APPLICATIONS TO FINANCE

CHAPTER 3 SUPPLEMENTARY EXERCISES

4.6 APPLICATIONS TO FINANCE

APPLICATIONS TO FINANCE 129

Annual Compound Interest

A= (1+r)^tP₀=P_t t=1,2,… A=P_t=the total amount accumulated P₀= principal t= time

r= interest rate(expressed as a decimal)

Example 4.6.1 Compound Interest

a) What compound interest accrues on $5000 at 6% annually for 3 years?

b) On $10,000 at 5% annually for 48 months?

Solution:

a) After 3 years, the total is 5000(1.06)³=$5955.08. Subtract the original principal,

$5000, to yield interest of $955.08. Notice that this exceeds the $900 simple interest that would have been earned as compound interest always yields more than simple interest.

b) After 4 years, the total amount is 10,000(1.05)⁴ =$12,155.06 and the interest is

$2155.06.

The example above assumes that interest is compounded annually. However, sometimes, interest is compounded quarterly or monthly. LetP₀be the original principal; r, the annual interest rate (as a decimal); n, the number of periods per year that interest is compounded, the time,t, in years. Then, the amount accumulated isP_t(=A)isP₀(1+r∕n)^nt.

Whenn=1, the above-mentioned formula matches the annual compound interest for- mula. Note that if there arenperiods per year, then the relevant interest per period isr∕n and the number of compoundings intyears isnt. With these alterations the expression for P_tfollows.

Compound Interest Formula

The principalA=P_t(the sum ofP_𝟎and earned interest) aftertyears when interest is compoundedntimes a year is

A=P_t=P₀(1+r∕n)^nt t=0,1,… r=interest rate (expressed as a decimal)

APPLICATIONS TO FINANCE 131

Example 4.6.2 Effects of Compounding Periods

a) What interest accrues on $5000 at 6% interest for 3 years when compounded quarterly?

b) When compounded monthly?

Solution:

a) Here, n=4 and

P_t=5000 (1+0.06∕4)^4×3 =5000(1.015)¹²=$5978.09.

Therefore, the accrued interest is $978.09.

b) Now, n=12 so

P_t=5000 (1+0.06∕12)^12×3=5000(1.005)³⁶=$5983.40.

Therefore, the accrued interest is $983.40.

Notice that both amounts exceed the interest accrued in the previous examples when interest was compounded annually. In general, the more frequent the compounding, the greater the accrued interest.

The previous example illustrates the effect of increased compounding frequency. The amount realized increases as the frequency of compounding increases. That is, as the period between interest payments decreases. Clearly, this is due to the additional sums earned by interest on interest already accrued. That is, by compounding interest.

The “Rule of 72” is well known and widely used in financial circles. The rule permits one to approximately calculate either the time or the interest rate for a sum to double. At a 6% rate, for example, about 72∕6=12 years is required. The annual interest rate required to double a sum in 3 years is about 72∕3=24%, and so on.

The previous example raises the question of instantaneous or continuous com- pounding. It is easy to jump to the guess that continuous compounding might result in huge sums. Alas, that is not the case! The following formula applies for continuous compounding.

Continuous Compounding Formula

The principalA=P_t(P₀plus interest earned) at timet(years) when compounding continuously is

P_t=P_oe^rt

r= interest rate (expressed as a decimal) e≈2.71828 (see Chapter 1)

↑Students familiar with the curious limit lim

n→∞

( 1+ x

n )n

=e^xwill recognize the compound intereste^rtas the limit ofP_tasn→∞.

Example 4.6.3 Compounding Continuously

What is the interest on $5000 at 6% for 3 years compounded continuously?

Solution:

Using the continuous compounding formula, we have

P_t =5000 e^0.06×3=5000 e^0.18=$5986.09.

As noted earlier, compounding continuously doesn’t yield an infinite sum. However, it does realize a larger return than any other compounding period.

Most variables in economic theory vary continuously with time. For small time changes, the derivative is an estimate of the variables’ time rate of change. The percentage change in the GDP between reporting periods is a simple notion of a growth rate.

↓ A logarithmic scale often provides a clearer view of long-term growth. The derivative of a logarithmic function with respect to time is the growth rate.

Recall that the derivative of a natural logarithmic function,f(t), isf^′(t)∕f(t). This is the relative rate of changeoff(t) per unit change oft. Thepercentage rate of changeis the relative rate of change multiplied by 100%.

Percentage Rate of Change

The percentage rate of change is f^′(t)

f(t) ×100%

Example 4.6.4 Percentage Rate of Change

A country’s GDP is f(t) =3.6+0.03t+0.02t². What is the percentage rate of change of GDP at t=2 years?

APPLICATIONS TO FINANCE 133 Solution:

f^′(t)

f(t) ×100% = 0.03+0.04t

3.6+0.03t+0.02t² ×100%

At t=2,

0.03+0.04(2)

3.6+0.03(2) +0.02(2)² ×100% = 0.11

3.74×100% =2.94%

Economists express demand,q, as a function of price,p, by thedemand function q=f(p)in whichq decreases aspincreases. The ratio of the relative rate of change of demand to the relative rate of change of price is theelasticity of demand.

Elasticity of Demand

The elasticity of demand at a pricep, E(p),is E(p) = −pf^′(p)

f(p) ×100% WhenE(p)<1, demand is inelastic

WhenE(p)>1, demand is elastic WhenE(p) =1, there is unitary elasticity

When demand iselasticat pricep,the change in demand is opposite to the change in price (hence the negative sign in the definition). When demand isinelastic,the change in demand is in the same direction as the change in price.

Example 4.6.5 Elasticity of Demand

Show that the demand function f(p) =50 p⁻²is elastic with respect to price.

Solution:

E(p) = −p(−100 p⁻³)

50 p⁻² = 100 p⁻²

50 p⁻² =2. As E(p)> 1, demand is elastic. As E(p) does not depend on the value of p, demand remains elastic regardless of the price.

Another application of exponential and logarithmic functions deals with sales decay.

Some marketing studies have demonstrated that if a product ceases to be promoted that sales for the product will decrease in proportion to current sales. The following example illustrates the concept.

Example 4.6.6 Sales Decay

A corporation finds that daily sales, S(t), are falling as S(t) =50,000(3^−0.10t)items, where t is the number of days since January 1^st.

a) What is a sales estimate for January 11?

b) When are sales expected to fall below 5000 items?

Solution:

a) January 11 implies that t=10. So

50,000(3−(0.10)(10)) =50,000(3⁻¹) ≈16,667 items.

b)

50,000(3^−(0.10t)) =5000 3^−0.10t= (1∕10)

−0.10t ln3=ln(0.10) t≈21 days On January 22, sales are expected to be below 5000.

⧫ In one marketing study, the sales S(t) at time t for an unpromoted product is S(t) =S(0)e^−𝜆t, where 𝜆is referred to as the exponential sales decay constant. There are variations of this model.

M. Vidale and B. Wolfe, “An Operations-Research Study of Sales Response to Adver- tising”, Operations Research, Volume 5, Issue 3, June 1957, page 371.

EXERCISES 4.6

1. If $50,000 is compounded quarterly at 8% for 4 years, what is the sum at maturity?

2. If $250,000 is compounded monthly at 6% for 3 years, what is the sum at maturity?

3. If $500,000 is compounded continuously at 7% for 6 years, what is the sum at maturity?

4. Which is greater: $75,000 at 10% compounded quarterly for 4 years or $60,000 com- pounded continuously at 4% for 8 years?

5. Which is greater: $250,000 at 3% compounded annually for 5 years or $225,000 com- pounded continuously at 5% for 4 years?

HISTORICAL NOTES 135 6. For $20,000 to double in 7 years with continuous compounding, what interest rate is

required?

7. When will $25,000 compounded continuously at 4% increase to $100,000?

8. When will $50,000 compounded continuously at 5% triple?

9. A sales decay model isS(t) =S₀e^−𝛼t where𝛼=0.2. When will sales drop to 30% of initial value? (tis in weeks)

In Exercises 10–15, determine the percentage rate of change of the functions at the indicated point .

10. f(t) =2t²att=2

11. f(t) =3t²+2t+1 att=3 12. f(t) =3e^0.2tatt=6

13. f(t) =e^0.4tatt=5 14. f(t) =2∕(3t+8)att=4 15. f(t) =1∕(t+3)att=2 16. f(p) =600−4pelastic atp=75?

17. In what interval is demandf(p) =500e^−0.25pelastic?

18. Consider a demand function f(p) =450p−p²,0<p<450. Verify that demand is elastic for 0<p<300 and inelastic for 300<p<450.

HISTORICAL NOTES

Leonhard Euler (1707–1783)– Perhaps the most prolific of mathematicians, he authored 866 books and papers and won the Paris Academy Prize 12 times. Born in Switzerland, he began as a theology student and changed to mathematics under the influence of Johann Bernoulli. Euler made significant contributions in differential calculus, mathematical anal- ysis, and number theory. He introduced the symbols e,i,f(x),𝜋, and the sigma summation sign. Remarkably, most of Euler’s publications appeared in the last 20 years of his life when he was totally blind.

John Napier (1550–1617)– Napier was born in Edinburgh, Scotland and little is known of his early years. He entered St. Andrews University at 13 years. Napier’s study of mathe- matics was only a hobby. He is best known for inventing logarithms and introducing decimal notation. He also invented “Napier’s Bones,” which are a mechanical means to multiply, divide, and take square roots and cube roots. (Also, Section 4.2 diamond box.)

CHAPTER 4 SUPPLEMENTARY EXERCISES