4Compound InterestAnnuitiesAmortization and Sinking FundsArithmetic and Geometric ProgressionsMathematics of Finance

4.1Compound Interest

Simple Interest FormulasSimple interest is the interest that is computed on the original principal only.If I denotes the interest on a principal P (in dollars) at an interest rate of r per year for t years, then we haveI = PrtThe accumulated amount A, the sum of the principal and interest after t years is given byA= P + I = P + Prt= P(1 + rt)and is a linear function of t.

ExampleA bank pays simple interest at the rate of 8% per year for certain deposits. If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years?What is the interest earned in that period?SolutionUsing the accumulated amount formula with P = 1000, r = 0.08, and t = 3, we see that the total amount on deposit at the end of 3 years is given by

or $1240.

ExampleA bank pays simple interest at the rate of 8% per year for certain deposits. If a customer deposits $1000 and makes no withdrawals for 3 years, what is the total amount on deposit at the end of three years?What is the interest earned in that period?SolutionThe interest earned over the three year period is given by

or $240.

Applied Example: Trust FundsAn amount of $2000 is invested in a 10-year trust fund that pays 6% annual simple interest.What is the total amount of the trust fund at the end of 10 years?SolutionThe total amount is given by

or $3200.

Compound InterestFrequently, interest earned is periodically added to the principal and thereafter earns interest itself at the same rate. This is called compound interest.Suppose $1000 (the principal) is deposited in a bank for a term of 3 years, earning interest at the rate of 8% per year compounded annually.Using the simple interest formula we see that the accumulated amount after the first year is

or $1080.

Compound InterestTo find the accumulated amount A2 at the end of the second year, we use the simple interest formula again, this time with P = A1, obtaining:

or approximately $1166.40.

Compound InterestWe can use the simple interest formula yet again to find the accumulated amount A3 at the end of the third year:

or approximately $1259.71.

Compound InterestNote that the accumulated amounts at the end of each year have the following form:

These observations suggest the following general rule:If P dollars are invested over a term of t years earning interest at the rate of r per year compounded annually, then the accumulated amount isor:

Compounding More Than Once a YearThe formula

was derived under the assumption that interest was compounded annually.In practice, however, interest is usually compounded more than once a year.The interval of time between successive interest calculations is called the conversion period.

Compounding More Than Once a YearIf interest at a nominal rate of r per year is compounded m times a year on a principal of P dollars, then the simple interest rate per conversion period is

For example, if the nominal interest rate is 8% per year, and interest is compounded quarterly, then

or 2% per period.

Compounding More Than Once a YearTo find a general formula for the accumulated amount, we applyrepeatedly with the interest rate i = r/m.We see that the accumulated amount at the end of each period is as follows:

Compound Interest FormulaWhere n = mt, andA= Accumulated amount at the end of t yearsP= Principalr= Nominal interest rate per yearm= Number of conversion periods per yeart= Term (number of years)There are n = mt periods in t years, so the accumulated amount at the end of t years is given by

ExampleFind the accumulated amount after 3 years if $1000 is invested at 8% per year compoundedAnnuallySemiannuallyQuarterlyMonthlyDaily

ExampleSolutionAnnually. Here, P = 1000, r = 0.08, and m = 1. Thus, i = r = 0.08 and n = 3, so

or $1259.71.

ExampleSolutionSemiannually. Here, P = 1000, r = 0.08, and m = 2. Thus, and n = (3)(2) = 6, so

or $1265.32.

ExampleSolutionQuarterly. Here, P = 1000, r = 0.08, and m = 4.Thus, and n = (3)(4) = 12, so

or $1268.24.

ExampleSolutionMonthly. Here, P = 1000, r = 0.08, and m = 12.Thus, and n = (3)(12) = 36, so

or $1270.24.

ExampleSolutionDaily. Here, P = 1000, r = 0.08, and m = 365.Thus, and n = (3)(365) = 1095, so

or $1271.22.

Continuous Compounding of InterestOne question arises on compound interest: What happens to the accumulated amount over a fixed period of time if the interest is compounded more and more frequently?Weve seen that the more often interest is compounded, the larger the accumulated amount.But does the accumulated amount approach a limit when interest is computed more and more frequently?

Continuous Compounding of InterestRecall that in the compound interest formula

the number of conversion periods is m.So, we should let m get larger and larger (approach infinity) and see what happens to the accumulated amount A.

Continuous Compounding of InterestIf we let u = m/r so that m = ru, then the above formula becomesThe table shows us that when u gets larger and larger the expression

approaches 2.71828 (rounding to five decimal places).It can be shown that as u gets larger and larger, the value of the expression approaches the irrational number 2.71828 which we denote by e.

u

10 2.593741002.7048110002.7169210,0002.71815100,0002.718271,000,0002.71828

Continuous Compounding of InterestContinuous Compound Interest FormulaA = PertwhereP= Principalr= Annual interest rate compounded continuouslyt= Time in yearsA= Accumulated amount at the end of t years

ExamplesFind the accumulated amount after 3 years if $1000 is invested at 8% per year compounded (a) daily, and (b) continuously.SolutionUsing the compound interest formula with P = 1000, r = 0.08, m = 365, and t = 3, we find

Using the continuous compound interest formula with P = 1000, r = 0.08, and t = 3, we find A = Pert = 1000e(0.08)(3) 1271.25Note that the two solutions are very close to each other.

Effective Rate of InterestThe last example demonstrates that the interest actually earned on an investment depends on the frequency with which the interest is compounded.For clarity when comparing interest rates, we can use what is called the effective rate (also called the annual percentage yield): This is the simple interest rate that would produce the same accumulated amount in 1 year as the nominal rate compounded m times a year.We want to derive a relation between the nominal compounded rate and the effective rate.

Effective Rate of InterestThe accumulated amount after 1 year at a simple interest rate R per year is

The accumulated amount after 1 year at a nominal interest rate r per year compounded m times a year is

Equating the two expressions gives

Effective Rate of Interest FormulaSolving the last equation for R we obtain the formula for computing the effective rate of interest:wherereff = Effective rate of interestr= Nominal interest rate per yearm= Number of conversion periods per year

ExampleFind the effective rate of interest corresponding to a nominal rate of 8% per year compoundedAnnuallySemiannuallyQuarterlyMonthlyDaily

ExampleSolutionAnnually. Let r = 0.08 and m = 1. Then

or 8%.

ExampleSolutionSemiannually. Let r = 0.08 and m = 2. Then

or 8.16%.

ExampleSolutionQuarterly. Let r = 0.08 and m = 4. Then

or 8.243%.

ExampleSolutionMonthly. Let r = 0.08 and m = 12. Then

or 8.300%.

ExampleSolutionDaily. Let r = 0.08 and m = 365. Then

or 8.328%.

Effective Rate Over Several YearsIf the effective rate of interest reff is known, then the accumulated amount after t years on an investment of P dollars can be more readily computed by using the formula

Present ValueConsider the compound interest formula:

The principal P is often referred to as the present value, and the accumulated value A is called the future value, since it is realized at a future date.On occasion, investors may wish to determine how much money they should invest now, at a fixed rate of interest, so that they will realize a certain sum at some future date.This problem may be solved by expressing P in terms of A.

Present ValuePresent value formula for compound interest

Where and

ExamplesHow much money should be deposited in a bank paying a yearly interest rate of 6% compounded monthly so that after 3 years the accumulated amount will be $20,000?SolutionHere, A = 20,000, r = 0.06, m = 12, and t = 3.Using the present value formula we get

ExamplesFind the present value of $49,158.60 due in 5 years at an interest rate of 10% per year compounded quarterly.SolutionHere, A = 49,158.60, r = 0.1, m = 4, and t = 5.Using the present value formula we get

Present Value with Continuously Compounded InterestIf we solve the continuous compound interest formulaA = Pertfor P, we getP = Aert This formula gives the present value in terms of the future (accumulated) value for the case of continuous compounding.

Applied Example: Real Estate InvestmentBlakely Investment Company owns an office building located in the commercial district of a city.As a result of the continued success of an urban renewal program, local business is enjoying a mini-boom.The market value of Blakelys property is where V(t) is measured in dollars and t is the time in years from the present.If the expected rate of appreciation is 9% compounded continuously for the next 10 years, find an expression for the present value P(t) of the market price of the property that will be valid for the next 10 years.Compute P(7), P(8), and P(9), and then interpret your results.

Applied Example: Real Estate InvestmentSolution Using the present value formula for continuous compoundingP = Aert with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is

Letting t = 7, we find that

or $599,837.

Applied Example: Real Estate InvestmentSolution Using the present value formula for continuous compoundingP = Aert with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is

Letting t = 8, we find that

or $600,640.

Applied Example: Real Estate InvestmentSolution Using the present value formula for continuous compoundingP = Aert with A = V(t) and r = 0.09, we find that the present value of the market price of the property t years from now is

Letting t = 9, we find that

or $598,115.

Applied Example: Real Estate InvestmentSolution From these results, we see that the present value of the propertys market price seems to decrease after a certain period of growth.This suggests that there is an optimal time for the owners to sell.You can show that the highest present value of the propertys market value is $600,779, and that it occurs at time t 7.72 years, by sketching the graph of the function P.

Example: Using Logarithms in Financial ProblemsHow long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly?SolutionUsing the compound interest formula

with A = 15,000, P = 10,000, r = 0.12, and m = 4, we obtain

Example: Using Logarithms in Financial ProblemsHow long will it take $10,000 to grow to $15,000 if the investment earns an interest rate of 12% per year compounded quarterly?SolutionWeve gotSo, it will take about 3.4 years for the investment to grow from $10,000 to $15,000.Taking logarithms on both sideslogbmn = nlogbm

4.2Annuities

Future Value of an AnnuityThe future value S of an annuity of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period, is

ExampleFind the amount of an ordinary annuity consisting of 12 monthly payments of $100 that earn interest at 12% per year compounded monthly.SolutionSince i is the interest rate per period and since interest is compounded monthly in this case, we have

Using the future value of an annuity formula, with R = 100, n = 12, and i = 0.01, we have

or $1268.25.

Present Value of an AnnuityThe present value P of an annuity consisting of n payments of R dollars each, paid at the end of each investment period into an account that earns interest at the rate of i per period, is

ExampleFind the present value of an ordinary annuity consisting of 24 monthly payments of $100 each and earning interest of 9% per year compounded monthly.SolutionHere, R = 100, i = r/m = 0.09/12 = 0.0075, and n = 24, so

or $2188.91.

Applied Example: Saving for a College EducationAs a savings program towards Albertos college education, his parents decide to deposit $100 at the end of every month into a bank account paying interest at the rate of 6% per year compounded monthly.If the savings program began when Alberto was 6 years old, how much money would have accumulated by the time he turns 18?

Applied Example: Saving for a College EducationSolutionBy the time the child turns 18, the parents would have made

deposits into the account, so n = 144.Furthermore, we have R = 100, r = 0.06, and m = 12, so

Using the future value of an annuity formula, we get

or $21,015.

Applied Example: Financing a CarAfter making a down payment of $4000 for an automobile, Murphy paid $400 per month for 36 months with interest charged at 12% per year compounded monthly on the unpaid balance.What was the original cost of the car?What portion of Murphys total car payments went toward interest charges?

Applied Example: Financing a CarSolutionThe loan taken up by Murphy is given by the present value of the annuity formula

or $12,043.Therefore, the original cost of the automobile is $16,043 ($12,043 plus the $4000 down payment).The interest charges paid by Murphy are given by(36)(400) 12,043 = 2,357or $2,357.

4.3Amortization and Sinking Funds

Amortization FormulaThe periodic payment R on a loan of P dollars to be amortized over n periods with interest charged at the rate of i per period is

Applied Example: Home Mortgage PaymentThe Blakelys borrowed $120,000 from a bank to help finance the purchase of a house.The bank charges interest at a rate of 9% per year on the unpaid balance, with interest computations made at the end of each month.The Blakelys have agreed to repay the loan in equal monthly installments over 30 years.How much should each payment be if the loan is to be amortized at the end of term?

Applied Example: Home Mortgage PaymentSolutionHere, P = 120,000, i = r/m = 0.09/12 = 0.0075, and n = (30)(12) = 360.Using the amortization formula we find that the size of each monthly installment required is given by

or $965.55.

Applied Example: Home AffordabilityThe Jacksons have determined that, after making a down payment, they could afford at most $2000 for a monthly house payment.The bank charges interest at a rate of 7.2% per year on the unpaid balance, with interest computations made at the end of each month.If the loan is to be amortized in equal monthly installments over 30 years, what is the maximum amount that the Jacksons can borrow from the bank?

Applied Example: Home AffordabilitySolutionWe are required to find P, given i = r/m = 0.072/12 = 0.006, n = (30)(12) = 360, and R = 2000.We first solve for P in the amortization formula

Substituting the numerical values for R, n, and i we obtain

Therefore, the Jacksons can borrow at most $294,643.

Sinking Fund PaymentThe periodic payment R required to accumulate a sum of S dollars over n periods with interest charged at the rate of i per period is

Applied Example: Sinking FundThe proprietor of Carson Hardware has decided to set up a sinking fund for the purpose of purchasing a truck in 2 years time.It is expected that the truck will cost $30,000.If the fund earns 10% interest per year compounded quarterly, determine the size of each (equal) quarterly installment the proprietor should pay into the fund.

Applied Example: Sinking FundSolutionHere, S = 30,000, i = r/m = 0.1/4 = 0.025, and n = (2)(4) = 8.Using the sinking fund payment formula we find that the required size of each quarterly payment is given by

or $3434.02.

4.4Arithmetic and Geometric Progressions

Arithmetic ProgressionsAn arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant d to the preceding term.The constant d is called the common difference.An arithmetic progression is completely determined if the first term and the common difference are known.

nth Term of an Arithmetic ProgressionThe nth term of an arithmetic progression with first term a and common difference d is given byan = a + (n 1)d

ExampleFind the twelfth term of the arithmetic progression2, 7, 12, 17, 22, SolutionThe first term of the arithmetic progression is a1 = a = 2, and the common difference is d = 5.So, upon setting n = 12 in the arithmetic progression formula, we findan= a + (n 1)d a12= 2 + (12 1)5 = 57

Sum of Terms in an Arithmetic ProgressionThe sum of the first n terms of an arithmetic progression with first term a and common difference d is given by

ExampleFind the sum of the first 20 terms of the arithmetic progression 2, 7, 12, 17, 22, SolutionLetting a = 2, d = 5, and n = 20 in the sum of terms formula, we get

Applied Example: Company SalesMadison Electric Company had sales of $200,000 in its first year of operation.If the sales increased by $30,000 per year thereafter, find Madisons sales in the fifth year and its total sales over the first 5 years of operation.SolutionMadisons yearly sales follow an arithmetic progression, with a = 200,000 and d = 30,000.The sales in the fifth year are found by using the nth term formula with n = 5. Thus,

or $320,000.an= a + (n 1)d a5= 200,000 + (5 1)30,000 = 320,000

Applied Example: Company SalesMadison Electric Company had sales of $200,000 in its first year of operation.If the sales increased by $30,000 per year thereafter, find Madisons sales in the fifth year and its total sales over the first 5 years of operation.SolutionMadisons total sales over the first 5 years of operation are found by using the sum of terms formula.Thus,

or $1,300,000.

Geometric ProgressionsA geometric progression is a sequence of numbers in which each term after the first is obtained by multiplying the preceding term by a constant.The constant is called the common ratio.A geometric progression is completely determined if the first term a and the common ratio r are given.

nth Term of a Geometric ProgressionThe nth term of a geometric progression with first term a and common ratio r is given by

ExampleFind the eighth term of a geometric progression whose first five terms are 162, 54, 18, 6, and 2.SolutionThe common ratio is found by taking the ratio of any term other than the first to the preceding term.Taking the ratio of the fourth term to the third term, for example, gives

Using the nth term formula to find the eighth term gives

Sum of Terms in a Geometric ProgressionThe sum of the first n terms of a geometric progression with first term a and common ratio r is given by

ExampleFind the sum of the first six terms of the geometric progression 3, 6, 12, 24, SolutionUsing the sum of terms formula

with a = 3, r = 6/3 = 2, and n = 6, gives

Applied Example: Company SalesMichaelson Land Development Company had sales of $1 million in its first year of operation.If sales increased by 10% per year thereafter, find Michaelsons sales in the fifth year and its total sales over the first 5 years of operation.SolutionThe sales follow a geometric progression, with first term a = 1,000,000 and common ratio r = 1.1.The sales in the fifth year are thus

or $1, 464,100.

Applied Example: Company SalesMichaelson Land Development Company had sales of $1 million in its first year of operation.If sales increased by 10% per year thereafter, find Michaelsons sales in the fifth year and its total sales over the first 5 years of operation.SolutionThe sales follow a geometric progression, with first term a = 100,000,000 and common ratio r = 1.1.The total sales over the first 5 years of operation are

or $6,105,100.

End of Chapter