BQT 133-Business Mathematics Teaching Module

2.1Introduction

The objective of this chapter is to examine those procedures which used to answer questions associated with major financial transactions involving the interest problem, annuity and depreciation. These procedures are part of what is usually financial mathematics. Most important financial transactions, such as repaying housing loan, involve a series of repayments. In section 2.2 and 2.3, both simple and compound interest will be discussed. Next, we will define the effective interest rate and present value. To allows customers greater flexibility in the way in which they repay loans, financial institutions use a procedure called continuous compounding to calculate interest payments.

In section 2.4 we have a series of regular payments made at the end of each period and whose compounding and payment periods coincide, this called an ordinary annuity. In this section, the formulas for the future and present value of an ordinary annuity have been discussed.

Section 2.5 examines the three methods commonly used to calculate the depreciation on particular capital assets.

2.2Simple Interest

The study of interest is very important and fundamental to the understanding of the economy of a country. Interest is money earned when money is invested or interest is charge incurred when a loan or credit is obtained.

If you deposit a sum of money P in a savings account or if you borrow a sum of money P from a lending agent such as financial agency or bank, then P is call principal. Usually we have to repay this amount P plus an extra amount. These extra amounts, which pay to the lender for the convenience of using lender money is called interest. Simple interest formula is given by the following formula:

In general, if the principal P is borrowed at a rate r after t years, the borrower will owe the lender an amount A that will include the principal P plus interest I. Since P is the amount that is borrowed now and A is the amount that must be paid back in the future, P is often referred to as the present value and A as the future value. The formula relating A and P is as follows:

Example 2.2.1

RM1000 is invested for two years in a bank, earning a simple interest rate of 8% per annum. Find the simple interest earnedSolution

Exercise 1

RM5000 is invested for 6 years in a bank, earning a simple interest rate of 5.7 % per annum. Find the simple interest earned

AnswerI=RM 1710

Example 2.2.2

RM10000 is invested for 4 years 9 month in a bank earning a simple interest rate of 10% per annum. Find the simple amount at the end of the investment period.

Solution

Exercise 2If Bank A offers a simple interest rate of 8 % per annum, Ahmad invested RM 9000 for 4 years 6 months in a bank earning. Find the future value obtain by Ahmad at the end of the investment period.

Anwer

Example 3.2.3Find the present value at 8% simple interest of a debt amount RM3000 due in ten months.

Solution

Example 3

Find the present value at 6% simple interest with total amount of debt RM 40000 due in 15 years.Anwer

2.3Compound InterestCompound interest is based on the principal which interest changes from time to time. Interest that is earned is compounded or converted into principal and earns thereafter. Hence the principal increases from time to time. The formula to compute the amount of compound interest is given below

Some important terms are best explained with the following example. Suppose RM9000 is invested for 7 years at 12% compounded quarterly

Principal, P

The original principal, denoted by P is the original amount invested. Here the principal is P = RM 9000

Annual nominal rate, rAnnual nominal rate denoted by r is the interest rate for a year together with the frequency in which interest is calculated in a year. Thus the annual nominal rate is given by r = 12% compounded quarterly, that is four times a year.Frequency of conversions/ number of compounding periods per

year, mFrequency of conversion denoted by m is the number of times interest is calculated in a year. The annual nominal rate is given by r = 12% compounded quarterly, that is four times a year. In this case, m=4Interest periodInterest period is the length of time in which interest is calculated. Thus, the interest period is three month.

Periodic interest rate, iPeriodic interest rate denoted by i is the interest rate for each interest period. Thus, the periodic interest rate in this case is given by

Number of interest periods in the investment period/ total number of compounding periods, nThe number of interest periods during the investment period denoted by n is the number of times interest is calculated. The number of interest periods is given by . Thus, .Example 2.3.1

Find the accumulated amount after 3 years if RM1000 is invested at 8% per year compounded

a) Annually

b) Semi-annually

c) Quarterly

d) Monthly

e) Daily

Solution

Example 2.3.2RM9000 is invested for 7 years 3 months. This investment is offered 12% compounded monthly for the first 4 years and 12% compounded quarterly for the rest of the period. Calculate the future value of this investment.

Solution

Example 3.3.3What is the annual nominal rate compounded monthly that will make RM1000 become RM2000 in five years?

Solution

Example 4Determine the annual nominal rate compounded quarterly that will make RM10000 increase to RM25000 in 10 years?

Answer

r=9.27%

2.3.1Effective Rate of Interest

Interest actually earned on an investment depends on the frequency with which the interest is compounded. One such way of comparing interest rates is provided by the use of the effective rate. The effective rate is the simple interest rate that would produce same accumulated amount in 1 year as the nominal rate compounded m times a year. The effective rate also called the effective annual yield. The formula for computing the effective rate of interest is given below

Example 2.3.4Find the effective rate of interest corresponding to a nominal rate of 8% per year compounded

a) Annually

b) Semi-annually

c) Quarterly

d) Monthly

e) DailySolution

Example 2.3.5 (Application Effective rate of Interest)Kamal wishes to borrow some money to finance some business expansion. He has received two different quotes:

Bank A: charges 15.2% compounded annually

Bank B: charges 14.5% compounded monthly

Which bank provides a better deal?Solution

BANKAnnual Nominal rateEffective rate

A15.2% 15.2%

B14.5% 15.5%

2.3.2Present Value

Recall:

The process for finding the present value is called discounting.

Example 3.3.6 How much money should be deposited in a bank paying interest at the rate of 6% per year compounded monthly so that, at the end of 3 years, the accumulated amount will be RM20000?

Solution

RM 16713 should be invested so that we get accumulated RM 20000 after 3 yearsExercise 5

Find the present value of RM 56 500 due in 7 years at an interest of 8% per year compounded semi-annually.

AnswerP= RM 32, 627.34Example 2.3.7 Find the present value of RM49, 158.60 due in 5 years at an interest of 10% per year compounded quarterly.

Solution

Example 2.3.8 (HOMEWORK1)A debt of RM3000 will mature in three years time. Find

a) The present value of this debt

b) The value of this debt at the end of the first year

c) The value of this debt after four years

Assuming money is worth 14% compounded semi-annually

Solution

2.3.3Continuous Compounding of Interest

Thus far, we have been discussing compounding of interest on discrete time intervals (daily, monthly, etc). If compounding of interest is done on a continuous basis the formula is

Example 2.3.9Find the accumulated value of RM1000 for six months at 10% compounded continuously.

Solution

Exercise 6

Find the accumulated value of RM8000 in 1 year 4 months at 8% compounded continuously.AnswerA=8900.51Example 2.3.10 (HOMEWORK2)Find the amount to be deposited now so as to accumulate RM1000 in eighteen months at 10% compounded continuously.

Solution

2.4Annuities

Annuity is a sequence of equal payments made at equal intervals of time. Examples of annuity are shop rentals, insurance policy premiums, instalment payment, etc. In this section we shall discuss ordinary annuity certain. An annuity in which the payments are made at the end of each payment period is call ordinary annuity certain.

2.4.1Future Value of Ordinary Annuity Certain

Future value of an ordinary annuity certain is the sum of all the future values of the periodic payments. Formula for calculate the future value of ordinary annuity certain is

Example 2.4.1

Find the amount of an ordinary annuity consisting of 12 monthly payments of RM100 that earn interest at 12% per year compounded monthly.

Solution

sum of all the future valuesExample 2.4.2 (D)RM 100 is deposited every month for 2 years 7 month at 12% compounded monthly. What is the future value of this annuity at the end of the investment period?

Solution

Example 2.4.3 (Try verify given answer)RM100 was invested every month in an account that pays 12% compounded monthly for two years. After the two years, no more deposit was made. Determine the future value of this annuity at the end of the investment period?AnwserS=2697.35Example 2.4.4 (D)Lily invested RM100 every month for five years in an investment scheme. She was offered 5% compounded monthly for the first three years and 9% compounded monthly for the rest of the period. Determine the future value of this annuity at the end of five years and total amount money after 5 years.Solution

2.4.2Present Value of Ordinary Annuity Certain

In certain instances, we may try to determine the current value P of a sequence of equal periodic payment that will be made over a certain period of time. After each payment is made, the new balance continues to earn interest at some nominal rate. The amount P is referred to as the present value of ordinary annuity certain. Formula to calculate this present value is given below.

Example 2.4.5 After making a down payment of RM4000 for an automobile, Maidin paid RM400 per month for 36 month with interest charged at 12% per year compounded monthly on the unpaid balance. What was the original cost of the car?

Solution

Example 2.4.6

As a savings program toward Alfian college education, his parents decide to deposit RM100 at the end of every month into a bank account paying interest at the rate of 6% per year compounded monthly. If the saving program began when Alfian was 6 years old, how much money would have accumulated by the time he turn 18?

AnswerP= RM 10247.47Example 2.4.7 (HOMEWORK3)Ray has to pay RM300 every month for twenty-four months to settle loan at 12% compounded monthly.

a) What is the original value of the loan?

b) What is the total interest that he has to pay?

Answer

2.4.3AmortizationAn interest bearing debt is said to be amortized when all the principal and interest are discharged by a sequence of equal payments at equal intervals of time.

Example 2.4.8 A sum of RM50000 is to be repaid over a 5 year period through equal instalments made at the end of each year. If an interest rate of 8% per year is charged on the unpaid balance and interest calculations are made at the end of each year, determine the size of each instalment so that the loan is amortized at the end of 5 years.

Solution

Have to pay RM 12522.82 per year within 5 yearsExample 2.4.9Andy borrowed RM120, 000 from a bank to help finance the purchase of a house. The bank charges interest at a rate 9% per year in the unpaid balance, with interest computations made at the end of each month. Andy has agreed to repay the loan in equal monthly instalments over 30 years. How much should each payment be if the loan is to be amortized at the end of the term?

Solution

Have to pay RM 965.55 per month within 30 years

2.4.4Sinking Fund

Sinking funds are another important application of the annuity formula. Simply stated, a sinking fund is an account that is set up for a specific purpose at some future date. For example, an individual might establish a sinking fund for the purpose of discharging a debt at a future date. A corporation might establish a sinking fund in order to accumulate sufficient capital to replace equipment that is expected to be obsolete at some future date. The formula for finding the sinking fund is given below

Example 2.4.10 DA debt of RM1000 bearing interest at 10% compounded annually is to be discharged by the sinking fund method. If five annual deposits are made into a fund which pays 8% compounded annually,

a) Find the annual interest payment

b) Find the size of the annual deposit into the sinking fund

c) What is the annual cost of this debt?

Solution

Pay RM 270.46 annually for 5 yearsExample 2.4.11

The proprietor of Carling 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 RM 30000. If the fund earns 10% interest per year compounded quarterly, determine the size of each instalment the proprietor should pay.Solution

per month for 2 yearsExample 2.4.12

Harris, a self employed individual who is 46 years old, is setting up a defined benefit retirement plan. If he wishes to have RM250000 in this retirement account by age 65, what is the size of each yearly instalment he will be required to make into a saving account earning interest at % yr?

m=1Solution

2.5DepreciationDepreciation is an accounting procedure for allocating the cost of capital assets, such as buildings, machinery tools and vehicles over their useful life. It is important to note that depreciation amount are estimates and no one can estimate such amounts with certainty. Depreciation expenses allow firms to recapture the amount of money to provide for replacement of the assets and to recover the original investments. Depreciation can also be viewed as decline in value of assets because of age, wear or decreasing efficiency. Many properties such as buildings, machinery, vehicles and equipment depreciate in value as they get older.Several terms are commonly used in calculation relating to depreciation. The terms are

Original costThe original cost of an asset is the amount of money paid for an asset plus any sales taxes, delivery charges, installation charges and other cost.

Salvage valueThe salvage value (scrap value or trade in value) is the value of an asset at the end of its useful life. If a company purchases a new machine and sells it for RM600 at the end of its useful life, then the salvage value is RM600. The salvage value of an asset is an estimate that is usually based on previous salvage value of a similar asset. Useful lifeThe useful life an asset is the life expectancy of the asset or the number of years the asset is expected to last. For example, if a company expects to use machinery for 10 years, then its useful life is 10 years. Total depreciationThe total depreciation or the wearing value of an asset is the difference between cost and scrap value. Annual depreciationThe annual depreciation is the amount of depreciation in a year. It may or may not be equal from year to year. Accumulated depreciationThe accumulated depreciation is the total depreciation to date. If the depreciation for the first year is RM2000 and for the second year RM1000 then the accumulated depreciation at the end of the second year is RM3000

Book valueThe book value or carrying value of an asset is the value of the asset as shown in the accounting record. It is the difference between the original cost and the accumulated depreciation charged to that date. For example a car which was purchased for RM40000 two years ago, will have a book value of RM34000 if its accumulated depreciation for two years is RM6000.

Three method of depreciation are commonly used. These methods are

1. Straight line method

2. Declining balance method

3. Sum of years digits method

2.5.1 Straight Line Method

The straight line method of computing depreciation is the simplest of the three methods and probably the most common method used. Under the straight line method, the total amount of depreciation is spread evenly to each accounting period through the useful life of the asset. The formula for finding the annual depreciation, annual rate of depreciation and book value are given below

Example 2.5.1

The book values of an asset after the third year and fifth year using the straight line method are RM7000 and RM5000 respectively. What is the annual depreciation of the asset?

Solution

Example 2.5.2John Company bought a lorry for RM38000. The lorry is expected to last 5 years and its salvage value at the end of 5 years is RM8000. Using the straight line method to;a) Calculate the annual depreciation

b) Calculate the annual rate of depreciation

c) Calculate the book value of the lorry at the end of third year

d) Prepare a depreciation scheduleSolution

(d)End of

YearAnnual

Depreciation(RM)Accumulated Depreciation (RM)Book value at end

Of year(RM)

00038000

16000600032000

260001200026000

360001800020000

460002400014000

56000300008000

2.5.2Declining Balance MethodDeclining balance method is an accelerated method in which higher depreciation charged are deducted in the early life of the asset. Formula for finding book values, annual rate of depreciation and accumulated depreciation is given below

Example 2.5.3

The cost of fishing boats is RM150000. Declining balance method is used for computing the depreciation. If the depreciation rate is 15%, compute the book value and accumulated depreciation of the boat at the end of 5 years.

SolutionBV=RM 66555.80

Example 2.5.4

Given

Cost of the asset = RM15000

Useful life = 4 years

Scrap value = RM3000

a) Find the annual rate of depreciation

b) Construct the depreciation schedule

Using the declining balance method

Solution

YearAnnual

Depreciation(RM)Accumulated

Depreciation (RM)Book Value (RM)

00015000

14969.504969.5010030.50

23323.108292.606707.40

32222.1610514.764485.24

41485.24120003000

2.5.3 Sum of Years Digits Method

The sum of years digits method is another accelerated method. In this method, the rate of depreciation is based on the sum of the digits representing the number of years of useful life of the asset. If an asset has a useful life of 3 years, the sum of digits is S = 1+2+3=6, while for an asset with a useful life of 5 years, the sum of digits is S = 1+2+3+4+5=15. Since S is an arithmetic progression, S can be calculated with the formula

The depreciation in the first year is X the depreciable value of the asset, Depreciation in the the second is X the depreciable value of the asset, Depreciation in the the third is X the depreciable value of the asset and so on

Example 2.5.5

A machine is purchased for RM45000. Its life expectancy is 5 years with a zero trade in value. Prepare a depreciation schedule using the sum of the years digits method.Solution

Amount of depreciation for each year is calculated as follows

YearAnnual Depreciation

1

2

3

4

5

The depreciation schedule is as follows

YearAnnual

Depreciation(RM)Accumulated

Depreciation (RM)Book Value (RM)

00045000

1150001500030000

2120002700018000

39000360009000

46000420003000

53000450000

Example 2.5.6

A computer is purchased for RM3600. It is estimated that its salvage value at the end of 8 years will be RM600. Find the depreciation and the book value of the computer for third year using the sum of the years digits method.Answer

Solution

CHAPTER 2 : FINANCIAL MATHEMATICS

Simple Interest

EMBED Equation.3

Where

I = simple interest

P = principal

r = rate of simple interest

t = time in years

Amount Simple Interest/Simple Amount/ Future Value

EMBED Equation.3

Where

A = amount or future value

P = principal or present value

r = rate of simple interest

t = time in years

Amount: Compound Interest/Future Value

EMBED Equation.3

Where EMBED Equation.3 and EMBED Equation.3

A = amount or future value at the end of n periods

P = principal or present value

r = annual nominal rate

m = number of compounding periods per year

i = rate per compounding period

n = total number of compounding periods

t= time in year

Effective Rate of Interest

EMBED Equation.3

Where

EMBED Equation.DSMT4 = effective rate of interest

r = annual nominal interest rate

m = number of compounding periods per year

Amount: Compound Interest/Future Value

EMBED Equation.3

Where EMBED Equation.3 and EMBED Equation.3

Present Value of Compound Interest

EMBED Equation.3

Where EMBED Equation.3 and EMBED Equation.3

A = amount or future value at the end of n periods

P = principal or present value

r = annual nominal rate

m = number of compounding periods per year

i = rate per compounding period

n = total number of compounding periods

Amount: Compound Interest

EMBED Equation.3

A = amount or future value at the end of n periods

P = principal or present value

e = 2.718282

i = rate per compounding period

t = time in years

Future Value of Ordinary Annuity Certain

EMBED Equation.3

Where EMBED Equation.3 and EMBED Equation.3

S = future value of annuity at the end of n periods

R = periodic payments

i = interest rate per interest period

n = term of investment

Present Value of Ordinary Annuity Certain

EMBED Equation.3

Where

P = present value of annuity at the end of n periods

R = periodic payments

i = interest rate per interest period

n = term of investment

Amortization

EMBED Equation.3

Where

P = present value of annuity at the end of n periods

R = periodic payments

i = interest rate per interest period

n = term of investment

Sinking Fund

EMBED Equation.3

Where

S = future value of annuity at the end of n periods

R = periodic payments

i = interest rate per interest period

n = term of investment

Straight Line Method

EMBED Equation.3

Declining Balance Method

EMBED Equation.3

Sum of Years Digits Method

EMBED Equation.3

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Universiti Malaysia Perlis 2013PAGE 2 __________________________________________________________________

Universiti Malaysia Perlis 2013

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