000 cash payment
b) Or Note its normal value $ 20000 due in 5 years
Find:
1-
The annual rate of compound interest using in
settlement.
2-
The normal value of the three debts.
Solution:
We can find the rate of interest from the two
alternatives, $ 13000 represent the present value of a
note its normal value $ 20000 due in 5 years,
310
13000 = 20000 (V)5
(V)5 = 13000 ÷ 20000 = 0.65
By searching in table (2) in front of 5 period about the
value 0.65,
We find it under 9%
To find the normal value of the three debts we
can use the following equation:
The present value of old debts = the present
value of new debts.
Suppose the normal value of each debt = S
The old debts:
S
S
S
3
4
7
3
The present value of old debts = S (V) 9% + S (V)49% +
S (V)79%
= S (1.09)-3 + S (1.09) -4 + S (1.09) -7
= S [ 0.7721835 + 0.708425 + 0.547034 ]
= S [ 2.027643 ] …………………… (1)
The new debts: 13000
S
S
S
5
6
9
The present value of new debts = 13000 + S(V) 59% + S
(V)69% + S (V)99%
=13000 + S (1.09)-5 + S (1.09)-6 + S (1.09)-9
311
= 13000 + S [ 0.649931 + 0.596267 +
0.4604278 ]
= 13000 + S (1.70662578)………… (2)
From (1), (2) we conclude that:
S [ 2.027643 ] = 13000 + S (1.70662578)
S [ 2.027643 - 1.70662578 ] = 13000
S (0.3210172) = 13000
S = 13000 ÷ 0.3210172 = $ 40496.27
So the first debt = the second debt = the third debt = $
40496,27.
Fourthly - Repayment of old debts with one debt
whose nominal value equals the total nominal value of
the old debts (Medium Maturity).
Example (7):
If the El Araby Company owes the following debts:
.$ 50000 due in 4 years
.$ 70000 due in 6 years
If the company agrees to the settlement of both
obligations with debt its nominal value equal the total
nominal value of the two debts, at compound interest
8% annually, Find the maturity date of the new debt.
312
Solution:
In this case, the maturity date of the new debt
known as Medium maturity, and the duration of the
new debt is called a equivalent period.
We can
determine the maturity date of the new debt by using
present value equation in the focal date:
The present value of old debts = The present value
of new debts
50000
70000
4
Δ
6
4
50000 (V) 8% + 70000 (V)6 8% = 120000 (V)n8%
50000 (1.08)-4 + 70000 (1.08)-6 = 120000 (V)n8%
50000 (0.7350298) + 70000 (0.6301696) = 120000
(V)n8%
80863,36 = 12000 (V)n8%
(V)n8% = 80863,36 ÷ 120000 = 0.673861
By searching in table (2) under 8% about the value
0.673861,
We find it between 5, 6 periods.
This means that, n = 5 years + fraction (Δ)
(V)58% = 0.680583 ……………………..(1)
(V)n8% = 0.673861 ……………………..(2)
313
(V)68% = 0.63017 ………………………(3)
By subtracting (3) from (1) = 0.050413, this
difference in the present value equal the Difference in
the time its value one period
By subtracting (2) from (1) = 0.006722, this
difference in the present value equal the Difference in
the time its value Δ
Δ = 0.006722
÷ 0.050413 = 0.13333333
So the duration of the new debt = 5,13333333 years
To return the fraction into months and days:0.1333333 × 12 = 1,6 months
0.6 × 30 = 18 days
So: n = 5 years, 1 month and 18 days.
By calculator: _
(V)n8% = 0.673861
(1.08)-n = 0.673861
-n log (1.08) = log 0.673861
-n = log 0.673861 ÷ log (1.08) = - 5,13 periods
n = 5,13 years.
314
Example (8):
Amr owes $ 100,000 due in 4 years with interest
at 8% compound quarterly, and $200,000 due in 5
years with interest at 9% annually. If money is worth
10%, what single payment due in 6 years hence will be
equivalent to the original obligations?
Solution:
100,000
200,000
S
4
5
6
Amount of $100, 000 invested at 8% compound
quarterly for 4 years = 100,000 (1.02)16 = 100,000
(1.372786) = $ 137278,6.
Since: m = 4, r = 0.08/4 = 0.02, n = 4×4 = 16 periods
Amount of $ 200,000 invested at 9% annually
for 5 years
= 200,000 (1.09)5 = 200,000 (1.538624) = $
307724,79.
137278,6
307724,79.
S
4
5
6
2
S = 137278,6 (1.10)
+ 307724,79 (1.10) = $
504,604.38
Another solution:
315
S (V)610% = 137278,6(V)410% + 307724,79(V)5 10%
S (1.1)-6 = 137278,6 (1.1)-4 + 307724,79 (1.1)-5
S (0.564474) = 137278,6 (0.683013) + 307724,79
(0.620921)
S (0.564474) = 284835,95.
S = 284835,95 ÷ 0.564474 = $ 504,604.2
Example (9):
A piece of property is sold for $ 10,000. The
buyer pays $ 2000 cash and signs a non- interest –
bearing note for $ 4000 due in 1 year. If the seller
charges 10% converted quarterly for credit, what noninterest – bearing note due in 2 years will pay off the
debt?
Solution:
m = 4,
r = 0.10/4 = 0.025
10,000
2000
4000
S
4
8
10,000 = 2000 + 4000 (V)40.025% + S (V)80.025%
8000 = 4000 (1.025)-4 + S (1.025)-8
8000 – 4000 (0.9059506) = S (0.82074657)
8000 – 3623,8 = S (0.82074657)
316
4376,2 = S (0.82074657).
S = 4376,2 ÷ 0.82074657 = $ 5331,97.
317
Exercises
1-
Alli –El Den owes $ 100,000 due now. The
lender agrees to settle this obligation with 2
equal payments in 1 and 2 years, respectively.
Find the size of the payments if the settlement is
based
on
8%
annually.
($ 56,076.92).
2-
Mohamed owes $ 40,000 due in 2 year and $
50,000 due in 3 year. The lender agrees to the
settlement of both obligations with a cash
payment at 10% compounded semiannually,
Determine the size of the cash payment.($
70,218.87).
3-
A piece of property is sold for $ 500,000. The
buyer pays $ 200,000 cash, and signs a note for
$ 100,000 due in 1 year and a second note for $
100,000 due in 2 years.
If the seller charges 10% compound annually,
What a note due in 3 years will pay off the debt?
4-
Wessam Omar owes $ 200,000 due in 3 years
with interest at 8% compound quarterly, and
318
$100,000 due in 5 years with interest at 8%. If
money is worth 9%, what single payment due in
6 years hence will be equivalent to the original
obligations?
5-
Hesham owes $ 40,000 due in 2 years and
$50,000 due in 4 years. If money is worth 8%
converted semiannually, what single payment
due in 3 years hence will settle the obligations?.
6-
On March 1, 2016, Reham owes $ 50000,
Which she is unable to repay. Her creditor
charges her 10% converted semiannually from
that date. If the debtor pays $ 25,000 on March
1, 2017, how much would she have to pay on
March 1, 2018, to discharge the rest of her
obligation? ($33,212.81)
7-
A mother had left each child $5000 with the
provision that the money be invested at 12%
compound quarterly and the amount be given to
each child upon reaching age 21, If the children
are 12 and 16, How much would each get?
319
8-
Suppose that women in problem 7 had left $
10,000 to her children with the provision that
they are to get equal amount when they reach
age 21 at 12% compound quarterly, If the
children are 12 and 16 when the women dies,
How much will each receive?($ 11127.08)
9-
On January 1, 2016, Mohamed Omar was owed
the following debts:
$ 10000 due in 3 years.
$ 20000 due in 4 years.
$ 30000 due in 5 years.
On January 1, 2018, He agreed with the creditor
to settle the old debts with one debt due in one year,
Find the nominal value of the new debt at compound
interest 12% annually?
10-
If the flours company was owed the following
debts:
$ 20000 due in 1 year.
$ 30000 due in 2 years.
$ 40000 due in 4 years.
320
If the company wanted to settle this debts with
three equal debts due in the same maturity dates. Find
the nominal value of the new debts at compound
interest 6% semiannually?
11-
On January 1, 2018, Hesham was owed the
following debts:
$ 5000 due in 2 years.
$ 6000 due in 3 years.
$ 8000 due in 4 years.
If he wanted to settle this debts as follow:
a) $ 5000 cash payment
b) $ 5000 due On January 1, 2019
c) The rest will paid in On January 1, 2020
Find the balance remaining at compound interest
9% annually.
12-
A dealer owed the following debts:
$ 10,000 due in 4 years.
$ 20,000 due in 6 years.
$ 30,ooo due in 8 years.
$ 40,000 due in 10 years.
321
If the dealer agrees with the creditor to settle this
debts with two debts, the first due in 2 years and the
second due in 5 years, Find the new debts if the
nominal value of the first debt is twice the nominal
value of the second debt at compound interest 8%
converted quarterly?
322
CHAPTER ( 5 )
Equal Annuities
323
324
CHAPTER ( 5)
Equal Annuities
Introduction:
General definition of Annuities:
Any set of equal payments made at equal
intervals of time, form an Annuity.
Another definition:
We
can
define
Annuities
as
equal
payments with equal values pay or investment
with a regular and periodic forms such as
periodic interests, salaries, rents and revenues.
325
Types of Annuities
We
can
divide
Annuities
to
many
divisions according to the basis that used in
division.
We will present the most divisions to the
Annuities as following:-
The first division:
We can divide Annuities according to the
beginning and ending the period of payment in
two kinds.
A- Certain Annuities:
That must or sure to happen. We notice
that payments begin and end at a fixed times
such as the following example:-
326
In case of purchasing a home, we can find
a certain payments that form an Annuity
because the payments start on a fixed date and
continue until the required number has been
made. Even if the buyer of the home dies, any
outstanding debt on the home must be paid.
B-
Uncertain
Annuities
or
Contingent
Annuities:
In this kind we can not determine exactly
the beginning and ending of these Annuities
because the first or last payment or both,
depends on the occurrence of special event and
we cannot expect when it may happen or not,
such as pensions, social security and many life
insurance policies.
327
The second division:
We can divide Annuities according to its
time of payment if now or later so we will find
two kinds:
a-
Immediate Annuities: Which first
payment pay now and it may pay at the
beginning or ending the period.
b-
Deferred Annuities: In which the first
payment is made not at the beginning or
ending of the first period, but at some later
date, called deferment period.
The third division:
We can divide Annuities from the aspect
of equality or inequality to the following.
328
a-
Equal Annuities: In which all values
are equal or the same in amounts.
b-
Changeable Annuities: In which all
values of Annuities are unequal and they
may increase or decrease.
The fourth division:
We can divide Annuities from the
continuously or the act of stopping its payments
to the following:
A-
Term
or
temporary
of
limited
Annuities: In which its payments pay
during limited period.
B-
Permanent Annuities: In which its
payments continue without stopping.
The fifth division:
329
We will divide Annuities from the time of
payment in two kinds.
A-
Ordinary
Annuities:
In
which
the
periodic payments are made at the end of
each period.
B-
Due Annuities: In which the periodic
payments are made at the beginning of
each period.
Important note:
We will focus in this study only on the
certain Annuities.
So we can say that the kinds of certain
equal Annuities are eight kinds as following in
this diagram:
333
The certain Equal Annuities
Temporary or (term)
Annuities
Immediate
Permanent
Annuities
Deferred
Ordinary
Due
Ordinary
Due
(1)
(2)
(3)
(4)
Immediate
deferred
Ordinary
Due
(5)
(6)
Ordinary
Due
(7)
(8)
Very important note:
We will focus in this chapter on how we
can find the amounts and present values to all
these eight previous Annuities as following:333
First: The amount of temporary equal
Annuities:
Question:
What do we mean by the amount of
temporary equal Annuities?
The answer:
It is the sum of all periodic payments and
the accumulated compound interest to the end of
the term.
Then:
The equations of the amount of Annuities =
The value of Annuity multiply in the
amount of a temporary equal Annuity to n years
and for one pound.
332
Important Note:
Now we will present the symbols that will
use in obtaining the amount of Annuities where
the value of Annuity is one pound and for n
years.
sn
Refer to the amount of a temporary
immediate ordinary Annuity to n
years and for one pound.
sn
Refer to the amount of a temporary
immediate due Annuity to n years and
for one pound.
m/sn
Refer to the amount of Annuity which
is deferred to m years and temporary,
immediate and ordinary to n years and
for one pound.
m/sn
Refer to the amount of Annuity
deferred to m years and temporary,
333
immediate, and due to n years and for
one pound.
Note:
We will explain how we can find the
amount of equal Annuities as following.
1-
The amount of a temporary immediate
ordinary Annuity to n payments and for
one pound.
The used symbol is sn
Note: At the first we will present this numerical
example.
Dr. Hossam deposits $ 1000 a year in an
account paying 10% compounded annually.
334
Question:
What amounts that accumulated just after
the fifth deposit is made?
now
solution
1000
1000
1000
focal date
1000
5 year
only 1000
4
3
1000 (1.1)
1000 (1.1)2
1000 (1.1)3
1000 (1.1)4
2
1
We find:
5 th payment
Still fixed
$ 1000
4 th payment
Invested to one year
(1000x1.1)
1100
3 rd payment
Invested to two year
(1000x1.1)2
1210
2 nd payment
Invested to 3 years
(1000x1.1)3
1331
1 st payment
Invested to 4 years
(1000x1.1)4
The amount of Annuity
335
=
1464.1
6105.1
Note:
Nc n nfhfgghm4rh $1
1
2
Beginning
$1
n–2
n–1
Annuity ends
n
$1
$1
payments
$1
1(1+i)1
1(1+i)2
sn
1(1+i)n-2
1(1+i)n-1
The main hypotheses to deduce the formula:
1-
We have an ordinary Annuity of n
payments of $ 1 each, and rate of i per
period.
2-
We accumulate each payment to the end
of the term.
3-
The last payment will be $ 1 because it has
had no time to earn investment.
336
4-
The next to the last payment will be 1(1+i)
or simply (1+i) because it has earned
interest for 1 period.
5-
The payment before this will amount to
(1+i)2.
6-
The first payment will amount to (1+i)n-1
because it has earned interest for 1 period
less than the number of payments
Important note:
In the previous diagram we can conclude
the amount of Annuities = the total of all
periodic payments at the end of n periods.
= amount of first payment + the amount of second
payment + the amount of third payment + …… +
the amount of last payment + 1.
337
So we have sn = (1+i)n–1 + (1+i)n–2 + (1+i)n–3 +
…… + (1+i) + 1
We will rearrange the previous terms
using ascending order to become as ascending
geometric progressing as following
Sn = 1 + (1+i) + (1+i)2 + …… + (1+i)n–2 +
(1+i)n–1
We obtain the sum of the terms on the
right side.
An observation:
These
terms
represent
geometric progression
So we find:

The first term is ( 1 ).

The common ratio is ( 1 + i ).

The number of terms is n.
338
ascending
Notice:
In algebra the sum of ascending geometric
progression is:
S=Ax
rn – 1
r–1
Hence:
A
first term
substitute by ( 1 )
r
common ratio
substitute by (1+i)
n
number of terms
In substituting values in formula of
ascending geometric progression
Sn = 1 *
( 1 + i )n – 1
(1+i)–1
hence:
Sn =
( 1 + i )n – 1
i
339
At the end:
Sn = R * sn = R *
( 1 + i )n – 1
i
Hence:
Sn
= Refer to amount of an ordinary Annuity
of n payments.
R
= Refer to periodic payment or rent.
sn
= Refer to amount of $ 1 per period for n
periods at the rate i per period.
Question:
How we can obtain the value of s n ?
Answer: By using two methods:
The first:
Using the tables of compound interest that
there is a numerical values of s n in the third
column.
343
The second:
Using calculators with exponential capability.
2-
The amount of a temporary immediate
due Annuity to n payments and for one
pound.
The used symbol is sn
We will clear the main idea through this
diagram.
Beginning
1
2
N periods
n–2
Ending
n–1
(1+i)
(1+i)2
sn
n-1
(1+i)
(1+i)n
From the previous diagram we can notice:

The first payment invests to all period and
the amount of it equals = 1 ( 1 + i )n.
343

The second payment invests from the
beginning of second period to the end of
period, so it invests to ( n – 1 ) and its
amount S = 1 ( 1 + i )n–1.

The third payment invests from the
beginning of third period to the end of
period, so its amount S = 1 ( 1 + i )n–2 and
so on.

The payment before the last invests to two
years and its amount S = 1 ( 1 + i )2.

The last payment invests to one year and
its amount S = 1 ( 1 + i ).
So the amount of Annuities = the amount
of first payment + the amount of second
payment + …… the amount of last payment.
342
So: S n = (1 + i)n + (1 + i)n–1 + (1 + i)n–2 +
…… + (1 + i)n + (1 + i)1
Then: We apply the formula of ascending
geometric progression S = A *
rn – 1
r–1
So:
a
First term equal ( 1 + i ).
r
Common ratio equal ( 1 + i ).
n
Number of terms.
Then:
Sn =(1+i)*
=(1+i)*
(1+i)n – 1
(1+i) – 1
(1+i)n – 1
i
The final formula:
Sn =(1+i)*
(1+i)n – 1
i
Question:
How we can option the amount of Sn .
343
Answer: Using two methods.
The first:
Using the tables of compound interest that
there is a numerical values of S n in the third
column and the relationship between the amount
of ordinary and due Annuity as following.
The amount of due immediate Annuity
= S n i for A mount R.
Sn i=R*(1+i)*
(1+i)n – 1
i
Sn =R*(1+i)*Sn
In this case we can use the tables of
compound interest from third column under the
rate i and n periods.
Also: We can use the relationship between the
due and ordinary Annuity as following:
344
Sn i=R*{Sn+1 –1}
The second method:
Using calculators with exponential capability.
So the final result:
How we can a obtain : S n
The amount of due Annuity by:
=R*Sn
Or
=R*(1+i)*Sn
Or
=R*{Sn+1 –1}
Or
= R * ( 1 + i )*
(1+i)n – 1
i
345
3- The amount of a temporary deferred,
ordinary Annuity.
The used symbol is m/s n is the same Sn
we will display the following diagram.
Deferred period
m
1
Temporary period
2
n
1
(1+i)
(1+i)2
(1+i)n-2
(1+i)n-1
Important notice:
We notice that the interval of deferment
(m) period ends before the first payment.
So we can decide that the deferred period
has no effect in calculation the amount of
Annuity because we find from the pervious
346
diagram that the deferred period ends before the
first payment.
So we can say that m/s n
is the same s n
4- The amount of a temporary defered,
due, Annuity:
The used symbol is m/s n is the same
s m because the deferred period has no effect in
calculation the amount of Annuity.
Important notice:
We can understand and deduce this fact
from the previous diagram.
5-
The amount of a permanent equal Annuities:
Question: What we mean by a permanent
Annuity?
347
The answer:
It is define as the Annuity that in which its
payments continue without stopping such as the
rent of land, so its numbers are unlimited.
So we can not calculate or determine
the amount of a permanent Annuities:
At the end of this part from this chapter
we can summarize all the equations that we will
use subsequently in the solved examples.
The amount of a temporary Annuities:
1-
The
amount
of
a
temporary,
immediate ordinary Annuity.
The used symbol is s n i
The amount of Annuities is = R x S n i
348
We can obtain s n using this equation
sn =
(1+i)n – 1
i
In this case you can use calculators with
exponential capability.
Or
Using the table of compound interest.
2-
The amount of a temporary, immediate,
due, Annuity.
The used symbol is s n i
We can obtain it using calculator through
applying this equation
sn =R*(1+i)*
Or
(1+i)n – 1
i
Using the table of compound interest
through these equations:
The amount of Annuity = R * ( 1 + i ) * s n
349
Or
=R*{sn+1i–1}
3-
The amount of m/s n is the same s n
4-
The amount of m/s n is the same s n
5-
There is no amount of permanent Annuities.
Solved Examples
Example 1:
Engineer Hesham deposits $ 2000 in a
bank every year with a compound interest
annually 12% for 10 years.
Find the amount of deposit in the
following cases.
A-
If he deposits the money at the end of
every year.
353
Solution
Important observation:
Any person who will study this subject
must basically, determine the type of Annuity
before solution to prove that he understands the
problem.
The required:
Is to obtain the amount of Annuity,
ordinary and temporary to 10 years, its payment
is $ 2000 .
The amount = R * s n i = 2000 S 10 12%
=R*
(1+i)n – 1
i
= 2000 *
(1+0.12)10 – 1
0.12
= 2000 *
3.108482 – 1
0.12
353
= 2000 *
2.108482
0.12
= 2000 * 17.548735
= 35097.47
Another solution:
We can search in the table of compound
interest at the column s n under the rat 12%.
In front of 10 years we will find the value
17.548735.
So the amount of Annuity
= 2000 * 17.548735 = 35094.47
The same answer.
B-
If the Engineer deposits the money at
beginning of every year.
352
So the required is:
To obtain the amount of due and a
temporary Annuity as following:
The first method: By calculator.
The amount is =
= R * s n i, R = 2000, i = 12%, n = 10
= 2000 S 10 12%
= 2000 * ( 1 + i ) *
(1+i)n – 1
i
= 2000 * ( 1 + 0.12 ) *
(1+0.12) – 1
0.12
= 2000 * 1.12 * 17.548735 = 39309.17
Note:
Try to determine the reasons of the
difference between the final results in A and B.
353
The second method:
Using the table of compound interest.
A-
The amount of Annuity = R * s n
= 2000 * s 10 12%.
= 2000 * ( 1 + 0.12 ) * s 10 12%
= 2000 * 1.12 * 17.548735 = 39309.17
B-
We can transfer due Annuity to ordinary
Annuity by applying this equation.
The amount = R * s n i
=R*{sn+1 –1}
= 2000 * { s 11 12% – 1 }
354
By searching in the table of compound
interest under the rate 12% and in the front of
11 years.
= 2000 { 20.6545800 – 1 }
= 2000 * 19.65458 = 39309.17
C-
If the Engineer deposits money at the end
of every year after deferred period equal 5
years.
m=5
n = 10
S
m/sn
5/s 10 12%
Determination the type of Annuity:
A deferred, ordinary, and temporary
Annuity.
Then: m = 5
n = 10
i = 12%
The amount = R * m / S n i
= 2000 * 5 / S 10 12%
355
R = 2000
We know previously that the deferred
period has no effect on the amount of Annuity
So the amount = 2000 * 5 / s 10 12%
= 2000 * 5 10 12%
E-
If the Engineer deposits money at the
beginning of every year after deferred
period equal 5 years.
Determination the type of Annuity:
A deferred, due, and temporary Annuity
Then: m = 5
n = 10
i = 12%
The amount = R * m / s n i
= 2000 * 5 / S 10 12%
= 2000 * S 10 12%
= 2000 { s 11 – 1 } 12%
356
R = 2000
= 2000 { 20.654583 – 1 } = 39309.17
Or
= 2000 * (1.12 ) *
(1+0.12)10 – 1
0.12
= 39309.17
Example 2:
Wessam deposits $ 500 at the end of each
year in an account paying 6% provided that
interest compounded annually.
What amount in her account after 4 years?
Solution
Determination the kind of Annuity:
Temporary to 4 years and ordinary.
Then: R = 500 n = 4
i = 6%
By substituting in the formula of ordinary,
temporary Annuity.
357
Sn = R * S n i = R *
= 500 *
(1+i)n – 1
i
(1+0.06)4 – 1
0.06
= 500 * S 4 6%
= 2187.308
Question: In the previous example.
What is the amount of a compound
interest she achieved?
Solution
The interest = the amount of ordinary Annuity –
the original payments
= 2187.308 – 500 * 4 = 187.308
Example 3:
Soso deposits $ 200 at the end of each 3
months in a bank that pays 5% converted
quarterly.
How much will she has to her credit at the end
of 10 years?
358
Solution
Notice:
We find that the bank adds the interest
quarterly.
So we must transfer the rate and period as
following.
i=
5%
4
=
0.05
4
= 1 1%
4
or = 0.0125
N = 10 * 4 = 40 – then substituting.
In the formula of temporary, ordinary Annuity.
The amount of Annuity = R * S n i
=R*
(1+i)n – 1
i
= 200 * 5 40 1
1
4
%
The amount of Annuity
= 200 *
(1+0.0125)40 – 1
0.0125
359
= 10297.91
The total compound interest:
= the amount of Annuity – the total deposits
= 10297.91 – 200 * 40 = 2297.91
Example 4:
Find the amount of an Annuity of
$ 5000 per year for 10 years at: (a) 6%, (b) 7%
if the interest is compounded annually.
Solution
Very important remark:
If the example does not mention the type
of the Annuity frankly, if it is ordinary or due
we must consider it as an ordinary Annuity
continuously.
So in the previous example, we will
consider the Annuity as ordinary Annuity.
363
When the rate 6%: R = 500, n = 10, i = 6%
By substituting in the formula of the
amount of an ordinary and temporary Annuity.
The amount of Annuity = R * S n i
=R*
(1+i)n – 1
i
= 5000 x
= 5000 * 5 10 6%
(1+0.06)10 – 1
0.06
= 65903.9747
When the rate 7%: R = 5000, i = 7%, n = 10
The amount of Annuity = 5000 * S 10 7%
= 5000 x
(1+0.07)10 – 1
0.07
= 69082.239
Example 5:
Find the amount of an Annuity of
$ 1200 at the end of 6 months for 5 years. if
money is worth (a) 5%, (b) 6% all rates are
converted semiannually.
363
Solution
Because of the interest is added at the end
of each 6 months.
So we must change the rate and the period.
When the rate 5%: then: i =
or
0.025
5%
=
2
2
1
2
%
n = 5 * 2 = 10
Then the amount of Annuity:
=R*Sn =R*
= 1200 *
(1+i)n – 1
i
(1+0.0 25)10 – 1
0.025
When the rate 6%: The i =
= 13444.06
6%
2
= 3%
0.03
N = 5 * 2 = 10
The amount of Annuity= 1200 *
= 13756.65
362
(1+0.03)10 – 1
0.03
Example 6:
Noorhan deposits $ 1000 as Annuity at the
end of each month for 6 years.
What the amount of the Annuity if money is
worth 6% compounded monthly?
Solution
We must transfer the period and rate.
N = 6 * 12 = 72, i =
6%
12
= 12 %
The amount of Annuity = R *
R = 1000
(1+0.005)72 – 1
0.005
= 86408.8557
Example 7:
Find the amount of an Annuity of $ 750 at
the end of each month for 12 years if money is
worth 7% converted monthly.
363
Solution
Number of periods become = 12 * 12 = 144
i=
7%
12
=
0.07
12
% = 0.0058333
Apply the Equation = the amount of Annuity
= 750 *
(1+0.005833)144 – 1
0.005833
= 168616.1729
Example 8:
Pro. Dr. Eieed try to provide for his son’s
Ahmed education, he deposits $ 1200 at the end
of each year for 18 years. If the money draws
8% interest.
How much does the fund contains just after the
eighteenth deposit is made?
364
Solution
Notice: The main idea in this example is to
obtain the amount of an ordinary Annuity.
So
n = 18 , i = 8% , R = 1200
Then we apply the formula.
The amount of Annuity = R * S n
= 1200 *
(1+0.08)18 – 1
0.08
= 44940.29
Example 9:
In the previous example if there is no more
deposits are made, but the amount in the fund is
allowed to accumulated at the same interest.
Required
How much will the fund contain in 3 more
years?
365
The idea:
Is to obtain the amount after 3 years using
compound interest as, one, payment.
So: P = 44940.29 , i = 8% , n = 3
Apply the formula: S = P ( 1 + i )n
= 44940.29 * ( 1 + 0.08 )3 = 56611.83
Example 10:
Engineer Hesham purchased a real estate
and paid a deposit and agreed with the seller or
vendor to pay the rest of the price with annual
premiums, pay at the end of every year for 15
years provided that the premiums will double
every 5 years, using compound annual rate
equal 8%.
366
Required
Calculate every premium if you know that
the accumulated capital from all the premiums
at the seller equal 53371.818 at the end of all the
periods. Using two methods in solution.
The first solution:
Important notice:
We find that the premiums will double
every 5 years through 15 years.
So we will suppose the value of every
premium through the first 5 years = X pound
and the value of the premiums at the second
5 years = 2 X Pound and the third =
4 X pound.
367
The following, diagram, will, clear, the
example.
5 years
x
x
x
5 years
x
5 years
x
2x
2x
2x
2x
2x
4x
4x
4x
4x
4x
So:
The total accumulated capital at the end of
15 years.
= X * S 15 + X * S 10 + 2 X * S 5
8%
8%
8%
= X { S 15 + S 10 + S 5 } = 53371.818
=X
(1+0.08)15 – 1
0.08
+
(1+0.08)10 – 1
+
0.08
2*(1+0.08)5 – 1
0.08
= 53371.818
= X { 27.152114 +14.486562 + 2 * 5.866601}
= 53371.818
368
= 53.371818 x = 53371.818
X=
53371.818
53.371819
= 1000
So: The value of premium through the first
five years = 1000
The second = 2000 – the third = 4000
The second solution:
8%
53371.818 = X * S 5 * (1 + 0.08)10
8%
8%
5
+ 2 X * S 5 * (1+0.08) + 4 X * S 5
53371.818 = { 5.8666096 * 2.158984997 } X
{ 12.66557011 } X
+ 2 X * { 5.8666096 * 1.469328077 }
X { 17.2399489 } + 4 X { 5.8666096 }
= [ 23.444384 ] X = 53.371818
5337.818 = 53.371818 X
369
= 1000
5
1
2000
5
2
4000
5
3
Periodic payment of an Annuity:
Hence
Sn=R*Sn =
(1+i)n – 1
i
So: R =
Sn
Sn
=Sn*
=Sn*
1
Sn
*R
i
(1+i)n - 1
Hence:
R
= Periodic payment or rent.
S n = Amount of Annuity of n payments.
1
Sn
Periodic deposit that will grow to $ 1 in n
payments.
373
Example:
How much must a person save every 6
months to accumulate $ 3000 in 4 years if
money is worth 5% compounded semiannually.
1
R
2
R
3
R
4
R
5
R
6
R
Amount
$ 3000
8 payments
8
R
7
R
A time diagram shows that the 3000 is an
amount in the future
By substituting
S 8 = 3000 , n = 8 and i = 2
R=
Sn
Sn
=
Or = S n *
3000
8.7361159004
3000
S 8 2 12 %
=
i
(1+i)n - 1
= 3000 *
1
2
%
= $ 343.4
0.025
(1.025)8 – 1 =
343.4
Note that: The gross of 8 payments equal 343.4
has total 2747.2
373
The balance needed to produce an amount
of $ 3000 comes from the accumulated interest
on each payment from the time it is made to the
end of the 4 years.
To make sure that your solution is right.
Sn=R*Sn =R*
= 343.4 *
(1+0.025)8 – 1
0.025
(1+i)n – 1
i
= 3000
General example:
Example 1:
Dr. Ahmed Eeid deposits $ 100 at the end
of two months for two years with rate 6%
annually. After that he deposits $ 300 at the end
of 3 months for 3 years with rate 8% annually
and he left all his balance at the bank to invest
for 18 month with rate $ 10% semi annually.
372
Required: Find the final balance:
Steps of solution:
1- Draw the following diagram.
N=
100
i=
24
2
3 years
2
3 3
= 12
6% = 1%
6
100
18 months
N = 36 = 12
3
i = 8% = 2%
4
300 300
3
n=
i=
300
18 = 3
6
10 = 5%
2
300
The first payment:
We consider it as an ordinary, temporary
Annuity.
Then: n =
So
24
2
= 12 ,
i=
6%
6
= 1%
The amount of first Annuities.
= 100 * S 12 1%
= 100 *
(1+0.01)12 – 1
0.01
373
= 1268.25
The balance
Two years
2 2
Note:
This previous amount will invest two
times so the amount consider as one payment
then we apply the following formula.
S = P ( 1 + i )n
So
S = 1268.25 * (1 + 0.02)12 * (1 + 0.05)3
= 1861.97
Note : We change n and i.
The second Annuity:
The amount of Annuities
= 300 *
(1+0.02)12 – 1
0.02
= 4023.63
Note:
This amount will invest to another period.
374
S = P ( 1 + i )n
Then
= 4023.63 ( 1 + 0.05 )3 = 4657.85
So the final balance = 1861.97 + 4657.85
= 6519.82
In the previous example if the rate still
fixed during all the invested period equal 6%.
Find the final balance
n = 12
i = 66 = 1%
n = 12
i = 64 = 1.5%
n=3
i = 62 = 3%
The first Annuity:
= 100 * S 12 1% (1 + 0.015)12 * (1 + 0.03)3
= 1656.95
The second Annuity:
= 300 *
(1+0.015)12 – 1
0.015
* (1+0.03)3 = 4275.15
375
= 4275.15 + 1656.95
The final balance = 5932.1
Please try to make a comparison between
the final result in two cases and write a
comment from you own mind.
Example 2:
Dr. Ammr, Nadi, Ezat deposits $ 1000 at
the end of every year for 7 years, after that he
reduce the deposit to $ 500 for another 3 years.
Find:
The final balance to him if the compound
interest rate 9% annually.
In two cases:
a-
At the end of 10 years.
b-
At the end of 12 years.
376
A- At the end of 10 years:
7 years
3 years
1000
500
10
The final balance = 1000 s 7 9% * (1+0.09)3
+ 500 * S 3 9% = ???
Try to find the final result by your self.
B- At the end of 12 years:
12
7 years
3 years
1000
2 years
500
The final balance:
= 1000 s 7 (1+0.09)5 + 500 x S 3 (1+0.9)2=???
Try to find the final result by your self.
377
Exercises on the amount of an ordinary
Annuity:
1-
Dr. Hossam deposits $ 10000 at the end of
each year in an account paying 10%
provided
that
interest
compounded
annually what amount in his account after
10 years?
2-
Wessam deposits $ 2000 at the end of each
3 months in a bank that pay 8% converted
quarterly how much will she has to her
credit at the end of 8 years?
3-
Dr. Mohamed Omar, deposits an Annuity
of $ 7000 per year for 10 years at:
a- 10%. b- 13.5%.
c- 14.25%.
If the interest is compounded annually.
979
Find the amount of the annuity?
4-
Engineer Hesham deposits $ 5000 at the
end of each 6 months for 6 years if money
is worth:
a- 6%.
b- 8%.
c- 10%.
All rates are converted semiannually.
Find the amount of annuity.
5-
Noorhan puts $ 300 every three months in
A saving account that pay 12% compounded quarterly, if the first deposit was
made on June, 1, 2018. How much will be
in her account Just after the deposit made
on December, 1, 2027 ?
983
Second:
The present value of temporary
equal Annuities.
Question:
What do we mean by the present value of
A temporary, ordinary Annuity?
It is the sum of the present values of all
periodic payments of the Annuity at the
beginning of A period or any time before the
beginning.
So: The present value of Annuities:
= The value of payment * the present
value of Annuity for one pound.
983
Question:
What are the kinds of symbols that we will
use in calculating the present value of
Annuities.
They are :
1- A n
Refer to the present value of A
temporary,
immediate,
and
ordinary
Annuity for n period and for one pound.
2- Ä n
Refer to the present value of A
temporary, immediate, and due Annuity
for n period and for one pound.
3- m/A n
Refer to the present value of
Annuity
deferred
to
m
years
also
temporary to n years, and ordinary to one
pound.
983
4- m/Ä n
Refer to the present value of
Annuity
deferred
to
m
years
also
temporary to n years and due to one
pound.
5- A ∞
Refer to the present value of A
permanent, immediate, ordinary Annuity
for one pound.
6- Ä ∞
Refer to the present value of A
permanent, immediate, due Annuity for
one pound.
7- m/A ∞
Refer to present value of A
deferred
Annuity
to
m
years
also
permanent and ordinary for one pound.
8- m/Ä ∞
Refer to the present value of A
deferred
Annuity
to
m
years
permanent and due for one pound.
989
also
Question:
How we can find the present value of A
certain equal Annuities for one pound?
First:
The present value of A temporary Annuities:
1-
The present value of
A temporary
immediate, ordinary Annuity.
The used symbol is A n . To obtain the
present value of A n , we must follow these
steps:
a- We assume an Annuity of n payments of
$ 1 each and A rate of i per period.
b- We
discount
each
payment
beginning of the Annuity.
983
to
the
c- The sum of these discounted values is
designated by the symbol A n .
d- We will clear these steps in the following
figure.
Present value of A temporary, ordinary
Annuity of n payments of $ 1.
Annuity begins
An
V1
V2
V3
Annuity ends
n–1
n
$
Vn-1
Vn
From the previous diagram we notice:
*
The first payment mature after one period
so present value = 1 * V1 .
*
The second payment mature after two
period so present value = 2 * V2 .
983
*
The last payment mature after n period so
present value = 1 * Vn .
Then: The present value to Annuities =
A n = V1 i% + V2 i% + V3 i% + …… Vn i%
We want to find the sum of the terms in
the right side, also we notice that these terms
represent A decreasing geometric progression
hence:
 The first term is v .
 The common ratio is v .
 The number of terms is n .
Now: We can apply the formula of the sum of
decreasing geometric progression.
S= A *
1–rn
1–r
983
By substituting the previous terms in the
formula of decreasing geometric progression.
We find:
An =V*
1–Vn
1–V
We try to remember our students to this
subject that:
V=
1
1+i
So: A n = V *
=V*
1–Vn
1Xi
1+i
Hence:
1 – Vn
1– 1
1+i
=V*
1–Vn
V Xi
=V*
Vn =
1
(1+ i)n
Then: A n =
1–Vn
i
Or: A n =
1 - (1 + i)
i
1 – Vn
1 + i – 1=
1+i
=
1–Vn
i
= ( 1 + i )-n
–n
987
V*
1 – Vn
i
1+i
Question:
How we can obtain the value of A n ?
By using two methods:
The first:
From the table of compound interest under
the rate i and in the front of n in the column of
present value of ordinary Annuity for one
pound.
The second:
Using calculator
The present value of Annuities =
=R*An=R*
Or: = R *
Hence: Vn =
1 – (1 + i)–n
i
1–Vn
i
1
(1 + i)n
Or: = ( 1 + i )-n
988
2-
The present value of A temporary, immediate, due Annuity.
The used symbol is Ä n :
We will present this diagram.
Annuity begins
Än
Annuity ends
n-1
n
1 1
V1
V2
Vn-1
Ä n = 1 + V1 + V2 + V3 + …… Vn-1
Än =1*
=
1–V
1–V
1–Vn
i
1+i
=
n
=
1–Vn
1– 1
1+i
1–Vn
i* 1
1+i
Ä n = ( 1 + i )*
=
1–Vn
i
Or: Ä n = ( 1 + i ) * A n
989
=
1–Vn
1+ i - 1
1+i
1–Vn
iXV
Question: How we can obtain Ä n
1-
By using tables of compound interest but
we can not obtain the present value of due
Annuity frankly, so we transfer it to
ordinary Annuity by two methods.
The first: Ä n = R * Ä n
Än=R*(1+i)An
Or = R { A n – 1 + 1 }
The second method using this formula:
Än=R*Än
2-
=R*{An–1+1}
Using calculator.
Än=R*Än =R*(1+i)*An
=R*(1+i)*
1
(1+ i)n
1 – Vn
i
As
Vn =
= ( 1 + i )-n
So
Än=R*(1+i)*
993
(1 – i)-n
i
3-
The present value of deferred, temporary,
ordinary Annuity.
The used symbol is m / A n
Deferred period
m
n payments
1XVm+1
1XVm+2
m+n
V
So
m+n
m / A n = Vm+1 + Vm+2 + …… Vm+n
= Vm { V1 + V2 + V3 + …… Vn }
m / A n = Vm * A n i
Question:
How we can obtain the value m/A n :
by:
993
A- Using the table of compound interest from
the column of Vm and A n directly.
Or
Using the following formula.
m/An=R*{Am+n –Am }
B- Using calculator.
m / A n = R * Vm i * A n i
Hence:
Vm i =
1
(1 + i)m
Also
Ani=
1 – (1+i)-n
i
4-
The
present
value
Vn =
of
A
1
(1 + i)n
deferred,
temporary, due, Annuity.
The used symbol is m / Ä n
Deferred period
m
n-1
1XVm
1XVm+1
Vm+n–1
993
n
The present value
m/Ä n = Vm + Vm+1 +Vm+2 + Vm+3 + … Vm+n-1
m / Ä n = Vm { 1 + V + V2 + V3 + …… Vn-1 }
m / Ä n = Vm * Ä n
In this case we must transfer due Annuity
to ordinary Annuity.
So: m / Ä n = R * V m–1 * A n
Or: m /Ä n = R { A m + n – 1 – A m – 1 }
Using calculator:
m/Ä n = R * m/Ä n
Or: R * Vm * Ä n
= R Vm–1 * A n i
Or: = R * Vm-1 *
1 – Vm
i
Or: m/Ä n = R { A m + n – 1 – A m – 1 }
Hence:
Vm =
1
(1+i)m
999
Vn =
1
(1+i)n
Solved examples:
1-
Dr. Islam Nadi Ezat deposits $ 1000 in A
bank to 25 years. The bank gives compound return 11% Annually.
Find the present value to this Annuity if:
a-
The deposit begins at the end of every
year.
Required:
Find the present value of A temporary,
immediate, ordinary Annuity. Then the used
symbol is { A n }:
1-
Using table of compound interest:
- So we search under the rate 11% and in
the front of 25 to the fourth column.
- So A n = 1000 * 8.421745 = 8421.745
993
2- Using calculator:
1 – (1+i)-n
i
Än=R*An =R*
= 1000 *
b-
1 – (1+0.11)-25
0.11
= 8421.745
The deposit begins at the beginning of
every year:
Required:
Find the present value of A temporary,
immediate, due Annuity.
The used symbol is {Ä n }
Using two methods:
1-
Än=R*Än
So
Än=R* (1+i)An
= 1000 ( 1 + 0.11 ) * A 25 11%
= 1000 * ( 1.11 ) *
1 – (1+0.11)-25
0.11
993
= 9348.137
2-
Ä=R{An–1 +1}
= 1000 { A 25 – 1 } + 1 = 1000 { A 24 + 1 }
= 1000 * { 8.348136 + 1 }
= 1000 * 9.348136 = 9348.136
C-
The deposit at the end of every year after
deferred period = 10 years
The used symbol is m / A n
In briefly:
m/An=R*m/An
= R Vm * A n
= 1000 V10 * A 25 11%
By using compound interest table:
= 1000 * 0.352184 * 8.421745 = 2966.008
993
Or
m/An=R*m/An
=R*{Am+n –Am }
= 1000 { A 35 – A 10 }
= 1000 { 8.855240 – 5.889232 }
= 1000 * 2.966008 = 2966.008
1 – (1+0.11)-25
0.11
Or
= 1000 * V10 *
D-
The deposit at the beginning of every year
= 2966.004
after deferred period equal 10 years.
The used symbol is m / Ä n
Then m / Ä n = R * Vm-1 * A n 11%
= 1000 * V9 * A 25 11%
= 1000 * 0.3909247 * 8.421745 = 3292.27
997
Or: m/ Ä n = R { A m + n – 1 – A m – 1 }
= 1000 { A 34 11% – A 9 } = 3292.27
Example:
In 1/1/2020, A trader agreed with the Arab
bank to pay annual Annuity instead of him
equal $ 3000 for 7 years, but the first Annuity
begins in 1/1/2026 the bank used compound
interest equal 7% converted Annually.
Find the deposit that the trader must pay now to
the bank in 1/1/2020.
Solution
The min idea of the example is to obtain
present value of Annuity.
998
We will present this diagram.
1/1/2020
1/1/2026
1
2
3
7
We notice that the example do not
determined if the Annuity may pay at beginning
or at ending of every year.
So we can consider it due or ordinary:
When we consider it due.
So the Annuity deferred to 6 years and
temporary to 7 years.
The present value of Annuity = R * m / Ä n
= 3000 * 6 / Ä 7
= 3000 * V5 * A 7 7% = 11527.46
We substitute symbols by using tables.
999
When we consider it ordinary:
The Annuity deferred to 5 years and
temporary to 7 years.
By using tables the present value
= 3000 * 5 / A 7 7%
= 3000 * V5 * A 7 7% = 11527.46
The present value of A permanent,
equal Annuities:
1-
The
present
value
of
an
immediate, permanent Annuity.
The used symbol is A ∞
The present value A ∞ =
Hence A ∞ = R *
1
i
333
1
i
=R*
1
i
ordinary
2-
The present value of A due, immediate,
permanent Annuity.
The used symbol is Ä ∞ = 1 +
1
i
So the present value Ä ∞ = R * Ä ∞
=R*{1+
3-
1
i
}
The present value of A deferred, permanent, ordinary Annuity.
1
i
The used symbol is m / A ∞ = Vm *
The present value m / A ∞ = R {
4-
1
i
–A m }
The present value of A deferred, permanent, due Annuity.
The used symbol m / Ä ∞
The present value m / Ä ∞ = Vm * { 1 +
So
m/ Ä ∞ = R * m / Ä ∞
333
1
i
}
m/ Ä n = R * Vm * { 1 +
Or
1
i
=R{
1
i
}
–Am–1 }
Important Notice:
In all previous solved examples, we refer
to the amount of annuity by the symbol ( R ) to
simplify the subject of annuities.
Solved general examples
Example 1:
Adham Elsharkawey purchased A real
estate and paid A deposit equal $ 18000. The
purchaser agreed with the vendor to pay the rest
of the price with A regular Annuities equal $
1500 at the end of every year for 30 years.
333
Required:
Calculate the price of purchase the real
estate using converted annual rate 15%.
Solution
Notice that:
The price of purchase the real estate =
The deposit of the price + present value of
regular Annuities
Steps of solution:
1-
Finding the present value of Annuities
using this formula.
An=R*An =R*
Notice: R = 1500 i = 15%
1 – ( 1 + i )-n
i
n = 30
The present value of Annuities:
339
= 1500 *
1 – (1 + 0.15)–30
0.15
= 1500 * 6.565979
= 9848.9694
The price of purchase the real estate=
The deposit of the price + the present value
of Annuities.
= 18000 + 9848.9694 = 27848.9694
Examples 2:
Farida Elkholy as A donor wants to
provide A 3000 scholarship every year for 4
years with the first to be awarded 1 year from
now. If the school can get 9% return on its
investment.
Required:
A.
How much money should the donor give
now?
333
B.
Show the table to clear what will happen
to the investment?
Solution
The steps:
Substituting in the formula of present
value to determine the money should the donor
give now.
An=R*An =R*
= 3000 * A 4 = 3000 *
2-
1 – ( 1 + i )–n
i
1 – ( 1 + 0.09 )–4
=
0.09
9719.16
Prepare A table to clear what will happen
to investment.
Question:
Do you believe that 3000 will actually be
paid from this gift?
333
Answer:
The following table will show what will
happen to the investment through 4 years and
wait to find the answer of previous question by
your self.
Original investment
Interest at 9% for 1 year
Amount after first year
First scholarship
Investment at beginning of second year
Interest at 9% for 1 year
Amount after second year
Second scholarship
Investment at beginning of third year
Interest at 9% for 1 year
Amount after third year
Third scholarship
Investment at beginning of fourth year
Interest at 9% for 1 year
Amount after fourth year
Fourth scholarship
$
(+)
$
(–)
$
(+)
$
(–)
$
(+)
$
(–)
$
(+)
(–)
9719.16
1874.72
10593.88
3000
7593.88
863.45
8277.33
3000
5277.33
474.96
5752.29
3000
2752.29
247.71
3000
3000
0000
Notice:
We find that 3000 paid to four years.
333
Please: Write a comment from your own mind
about the final result.
Question:
Design examples from you own mind
similar with the previous example. To have
more training.
Example 3:
Hosney Elsharkay as A businessman
wants to give A charity amount of money to
spend on sick people for 4 years. He intends to
pay 12679.5 now. Prove that this amount
enough to pay 4000 for 4 years to this organization provided that the first amount will pay
after one year, and the amount can get 10%
return on its investment Annually. And try to
make sure that your solution is right.
337
Solution
To verify that 1279.5 enough to pay 4000
for 4 year, we will prepare the following table to
show what will happen to the investment
through 4 years using the rate of investment
10% converted compounded Annually.
Original investment
Interest at 10% for 1 year
Amount after one year
First amount of donation
Investment at beginning of second year
Interest at 10% for one year
Amount after second year
Second amount of donation
Investment at beginning of third year
Interest at 10% for 1 year
Amount after third year
Third amount of donation
Investment at beginning of fourth year
Interest at 10% for 1 year
Amount after fourth year
Fourth amount of donation
338
$
(+)
$
(–)
$
(+)
$
(–)
$
(+)
$
(–)
$
(+)
(–)
12679.5
1267.95
13947.45
4000
9947.45
994.75
10942.20
4000
6942.22
694.22
7639.42
4000
3636.42
363.64
4000
4000
0000
Final result:
We find that the amount 12679.5 will
become enough to pay 4000 for 4 years using
the rate of investment 10% converted Annually.
To make sure that the solution is right we can
substitute in the formula of the present value of
A temporary, ordinary Annuity
An =R*A n
= 4000 *
1 – ( 1 + 0.1 )–4
0.1
= 12679.5
Example 4:
Prof. Dr. Eeid wants to buy A flat has an
annual rent equal 6000, the used compound
interest rate is 6% Annually.
Find:
The price of purchasing the flat if:
339
a- The rent pay at the end of every year.
The price of purchasing the flat = the
present value of A permanent annual Annuity
( rent ) that equal $ 6000.
If the rent pay at the end of every year
( ordinary Annuity ) the price of purchasing the
flat
=R*A∞ =R *
= 60000 *
1
0.06
1
i
= 1000000
b- The rent pay at the beginning of every
year ( due Annuity ).
The price of purchasing the flat
=R*{1+
1
i
=R*Ä∞
}
= 60000 * { 1 +
1
0.06
333
} = 1060000
Q:
Try to make A comparison between a, b
from your own mind and write A
comment .
c- The rent pay after every year but after A
deferred period equal 5 years.
In this case:
We can consider the rent as A permanent,
deferred, ordinary Annuity.
So the price of purchasing the flat equal.
= R * m/ A ∞
= R * Vm *
1
i
= 60000 * V5 *
= 60000 * 0.747258 *
Note: V5 =
1
(1+0.06)5
333
1
0.06
1
0.06
= 747258
d- The rent pay at the beginning of every
year but after A deferred period equal 5
years.
In this case: We can consider the rent as A
permanent deferred, due Annuity.
The price of purchasing the flat
= R * m/ Ä ∞ i%
= 60000 * Vm * { 1 +
1
i
= 60000 * V5 * { 1 +
1
0.06
= 60000 * 0.747258 * { 1 +
}
1
0.06
}
} = 792094
Example 5:
An area of agricultural land gives an
annual revenue equal 5000 if the compound
interest rat equal 5% converted Annually.
Calculate the price of this land if:
333
a- The revenue pay at the end of every year.
Solution
In this case we can consider the revenue of
this land as an ordinary, permanent, Annuity.
So the price of the area of agricultural land
equal the present value of A permanent, ordinary, Annuity.
= R * A ∞ = 50000 *
1
0.05
= 1000000
b- If the revenue pay at the beginning of
every year.
The price = R * Ä ∞ = R * { 1 +
= 5000 * { 1 +
1
0.05
339
1
i
}
} = 1050000
Example 6:
Abo Ahamed aged now 40 years, he plans
to own A bulk of money equal 1500000 when
he reaches the age of retirements at 65, so he
deposited in A bank regularly at the end of
every year A periodic payment equal 12000 for
15 years until he reached the age 55. but at this
age he withdrew apart of his balance from the
bank equal 400000 to construct A villa for him,
if you know that the bank calculate A compound interest equal 11% converted Annually.
Determine the periodic payment that he must
deposit Annually through the following ten
years to achieve the expected bulk that he
planned previously that equal 1500000 when he
reaches the age 65.
333
Solution
We will present this diagram.
4000000
Withdrawal
12000
12000
12000
X
X
X
1500000
age
40
Age
55
i = 11%
Steps of solution:
1-
Finding the total of deposits at the end of
15 years.
S 15 = R * S 15
= 1200 *
2-
(1+0.11)15 – 1
0.11
= 412864.31
Finding the remainder balance = ( the total
of deposits ) – ( the withdrawal amount )
= 412864.31 – 400000 = 12864.31
333
3-
Determine the periodic payment that he
must deposit through 10 years, to obtain
1500000.
1500000 = the amount of remainder balance +
the amount of ordinary temporary
Annually for 10 years.
1500000 = 12864.31 (1 + 0.11)10 + R S 10 11%
10
–1
1500000 = 12864.31 * 2.1839421 + R *(1.11)
0.11
R * 16.722009 = 1500000 – 36527.19
R * 16.722009 = 1463472.81
R=
1463472.81
16722009
= 87517.76
The final result:
Abo Ahemed must deposit 87517.76
Annually for ten years to achieve his plan.
333
Example 7:
Dr. Hossam Elkholy A Businessman
wants to construct A charity hospital, so he
demand
from
A
consultancy
office
to
implement A feasibility study to determine the
expected amounts of money to construct and
spending to this hospital. So the office told him
that he will need the following payments.
1-
The expected value of money to purchase
the land for building the hospital equal
1200000.
2-
The cost of construction the hospital will
equal 200000 and must pay at the
beginning of building.
3-
To complete the building, it will need at
the end of the following
337
5 years A
periodical payment equal 300000 for
spending on it
4-
After finishing the construction of the
building it will need amount of money for
purchasing machines and equipments with
value equal 500000.
5-
The experts decided that the hospital needs
A current periodical administrative expenses for spending on the building begins
with the starting of sixth year equal 50000.
And if you know that the compound
interest rate that may used in investment equal
11%.
Required:
Determine the expected gross amount of
money that the businessman need for constru338
ction, foundation, and continuous performance
to this hospital.
Solution
1200000
300000
200000
500000
500000
500000
The expected amount of money = the
present value of gross GOST
= 1200000 + 200000 + 300000 * A S 11%
+ 500000 * V5 11% + 50000 * 5 / Ä ∞ 11%
= 1200000 + 200000 + 300000 *
–
–
1– (0.11)–5
0.11
* (1.11) 5 + 50000 * (1.11) 5 * (1 +
1
0.11
+ 50000
)
= 1400000 + 300000 * 3.695897 + 500000 *
0.593451 + 50000 * 0.593451 * 10.09091
= 3104917.6
339
General exercises on Annuities
1-
A person deposits $ 5000 in A bank every
year with compound interest Annually
15% for 15 years.
Find the amount of deposits in the following
cases:
a- If he deposits the money at the end of
every year.
b- If he deposits the money at the beginning
of every year.
c- If he deposits money at the end of every
year after deferred period equal 7 years.
d- If he deposits money at the beginning of
every year after deferred period equal 5
years.
333
2-
SoSo deposits $ 400 at the end of each 3
months in A bank that pay 6% converted
quarterly. How much will she has to her
credit at the end of 8 year.
3-
Find the amount of $ 2000 at the end of 6
months for 10 years if money is worth (a)
10%. (b) 12% all rates are converted semi
Annually.
4-
A person deposits $ 1000 as Annuity at the
end of each month for 5 years.
What is the amount of the Annuity if
money is worth 6% compounded monthly.
5-
Engineer Hesham purchased A real estate
and paid A deposit equal $ 50000 and
agreed with the seller to pay the rest of the
price with annual premium pay at the end
of every year for 10 years equal $ 10000.
333
Calculate the amount of accumulated
capital at the seller using compound annual rate
equal 14%.
6-
How much A person save every 6 months
to accumulate $ 10000 in 5 years if money
is worth 6% compounded semi Annually.
7-
A person deposits $ 200 at the end 3 of
months for 3 years, after that he deposit $
400 at the end of 6 months for 4 years.
With rate 10% Annually.
Required:
Find the final balance to the depositor at
the end of 10 years.
8-
A person deposits $ 500 in A bank to 10
years, the bank gives compound return
333
12% Annually. Find the present value to
this Annuity if:
a- The deposit begins at the end of every
year.
b- The deposit begins at beginning of every
year.
9-
A
donor
wants
to
provide
40000
scholarship every year for 5 years with the
first to be awarded 1 year from now. If the
school can get 10% return on its
investment.
Required:
A- How much money should the donor give
now?
B- Show the table to clear what will happen
to the investment through 5 years.
339
10- A businessman wants to construct A
hospital. The experts told him that he will
need the following payments.
1- The value of the land equal 2000000.
2- The cost of construction must pay now
equal 3000000.
3- The building needs 500000 at the end of
the following 6 years to complete it.
4- After construction the building, it needs
equipments with value equal 1000000.
5- The hospital needs A current periodical
administrative expenses for spending one
the building begins with starting with the
seventh year equal 80000, if you now that
the compound interest rate that may used
in investment equal 12%.
333
Required:
The expected gross amount of many for
construction, foundation and continuous performance to this hospital.
333
CHAPTER (6)
Amortization of Loans
427
428
CHAPTER (6)
Amortization of Loans
All the financial experts consider loans as
the best method to finance most projects all over
the world.
Definition of loan:
It is amount or debt of money due to pay
after a specific period.
Forms of loans:
1-
The loan may be as amount of money we
call it as loan.
2-
The loan may be as bond loan:
When a corporation or government needs
money for an extended period of time. The
429
amount required may be too large to obtain
from a single bank or other lenders. The situation can met by issuing bonds that are purchased
by individuals or insurance companies and other
investors. Thus the buyer of a bond lends money
to the organization that issued the bond.
Concept of Amortization loans:
It means that the borrower of the loan
think how he can repay or pay off the principal
loan and its interests or the extinction of the
debt by any satisfactory set of payments. In this
chapter, however when we say that a debt is
amortized we shall mean usually, that all liabilities as to principal and interest are discharged
by a sequence of equal payment due at the ends
of equal interval of time. In such case the
434
payments form an annuity whose present value
is the original principal of the debt.
Methods of Amortization loans:
There are many methods, but we will
present only two methods.
The first method:
1-
Is amortization the loan with equal
premiums from the principal loan and
repay the due interest on the decreasing
balance of the remainder of the loan.
The second method:
2-
Repay the loan with equal premiums
including the principal loan and the due
interest.
The first method of amortization loans:
434
Example:
A company borrowed L. E. 50000 from a
bank to buy beauty tools, and agreed with the
bank to repay only the principal loan by five
equal premiums and will repay the due interest
on the remainder, balance with the equal
premiums at the end of every year and the bank
will use compound interest equal 10% annually.
Find:
1-
Determine the equal annual premium from
only the principal loan.
The equal annual premium from the
principal loan =
=
The principal loan
Number of premiums
50000
= 10000 annually
5
432
=P L
n
2-
Determine the total payments that the
company will pay annually at the end of
every year.
The used equation is:
The due interest on the balance of the loan
+ The equal annual premium from the principal
loan.
The first year:
The interest of first year = The balance of
first year X the rate X period
I1 = L X i X n
10
= 50000 X 100 X 1 = 5000
The payable premium at the end of first
year ( P1 )
= P + i1 = 10000 + 5000 = 15000
433
The balance at the end of first year =
The balance at the beginning of first year –
the equal annual premium.
= 50000 – 100000 = 40000
The second year:
The balance at beginning of second year
= The balance at the end of first year = 40000
The interest of second year =
10
I2 = 40000 X 100 X 1
The payable premium at the end of second
year
P + i2 = 10000 + 4000 = 14000
The balance at the end of second year =
the balance at the beginning of second year –
the equal premium
434
= 40000 – 10000 = 30000
And so on. The following year.
3-
Construct an amortization schedule of
the loan:
The following schedule represent an
amortization schedule of the loan.
Years or
interval
Outstanding
Equal
Due interest on
Total
Outstanding
principal at
premium
the outstanding
payment at
principal at
the beginning
principal at the
the end of
the end of
of the year
end of every
every year
every year
year
1
50000
10000
5000
15000
40000
2
40000
10000
4000
14000
30000
3
30000
10000
3000
13000
20000
4
20000
10000
2000
12000
10000
5
10000
10000
1000
11000
Zero
Now we will clear some important notices:
1-
The schedule consists of six columns.
2-
The first column refer to the year or
intervals.
435
3-
The second column refer to the original
principal at the beginning of every year
and we notice that this balance decreasing
continuously
through
the
following
intervals.
4-
The last outstanding principal of the loan
at the last year must equal with the equal
premium.
5-
The third column refer to equal premium
and its equation =
=
6-
The principal loan
Number of premiums
The fourth column refer to the due interest
on the outstanding principal at the end of
every year and its equation: I = L . i . n
436
In wording:
Interest = the outstanding principal at the
beginning of every year X the rate of interest X
the period.
7-
The fifth column:
Refer to the total payment at the end of
every year and its equation is =
The equal premium + the due interest on
the outstanding principal at the end of every
year.
So: The fifth column = the values in third
column + the values in fourth column.
8-
The sixth column refer to the outstanding
principal at the end of every year and its
equation is the outstanding principal at the
beginning of the year ( – ) equal premium.
437
Or:
The values in the second column – the
values in the third column.
9-
The total paid interest of the loan = i 1 + i2
+ i3 + i4 + i5 = 5000 + 4000 + 3000 + 2000
+ 1000 = 15000
10- We can obtain the total paid interest by
using equation of arithmetic progression.
Hence:
The first term = 5000, the last term = 1000
and the number of terms = 5.
n
By substitution S = 2 * { t1 + tn }
Then the total paid interest =
S = 5 * { 5000 + 1000 } = 15000
2
438
Example 2:
Dr. Hossam Elkholy borrowed a loan
equal 35000 from a bank, and he agreed with
the bank to repay this loan through seven
semiannual equal premi-ums from the principal
loan and will repay the due interest on the
decreasing balance of the remainder of the loan
using interest rate equal 8% annually.
Find:
1-
The value of semiannual equal premium
from the principal only.
2-
Determine the remainder balance of the
loan at the end of fourth and sixth interval.
3-
Construct an amortization schedule to the
four intervals only.
439
4-
Determine the total paid interests to the
bank
Solution
1-
The semiannual equal premium =
=
2-
The principal of the loan
Number of premiums
=
35000
= 5000
7
The remainder balance from the loan at
the end of any interval = The principal of
the loan – ( number of paid premium X
The value of equal premium )
So:
The balance of the loan at the end of
fourth interval = 35000 – ( 4 X 5000 ) = 15000
Also: The balance of the loan at the end of sixth
interval = 35000 – (6 X 5000) = 5000
444
4-
Construct an amortization schedule of the
loan from the first interval to the fourth
interval :
Years or
intervals
Outstanding
Equal
Due interest on
Total
Outstanding
principal at
premium
the outstanding
payment at
principal at
the beginning
principal at the
the end of
the end of
of the interval
end of every
every
every
interval
interval
interval
1
35000
5000
1400
6400
30000
2
30000
5000
1200
6200
25000
3
25000
5000
1000
6000
20000
4
20000
5000
800
5800
15000
4-
Calculate the total paid interests:
Hence the balance at the beginning of
seventh or last interval equal with equal
premium = 5000
So:
The interest of the seventh or last interval
= The balance at the beginning of the interval X
the rate of interest X the period.
444
= 5000 X
8 X
100
6 = 200
12
Then the total of interests represent an
arithmetic progression.
n
The equation is ∑ i = 2 { i1 + i7 }
S=
7
2
{ 1400 + 200 } = 5600
Example 3:
The engineer Hesham Elkholy borrowed a
loan from a bank and agreed to repay it through
four equal premiums from only the principal,
also if you know that the balance at the
beginning of fourth interval equal L. E. 5000
and the interest of third interval equal L. E.
1200.
442
Find the following:
1-
The value of equal premium that will
repay only from the principal.
2-
The value of total loan.
3-
The used rate of interest.
4-
The total paid payments.
5-
The total paid interests.
6-
Construct an amortization schedule of the
loan.
Solution
We know from the example that the
balance of the beginning of fourth interval =
5000, and we know previously that this balance
must equal, the equal premium at the last year
of the loan.
443
SO:
1-
The value of equal premium = 5000 that
will repay only from the principal.
2-
The value of total loan = the total of equal
premium
=
The
number
of
equal
premiums X the value of equal premium =
4 X 5000 = 20000
3-
The balance of the loan at any interval =
The balance of the loan at the previous
interval – ( equal premium ).
The balance at the beginning of second
interval = 20000 – 5000 = 15000
The balance at the beginning of third
interval = 15000 – 5000 = 10000
The balance at the beginning of fourth
interval = 10000 – 5000 = 5000
444
Then:
To obtain the used rate of interest, we will
apply this equation.
I3 = The balance at the beginning of the interval
X The rate X The period
1200 = 10000 X The rate X 1
Then the rate will become 12%
Then the interest of all intervals:
12
I1
= 20000 X 100 X 1 = 2400
I2
= 15000 X 100 X 1 = 1800
I3
= 10000 X 100 X 1 = 1200
I4
= 5000 X 100 X 1 = 600
12
12
12
SO:
445
We can construct an amortization schedule
from the previous data as following:intervals
Outstanding
Equal
Due interest on
Total
Outstanding
principal at
premium
the outstanding
payment at
principal at
the beginning
principal at the
the end of
the end of
of the interval
end of every
every
every
interval
interval
interval
1
20000
5000
2400
7400
15000
2
15000
5000
1800
6800
10000
3
10000
5000
1200
6200
5000
4
5000
5000
600
5600
Zero
Then the total paid payment =
=
4 { 7400 + 5600 } = 26000
2
The total paid interest =
=
4 { 2400 + 600 } = 6000
2
The second Method of Amortization loans:
And also we can say extinction of debts by
periodic payments.
446
In current language the amortization of a
debt means the extinction of the debts by any
satisfactory set of payments. When we say that a
debt is amortized we shall mean usually, that all
liabilities as to principal and interest are discharged by a sequence of equal payments due at the
ends of equal intervals of time. In such case the
payments form an annuity whose present value
is the original principal of the loan.
In the following example we will explain
how we can repay the loan with equal premiums
including the principal loan and the due interest.
Example:
Toto borrows $ 10000 with the agreement
that money is worth 13% compounded annually.
The debt is to be paid interest included, by equal
instalments at the end of each year for 5 years.
447
Find:
1-
The annual payment or the annual equal
premium from the principal loan and the
due interest together.
2-
Construct an amortization schedule.
Solution
The annual equal premium =
P=L*
1
an i
1–
Hence a n i =
1
(1+i)n
i
Substituting in previous formula we have:
1–
a 5 13% =
1
(1.13)5
0.13
P = 10000 *
1
3.51723
= 3.517231
= 10000 * 0.2843145 = 2843.145
448
Also we can obtain the equal annual
premium using this formula.
P=L*
= 10000 *
i
1– 1
( 1+i )n
0.13
1– 1
( 1.13 )5
= 2843.145
Construct an amortization schedule.
We will present some, equation as
following:
1-
The equal premium = depreciation from
the loan at any year + Interest of this year.
P=d+i
2-
Balance of loan at first interval = principal
loan – depreciation of first year.
=L–d
449
The interest of first year I1 = the balance
of the loan at the beginning of the interval X the
rate X the interval.
i1 = L * i * 1
At the second year:
The balance at the last first year = the balance at
the beginning of second year.
i2 = ( L – d 2 ) * i
d2 = P – i2
The relation between depreciations of the
loan:
If we know the depreciation of first year
and rate of interest we can obtain the following
depreciations using these formulas:
d2 = d1 ( 1 + i )
d3 = d1 ( 1 + i )2 OR
454
d3 = d2 ( 1 + i )
d4 = d1 ( 1 + i )3 OR
d4 = d3 ( 1 + i )
d5 = d1 ( 1 + i )4 OR
d5 = d4 ( 1 + i ) and so on
The total of depreciations equal the loan
So: L = d1 + d2 + d3 …… dn
If we know the equal premium and the
following depreciation we can obtain the due
interests:
i1 = P – d 1
i2 = P – d 2
i3 = P – d 3
i4 = P – d 4
454
Construct an amortization schedule:
Payment
Balance at
Equal
Depreciation
Interest
number
the beginning
premium
d
(i)
or the
of the year
(P)
Balance at
the end of
every year
year
1
10000.00 2843.145 1543.145
1300.00
8456.855
2
8456.855 2843.145 1743.754 1099.391 6716.100
3
6746.100 2843.145 1970.442
872.700
4742.658
4
4742.658 2843.145 2226.600
616.545
2516.060
5
2516.060 2843.145 2516.060
327.085
____
Example:
Ashrakat Alkholy borrowed a loan from a
bank, and she agreed to repay it by five equal
premiums from the principal and interests using
compound interest 8%, and if you know that the
third depreciation from the loan equal $
5979.68.
Find:
1-
The
annual
depreciations
principal loan.
452
from
the
2-
Determine the equal annual premium from
the principal loan and the due interest
together.
3-
Determine the principal loan.
4-
Construct the amortization schedule of the
loan.
Solution
We know from the example that d3 = $
5979.68 and the rate of interest = 8%.
So we can determine the following
depreciations using this formula.
d4 = d3 ( 1 + i )
d4 = 5979.68 ( 1 + 0.08 ) = 6130.81
d5 = d4 ( 1 + i )
= 6130.81 ( 1 + 0.08 ) = 66210.28
453
Also we can find d2, d1 as following:
d 2:
d3 = d2 ( 1 + i )
d2 =
d 1:
d3
= 5979.68 = 5536.74
(1+i)
1.08
d2 = d1 ( 1 + i )
d1 =
Hence:
d2 = 5536.74 = 5126.61
1.08
(1 + i)
d1 = 5126.61
d2 = 5536.74
d3 = 5979.68
d4 = 6130.81
d5 = 6621.28
So the principal loan equal the total of
depreciations:
L = d1 + d2 + d3 + d4 + d5
= 5126.61 + 5536.74 + 5979.68 +
6130.81 + 6621.28 = 29395.12
454
Finding the equal annual premium:
1
An i
P=L*
Hence: A n i% =
1–
A5 8% =
1
(1+0.08)5
0.08
= 29395.12 *
1–
1
a 5 8%
1
(1.i)n
i
= 3.99271
1
P = 29395.12 * 3.99271 = 7362.2
Construct amortization schedule:

P=d+i
i=P–d
The balance at the end of any interval =
the balance at the beginning of any interval – d
the year
Balance at
Equal
Depreciation
Interest
the beginning
premium
d
(i)
Balance at
of the year
(P)
1
29395.12
7362.2
2126.61
2235.59
24268.51
2
24268.51
7362.2
5536.74
1825.46
18731.77
3
18731.77
7362.2
5979.68
1382.52
12752.09
4
12752.09
7362.2
6130.81
1231.39
6621.28
5
6621.28
7362.2
6621.28
740.92
____
the end of
every year
455
Example:
Elkholy company borrowed a loan and
agreed to repay it by equal annual premium
from the principal loan and the due interest
using compound interest 6%, and if you know
that the second depreciation equal $ 56412.1
and the third depreciation equal 59796.7.
Find: Construct an amortization schedule of the
loan.
Solution
Hence:
d3 = 59796.7
d2 = 56412.1
i = 6%
So: From the relation between the depreciations we can determine the following
depreciations.
456
d 2 = d1 ( 1 + i )
d1 =
d2
56412.1
=
= 53218.9
(1+i)
1 + 0.06
d 4 = d3 ( 1 + i )
= 59796.7 ( 1.06 ) = 63384.6
d 5 = d4 ( 1 + i )
= 63384.50 ( 1.06 ) = 67187.7
We know previously that.
L = d1 + d 2 + d3 + d 4 + d 5
L = 53218.9 + 56412.1 + 59796.7 +
So:
63384.6 + 67187.7 = 300000
Determine the equal premium:
P=L*
1
A 5 6%
= 300000 *
1
A 5 6%
457
= 71218.9
Also we can determine the following interests:
Hence:
P=d+i
So: i1 = P – d1
i2 = P – d 2
i3 = P – d 3
Also: The balance at the last of any year = the
balance at the beginning of the year –
depreciation.
Also: the balance at beginning of any year = the
balance at the last of previously year.
Amortization schedule of the loan:
Payment
Balance at
number the the beginning
Equal
Depreciation
Interest
premium
d
(i)
Balance at
the end of
years
of the year
(P)
every year
1
300000
71218.9
53218.9
18000
246789.9
2
246789.9
71218.9
56412.1
14906.8
190369
3
190369
71218.9
59796.7
11422.2
130572.2
4
130572.2
71218.9
63384.6
7834.3
67187.7
5
67187.7
71218.9
67187.7
4031.2
____
458
Exercises:
1-
A company borrowed L. E. 100000 from a
bank, and agreed to repay only the
principal loan by five equal premiums and
will repay the due interest on the
remainder balance with the equal premiums at the end of every year and the bank
will use compound interest equal 8%
annually.
Find:
a- Determine the equal annual premium from
only the principal loan.
b- Determine the total payments that the
company will pay annually at the end of
every year.
459
c- Construct on amortization schedule of the
loan.
2-
A man borrowed $ 20000 with the
agreement that money is worth 10%
compounded annually. The debt is to be
paid interest included, by equal instalment
at the end of each year for 6 years.
Find:
a- The annual premium or the annual equal
premium.
b- Construct an amortization schedule
3-
A person borrowed loan from a bank and
he agreed to repay it by six equal premium
from the principal and interests using
compound interest 10% and if you know
464
that the third depreciation from the loan
equal $ 5979.67.
Find:
a- The
annual
depreciations
from
the
principal loan.
b- Determine the equal annual premium.
c- Determine the principal loan.
d- Construct the amortization schedule of the
loan.
464