Mathematics of Finance

Transcript Mathematics of Finance

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 A 1 after 1 year: A 1 = p + pr(1) = p (1 + r) The accumulated amount A 2 A 2 = A 1 + A 1 r(1) = A 1 after 2 years: (1 + r) = p(1 + r) (1 + r)= p(1 + r) 2

The accumulated amount A 3 A 3 = A 2 + A 2 r(1) = A 2 after 3 years: (1 + r) = p(1 + r) 2 (1 + r)=p (1 + r) 3 The accumulated amount A after 4 years: A 4 = A 3 + A 3 r(1) = A 3 (1 + r) = p(1 + r) 3 (1 + r)= p(1 + r) 4 We conclude that the accumulated amount after t years A t = 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 A 1 A 1 = p + pi = p (1 + i) after 1 period: The accumulated amount A 2 A 2 = A 1 + A 1 i = A 1 after 2 periods: (1 + i) = p(1 + i) (1 + i)= p(1 + i) 2

The accumulated amount A 3 A 3 = A 2 + A 2 i = A 2 after 3 periods: (1 + i) = p(1 + i) 2 (1 + i)=p (1 + i) 3 The accumulated amount A after 4 periods: A 4 = A 3 + A 3 i = A 3 (1 + i) = p(1 + i) 3 (1 + i)= p(1 + i) 4 We conclude that the accumulated amount after n periods A n = 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 A n = 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

Riyals at interest rate of

years on a deposit of 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

Riyals at interest rate of

years on a deposit of 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 e rt Thus the amount accumulated if the interest is compounded continuously is:

A = p e rt

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 e rt

, 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

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) r

eff

= [1 + (r/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

r We use the formula eff = [1 + (r/m)] m – 1, Where r = 10/100 = 1/10 = 0.1

r 1. quarterly → m = 4 eff = [1 + (0.1)/4)] 4 – 1 r 2. monthly → m = 12 eff = [1 + (0.1)/12)] 12 – 1 r 3. annually → m = 1 eff = [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+ r

eff

) = p[1+ (r/m)]

Where, r = the nominal yearly interest rate m= the number of conversion periods per year →1+ r

eff =

[1+ (r/m)]

→r

eff =

[1+ (r/m)]

- 1

Homework

Exercises 4.1 Page: 197 Exercises 21 and 23

Homework

What is the effective rate of interest r eff corresponding to the following nominal interest rate: 1. 10% per year compounded semiannually

Answer: r eff = 10.25 % per year

2. 8% per year compounded monthly

Answer: r eff = 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) ]

→ p = A / [ 1 +(r/m) ]

→ 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.

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 a 2 =a 1 + d = 5 + 2 = 7 a 3 = a 2 + d = 7 + 2 = 9 = 5 + 2(2) = a + 2d a 4 = a 3 + d = 9 + 2 = 11= 5 + 3(2) = a + 3d a 5 = a 4 + d = 11 + 2 = 13= 5 + 4(2) = a + 4d In general:

a n = a n-1 + d = a + (n-1)d For this progression: a n For example: = 5 + 2 (n-1) a 21 = 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 a 2 =a 1 + d = 3 + 10 = 13 a 3 = a 2 + d = 13 + 10 = 23 = 3 + 2(10) = a + 2d a 4 = a 3 + d = 23 + 10 = 33= 3 + 3(10) = a + 3d a 5 = a 4 + d = 33 + 10 = 43= 3 + 4(10) = a + 4d In general: The n-th term Formula

a n = a + (n-1)d For this progression: a n For example: = 3 + 10 (n-1) a 21 = 3 + (21-1)(10) = 3 + 200 = 203 Textbook: Example 1 page 229

Example (10)

Find the arithmetic progression with the 21 st term is being 45 and the 23 rd 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 + a n s = a n + (a n -d) + (a → 2s = ( a + a n n -2d) + (a ) + ( a + a n n -3d)+ …….+[a n ) + ( a + a n - (n-2)d] + a ) + ……+ ( a + a n ) = n ( a + a n ) = 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: s n = 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 S n = n [ 2a + (n-1)d] / 2 From example (8), we have: a=5 and d=2 Thus: S 10 And S 5 = 10 [ 2(5) + (10 -1)2] / 2 = 140 = 5[ 2(5) + (5 -1)2] / 2 = 45 2. The sum of the terms from a = 140 – 45 = 95 6 to a 10 = S 10 – S 5

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 31 st sum of the first 31 terms: a. 2, 5, 8, 11, 13….

term and the

Answers: a 31 = 94 and s 31 = 1457

b. 1, 2, 3, 4, 5, …

Answers: a 31 = 32 and s 31 = 496

c. -5, 4 ½ , -4, -3 ½ , -3

Answers: a 31 = 5 and s 31 = 77.5

d. -5, -7, -9, 11,…

Answers: a 31 = - 70 and s 31 = - 1085

e. 5, 5 ½ , -6, -6½ , -7,…

Answers: a 31 = - 25 and s 31 = - 387.5

f. -5, -3, -1, 1,3, ….

Answers: a 31 = 50 and s 31 = 775

g. 22, 32, 42, 52, …

Answers: a 31 = 344 and s 31 = 5332

h.* x, x + 2y, x + 4y, x + 6y, x + 8y ….

Answers: a 31 = x + 60y and s 31 =31 x + 930y

i* . x + 2y, x + y, x, x – y, x - 2y,..

Answers: a 31 = x - 28y and s 31 = 31x – 403y

2. Repeat Ex(1) for n =33 3. Find the arithmetic progressions, satisfying the two given conditions (a n = the n-th term and s n = the sum of the first n terms)

a. a 31 = 94 and a 34 = 98 b. a 31 = 94 and a 1 = 2 b. a 31 = 94 and d= 3 d. a 4 = 11 and s 4 = 26 Answers: 2, 5, 8, 11, …..

I-2-02: Geometric Progression

A sequence of numbers in the form: a, ar, ar 2 , ar 3 , …….

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 a 2 =a r = 5 ( 2 ) = 10 a 3 = a 2 r = 10(2) = 20 = 5 (2) 2 = a r 2 a 4 = a 3 r = 20(2) = 40 = 5 (2) 3 = a r 3 a 5 = a 4 r = 40(2) = 80 = 5 (2) 4 = a r 4 In general: a n = a n-1 r = a r n-1

The formula for the n-th term of the geometric progression: a n = a r n-1

a 7 = a r 7-1 = 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 = ar 3-1= ar 2 and 160 = ar 6-1 → 160/20 = ar 5 / ar 2 = ar 5 → 8 = r 3 → r = 2 Thus, 20 = a (2) 2 = 4a → a= 20/4 = 5

Textbook: Example 6page 231

Sum of the first n terms of an geometric progression

Consider the geometric progression a, ar, ar 2 , ar 3 , ar 4 , ….ar

5 , ar 6 , …….

Where r is not equal to 1.

Let s be the sum of the first n terms of the progression.

Then: s = a + ar +ar 2 +ar 3 + ar 4 + …….. …….+ ar n-2 + ar n-1 (1) Multiplying both sides of equation (1) by r, we get: sr = ar +ar 2 +ar 3 + ar 4 + …….. + ar n-1 + ar n (2) Subtracting equation (2) from (1), we get: s - sr = a – ar n → s ( 1 – r ) = a ( 1 - r n ) → s = a ( 1 - r n ) / ( 1 – r )

The sum formula for the first n terms of a geometric progression with ratio distinct from 1

s n = a ( 1 - r n ) / ( 1 – r )

If r = 1, then, obviously all the terms are equal to a, and the sum of n such term will be na.

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 s n = a ( 1 - r n ) / ( 1 From example (12), we have: a=5 and d=2 – r ) Thus: S And S 6 3 = 5( 1 - 2 = 5( 1 - 2 6 = 315 – 35 = 280 3 ) / ( 1 – 2 ) = 5(-7)/(-1) = 35 ) / ( 1 – 2 ) = 5(-63)/(-1) = 315 2. The sum of the terms from a 4 to a 6 = s 6 – s 3

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 7 th term and the sum of the first 7 terms: a. 2, 6, 18, 54, ….

Answers: a 7 = 1458 and s 7 = 2186

b. 3, 6, 12, 24, 48, …

Answers: a 7 = 192 and s 7 = 381

2. For the following geometric progressions, find the 5 th term and the sum of the first 5 terms: 1, 10, 100, 1000, 10000, ….

Answers: a 5 = 10000 and s 5 = 11111

2. Find the geometric progressions, satisfying the two given conditions (a n = the n-th term and s n = the sum of the first n terms)

a. a 2 = 6 and a 4 = 54

Answer:: 2, 6, 18, 54, …..

b. a 2 = 6 and a 4 = 24

Answer: 3, 6, 12, 24, 48, …:

c. a 2 = 100 and a 3 = 1000

Answer: 1, 10, 100, 1000, …

d. a 2 = 6 and r = 3

Answer:: 2, 6, 18, 54

e. a 2 = 6 and a 1 = 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.

. 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 A n = 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 A n-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 A n-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 A 1 = R

Thus, the accumulated amount S of the annuity is: = A 1 + A 2 + A 3 + ……………. + A n-3 + A n-2 + A n-1 + A 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

S

n ┐ 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 Where; - 1 ] / i 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)

= R [ [(1+i)

- 1 ] / i ]

Where n is the number of equal periodic payments R made at the end of each period.

→ P = R [ [(1+i)

- 1 ] / i ] / (1+i)

= R [ [(1+i)

- 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

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)

- 1] / i → R = Si / [ (1+i)

- 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 formulas for

: 1 * .

Compound

int

erest

(

The accumulate d amount



Future value

) :



( 1 

)

2 * .

Sum of

geometric progressio n

S n



(

r n r

 1 )  1 ,

where r here is the ratio and a the We can deduce all the ther

1 .

Compound

int

erest

(

The formulas for

present value



P r in cipal

) 

( 1 

) 

n first term

2 .

Future value of annuity





( 1 

) 

( 1 

) 2 

( 1 

) 3 (

[( 1 ( 1  

i i

) )

  1 1 ]  

[( 1 

)

 1 ]

i by substituti ng r

 1 

i and a



R in

2 *      

( 1 

)

 1

3 .

The present value of P



[ 1  ( 1 

) 

]

i annuity

is deduced from the future value of annuity S



[( 1 

)

 1 ]

i by letting P

( 1 

)



[( 1 

)

 1 ]

4 .

The amount amortized P



[ 1  ( 1 

) 

]

i same as for the present value of annuity

4 .

The periodic payment needed to amortized a loan with an original amount P R

 1 

( 1 

) 

n deduced from

( 4 ) 4 .

Periodic payment towards a Si n king Fund of amount S deduced from formula

( 2 )

for the future value of annuity R



( 1 

)

 1

Using Excel

Arithmetic Progression The n-th Term & The Sum of the First n Terms

2 1 -5 -5 -5 -5 22 a 2 1 -5 -5 -5 -5 22 3 1 0.5

-2 -0.5

2 10 d 3 1 0.5

-2 -0.5

2 10

a d an sn First Term Difference The n-th term Sum of the first n term s Formula For the n-th term a n = a+(n-1)*d Formula s n = For the sum of the first n terms n*(2*a+(n-1)*d)/2

33 33 33 33 33 33 33 n 31 31 31 31 31 31 31 a n s n 94 32 1457 496 5 77.5

-70 -1085 -25 -387.5

50 775 344 5332 100 1650 34 6 561 99 -74 -1221 -26 -429 54 891 364 6006 Ex 1 b c d e f g d e f g Ex 2 a b c

Geometric Progression The n-th Term & The Sum of the First n Terms

a 2 3 1 r 3 2 10 a d a n s n First Term Ratio The n-th term Sum of the first n terms Formula For the n-th term a n = a*r^n Formula s n = Also s n = For the sum of the first n terms 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.

P 20000 100000 5000 25000 80000 100000 12000 12000 240000 240000 240000 160000 r 0.1

0.08

0.04

0.03

0.105

0.1

0.08

0.09

R = Pi / [ 1 - (1+i) -n ] m 1 1 4 4 12 1 12 12 12 12 12 12 t 10 10 3 12 20 10 3 4 30 30 30 30 i n 0.1

0.08

0.01

0.0075

0.00875

0.1

10 10 12 48 240 10 0.008333 36 0.008333 48 0.006667 360 0.006667 360 0.006667 360 0.0075

360 R 3254.908

14902.95

444.2439

622.1261

798.7039

16274.54

387.2062

304.351

1761.035

1287.396

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 P r m t n = mt I = r / m R 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 interest rate per period Periodic paym ent Formula R = i*P / (1-(1+i)^(-n))

→

Sinking Fund Periodic Payment Formula

Sinking Fund Periodic Payment Formula S 30000 20000 100000 250000 2500000 200000 250000 450000 r 0.1

0.04

0.045

0.105

0.07

0.09

0.085

0.1

m 4 2 6 12 1 4 12 12 t 2 6 20 25 20 10 25 30 i n 0.025 8 0.02

12 0.008 120 0.009 300 0.07

20 0.023 40 0.007 300 0.008 360 R 3434 Example (19) 1491.2 Ex 1 516.76 Ex 2 172.95 Ex 3 60982 Ex 4 3135.5 Ex 5 242.23 Ex 6 199.07 Ex 7 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)