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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
Arithmetic Sequences
and Series
Geometric 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
Application of Arithmetic and Geometric Series
Application of Arithmetic and Geometric Series
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
Simple Interest
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)
tt iPP )1(0 ..compounding once annually
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
necessary in t years
Three compoundings necessary in 1 year
32
How to compound m times annually
(rate = i% pa)
tt iPP )1(0 ..compounding once annually
mt
tm
iPP
10
..compounding m times annually
At each compoumding
use the annual rate,i, divided by m m x t compoundings
necessary in t years
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
• 6351.185 12 conversion periods
• 6356.079 365 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
Calculate the APR when interest is
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
Calculate the APR when interest is
compounded continuously
But Pt is the same whcihever method is used, hence
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)
But Pt is the same whcihever method is used, hence
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
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
Compound Interest for Fixed
Deposits at Regular Intervals of Time
Compound Interest for Fixed
Deposits at Regular Intervals of Time
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%.
Compound Interest for Fixed
Deposits at Regular Intervals of Time
Annuities
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
Annuities
Annuities
Annuities
Annuities
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
Debt Repayment
Debt Repayment
Sinking Funds
Sinking Funds
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)