Chapter I
Mathematics of Finance
I-1 Interest
I-1-01: Simple Interest
Let:
p = Principal in Riyals
r =Interest rate per year
t = number of years
→ The accumulated amount A after t years:
A = p + prt = p ( 1 + rt)
Graphing the straight line segment:
A(t) = p + prt ; t ≥ 0
Example (1)
What are the interest and the total accumulated amount
after 10 years on a deposit of 2000 Riyals at a simple
interest rate of 1% per year?
Solution:
The accumulated amount A = p + prt
The interest paid I = prt
where p = 2000, r = 1/100 = 0.01 and t = 10.
A = 2000 + 2000(1/100)(10) = 2000 + 200 = 2200
I = prt = 200
Textbook: Example 1 page 186
Graphing the straight line:
A(t) = 2000 + 2000(0.01)t ; t ≥ 0
Homework
Exercises 4.1 Page: 197
All odd numbered exercises from 1 to 9
Homework
1.a. What are the interest I and the total accumulated amount A
after 2 years on a deposit of 500 Riyals at a simple interest rate of 8% per
year?
Answer: I = 80 Riyals and A = 580 Riyals
2.b.c. If the accumulated amount after 2 years on a deposit at a simple
interest rate of 8% per year is 580, then what’s the principal P (the
deposit)?
Answer: P = 500 Riyals
3.d. How many years t would it take a deposit of 500 Riyals at a simple interest
rate of 8% per year to grow to an accumulated amount of 580 Riyals?
Answer: t = 2 years
3.f. If a deposit of 500 Riyals grows to an accumulated amount of 580 Riyals in
2 years, then what’s the simple interest rate?
Answer: r = 0.08 (8% per year )
Homework
2.a. What are the interest I and the total accumulated amount A
after 9 months on a deposit of 800 Riyals at a simple interest rate of 6% per
year?
Answer: I = 36 Riyals and A = 836 Riyals
2.b. If the accumulated amount after 9 months on a deposit at a simple
interest rate of 6% per year is 836, then what’s the principal P (the
deposit)?
Answer: P = 800 Riyals
3.c. How long would it take a deposit of 800 Riyals at a simple interest
rate of 6% per year to grow to an accumulated amount of 836 Riyals?
Answer: t = 0.75 year = (3/4)(12 months) = 9 months
3.d. If a deposit of 800 Riyals grows to an accumulated amount of 836 Riyals in
9 months, then what’s the simple interest rate?
Answer: r = 0.06 = 6% per year
I-1-02: Compound Interest
Let:
p = Principal in Riyals
r =Interest rate per year
t = number of years
→
The accumulated amount A1 after 1 year:
A1 = p + pr(1) = p (1 + r)
The accumulated amount A2 after 2 years:
A2 = A1 + A1 r(1) = A1 (1 + r) = p(1 + r) (1 + r)= p(1 + r)2
The accumulated amount A3 after 3 years:
A3 = A2 + A2 r(1) = A2 (1 + r) = p(1 + r)2 (1 + r)=p (1 + r)3
The accumulated amount A after 4 years:
A4 = A3 + A3 r(1) = A3 (1 + r) = p(1 + r)3 (1 + r)= p(1 + r)4
We conclude that the accumulated amount after t years
At = p(1 + r)t
Example (2)
What are the interest and the total accumulated amount
after 2 years on a deposit of 2000 Riyals at a compound
interest rate of 10% per year?
Solution:
The accumulated amount A = p(1+ r)t
The interest paid I = A – p
Where, p = 2000, r = 10/100 = 1 /10 = 0.1 and t = 2
A = 2000[1 + (1/10)]2
The interest paid = 2420 – 2000 = 420
I-1-03: Interest compounded m times a year
Let:
p = Principal in Riyals
r = Interest rate per year
m = number of times a year the interest is compounded
Conversion period = the period of time between successive
interest calculations
The interest rate per conversion period = i = r / m
t = the number of years (term)
n = Number of periods in t years = mt
→
The accumulated amount A1 after 1 period:
A1 = p + pi = p (1 + i)
The accumulated amount A2 after 2 periods:
A2 = A1 + A1 i = A1 (1 + i) = p(1 + i) (1 + i)= p(1 + i)2
The accumulated amount A3 after 3 periods:
A3 = A2 + A2 i = A2 (1 + i) = p(1 + i)2 (1 + i)=p (1 + i)3
The accumulated amount A after 4 periods:
A4 = A3 + A3 i = A3 (1 + i) = p(1 + i)3 (1 + i)= p(1 + i)4
We conclude that the accumulated amount after n periods
An = p(1 + i)n = p (1+i)mt = p (1 + r/m )mt
Where,
p = the principal
r = the interest per year
m = the number of times (periods) in a year the interest is
compounded
Example (3)
What is the total accumulated amount after 3 years on a
deposit of1000 Riyals at interest rate of 10% per year
compounded:
1. semiannually ( 2 periods in a year)
2. quarterly ( 4 periods in a year )
3. monthly ( 12 periods in a year)
4. daily ( 365 periods in a year)
5. every 4 months ( 3 periods in a year )
6. every two months ( 6 periods in a year )
7. annually
Solution:
In all of these cases, we use the formula
An = p(1 + i)n = p (1+i)mt = p (1 + r/m )mt
1. semiannually ( 2 periods in a year)
A = 1000[1 + (0.1)/2 ]2(3
2. quarterly ( 4 periods in a year )
A = 1000[1 + (0.1)/4 ]4(3)
3. monthly ( 12 periods in a year)
A = 1000[1 + (0.1)/12 ]12(3)
4. daily ( 365 periods in a year)
A = 1000[1 + (0.1)/365 ]365(3)
5. every 4 months ( 3 periods in a year )
A = 1000[1 + (0.1)/3 ]3(3)
6. every two months ( 6 periods in a year )
A = 1000[1 + (0.1)/6 ]6(3)
7. annually ( 1 periods in a year )
A = 1000[1 + 0.1 ]3
Textbook: Example 3 page 189
Homework
Exercises 4.1 Page: 197
All odd numbered exercises from 11 to 19
Homework
What is the total accumulated amount after t years on a deposit of P
Riyals at interest rate of r per year, compounded:
1. annually, if P= QR1000, t = 8 years and r = 7%
Answer: A = QR 1718.19
2. quarterly , if P= QR12000, t = 10.5 years and r = 8%
Answer: A = QR 27566.93
3. monthly , if P= QR150000, t = 4 years and r = 14%
Answer: A = QR 261751.04
4. daily , if P= QR150000, t = 3 years and r = 12%
Answer: A = QR 214986.69
4. semiannually, if P= QR2500, t = 10 years and r = 7%
Answer: A = QR 4974.47
Homework
What is the total accumulated amount after t years on a deposit of P
Riyals at interest rate of r per year, compounded:
1. annually, if P= QR1000, t = 8 years and r = 7%
Answer: A = QR 1718.19
2. quarterly , if P= QR12000, t = 10.5 years and r = 8%
Answer: A = QR 27566.93
3. monthly , if P= QR150000, t = 4 years and r = 14%
Answer: A = QR 261751.04
4. daily , if P= QR150000, t = 3 years and r = 12%
Answer: A = QR 214986.69
4. semiannually, if P= QR2500, t = 10 years and r = 7%
Answer: A = QR 4974.47
I-1-04: Continuous Compounding of Interest
“ If the interest is compounded more and more frequently”
We have:
A = p (1 + r/m )mt
= p (1 + r/m )\(m/r) rt
= p [1 + (1/u)]u(rt)
, by letting m/r = u and so r/m = 1/u
= p [1 + (1/u)]u(rt) = p [[1 + (1/u)]u](rt)
Notice that u →0 as m→∞
and so as you are supposed to know:
[[1 + (1/u)]u → e
And thus,
p [[1 + (1/u)]u](rt) → p ert
Thus the amount accumulated if the interest is compounded continuously is:
A = p ert
Where p is the principal, r the interest rate and t the term ( number of years)
Example (4)
What is the total accumulated amount after 3 years on a deposit of1000
Riyals at interest rate of 10% compounded continuously?
Solution:
The amount accumulated:
A = p ert , where p=1000, r = 0.1 and t = 3.
Thus,
A = 1000 e(0.1)(3)
Assignment:
1. Find the answer approximated to 2 decimal digits
(Accuracy of one Dirham)
2. Use the appropriate formula to calculate the amount accumulated if
the interest is compounded daily ( with the same accuracy) and
compare the that with the answer to question (1).
Textbook: Example 4 page 191
Homework
Exercises 4.1 Page: 198
Exercise 29
Homework
Find, using the continuous compounding formula:
the total accumulated amount A after 4 years on a deposit
of 5000 Riyals at interest rate of 8% per year.
Answer: A = QR 6885.64
I-1-05: Effective Rate of Interest
( the annual percentage yield)
reff = [1 + (r/m)]m - 1
Example (5):
What’s the effective rate if the rate if the
nominal rate of interest is 10% per year
Compounded:
1. quarterly
2. monthly
3. annually
Solution
We use the formula
reff = [1 + (r/m)]m – 1,
Where r = 10/100 = 1/10 = 0.1
1. quarterly → m = 4
reff = [1 + (0.1)/4)]4 – 1
2. monthly→ m = 12
reff = [1 + (0.1)/12)]12 – 1
3. annually → m = 1
reff = [1 + (0.1)/1)]1– 1 = 0.1 = the nominal interest
Textbook: Example 5 page 192
Deducing the Formula of the Effective rate of
Interest
We let:
p(1+ reff ) = p[1+ (r/m)]m
Where,
r = the nominal yearly interest rate
m= the number of conversion periods per year
→1+ reff = [1+ (r/m)]m
→reff = [1+ (r/m)]m - 1
Homework
Exercises 4.1 Page: 197
Exercises 21 and 23
Homework
What is the effective rate of interest reff corresponding to the following nominal
interest rate:
1. 10% per year compounded semiannually
Answer: reff = 10.25 % per year
2. 8% per year compounded monthly
Answer: reff = 8.3 % per year
I-1-06: Present Value
What’s the principal (present value) that
should be deposited in a bank paying an
Interest at a rate of r compounded m times a
Year such that at the end of t years the
accumulated amount will be A.
We had A = p [ 1 +(r/m) ]mt
→ p = A / [ 1 +(r/m) ] mt
→ p = A [ 1 +(r/m) ] - mt
Example (6)
What’s the principal (present value) that
should be deposited in a bank paying an
Interest at a rate of 10% compounded 3 times a
year such that at the end of 4 years the
accumulated amount will be 5000.
Solution:
p = A [ 1 +(r/m) ] - mt
We have A=5000, r =10/100=1/10= 0.1, m=3 and
t=4
→ p = 5000 [ 1 +(0.1)/3 ] –3(4)
Textbook: Examples 6 & 7 page 194
Homework
Exercises 4.1 Page: 197 - 198
Exercises 25, 27
Homework
1. What’s the present value( money already in an account) in a
bank paying an Interest at a rate of 6% compounded 2 times a year
such that at the end of 4 years the accumulated amount will be QR 40000.
Answer: P = 31576.37 Riyals
2. What’s the present value( money already in an account) in a
bank paying an Interest at a rate of 7% compounded monthly such that at the
end of 4 years the accumulated amount will be QR 40000.
Answer: P = 30255.95 Riyals
I-1-07: using Logarithms to Solve
Problems in Finance
a. Finding the term number of years) t needed for a principal p to grow
to an accumulated amount A if invested at rate r(per year)
compounded m times a year.
We have:
A = p [ 1 + (r/m) ]mt
Hence,
A / p = [ 1 + (r/m) ]mt , dividing both sides of the equation by p
Taking the natural logarithm for both sides, we get:
ln(A/p) = ln [ 1 + (r/m) ]mt = mt ln [ 1 + (r/m) ]
Thus,
t = ln(A/p) / m ln [ 1 + (r/m) ]
Example (7)-a
How long it will take a principal of 1000 Riyals to
grow to 1270.24 if deposited in a bank paying an
Interest at a rate of 8% compounded monthly ?
Solution:
t = ln(A/p) / m ln [ 1 + (r/m) ]
We have: p=1000, A=1270.24, r=0.08 and m=12
We can either use the deduced formula or do the
calculation from scratch:
1. Using the deduced formula:
t = ln(A/p) / m ln [ 1 + (r/m) ]
→ t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ]
Assignment find t !
Textbook: Example 8 page 194
2. Solving from scratch:
We have t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ]
A = p [ 1 + (r/m) ]mt
Thus,
1270.24 = 1000[ 1 + (0.08/12) ]12t
→ 1270.24/1000 = [ 1 + (0.08/12) ]12t ( dividing both sides by 1000)
→ ln(1270.24/1000) = ln [ 1 + (0.08/12) ]12t ( applying the logarithm to both
sides)
→ ln(1270.24/1000) =12t ln [ 1 + (0.08/12) ] ( using a logarithmic property)
→12t = ln(1270.24/1000) / ln [ 1 + (0.08/12) ]
→t = ln(1270.24/1000) / 12 ln [ 1 + (0.08/12) ]
Textbook: Example 8 page 194
Homework
Exercises 4.1 Page: 198
Exercises: 31, 33 and 37
Homework
1. How long it will take an investment of P Riyals to grow to A Riyals in a
bank paying an Interest at a rate of r, compounded monthly:
a. If P = QR 5000, r= 0.12 per year and A = QR6500
Answer: 2.2 years
b. If P = QR 2000, r= 0.09 per year and A = QR4000
Answer: 7.7 years
2. How long it will take an investment of 6000 Riyals to grow to 7000 Riyals in
A bank paying an Interest at a rate of 7.5%, compounded continuously.
Answer: 2.06 years
b. Finding the interest rate needed for a principal p to grow
to an accumulated amount A if invested for t years and the
Interest is compounded m times a year.
We have:
A = p [ 1 + (r/m) ]mt
Hence,
A / p = [ 1 + (r/m) ]mt , dividing both sides of the equation by p
Taking the natural logarithm for both sides, we get:
ln(A/p) = ln [ 1 + (r/m) ]mt = mt ln [ 1 + (r/m) ]
Thus,
ln [ 1 + (r/m) ] = ln(A/p) / mt
→ 1 + (r/m) = e ln(A/p) / mt
→ r/m = e ln(A/p) / mt – 1
→ r = m [ e ln(A/p) / mt – 1]
Example (7)-b
Find the interest rate needed a principal of 1000 Riyals to
grow to 1270.24 if deposited in a bank for 3 years and the interest is
compounded monthly ?
Solution:
We have: p=1000, A=1270.24, t=3 and m=12
We can either use the deduced formula or do the
calculation from scratch:
1. Using the deduced formula:
r = m [ e ln(A/p) / mt – 1]
→ r = 12 [ e ln(1270.24/1000) / 12(3) – 1]
Assignment find r !
2. Solving from scratch:
We have:
A = p [ 1 + (r/m) ]mt
→1270.24= 1000 [ 1 + (r/12) ]12(3)
Hence,
1270.24 / 1000 = [ 1 + (r/12) ]36 , dividing both sides of the equation by p
Taking the natural logarithm for both sides, we get:
ln(1270.24 /1000) = ln [ 1 + (r/12) ]36 = 36ln [ 1 + (r/12) ]
Thus,
ln [ 1 + (r/12) ] = ln(1270.24 /1000) / 36
→ 1 + (r/12) = e ln(1270.24/1000) / 36
→ r/12= e ln(A/p) / 36 – 1
→ r = 12 [ e ln(A/p) / 36 – 1]
Textbook: Example 9 page 195
Homework
Exercises 4.1 Page: 198
Exercise 35
Homework
1. Find the interest rate needed for an investment of 5000 Riyals to
grow to 6500 Riyals in 2.2 years If interest is compounded monthly
Answer: 0.12 = 12%
1. Find the interest rate needed for an investment of 5000 Riyals to
grow to 6000 Riyals in 3 years If interest is compounded continuously
Answer: 0.0608 = 6.08%
I.2. Progressions
I-2-01: Arithmetic Progression
An arithmetic progression is a sequence of numbers of the form:
a, a+d, a+2d, a+3d,……
Where the difference between any term and the one preceding it is a constant d
(called the difference).
Examples (8)
Consider the following sequence:
5, 7, 9, 11, 13, 15,
Here the first term a = 5 and the difference d = 2
a2 =a1 + d = 5 + 2 = 7
a3 = a2 + d = 7 + 2 = 9 = 5 + 2(2) = a + 2d
a4 = a3 + d = 9 + 2 = 11= 5 + 3(2) = a + 3d
a5 = a4 + d = 11 + 2 = 13= 5 + 4(2) = a + 4d
In general:
an = an-1 + d = a + (n-1)d
For this progression: an = 5 + 2 (n-1)
For example:
a21 = 5 + (21-1)(2) = 5 + 40 = 45
Examples (9)
Consider the following sequence:
3, 13, 23, 33, 43, 53,
Here the first term a = 3 and the difference d =10
a2 =a1 + d = 3 + 10 = 13
a3 = a2 + d = 13 + 10 = 23 = 3 + 2(10) = a + 2d
a4 = a3 + d = 23 + 10 = 33= 3 + 3(10) = a + 3d
a5 = a4 + d = 33 + 10 = 43= 3 + 4(10) = a + 4d
In general: The n-th term Formula
an = a + (n-1)d
For this progression: an = 3 + 10 (n-1)
For example:
a21 = 3 + (21-1)(10) = 3 + 200 = 203
Textbook: Example 1 page 229
Example (10)
Find the arithmetic progression with the 21st
term is being 45 and the 23rd term 49.
Solution:
We have:
45 = a + 20d and 49 = a + 22d
Subtracting the first equation from the
second, we get: 4 = 2d → d =2
Substituting d = 2 n the first equation, we get:
45 = a + 20(2) → a = 45 – 40 = 5
Thus the progression is the one given in example (8),
namely:
5, 7, 9, 11, 13, 15,……., 5+2(n-1), ………..
Textbook: Example 2 page 229
Sum of the first n terms of an arithmetic
progression
Consider the arithmetic progression
a, a+d, a+2d, a+3d, a+4d, ….a+(n-2)d, a+(n-1)d, …….
Let s be the sum of the first n terms of the progression.
Then:
s = a + (a+d) + (a+2d) + (a+3d)+…….+[a + (n-2]d + an
s = an + (an -d) + (an -2d) + (an -3d)+…….+[an - (n-2)d] + a
→ 2s = ( a + an ) + ( a + an ) + ( a + an ) +……+ ( a + an )
= n ( a + an )
= n [a + a + (n-1)d]
= n [ 2a + (n-1)d]
→ s = n [ 2a + (n-1)d] / 2
The sum formula for the first n terms of the
arithmetic progression:
sn = n [ 2a + (n-1)d] / 2
Example (11)
Consider the arithmetic progression of example (8)
Find:
1. The the sum of its first five terms and the first ten terms
2. the sum of the term from the sixth to the tenths
Solution:
1. We use the formula Sn = n [ 2a + (n-1)d] / 2
From example (8), we have: a=5 and d=2
Thus: S10 = 10 [ 2(5) + (10 -1)2] / 2 = 140
And S5 = 5[ 2(5) + (5 -1)2] / 2 = 45
2. The sum of the terms from a6 to a10 = S10– S5
= 140 – 45 = 95
Textbook: Example 3 page 229, Applied Example (4)
Homework
Exercises 4.4 Page: 234 - 235
1. All odd numbered exercise from 1 to 11
And exercise 13*
2. Exercise from 17 and 19
Homework
1. For each of the following arithmetic progressions, find the 31st term and the
sum of the first 31 terms:
a. 2, 5, 8, 11, 13….
Answers: a31 = 94 and s31 = 1457
b. 1, 2, 3, 4, 5, …
Answers: a31 = 32
and
s31 = 496
c. -5, -4 ½ , -4, -3 ½ , -3
Answers: a31 = 5 and s31 = 77.5
d. -5, -7, -9, -11,…
Answers: a31 = - 70 and
s31 = - 1085
e. -5, 5 ½ , -6, -6½ , -7,…
Answers: a31 = - 25 and s31 = - 387.5
f. -5, -3, -1, 1,3,….
Answers: a31 = 50 and
g. 22, 32, 42, 52, …
Answers: a31 = 344 and
s31 = 775
s31 = 5332
h.* x, x + 2y, x + 4y, x + 6y, x + 8y….
Answers: a31 = x + 60y and s31 =31 x + 930y
i* . x + 2y, x + y, x, x – y, x - 2y,..
Answers: a31 = x - 28y and s31 = 31x – 403y
2. Repeat Ex(1) for n =33
3. Find the arithmetic progressions, satisfying the two given conditions
(an = the n-th term and sn= the sum of the first n terms)
a. a31 = 94
and
a34 = 98
b. a31 = 94
and
a1 = 2
b. a31 = 94
and
d= 3
d. a4 = 11
and
s4 = 26
Answers: 2, 5, 8, 11,…..
I-2-02: Geometric Progression
A sequence of numbers in the form:
a, ar, ar2, ar3, …….
Where the ratio between of any term to the one preceding it is a constant r (called the ratio).
Examples (12)
Consider the following sequence:
5, 10, 20, 40, 80, 160,
Here the first term a = 5 and the ratio r= 2
a2 =a r = 5 ( 2 ) = 10
a3 = a2 r = 10(2) = 20 = 5 (2)2 = a r2
a4 = a3 r = 20(2) = 40 = 5 (2)3 = a r3
a5 = a4r = 40(2) = 80 = 5 (2)4 = a r4
In general:
an = an-1 r = a rn-1
The formula for the n-th term of the geometric progression:
an = a rn-1
a7 = a r7-1= 5(2)6 = 5(64) = 320
Textbook: Example 5 page 230
Examples (13)
Find the geometric progression with the with the third term equal 20 and the sixth term
equal 160.
Solution;
We have:
20 = ar3-1= ar2 and 160 = ar6-1 = ar5
→ 160/20 = ar5 / ar2
→ 8 = r3
→
Thus,
20 = a (2)2 = 4a
r=2
→
a= 20/4 = 5
Textbook: Example 6page 231
Sum of the first n terms of an geometric
progression
Consider the geometric progression
a, ar, ar2, ar3, ar4, ….ar5, ar6, …….
Where r is not equal to 1.
Let s be the sum of the first n terms of the progression.
Then:
s = a + ar +ar2 +ar3 + ar4 +…….. …….+ arn-2 + arn-1
(1)
Multiplying both sides of equation (1) by r, we get:
sr = ar +ar2 +ar3 + ar4 +…….. + arn-1 + arn
(2)
Subtracting equation (2) from (1), we get:
s - sr = a – arn
→ s ( 1 – r ) = a ( 1 - rn )
→ s = a ( 1 - rn ) / ( 1 – r )
The sum formula for the first n terms of a geometric progression
with ratio distinct from 1
s n = a ( 1 - rn ) / ( 1 – r )
If r = 1, then, obviously all the terms are equal to a, and the sum of n such term will
Example (14)
Consider the geometric progression of example (12)
Find:
1. The the sum of its first three terms and the first six terms
2. the sum of the term from the fourth to the sixth term
Solution:
1. We use the formula sn = a ( 1 - rn ) / ( 1 – r )
From example (12), we have: a=5 and d=2
Thus: S3 = 5( 1 - 23 ) / ( 1 – 2 ) = 5(-7)/(-1) = 35
And S6 = 5( 1 - 26 ) / ( 1 – 2 ) = 5(-63)/(-1) = 315
2. The sum of the terms from a4 to a6 = s6 – s3
= 315 – 35 = 280
Textbook: Example 7 page 229, Applied Examples (8)
Homework
Exercises 4.4 Page: 235
All odd numbered exercise from 23 to 31
Homework
1. For each of the following geometric progressions, find the 7th term
and the sum of the first 7 terms:
a. 2, 6, 18, 54, ….
Answers: a7 = 1458 and s7= 2186
b. 3, 6, 12, 24, 48, …
Answers: a7 = 192 and
s7 = 381
2. For the following geometric progressions, find the 5th term and the
sum of the first 5 terms:
1, 10, 100, 1000, 10000,….
Answers: a5 = 10000 and s5 = 11111
2. Find the geometric progressions, satisfying the two given conditions
(an = the n-th term and sn= the sum of the first n terms)
a. a2 = 6
and
a4 = 54
Answer:: 2, 6, 18, 54,…..
b. a2 = 6
and
a4 = 24
Answer: 3, 6, 12, 24, 48,…:
c. a2 = 100 and
a3 = 1000
Answer: 1, 10, 100, 1000,…
d. a2 = 6 and
r= 3
Answer:: 2, 6, 18, 54
e. a2 = 6 and a1 = 2
Answer:: 2, 6, 18, 54
I-3-01: Annuity
Definition
1. An annuity is a sequence of payment made at regular time interval.
2. The term of an annuity is the time period in which these payment are
made.
Types of Annuity
A. Classification by Payment Date
1. Ordinary Annuity: If the payments are made at the end of each
payment period
2. An Annuity due: If the payments are made at the beginning of each
payment period.
3. Simple Annuity: If the payment period coincides with the interest
conversion period.
4. Complex Annuity: If the payment period differs from the interest
conversion period.
B. Classification by the start and the end of the
Term (time period in which the payments are paid)
1. An Annuity Certain: If the term is given by a fixed time interval.
2. Perpetuity: If the term is given by a fixed time interval but extends
indefinitely.
3. Contingent Annuity: If the term is not fixed in advance
Suppose that R Riyals are to be paid into an account at the end of each
period for n periods and that the account earns interest at the rate of i
per conversion period ( i = r / m , where r is the nominal interest and m is the number of
conversion periods in a
year)
Discussion:
1. The first payment is paid at the end of the first period and earns
interest at the rate i over the remaining (n-1) periods
And hence it has an accumulated amount of An = R ( 1 + i)n-1
2. The second payment is paid at the end of the second period and
earns interest at the rate i over the remaining (n-2) periods and hence it
has an accumulated amount of An-1 = R ( 1 + i )n-2
3. The third payment is paid at the end of the third period and
earns interest at the rate i over the remaining (n-3) periods and hence it
has an accumulated amount of An-2 = R ( 1 + i)n-3
.
.
.
.
Lastly, the last payment ( the n-th) payment is paid at the end of the last period and it
earns no
interest. an earns interest at the rate i over the remaining (n-3) periods and hence it
has an accumulated amount of A1 = R
Thus, the accumulated amount S of the annuity is:
= A + A + A + ……………. + A + A + A + A
1
2
3
n-3
n-2
n-1
n
= R + R(1+i) + R(1+i)2 + R(1+i)3 +…………… R(1+i)n-1
= R [ [(1+i)n - 1 ] / [(1+i) - 1] , the sum of n terms of a
geometric progression
With: ratio r = 1 + I and first term a = R
The formula for (Future) Value of an Annuity:
S = R [ [(1+i)n - 1 ] / i]
Definition:
The expression [ [(1+i)n - 1 ] / i ] is called the compound-amount
factor, and denoted by Sn┐I
( pronounced S angle m @ i)
Example (15)
Suppose that 100 Riyals be paid into an account at the end of each
quarter for 3 years and that the account earns interest at the rate of
8% per year compounded quarterly. Find the accumulated S of this
annuity (called future value of the annuity)?
Solution:
We have : S = R [(1+i)n - 1 ] / i
Where;
R = 100, n = the number of periods = 4(3) = 12
i = The interest per period = (0.08) / 4 = 0.02
Thus,
S = 100 [(1+ 0.02)12 - 1 ] / 0.02
Example (16)
Suppose that 100 Riyals be paid into an account at the end of each
month for two years and that the account earns interest at the rate of
12% per year compounded monthly. Find the accumulated S of this
annuity?
Solution:
We have : S = R [(1+i)n - 1 ] / i
Where;
R = 100, n = the number of periods = 12(2) = 24
i = The interest per period = (0.12) / 12 = 0.01
Thus,
S = 100 [(1+ 0.01)24 - 1 ] / 0.02
Textbook: Example 1 page 206. Please notice that the payment here is made for
one year (the question says “12 monthly payment” ) and the conversion periods are also 12
( the question says “ 12% compounded monthly” ). Thus here both n and m are equal to 12.
Homework
Exercises 4.2 Page: 211
All odd numbered exercise from 1 to 7 and
exercise 17.
Homework
1. If a payment of 1800 Riyals is paid into an account at the end of
each quarter for 6 years and that the account earns interest at the
rate of 8% per year compounded quarterly. Find the amount S of this
annuity (future value of the annuity)?
Answer: 54759.35
1. If a payment of 200 Riyals is paid into an account at the end of each
month for 20.25 years and that the account earns interest at the rate
of 9% per year compounded monthly. Find the amount S of this
annuity (future value of the annuity)?
Answer: 137209.97
1. If a payment of 1000 Riyals is paid into an account at the end of
each year for 10 years and that the account earns interest at the
rate of 10% per year compounded anually. Find the amount S of this
annuity (future value of the annuity)?
Answer: 15937.42
Present Value of An Annuity
To find the amount P that if deposited will grow to be equal
to the future value of an annuity, let:
P(1+i)n = R [ [(1+i)n - 1 ] / i ]
Where n is the number of equal periodic payments R made
at the end of each period.
→ P = R [ [(1+i)n - 1 ] / i ] / (1+i)n
= R [ [(1+i)n - 1 ] / i ] (1+i)-n
= R [ 1 - (1+i)-n ] / i
Example (17)
Find the present value of an ordinary annuity consisting of
24 monthly payments of 100 Riyals each and earning
interest at 9% per year compounded monthly?
Solution:
We have : P = R [ 1 - (1+i)-n ] / i ]
Where;
R = 100, n = 24, i = r/m = 0.09 / 12 = 3/400 =0.0075
Thus,
P = 100 [ 1 - (1+0.0075)-24 ] / 0.0075
Textbook: Example 2 page 207 and Applied Examples (3) to (6) pages: 208
to 209
Assignment
Go through the following applied examples in the textbook:
1. Example 3, Saving for college education
2. Example 4, Financing a car
3. Example 5, IRAs
4. Example 6, Investment Analysis.
Note:
1. In all of these examples, we use the formula for “ Present value for an
annuity”
2. In Examples(5) i = r/ 1 = r
3. In Examples(6)
For Clarck’s: i = 0.8/1 = R [ [(1+i)n - 1 ] / i ] (1+i)-n
= R [ 1 - (1+i)-n ] / i
Homework
Exercises 4.2 Page: 211
Exercise from 9, 11 and13
Homework
1. Find the present value of an ordinary simple annuity of 5000 Riyals
for 8 years @ 6% per year compounded annually.
Answer: QR 31048.97
2. Find the present value of an ordinary annuity of 4000 Riyals for 5
years @ 9% per year compounded annually.
Answer: QR 15558.61
3. Find the present value of an ordinary annuity of 800 Riyals for 7
years @ 12% per year compounded quarterly.
Answer: QR 15011.29
I-3-01: Amortization & Sinking Funds
Definition
Amortization means essentially, finishing repaying back the
Loan, consisting of the principal (the amount borrowed) and
the interest, at the end of a fixed term in periodic
installment (usually of the of same size).
We will look at the case when the number of interest
conversion periods in year. is equal to the number of
payments per year.
.
Discussion:
We can think of the periodic loan repayment R as the payment on a simple
annuity and the original amount borrowed as the present value of the
annuity. Thus we can use the formula :
P = R [ 1 - (1+i)-n ] / i
→ R = Pi / [ 1 - (1+i)-n ]
This is the formula of the periodic payment R on a loan
P to be repaid (amortized) over n periods with interest
rate of i per (interest) conversion period.
Notice that
1. i =r/m,, where r is the nominal interest and m is the
number of periodic conversions per year and not the number of
Installments.
2. n is the number of all installments paid. If there are monthly
installments over 3 years, then n = 12(3) = 36.
Please, do not confuse m with n.
Example (18)
Determine the size of each of equal installments paid at the end of the
year in order to repay (amortize) at the end of 10 years a loan of
borrowed 20000 plus the interest, if the interest rate is 10%, calculated
at the end of each year
Solution:
We use the formula:
R = Pi / [ 1 - (1+i)-n ]
Where; m = 1 ( only one conversion period per year).
n = 10 ( one installment per year for 10 years = 10 installments)
P = 20000, i = r/m = 0.1 / 1 = 0.1
Thus,
R = 20000(0.1) / [ 1 - (1 + 0.1)-10 ]
Textbook: Example 2 page 207 and Applied Examples (1) to (5) pages: 216 to 219
Homework
Exercises 4.3 Page: 222 - 223
1. All odd numbered exercise from 1 to 7
1. Exercise from 19, 23 and 25.
Homework
1. Determine the size of each of equal installments paid at the end of
the year in order to repay (amortize) at the end of 10 years a loan of
borrowed 100000 plus the interest, if the interest rate is 8%,
calculated at the end of each year
Answer: 14902.95
2. Determine the size of each of equal installments paid at the end of
each quarter in order to repay (amortize) at the end of 3 years a loan
of borrowed 5000 plus the interest, if the interest rate is 4%,
calculated at the end of each quarter.
Answer: 444.24
3. Determine the size of each of equal installments paid at the end of
each quarter in order to repay (amortize) at the end of 12 years a
loan of borrowed 25000plus the interest, if the interest rate is 3%,
calculated at the end of each quarter.
Answer: 622.13
4. Determine the size of each of equal installments paid at the end of
each month in order to repay (amortize) at the end of 20 years a
loan of borrowed 80000 plus the interest, if the interest rate is 10%,
calculated at the end of each month.
Answer: 798.70
5. Determine the size of each of equal installments paid at the end of
the year in order to repay (amortize) at the end of 10 years a loan of
borrowed 100000 plus the interest, if the interest rate is 10%,
calculated at the end of each year
Answer: 16274.54
6. Determine the size of each of equal installments paid at the end of
each month in order to repay (amortize) at the end of 3 years a loan
plus the interest borrowed to buy a car costing QR 16000, if the
down payment is quarter of the price of the car and the interest rate
is 4%, calculated at the end of each month. What will be the interest
paid?
Answer: a. 387.21 , b. The interest paid = 12000 – 36(387.21) =
1939.42
Note that the down payment = 16000 – (16000)(1/4) = 12000
7. Change the number of years in Ex (6) t0 4 and aswer the same
questions
Answer: a. 304.35 , b. The interest paid = 2608.85
8. Determine the size of each of equal installments paid at the end of
each month in order to repay (amortize) at the end of 30 years a
loan plus the interest borrowed to buy a house costing QR 270000,
if the down payment is QR 30000 and the interest rate is 8%,
calculated at the end of each month. What will be the equity after 5
years, 10 years and 20 years ? ( ignore appreciation)
Answer: a. 1761.03
b.
b1. 41833.34
b2. 59460.71
b3. 124852.9
Sinking Funds
Definition
Sinking funds means essentially, an account set up for a
particular goal at some date in the future.
Examples:
1. An account set up by a company to purchase new equipments
at some future date.
2. An account set up by a person to pay a debt at some future date.
3. . An account set up by a corporation to hire some expertise at some future
date.
Discussion:
We can think of the amount to be accumulated by a certain date in the
future as the future value S of an annuity. Thus, we will be able to
calculate the size R of an equal number n of installments that should be
paid into the fund, taking into consideration the nominal interest rate
and the number of interest conversion periods in a year. This allows us
to use the formulas:
S = R [ (1+i)n - 1] / i
→ R = Si / [ (1+i)n - 1] (Sinking Fund Payment Formula)
This is the formula of the periodic payment R necessary to reach
an accumulated amount S over n periods with interest rate of i
per( interest conversion) period.
Example (19)
Determine the size of each of equal quarterly installments that should be paid
into a (sinking) fund of 50000 to be collected in 3 years if the fund is earning
20% interest per year compounded quarterly.
Solution:
We use the formula:
R = Si / [ (1+i)n - 1]
Where; m = 4 ( 4 conversion period per year).
n = 4(3) = 12 ( 4 installments per year for 3 years)
S = 50000, i = r/m = 0.2 / 4 = 0.05
Thus,
R = 50000(0.05) / [ (1+ 0.05)12 - 1]
Textbook: Applied Examples 6 page: 220
Homework
Exercises 4.3 Page: 222 - 223
1. All odd numbered exercise from 9 to 13
2. All odd numbered exercise from 27 to 33
Homework
1.
Determine the size of each of equal semiannual installments that
should be paid into a (sinking) fund of QR 20000 to be collected in 6
years if the fund is earning 4% interest per year compounded
semiannually.
Answer: 1491.19
2. Determine the size of each of equal installments paid at the end of
every two month that should be paid into a (sinking) fund of QR
100000 to be collected in 20 years if the fund is earning 4.5%
interest per year compounded at the end of each two month.
Answer: 516.76
3. Determine the size of each of equal monthly installments that should
be paid into a (sinking) fund of QR 250000 to be collected in 25
years if the fund is earning 10.5% interest per year compounded
monthly
Answer: 172.95
4.
Determine the size of each of equal annual installments that should
be paid into a (sinking) fund of QR 2500000 to be collected in 20
years if the fund is earning 7% interest per year compounded
annually.
Answer: 60982.31
5.
Determine the size of each of equal quarterly installments that should
be paid into a (sinking) fund of QR 200000 to be collected in 10
years if the fund is earning 9% interest per year compounded
quarterly.
Answer: 3135.48
6. Determine the size of each of equal monthly installments that should
be paid into a (sinking) fund of QR 250000 to be collected in 25
years if the fund is earning 8.5% interest per year compounded
monthly.
Answer: 242.23
7. Determine the size of each of equal monthly installments that should
be paid into a (sinking) fund of QR 450000 to be collected in 30
years if the fund is earning 10% interest per year compounded
monthly.
Answer: 199.07
Summary Hint
Starting from the form ulas for :
1* . Com poundinterest(The accum ulated am ount Future value) : A  P (1  i ) n
&
a (r n  1)
2 * . Sum of geom etric progression : S n 
, where r here is the ratio and a the first term.
r 1
We can deduceall the ther form ulas for :
1. Com poundinterest(The present value  Principal)  A(1  i )  n
2. Future value of annuity:
S  R  R (1  i )  R (1  i ) 2  R (1  i ) 3     R (1  i ) n 1
R[(1  i ) n  1]
(
(1  i )  1
R[(1  i ) n  1]

i
by substituting r  1  i and a  R in 2 *
3. The present value of annuity:
R[1  (1  i )  n ]
P
i
R[(1  i ) n  1]
is deduced from the future value of annuity S 
i
R[(1  i ) n  1]
n
by letting P(1  i ) 
i
4. The am ountam ortized
R[1  (1  i )  n ]
P
i
sam e as for the present value of annuity
4. The periodic paym entneeded to am ortized a loan with an originalam ountP
iP
R
1  (1  i )  n
deduced from (4)
4. Periodic paym enttowards a Sinking Fund of am ountS
deduced from form ula(2) for the future value of annuityR 
iS
(1  i ) n  1
Using Excel
Arithmetic Progression
The n-th Term & The Sum of the First n Terms
a sn
a
2
1
-5
-5
-5
-5
22
d
3
1
0.5
-2
-0.5
2
10
n
31
31
31
31
31
31
31
n
94
32
5
-70
-25
50
344
1457
496
77.5
-1085
-387.5
775
5332
b
c
d
e
f
g
2
1
-5
-5
-5
-5
22
3
1
0.5
-2
-0.5
2
10
33
33
33
33
33
33
33
100
34
6
-74
-26
54
364
1650
561
99
-1221
-429
891
6006
Ex 2a
b
c
d
e
f
g
a
d
an
First Term
Difference
The n-th term
Sum of the first n
term s
sn
For the n-th
term
Formula
an
= a+(n-1)*d
For the sum of
the first n terms
Formula
sn
=
n*(2*a+(n-1)*d)/2
Ex 1-
Geometric Progression
The n-th Term & The Sum of the First n Terms
a
2
3
1
r
3
2
10
a
d
First Term
an
The n-th term
Ratio
Sum of the
first n terms
sn
For the n-th
term
Formula
an
= a*r^n
For the sum of
the first n terms
Formula
Also
sn
sn
=
a*(r^n-1)/(r-1)
=
a*(1-(r^n))/(1-r)
n
7
7
5
a
n
1458
192
10000
s
n
2186
381
11111
Ex 1 - a
Ex 1 - b
Ex 2
Amortization
Applying the formula of the periodic payment R on a loan
P to be repaid (amortized) over n periods with interest
rate of i per interest conversion period.
R = Pi / [ 1 - (1+i)-n ]
P
20000
100000
5000
25000
80000
100000
12000
12000
240000
240000
240000
160000
P
r
m
t
n = mt
r
m
t
i
n
R
0.1
0.08
0.04
0.03
0.105
0.1
0.1
0.1
0.08
0.08
0.08
0.09
1
1
4
4
12
1
12
12
12
12
12
12
10
10
3
12
20
10
3
4
30
30
30
30
0.1
0.08
0.01
0.0075
0.00875
0.1
0.008333
0.008333
0.006667
0.006667
0.006667
0.0075
10
10
12
48
240
10
36
48
360
360
360
360
3254.908
14902.95
444.2439
622.1261
798.7039
16274.54
387.2062
304.351
1761.035
1761.035
1761.035
1287.396
Original Loan
yearly interest rate
num ber of periods/yr
num ber of years
num ber of paym ents in t yrs
sam e as num ber of periods in t yrs
→
I = r / m interest rate per period
R
Periodic paym ent
Formula
R =
i*P / (1-(1+i)^(-n))
Example (18)
Ex 1
Ex 2
Ex 3
Ex 4
Ex 5
Ex 6
Ex 7
Ex 8
Ex 8-2
Ex 8-3
Example (18)-02
Sinking Fund
Periodic Payment Formula
Sinking Fund
Periodic Payment Formula
S
30000
20000
100000
250000
2500000
200000
250000
450000
r
m
t
i
n
R
0.1
0.04
0.045
0.105
0.07
0.09
0.085
0.1
4
2
6
12
1
4
12
12
2
6
20
25
20
10
25
30
0.025
0.02
0.008
0.009
0.07
0.023
0.007
0.008
8
12
120
300
20
40
300
360
3434
1491.2
516.76
172.95
60982
3135.5
242.23
199.07
S
r
m
The Sum to be accumulated
interest/yr
number of conversion periods/yr
equal number of payment/yr
t
number of years
I = r / m interest per period
n = mt number of periods in t years
→
R
The periodic Payment
The
R =
Formula
i*S/((1+i)^n-1)
Example (19)
Ex 1
Ex 2
Ex 3
Ex 4
Ex 5
Ex 6
Ex 7