1.3 Solution of Problems in Interest

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TOPIC 1: INTEREST RATE MEASUREMENT SOLUTION OF PROBLEMS IN INTEREST

After completing this topic, the students should be able to: ď ˝

ď ˝

ď ˝

Describe the fundamental concepts and key terms of the measurement of interest. Apply the key procedure of financial mathematics such as determining equivalent measures of interest, discounting, accumulating, determining yield rates, and estimating the rate of return on fund. Apply those concepts in calculating present and accumulated values for various streams of cash flows.

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EQUATIONS OF VALUE UNKNOWN INTEREST RATE UNKNOWN TIME DETERMINING TIME PERIOD

EQUATIONS OF VALUE •

An equation of value helps to solve problems by providing a relationship between cash inflows and cash outflows, which we can use to solve for an unknown variable.

•

To set up the equation of value, we simply equate two equivalent expressions and simplify them until we can solve for the unknown variable.

•

We use this equation of value in variety contexts.

•

We may solve an equation of value to find an unknown amount of money, a number of years or an interest rate.

•

Time diagram can be used to solve the equation of value.

EQUATIONS OF VALUE EXAMPLE 1: John owes Bill $3000, due in 5 years. Find the equivalent amount due if John wants to pay in three years, assuming the value of money grows at 5% interest compound annually. X 0

1

2

3

3000 4

j = 0.05 đ?‘ƒđ?‘‰ = (1 + đ?‘–)−đ?‘Ą = 3000đ?‘Ł 2

= 3000(1 + 0.05)−2 = $ 2721.09

5

EQUATIONS OF VALUE EXAMPLE 2: In return for a promise to receive $900 at the end of 9 years, Athiqah agrees to pay $200 at the end of 1 years, $300 at the end of 6 years, and to make a further payment at the end of 12 years. Find the payment at the end of 12 years if the nominal rate of interest is 9% convertible semiannually.

200

0

1

300

2

3

4

5

j = 0.045

900

6

7

8

m=2

X=?

đ?‘ƒđ?‘‰ = (1 + đ?‘–)−đ?‘Ą đ?‘‹đ?‘Ł 24 + 300đ?‘Ł 12 + 200đ?‘Ł 2 = 900đ?‘Ł 18

đ?‘‹= =

900đ?‘Ł 18 − 300đ?‘Ł 12 − 200đ?‘Ł 2 đ?‘Ł 24 900(1+0.045)−18 − 300(1+0.045)−12 − 200(1+0.045)−2 (1+0.045)−24

= $136.54

9

X

10

11

12

UNKNOWN INTEREST RATE In this section, the interest rate, that is unknown. Methods for solving i :  Direct Method  Analytical Method  Linear Interpolation  Successive Iterations Method

UNKNOWN INTEREST RATE Direct Method • When a single payment is involved, the method that works best is to solve for i directly in the equation of value (using exponential and logarithmic functions). • In this situation, the equation of value takes the form đ??´(đ?‘Ą) = đ??ž 1 + đ?‘– đ?‘Ą EXAMPLE 1: At what interest rate convertible semi-annually would $500 accumulate to $800 in 4 years?

Let đ?‘— =

đ?‘–2 2

500 1 + đ?‘—

8

= 800 1

đ?‘—=

8 8 5

−1

= 0.06051 Thus, đ?‘– 2 = 2đ?‘— = 0.121 = 12.1%

UNKNOWN INTEREST RATE Analytical Method • When multiple payments are involved, the equation of value takes the form � � = 0. • The problem is the classical problem of determining the nonnegative solutions to the equation � � = 0 with � � being a differentiable function. • If �(�) is a polynomial then the problem reduces to solving polynomial equations. Some known algebraic methods can be used such as the rational zero test, or the quadratic formula, etc. EXAMPLE 2: At what eective interest rate would the present value of $1000 at the end of 3 years plus $2,000 at the end of 6 years be equal to $2,700? �=0 2000� 6 + 1000� 3 = 2700 20� 6 + 10� 3 = 27 the fact that � 3 > 0 we obtain � 3 = 0.9385 from which we obtain v = 0.9791 1

đ?‘–= −1 0.9791 = 0.0213 = 2.13%

UNKNOWN INTEREST RATE Linear Interpolation Suppose that we know the values of a function đ?‘“(đ?‘Ľ) at distinct points đ?‘Ľ1 and at đ?‘Ľ2 , and that đ?‘“ đ?‘Ľ1 ≠đ?‘“ đ?‘Ľ2 . If│đ?‘Ľ1 − đ?‘Ľ2 │is small, it may be reasonable to assume that the graph of đ?‘“ is approximately linear between đ?‘Ľ1 and đ?‘Ľ2 . That is equivalent to assuming that

đ?‘“ đ?‘Ľ = đ?‘“ đ?‘Ľ1 + {

đ?‘“ đ?‘Ľ2 −đ?‘“ đ?‘Ľ1 đ?‘Ľ2−đ?‘Ľ1

}(đ?‘Ľ − đ?‘Ľ1 )

For đ?‘Ľ1 < đ?‘Ľ < đ?‘Ľ2 . If now we wish to determine an approximate value of đ?‘Ľ where đ?‘“(đ?‘Ľ) has a speciďŹ c value đ?‘Ś0 , we may use this equation for that purpose and ďŹ nd đ?‘Ś0 − đ?‘“ đ?‘Ľ1 đ?‘Ľ = đ?‘Ľ1 + đ?‘Ľ2 − đ?‘Ľ1 { } đ?‘“ đ?‘Ľ2 − đ?‘“ đ?‘Ľ1

UNKNOWN INTEREST RATE In particular, if we wish to ďŹ nd a zero of f near the given points (i.e. y0 = 0), and if f(x1) 6= f(x2), a good approximation could be

đ?‘Ľ ≈ đ?‘Ľ1 − đ?‘“ đ?‘Ľ1

đ?‘Ľ2 − đ?‘Ľ1 đ?‘“ đ?‘Ľ2 − đ?‘“(đ?‘Ľ1 )

When we apply these formulas for points between two points x1 and x2, we speak of linear interpolation.

UNKNOWN INTEREST RATE EXAMPLE 3: At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? The equation of value at time t = 10 years is 1000(1 + đ?‘—)20 + 2000(1 + đ?‘—)14 = 5000 đ?‘– (đ?‘Ľ)

Where đ?‘— = . We will use linear interpolation to solve for j. For that purpose, we deďŹ ne 2 đ?‘“ đ?‘— = 1000(1 + đ?‘—)20 + 2000(1 + đ?‘—)14 −5000 We want to ďŹ nd đ?‘— such that đ?‘“(đ?‘—) = 0. By trial and error we see that đ?‘“(0.03) = −168.71 and đ?‘“(0.035) = 227.17. Since đ?‘“ is continuous, đ?‘— is between these two values. Using linear interpolation we find đ?‘— ≈ 0.03 + 168.71 Ă— đ?‘“(0.0321) = −6.11.

Hence, đ?‘– 2 = 2đ?‘— = 0.0642 = 6.42%

0.005 227.17 + 168.71

≈ 0.0321

UNKNOWN INTEREST RATE Successive Iterations Methods Iteration methods can be used to achieve a higher level of accuracy than the one provided with linear interpolation. We will discuss two iteration methods: the bisection method and the Newton-Raphson method. The bisection method is based on the fact that a dierentiable function đ?‘“ that satisďŹ es đ?‘“(đ?›ź)đ?‘“(đ?›˝) < 0 must satisfy the equation đ?‘“(đ?‘Ľ0 ) = for some đ?‘Ľ0 between Îą and β. The ďŹ rst step in the methods consists of ďŹ nding two starting values đ?‘Ľ0 < đ?‘Ľ1 such that đ?‘“(đ?‘Ľ0 )đ?‘“(đ?‘Ľ1 ) < 0. Usually, these values are found by trial and error. đ?‘Ľ +đ?‘Ľ

Next, bisect the interval by means of the mid- point �2 = 0 1 . If �(�0 )�(�2 ) < 2 0 then apply the bisection process to the interval �0 ≤ � ≤ �2 .If �(�1 )�(�2 ) < 0 then apply the bisection process to the interval �2 ≤ � ≤ �1 . We continue the bisection process as many times as necessary to achieve the desired level of accuracy. We illustrate this method in the next example.

UNKNOWN INTEREST RATE EXAMPLE 4: At what interest rate convertible semiannually would an investment of $1,000 immediately and $2,000 three years from now accumulate to $5,000 ten years from now? Use the bisection method. đ?‘Ą = 10 1000 1 + đ?‘—

where đ?‘— =

20

+ 2000 1 + đ?‘—

14

= 5000

đ?‘–2

2

đ?‘“ đ?‘— = 1000 1 + đ?‘—

20

Thus, đ?‘— = 0.032178 Hence, đ?‘– 2 = 2đ?‘— = 0.06436 = 6.43%

+ 2000 1 + đ?‘—

14

− 5000

UNKNOWN TIME • • •

Logarithms can be used as the best method to solve unknown time involving a single payment. A situation in which several payments made at various point in time need to be replaced by one payment numerically equal to the sum of the other payments. The fundamental equation of value: t t t t – Exact method: s1  s 2  ...  s n v  s1v 1  s 2 v 2  ...  s n v n n

– Method of equated time: t  s1t1  s 2 t 2  ...  s n t n  s1  s 2  ...  s n

s t k 1 n

s k 1

•

k k

k

The rule of 72: – n can be approximated immediately by dividing 72 by the rate of interest expressed as a percentage (i.e. as 100i). .6931 1.0395 i .72  i

n

UNKNOWN TIME EXAMPLE 1: A single payment of $3,938.31 will pay o a debt whose original repayment plan was $1,000 due on January 1 of each of the next four years, beginning January 1, 2006. If the eective annual rate is 8%, on what date must the payment of $3,938.31 be paid? Equation of value 1000 1 + 1 + 0.08

−1

+ 1 + 0.08

−2

+ 1 + 0.08

−3

= 3938.31 1 + 0.08

−đ?‘Ą

Solving this equation for đ?‘Ą we find −1 −2 −3 đ?‘™đ?‘› 1000 3938.31 −1 1 + 1 + 0.08 + 1 + 0.08 + 1 + 0.08

��1.08

≈ 1.25. đ?‘Ą

Therefore, the single payment of 3938.31 must be paid 1.25 years after January 1, 2006, or on March 31, 2007.

UNKNOWN TIME

EXAMPLE 2: Find the length of time necessary for $1000 to accumulate to RM1500 if invested at 6% per year compounded semiannually by a direct method, i.e. using a calculator.

Let đ?‘Ą be the comparison date. Then we must have 1000 1.03

2đ?‘Ą

= 1000 or

1 2

đ?‘Ą = ( ){

��1.5 ��1.03

= 6.859 đ?‘Śđ?‘’đ?‘Žđ?‘&#x;đ?‘

1.03

}

2đ?‘Ą

= 1.5

DETERMINING TIME PERIOD •

Exact simple interest

đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™

– Exact number of days – refer Appendix A (page 587) for numbering the days of the years

•

Ordinary simple interest

30 360

– Every month has 30 days and every year has 360 days

360 (đ?‘Œ2 – đ?‘Œ1) + 30 (đ?‘€2 – đ?‘€1) + (đ??ˇ2 – đ??ˇ1) •

Banker’s Rule

đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ 360

– Hybrid between both method

DETERMINING TIME PERIOD

EXAMPLE 1 : Find the amount of interest that $4000 deposited on May 15 will earn, if the money is withdrawn on August 30 in the same year and if the rate of simple interest is 10%, using exact simple interest.

Refer Appendix A, May 15 is day 135 and August 30 is day 242.

242 − 135 4000 × 0.1 = $117.26 365

DETERMINING TIME PERIOD

EXAMPLE 2: Find the amount of interest that $4000 deposited on May 15 will earn, if the money is withdrawn on August 30 in the same year and if the rate of simple interest is 10%, using ordinary simple interest

360 0 + 30 8 − 5 + 30 − 15 = 105 4000 0.1

105 = $116.67 360

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