Introductory maths analysis chapter 05 official

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 5Chapter 5
Mathematics of FinanceMathematics of Finance

INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

• To solve interest problems which require
logarithms.
• To solve problems involving the time value of
money.
• To solve problems with interest is compounded
continuously.
• To introduce the notions of ordinary annuities
and annuities due.
• To learn how to amortize a loan and set up an
amortization schedule.
Chapter 5: Mathematics of Finance
Chapter ObjectivesChapter Objectives

Compound Interest
Present Value
Interest Compounded Continuously
Annuities
Amortization of Loans
5.1)
5.2)
5.3)
5.4)
Chapter OutlineChapter Outline
5.5)

5.1 Compound Interest5.1 Compound Interest
Example 1 – Compound Interest
• Compound amount S at the end of n interest
periods at the periodic rate of r is as
( )n
rPS += 1
Suppose that $500 amounted to $588.38 in a
savings account after three years. If interest was
compounded semiannually, find the nominal rate of
interest, compounded semiannually, that was
earned by the money.

Solution:
There are 2 × 3 = 6 interest periods.
The semiannual rate was 2.75%, so the nominal
rate was 5.5 % compounded semiannually.
5.1 Compound Interest
( )
( )
0275.01
500
38.588
500
38.588
1
500
38.588
1
38.5881500
6
6
6
6
≈−=
=+
=+
=+
r
r
r
r

How long will it take for $600 to amount to $900 at an
annual rate of 6% compounded quarterly?
Solution:
The periodic rate is r = 0.06/4 = 0.015.
It will take .
( )
( )
( )
233.27
015.1ln
5.1ln
5.1ln015.1ln
5.1ln015.1ln
5.1015.1
015.1600900
≈=
=
=
=
=
n
n
n
n
n
months9years,68083.6 2
1
4
233.27
=≈

Example 5 – Effective Rate
Effective Rate
• The effective rate re for a year is given by
11 −





+=
n
e
n
r
r
To what amount will $12,000 accumulate in 15 years
if it is invested at an effective rate of 5%?
Solution: ( ) 14.947,24$05.1000,12
15
≈=S

Example 7 – Comparing Interest Rates
If an investor has a choice of investing money at 6%
compounded daily or % compounded quarterly,
which is the better choice?
Solution:
Respective effective rates of interest are
The 2nd
choice gives a higher effective rate.
%27.61
4
06125.0
1
and%18.61
365
06.0
1
4
365
≈−





+=
≈−





+=
e
e
r
r
8
1
6

5.2 Present Value5.2 Present Value
Example 1 – Present Value
• P that must be invested at r for n interest
periods so that the present value, S is given by
Find the present value of $1000 due after three
years if the interest rate is 9% compounded
monthly.
Solution:
For interest rate, .
Principle value is .
( ) n
rSP
−
+= 1
( ) 15.764$0075.11000
36
≈=
−
P
0075.012/09.0 ==r

5.2 Present Value
Example 3 – Equation of Value
A debt of $3000 due six years from now is instead
to be paid off by three payments:
• $500 now,
• $1500 in three years, and
• a final payment at the end of five years.
What would this payment be if an interest rate of
6% compounded annually is assumed?

5.2 Present Value
Solution:
The equation of value is
( ) ( )
27.1257$
02.160002.11000
208
≈
+=
−−
x

5.2 Present Value
Example 5 – Net Present Value
You can invest $20,000 in a business that guarantees
you cash flows at the end of years 2, 3, and 5 as
indicated in the table.
Assume an interest rate of 7% compounded annually
and find the net present value of the cash flows.
Net Present Value
( ) investmentInitial-valuespresentofSumNPVValuePresentNet =
Year Cash Flow
2 $10,000
3 8000
5 6000

5.2 Present Value
Example 5 – Net Present Value
Solution:
( ) ( ) ( )
31.457$
000,2007.1600007.1800007.1000,10NPV
532
−≈
−++=
−−−

5.3 Interest Compounded Continuously5.3 Interest Compounded Continuously
Example 1 – Compound Amount
Compound Amount under Continuous Interest
• The compound amount S is defined as
If $100 is invested at an annual rate of 5%
compounded continuously, find the compound
amount at the end of
a. 1 year.
b. 5 years.
kt
k
r
PS 





+= 1
( )( )
13.105$100 105.0
≈== ePeS rt
( )( )
40.128$100100 25.0505.0
≈== eeS

5.3 Interest Compounded Continuously
Effective Rate under Continuous Interest
• Effective rate with annual r compounded
continuously is .1−= r
e er
Present Value under Continuous Interest
• Present value P at the end of t years at an
annual r compounded continuously is .rt
SeP −
=

5.3 Interest Compounded Continuously
Example 3 – Trust Fund
A trust fund is being set up by a single payment so
that at the end of 20 years there will be $25,000 in
the fund. If interest compounded continuously at an
annual rate of 7%, how much money should be paid
into the fund initially?
Solution:
We want the present value of $25,000 due in 20
years.
( )( )
6165$000,25
000,25
4.1
2007.0
≈=
==
−
−−
e
eSeP rt

5.4 Annuities5.4 Annuities
Example 1 – Geometric Sequences
Sequences and Geometric Series
• A geometric sequence with first term a and
common ratio r is defined as
a. The geometric sequence with a = 3, common
ratio 1/2 , and n = 5 is
0where,...,,,, 132
≠−
aarararara n
432
2
1
3,
2
1
3,
2
1
3,
2
1
3,3 
























5.4 Annuities
Example 1 – Geometric Sequences
b. Geometric sequence with a = 1, r = 0.1, and
n = 4.
c. Geometric sequence with a = Pe−kI
, r = e−kI
,
n = d.
001.0,01.0,1.0,1
dkIkIkI
PePePe −−−
,...,, 2
Sum of Geometric Series
• The sum of a geometric series of n terms, with
first term a, is given by
( ) 1rfor
1
11
0
≠
−
−
== ∑
−
= r
ra
ars
nn
i
i

5.4 Annuities
Example 3 – Sum of Geometric Series
Find the sum of the geometric series:
Solution: For a = 1, r = 1/2, and n = 7
62
2
1
...
2
1
2
1
1 





++





++
( )
64
127
2
1
1
2
1
11
1
1
2
1
128
127
7
==
−














−
=
−
−
=
r
ra
s
n
Present Value of an Annuity
• The present value of an annuity (A) is the sum
of the present values of all the payments.
( ) ( ) ( ) n
rRrRrRA
−−−
++++++= 1...11
21

5.4 Annuities
Example 5 – Present Value of Annuity
Find the present value of an annuity of $100 per
month for years at an interest rate of 6%
compounded monthly.
Solution:
For R = 100, r = 0.06/12 = 0.005, n = ( )(12) = 42
From Appendix B, .
Hence,
005.042
__100aA =
798300.37
005.042
__ =a
( ) 83.3779$798300.37100 =≈A
2
1
3
2
1
3

5.4 Annuities
Example 7 – Periodic Payment of Annuity
If $10,000 is used to purchase an annuity consisting
of equal payments at the end of each year for the
next four years and the interest rate is 6%
compounded annually, find the amount of each
payment.
Solution:
For A= $10,000, n = 4, r = 0.06,
91.2885$
465106.3
000,10000,10
000,10
06.04
06.04
____
__
=≈==
=
aa
A
R
Ra
rn

5.4 Annuities
Example 9 – Amount of Annuity
Amount of an Annuity
• The amount S of ordinary annuity of R for n
periods at r per period is
Find S consisting of payments of $50 at the end of
every 3 months for 3 years at 6% compounded
quarterly. Also, find the compound interest.
Solution: For R=50, n=4(3)=12, r=0.06/4=0.015,
( )
r
r
RS
n
11 −+
⋅=
( ) 06.652$041211.135050
015.012
__ ≈≈=S
( ) 06.52$501206.652InterestCompund =−=

5.4 Annuities
Example 11 – Sinking Fund
A sinking fund is a fund into which periodic
payments are made in order to satisfy a future
obligation. A machine costing $7000 is replaced at
the end of 8 years, at which time it will have a salvage
value of $700. A sinking fund is set up. The amount in
the fund at the end of 8 years is to be the difference
between the replacement cost and the salvage value.
If equal payments are placed in the fund at the end of
each quarter and the fund earns 8% compounded
quarterly, what should each payment be?

5.4 Annuities
Example 11 – Sinking Fund
Solution:
Amount needed after 8 years = 7000 − 700 = $6300.
For n = 4(8) = 32, r = 0.08/4 = 0.02, and S = 6300,
the periodic payment R of an annuity is
45.142$
6300
6300
02.032
02.032
________
____
≈==
=
ss
S
R
Rs
rn

5.5 Amortization of Loans5.5 Amortization of Loans
Amortization Formulas

5.5 Amortization of Loans
Example 1 – Amortizing a Loan
A person amortizes a loan of $170,000 by obtaining
a 20-year mortgage at 7.5% compounded monthly.
Find
a.monthly payment,
b.total interest charges, and
c.principal remaining after five years.

5.5 Amortization of Loans
Example 1 – Amortizing a Loan
Solution:
a. Monthly payment:
b. Total interest charge:
c. Principal value:
( )
51.1369$
00625.01
00625.0
000,170 240
≈







−
= −
R
( ) 40.682,158$000,17051.1369240 =−
( ) 74.733,147$
00625.0
00625.01
51.1369
180
≈






 −
−

Introductory maths analysis chapter 05 official

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