 # WWhat is financial math? - field of applied mathematics, concerned with financial markets. PProcedures which used to answer questions associated with.

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WWhat is financial math? - field of applied mathematics, concerned with financial markets. PProcedures which used to answer questions associated with major financial transactions EExample: Interest problem, annuity and depreciation

 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.

 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.

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 earned

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  Solution

Example 2.2.2 R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. SSolution

Exercise 2  If 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 2.2.3 Find 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.  Solution

 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.

 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, r  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, m  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=4 Interest period  Interest period is the length of time in which interest is calculated. Thus, the interest period is three month

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 2.3.2  RM9000 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 2.3.3 What is the annual nominal rate compounded monthly that will make RM1000 become RM2000 in five years? Solution

 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.

Example 2.3.4

Solution

Example 2.3.5 Solution

BANKAnnual Nominal rate Effective rate A15.2% B14.5%15.5%

The process for finding the present value is called discounting.

Solution RM 16713 should be invested so that we get accumulated RM 20000 after 3 years

Solution

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

 Sequence of equal payments made at equal intervals of time.  Examples of annuity are shop rentals, insurance policy premiums, instalment payment, etc.  An annuity in which the payments are made at the end of each payment period is call ordinary annuity certain.

 Future value of an ordinary annuity certain is the sum of all the future values of the periodic payments.  If you know how much you can invest per period for a certain time period, the future value of an ordinary annuity formula is useful for finding out how much you would have in the future by investing at your given interest rate.  If you are making payments on a loan, the future value is useful for determining the total cost of the loan.

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

Example 2.4.2 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 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

 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.

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 Original cost = RM4000 + RM 12 043 = RM16 043

 An 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

Example Andy 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 Andy has to pay RM 965.55 per month within 30 years

 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.

Example 2.4.10 A 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

 Depreciation 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.

Several terms are commonly used in calculation relating to depreciation. The terms are Original cost  The 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 value  The 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 life  The 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 depreciation  The total depreciation or the wearing value of an asset is the difference between cost and scrap value.

Annual depreciation  The annual depreciation is the amount of depreciation in a year. It may or may not be equal from year to year. Accumulated depreciation  The 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 value  The 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 it’s 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

Example 3.5.1 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 2.5.2 John 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 schedule

Solution

End of Year Annual Depreciation( RM) Accumulated Depreciation (RM) Book value at end Of year(RM) 00038000 16000 32000 260001200026000 360001800020000 460002400014000 56000300008000

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 14969.50 10030.50 23323.108292.606707.40 32222.1610514.764485.24 41485.24120003000

 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

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 (RM) Accumulate d Depreciation (RM) Book Value (RM) 00045000 115000 30000 2120002700018000 39000360009000 46000420003000 5 450000 The depreciation schedule is as follows

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.

Solution

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