LOGO
MATH 2040
Introduction to
Mathematical Finance
Instructor: Dr. Ken Tsang
1
Chapter 2: Solution of problems in interest
Introduction: the basic problem
Equations of value
Unknown time & interest rate
Determining time periods
Examples
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Introduction
• How to solve an interest problem?
–Use basic principle
– Develop a systematic approach
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Obtaining numerical results
• Use a calculator with exponential functions.
• Use the table of compound interest functions.
• Series expansion could also be used.
• Use software like MatLab or R
4
A common problem
• A common problem is to determine a(t), where t
is not an integer.
• Use compound interest for integral periods of time
and simple interest for fractional periods.
• This method is equivalent to the method of using
the first two terms of the binomial expansion
assuming that 0 < k < 1.
5
Linear interpolation
• The use of simple interest for a fractional period is
equivalent to performing a linear interpolation
between (1 + i)n and (1 + i)n+1, where n is the
integral part of the period.
• (1 + i) n+k  (1  k)(1 + i)n + k (1 + i) n+1
= (1 + i)n(1 + k i)
• (1 – d) n+k  (1  k)(1  d )n + k (1 – d) n+1
= (1 – d)n(1 – k d)
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Example
• Find the accumulated value of $5000 at the end of
30 years and 4 months at 6% p.a. converted
semiannually: (1) assuming compound interest
throughout, and (2) assuming simple interest during
the final fractional period.
• (1) By direct calculation,
5000(1.03)60.667 = 30044.27
• (2) By assuming simple interest in the final
fractional period, we have
5000(1.03)60(1.02) = 30047.18
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Exact simple interest
• There are three commonly used methods to count
the number of days for the period of investment.
• The first one is called the exact simple interest
method.
• In this method, the exact number of days are
counted, and a year contains 365 days.
• This method is often denoted by “actual/actual”.
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Ordinary simple interest
• The second one is called the ordinary simple
interest method.
• In this method, it is assumed that there are 30
days in each month and 360 days in a year.
• This method is often denoted by “30/360”.
• For simplicity, a formula for calculating the days in
the “30/360” method is:
360(Y2 – Y1) + 30(M2 – M1) + (D2 – D1)
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The banker’s rule
• The third one is called the banker’s rule.
• This method is a hybrid of the previous two.
• It used the exact number of days in each month
and 360 days in a year.
• This method is often denoted by “actual/360”.
• Bankers use this rule to charge interest for loans to
their clients, and the rule is more favorable to the
bankers.
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Remarks on Determining time periods
•
Leap years lead to complications.
– February 29 counted as one day and 366 days for the
year.
– February 29 counted as one day and 365 days for the
year.
– February 29 is not counted and 365 days for the year.
•
•
In counting the number of days, either the first
day or the final day is counted, but not both.
The above methods are also used for compound
interest calculation.
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Example
•
Find the amount of interest that $2000 deposited From June
17 to September 10 of the same year will earn, if the rate of
interest is 8%, on the following:
–
–
–
•
Exact simple interest (actual/actual)
Ordinary simple interest (30/360)
Banker’s rule (actual/360)
Answers for the three methods are:
–
–
–
85 days. Interest = 2000(0.08)(85/365) = 37.26
Number of days = 360(0) + 30(9 – 6) + (10 – 17) = 83.
therefore interest = 2000(0.08)(83/360) = 36.89
85 days. Interest = 2000(0.08)(85/360) = 37.78.
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The basic problem
• A typical problem on interest involves four
basic quantities:
–
–
–
–
The original investment – principal
The length of the investment period – the period
The rate of interest
The accumulated value at the end of the period
• In principle, if three of the four quantities are
known, the fourth can be determined.
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A fundamental principle in the theory of interest
• Value of an amount of money depends on the
time when it is payable.
• This time value of money reflects the effect of
interest.
• Two or more amounts of money payable at
different time cannot be compared directly.
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Equation of value
• The time value of money at any given point in time, t,
will be either a present value or a future value from the
payment time.
• To compare two or more amounts of money payable at
different time, they have to be accumulated or discounted
to a common date, the comparision date.
• The equation which accumulates or discountes each
payment to the comparision date is called the equation
of value.
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Time diagram
• It helps to draw out a time line and plot
the payments and withdrawals accordingly.
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Example
• A $600 payment due in 8 years is equivalent to
receive $100 now, $200 in 5 years and $ X
in 10 years. If i = 8% p.a., find $ X.
• The line diagram is
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Example – solution 1
• Compare the values at t = 0.
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Example – solution 2
• Compare the values at t = 5.
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Example – solution 3
• Compare the values at t = 10.
• All three solutions leads to the same answer, because
they all treat the values of the payments consistently at a
given point of time t.
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Unknown time
• As discussed before, if any of the four basic
quantities are known, then the fourth can be
determined.
• Consider the situation where the length of
investment is not known.
• The easiest approach is to use logarithm.
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Example - logarithm
• How long does it take money to double if
the interest rate i = 6%?.
Very simple evaluation using MATLAB
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Example – interest table + interpolation
• If logarithms are not available, then use
interest tables and perform a linear
interpolation.
• Using interest tables, we have
and
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Doubling a payment – rule of 72
• For doubling a single payment we have the rule
of 72
• This rule gives a good approximation for the
period to double the principal over a wide range
of interest rates.
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Comparing rule of 72 with exact values
Rate of interest Rule of 72 Exact value Error
4%
18
17.67
+2%
6
12
11.90
+1%
8
9
9.01 0.1%
10
7.2
7.27
1%
12
6
6.12
2%
18
4
4.19
5%
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Tripling a payment – rule of 114
• For tripling a single payment, we have the
approximate rule of 114.
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Multiple payments
• Let St represent a sequence of payments made at time
t for t = 0, 1, …, n.
• How to replace the multiple payments with a single
payment S, equal to the sum of all St, such that the
present value of S at time t is equal to the present
value of the multiple payments?
• Determine the value of t.
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Multiple payments - solution
• To find the true value of t, we form the
equation:
• Taking logarithm on both sides, we have
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Multiple payments – approximate solution
• Suppose we use t , the weighted average of time for the
multiple payments, to approximate the value of t.
• This method is called the method of equated time.
• If we can prove that t  t , then the present value using
the method of equated time will be less than the present
value using exact t.
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Weighted average of time > exact time 1
• The present value of the payment at time tk is
.
• So the arithmetic weighted mean of present values is
• And the geometric weighted mean of present values is
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Weighted average of time > exact time 2
• Since the geometric means are less than
arithmetic mean
•  Method of equated time > exact t
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Example
• Find the length of time necessary for $1000 to
accumulate to $1500 if invested at 6% per annum
compounded semiannually: (1) by use of logarithm,
and (2) by interpolating in the interest table.
• Let n be the number of half-years required. The
equation of time is
1000(1.03)n = 1500.
(1.03)n = 1.5
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Example – cont’d
• Using logarithm, we have
n loge 1.03 = loge 1.5.
• Because loge 1.03 = 0.405465 and loge 1.5 =
0.029559, we can determine n to be 13.717.
• So the number of years is 6.859
Using MATLAB
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Example – cont’d
• From the interest table, we have
(1.03)13 = 1.46853 and
(1.03)14 = 1.51259
• Performing a linear interpolation,
1.50000  1.46853
n  13 
 13.714
1.51259  1.46853
• So the number of years is 6.857. Note that
the two answers are very close.
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Another example
• Payments of $100, $200 and $500 are due at the ends
of years 2, 3 and 8 respectively. Assuming an
effective rate of interest of 5% per annum, find the
point in time at which a payment of $800 would be
equivalent: (1) by an exact method, and (2) by the
method of equated time.
• The exact time equation is:
800 vt = 100 v2 + 200 v3 + 500 v8,
which can be solved for t = 5.832.
• By method of equated time, we have
100  2  200  3  500  8
t
 6.
100  200  500
• As expected, the true value t is less.
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Unknown rate of interest – single payment
• It is quite common to have a financial transaction
where the rate of return needs to be determined.
• For example, suppose $1000 investment triples in
10 years at nominal rate of interest convertible
quarterly. Find i(4).
Using MATLAB
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Unknown rate of interest – multiple payments
• Interest can also be determined if there are only a small
number of payments and the equation of value can be
reduced to a polynomial that is not too difficult to solve.
• For example, at what effective interest rate will the present
value of $130 at the end of 5 years and $300 at the end of
10 years be equal to $360?
• The equation of value is
130 v5 + 300 v10 = 360.
• If we put y = v5, the equation becomes a quadratic equation
in y, which can be solved, and consequently i can be
determined, giving y = 0.9 and i = (0.9)0.2  1.
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Linear interpolation
• Unfortunately, most of the times, a quadratic
equation or low degree polynomial equation is not
available.
• In this case we can use linear interpolation.
• For example, at what effective interest rate will an
investment of $100 immediately and $500 at the
end of the 3rd year from now accumulate to
$1000 at the end of the 10th year from now?
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Linear interpolation – cont’d
• The equation to solve is 100(1 + i)10 + 500(1 + i)7 = 1000.
• Put f (i) = (1 + i)10 + 5(1 + i)7. The problem is solved if we
can determine what value of i will lead to f (i) = 10.
• By trial and error, we find that f (9%) = 9.68 and that
f
(10%) = 10.39. So by linear interpolation, we have
• The actual answer is 9.46%.
• The linear interpolation can be repeated until the desired
level of accuracy is attained.
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Worked example 1
• At what interest rate convertible quarterly
would $1000 accumulate to $1600 in six
years?
• We need to determine x = i(4)/4. So the
equation of value is 1000(1 + x)24 = 1600.
• Solving the equation, we get x = 0.019776.
• The answer is i(4) = 4 x = 7.91%
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Worked example 2
• At what effective rate of interest will sum of the
present value of $2000 at the end of the 2nd year
from now and $3000 at the end the 4th year from
now be equal to $4000?
• The equation of value is 2000v2 + 3000v4 = 4000.
• That is simplified and re-written as 2v2 + 3v4  4 =
0.
• Solving it as a quadratic equation in v2, we get the
meaningful root 9 v2 = 0.868517, which gives the
answer i = 7.30%.
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Worked example 3
• Suppose an investment of $1000 now plus another
investment of $2000 at the end of 3 years from now will
accumulate to $5000 at the end of 10 years from now.
What is the nominal interest rate convertible semiannually?
• We need to determine j = i(2) /2.
• So the equation of value is:
1000(1 + j)20 + 2000(1 + j)14 = 5000
• We cannot solve this equation easily, so we use linear
interpolation.
• We may also use the graphic capability of MATLAB to help.
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Worked example 4 – cont’d
• Put f (j) = 1000(1 + j)20 + 2000(1 + j)14  5000
• Then we need to find j so that f (j) = 0.
• By trial and error, we get
f (0.030) = 168.71 and f (0.035) = 227.17
• Performing one linear interpolation, we get
168 .71
j  0.0300  0.0050
 0.0321 .
227 .17  168 .71
• This implies that i (2) = 2 (0.321) = 0.0642, or 6.42%.
• A higher level of accuracy can be achieved if the linear
interpolation is repeated until the desired accuracy is
attained.
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A simple MATLAB plot routine
• Use the following command lines in MATLAB
–
–
–
–
x = 1:0.005:1.2
y = x.^20 + 2*x.^14 – 5
Z=0
plot(x,y,x,z)
• and we get the plot
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A simple plot by MATLAB
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A fancier MATLAB plot routine
x=1:0.005:1.2;
y=x.^20 + 2*x.^14-5;
z=0
plot(x,y,x,z);
title('y=(1+j)^2^0+2(1+j)^1^4-5');
xlabel('1+j');
ylabel('y')
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A fancier MATLAB plot
From this plot we can determine j approximately.
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