Business Mathematicscontents.kocw.net/KOCW/document/2013/gacheon/LEEYongju/9.pdfFinancial...

Lecture Note #9 Chapter 5

Business

Mathematics

1

Financial Mathematics

1. Arithmetic and Geometric Sequences and

Series

2. Simple Interest, Compound Interest and

Annual Percentage Rates

3. Depreciation

4. NPV and IRR

5. Annuities, Debt Repayments, Sinking

Funds

6. Interest Rates and Price of Bonds

Arithmetic and Geometric Sequences and Series

• Sequence : A list of numbers which follow a

definite pattern or rule.

1. Arithmetic Sequence : Each term, after the first,

is obtained by adding a constant, d, to the

previous term, where d is called the common

difference

2. Geometric Sequence : Each term, after the first,

is obtained by multiplying the previous term by a

constant, r, where r is called the common ratio

Arithmetic and Geometric Sequences and Series

• Series : The sum of the terms of a sequence.

– Finite series: the sum of a finite number of terms

of a sequence

– Infinite series: the sum of an infinite umber of

terms of a sequence

1. Arithmetic Series (Arithmetic Progression, AP) : the

sum of the terms of an arithmetic sequence.

2. Geometric Series (Geometric Progression, GP) : the

sum of the terms of a geometric sequence.

Arithmetic Sequences and Series

Arithmetic Sequences

and Series

Geometric Sequences and Series

Geometric Sequences

and Series

Application of Arithmetic and Geometric Series

A manufacturer produces 1,200 computers in the first

week. But after week 1, it increases production by:

i) scheme I: 80 computers each week

Ii) scheme II: 5% each week.

(a) Find out the production quantity in week 20 under each

scheme.

(b) Find out the total production quantity over the first 20

weeks under each scheme.

(c) Find the week in which the production quantity reaches

8,000 or more for the first time under each scheme.

Application of Arithmetic and Geometric Series

Assignment #2

• Problems 1, 9, 10, 11

of Progress Exercises 5.1

Due on 2013/05/02 (Thursday)

Simple Interest

Compound Interest

How compounding is carried out

(when annual interest rate i %)

The next slide demonstrates

….how interest is calculated at the end of each

year

….interest earned is added to the principal

….principal at the start of next year = (principal

+ interest) from previous year

21

iP2 P2 + iP2 = P2(1+ i) = P3

The table below will be filled in, row by row…

..to demonstrate the idea of compounding annually at an interest

rate i %

Amount at start of

year = principal

Interest earned

during year

Amount at end of Year

= principal + interest

iP1 P1 + iP1 = P1(1+ i) = P2

Year t Pt-1 iPt-1 Pt-1 + iPt-1 = Pt-1(1+ i) = Pt

In general, at the end of year t ….

Year 1 P0 iP0 P0 + iP0 = P0(1+ i) = P1

P1 Year 2

Year 3 P2

22

Compound interest formula (II)

principal + interest BUT, in terms of P0…

P0 + iP0 = P0(1+ i) = P1

In general….

Next year

P1 + iP1 = P1(1+ i) = P2

Next year

P2 + iP2 = P2(1+ i) = P3

But P1(1+ i) = P0(1+ i) (1+ i)

so… P2 = P0(1+ i)2

so… P3 = P0(1+ i)3

so… P1 = P0(1+ i)

But P2(1+ i) = P0(1+ i)2 (1+ i)

P4 = P0(1+ i)4

….Pt = P0(1+ i)t

..and so on…

23

Worked Example 5.5 (see text) Calculate the amount owed on a loan of £1000 at the end of three years, interest compounded annually, rate of 8%

..the compound interest formula

Method

Substitute the values given into the compound interest formula

• t = 3 years

• P0 = 1000

• i = = 0.08

Calculations

100

8

you will need.. P t = P0(1+ i)t

712.1259

)2597120.1(1000

)08.1(1000 3

303 )1( iPP

3)08.01(1000

24

Terminilogy: present value; future value

In the compound interest formula;

P t = P0(1+ i)t

Pt is called the future value of P0 at the end of t years when

interest at i% is compounded annually

P0 is called the present value of Pt when discounted at i%

annually

…see following examples

25

The present value formula is deduced from the compound

interest formula as follows:

tt iPP )1(0

t

t

tt

i

iP

i

P

)1(

)1(

)1(

0

0)1(

Pi

P

t

t

t

t

i

PP

)1(0

26

Worked Example 5.6 (a)(i)

£5000 is invested at an interest rate of 8% for three years

..the compound interest formula

Method

Substitute the values given into the compound interest formula

• t = 3 years

• P0 = 5000

• i = = 0.08

Calculations

100

8

You will need P t = P0(1+ i)t

5.6298

)2597120.1(5000

)08.1(5000 3

303 )1( iPP

3)08.01(5000

27

Revise terminilogy: present value; future

value • In the compound interest formula;

P t = P0(1+ i)t

future value present value

In Worked Example 5.6

Pt = 6298.5 is called the future value of P0 = 5000 at the end of

3 years when invested at 8% compounded annually

P0 = 5000 is called the present value of Pt= 6298.5

when discounted at 8% annually for 3 years

28

Worked Example 5.6(b)(i) Present value calculations

(£6298.5 discounted at 8% annually for three years)

..the present value formula will be required

Method

Substitute the values given into the present value formula

• t = 3 years

• Pt = 6298.5

• i = = 0.08

Calculations

100

8

t

t

i

PP

)1(0

33

0)1( i

PP

3)08.01(

5.6298

3)08.1(

5.6298

5000

29

Worked Example 5.6 (b)(ii) Present value calculations

(£15,000 discounted at 8% annually for three years)

..the present value formula will be required

Method

Substitute the values given into the present value formula

• t = 3 years

• Pt = 15,000

• i = = 0.08

Calculations

100

8

t

t

i

PP

)1(0

33

0)1( i

PP

3)08.01(

15000

3)08.1(

15000

48.11907

30

How to compound twice annually

(rate = i % pa)

tt iPP )1(0 ..compounding once annually

t

ti

PP

2

02

1

..compounding twice annually

At each compoumding

use the annual rate, i, divided by 2 2 x t compoundings

necessary in t years

Two compoundings necessary in 1 year

31

How to compound three times annually

(rate = i% pa)


t

ti

PP

3

03

1

..compounding three times annually

At each compoumding

use the annual, I, rate divided by 3 3 x t compoundings


Three compoundings necessary in 1 year

32

How to compound m times annually

(rate = i% pa)


mt

tm

iPP

10

..compounding m times annually

At each compoumding

use the annual rate,i, divided by m m x t compoundings


m compoundings necessary in 1 year

33

Compounding continuously

tt iPP )1(0 …compounding once annually

mt

tm

iPP

10

…compounding m times annually

tm

tm

iPP

10 …rearranging

ittit ePePP 00

me

m

i im

as 1

itt ePP 0

34

Calculations

Worked Example 5.8 (a). £5000 is invested at an interest

rate of 8% for three years compounded semiannually

Method

Substitute the values given in the question into the compound interest formula above

• m = 2

• t = 3 years

• P0 = 5000

• i = = 0.08 100

8

595.6326

)2653190.1(5000

)04.1(5000 6

6)04.01(5000

mt

tm

iPP

10 3

03 1

m

m

iPP

32

32

08.015000

P

you will need the formula..

35

Calculations

Worked Example 5.8 (c)(i). £5000 is invested at an

interest rate of 8% for three years compounded monthly

Method

Substitute the values given into the compound interest formula above

• m = 12

• t = 3 years

• P0 = 5000

• i = = 0.08 100

8

185.6351

)270237.1(5000

mt

tm

iPP

10 3

03 1

m

m

iPP

312

312

08.015000

P

you will need the formula..

36

Calculations

Worked Example 5.8 (c)(ii) £5000 is invested at an

interest rate of 8% for three years compounded daily (assume 365 days per year)

Method

Substitute the values given

into the compound interest

formula above

• m = 365

• t = 3 years

• P0 = 5000

• i = = 0.08 100

8

079.6356

)2712157.1(5000

)0002192.1(5000 1095

mt

tm

iPP

10

3

03 1

m

m

iPP

3365

3365

08.015000

P

you will need the formula...

37

Calculations

Worked Example 5.9

£5000 is invested at an interest rate of 8% for three years

compounded continuously

Method

Substitute the values given

into the compound interest

formula above

• t = 3 years

• P0 = 5000

• i = = 0.08

100

8246.6356

)2712492.1(5000

5000 24.0

e

itt ePP 0

303

iePP

308.03 5000 eP

you will need the formula...

38

How much do you gain when interest is compounded

more than once annually ?

Review results in Worked Examples 5.6, 5.7 and 5.9

£5000 is invested at a nominal interest rate of 8% for three

years but compounded at various intervals annually.

The future value at the end of 3 years was calculated:

• 6298.560 compounded once annually

• 6326.595 compounded twice annually

• 6351.185 compounded monthly

• 6356.079 compounded daily

• 6356.246 compounded continuously

39

How much do you gain when interest is compounded

more than once annually ?

Review results in Worked Examples 5.6, 5.7 and 5.9

£5000 is invested at a nominal interest rate of 8% for

three years but compounded at various intervals annually

• 6298.560 one conversion period

• 6326.595 2 conversion periods



• 6356.246 infinete conversion periods (continuous)

40

How much do you gain by compounding more than

once annually ?

Conversion

periods/year

Amount at

end of 3

years

Difference over annual

compounding

1 6298.560

2 6326.595 6326.595 - 6298.560 = 28.035

12 6351.185 6351.185 - 6298.560 = 52.625

365 6356.079 6356.079 - 6298.560 = 57.519

Infinitely many

(continuous)

6356.246

6356.246 - 6298.560 = 57.686

41

How do we make comparisons when different

conversions periods are used?

• Use Annual Percentage Rates: APR

• What is the APR?

• The APR is the interest rate, compounded annually that yeilds an amount Pt

• the same amount Pt would be yeilded when any other method of compounding is used, for example..

42

Annual Percentage Rates: APR

Pt calculated using the APR rate annually

is the same as

Pt calculated by any other method

itt ePP 0

tt APRPP )1(0

mt

tm

iPP

10

43

Calculate the APR when interest is

compounded m times annually

mt

tm

iPP

10

compounding m times annually at a nominal rate of i % p.a.

tt APRPP )1(0 compounding once annually at

APR% p.a.

But Pt is the same whcihever method is used, hence mt

t

m

iPAPRP

1)1( 00

Next slide 44


compounded m times annually

But Pt is the same whcihever method is used, hence

mtt

m

iPAPRP

1)1( 00

mtt

m

iAPR

1)1(

m

m

iAPR

1)1(

11

m

m

iAPR

45


compounded continuously


itt ePAPRP 00 )1(

itt eAPR )1(

ieAPR )1(

1 ieAPR

46

Calculate the APR:

Progress Exercises 5.4 no 11(a)

Pt is the same whcihever method is used, hence

323

2

06.015500)1(5500

APR

323

2

06.01)1(

APR

2

2

06.01)1(

APR

0609.0103.012

APR

47

Calculate the APR

Progress Exercises 5.4 no 11(d)


306.00

30 )1( ePAPRP

306.03)1( eAPR

06.0)1( eAPR

06184.0106.0 eAPR

48

Assignment #3

• Problems 4 and 6 of Progress Exercises 5.2

• Problems 5, 6, 15 of Progress Exercises 5.3

• Problems 5, 6 of Progress Exercises 5.4

Due on 2013/05/07 (Tuesday)

Depreciation • Depreciation: allowance made for the wear and tear

of equipment during the production process which

involves reduction of the asset value.

• There are two depreciation methods:

1. Straight-line depreciation subtracts equal amount

from the original asset value each year. This is the

converse of simple interest.

2. Reducing-balance depreciation subtracts equal

rate from the asset value of the previous year. This is

the converse of compound interest.

Depreciation

Worked Example 5.11

Worked Example 5.12

NPV and IRR

• NPV and IRR are the two techniques which are used

to appraise investment projects (or investment

alternatives).

• More specifically, NPV and IRR are used to

determine whether to invest in a certain investment

project, or are used to select one or a few among

many investment alternatives.

• NPV(Net Present Value)

• IRR(Internal Rate of Return)

Net Present Value(NPV)

• NPV is the sum of the present values of several

future cash flows discounted at a given rate.

• NPV uses present values to appraise the profitability

of investment projects.

• While calculating NPV, a given discount rate (i) is

used to convert all future cash flows into present

values.

• Each Cash flow is either cash inflow or cash outflow.

• Cash inflow is a return from the investment.

• Cash outflow is a cost or money to be invested.

Net Present Value(NPV)

Calculating NPV

Year

(t)

Cash flow

0 -400,000 1 -400,000

1 120,000 0.9259 111,111

2 130,000 0.8573 111,454

3 140,000 0.7938 111,137

4 150,000 0.7350 110,254

43,956

Internal Rate of Return(IRR)

• In the previous example of calculating NPV, discount rate

of 8% was used. As a result, NPV=$43,956 was obtained.

The value of NPV, however, changes as the discount rate

changes.

• If the discount rate increases, the NPV decreases.

• If the discount rate is increased to 12.6555%, the NPV

becomes zero.

• If the discount rate is further increased to 15%, the NPV

would result in a negative value. (See Table 5.4 on page

232)

• IRR is the discount rate at which the NPV is zero.

• For the previous example, the IRR is 12.6555%

Internal Rate of Return(IRR)

• Decision rule for using IRR:

Invest in the project, if IRR > market rate of interest

Do not invest in the project, if IRR < market rate of interest

Calculating IRR

(1) Graphical method

•Calculate NPV’s for several different discount rates so that

NPV’s range from positive to negative values. Then, plot the

points of (discount rate, NPV) on a graph where the

horizontal axis represents discount rate and the vertical axis

represents value of NPV. Then connect the points to get a

curve.

•The value of the discount rate of the point at which the curve

crosses the horizontal axis is the IRR.

Calculating IRR

Comparison of NPV and IRR

• When comparing the profitability of two or more projects,

the most profitable project would be the project with the

largest NPV which would be the project with the largest

IRR.

• Advantage of using NPV: results are given in cash terms

• Disadvantage of using NPV: results change when

discount rate is changed

• Advantage of using IRR: results are independent of any

external rates of interest

• Disadvantage of using IRR: does not differentiate between

the scale of projects. Higher IRR with smaller NPV due to

small scale of the project.

Comparison of NPV and IRR

Project I (discount rate = 10%) Project II (discount rate = 10%)

year Cash flow Discount

factor

PV year Cash flow Discount

factor

PV

0 -100,000 1 -100,000 0 -10 1 -10

1 120,000 109,091 1 50 45

NPV of Project I = 9,091 NPV of Project II = 35

Compound Interest for Fixed

Deposits at Regular Intervals of Time



Worked Example 5.15

•New members of a golf club are admitted at the start of

each year and pay a joining fee of $2,000. Henceforth

members pay the annual fee of $400, which is due at

the end of each year. How much does the club earn

from a new member over the first 10 years, assuming

annual compounding at an annual interest rate of 5.5%.

Annuities

Annuities Worked Example 5.16

•To provide for future education, a family considers various

methods of saving. Assume saving will continue for a period

of 10 years at an interest rate of 7.5% per annum.

(a) Calculate the value of the fund at the end of year 10

when equal deposit of $2,000 is made at the end of each

year.

(b) How much should be deposited each year if the final

value of the fund is $40,000?

(c) How much should be deposited each month if the final

value of the fund is $40,000?

Annuities

Debt Repayment

• We say a loan is amortized if both principal and interest are to

be paid back by a series of equal payments made at equal

intervals of time assuming a fixed rate of interest throughout.

• Mortgage is an amortized loan used to purchase a real estate

(house or building) by offering the real estate to be purchased as

a collateral.

• Mortgage repayment is one type of debt repayment(or loan

repayment).

Debt Repayment

Sinking Funds

Assignment #4

• Problems 3 of Progress Exercises 5.5

• Problems 2, 3, 5 of Progress Exercises

5.6

Due on 2013/05/09 (Thursday)

Business Mathematicscontents.kocw.net/KOCW/document/2013/gacheon/LEEYongju/9.pdfFinancial...

Documents

Transcript of Business Mathematicscontents.kocw.net/KOCW/document/2013/gacheon/LEEYongju/9.pdfFinancial...