AMB111F Financial Maths Notes

AMB111F Financial Maths Notes

Compound Interest: Interest computed on the current amount that increases at regular intervals.

Simple interest: Interest computed on the original fixed amount (principal). Suppose an amount of P rands is invested (in a savigs bank or insurance) at the rate of r % per annum compound interest, then after 1 year the new amount is A1 = P + P (r/100) = P (1 + r/100). After 2 years the new amount is A2 = P (1 + r/100) + P (1 + r/100)(r/100) = P (1 + r/100)2_{. After 3 years the} new amount is A3 = P (1 + r/100)2 + P (1 + r/100)2(r/100) = P (1 + r/100)3, and so forth. Thus the amount after n years at r % p.a. compound interest is

P (1 + r/100)n.

The initial amount invested, denoted by P, is usually called the P rincipal; r is the rate at which the money earns compound interest. The number n of years during which P is invested is called the period, which may also be given in months or any fraction of a year.

NOTE: Simple interest is computed on the same fixed principal every year (or fraction of the year) and the resulting amount is A = P (1 + nr/100), whereas compound interest is computed on the new amount.

Example : 1. A student invests R500 for 5 years at 12,5 % per year compound interest. Calculate how much money will he have at the end of 5 years.

Soln. Amount after n years is A = P (1 + r/100)n_{. Here P = R500; n = 5 and}

r = 12, 5. Thus A = 500(1 + 0, 125)5 = 500(1, 125)5 = R905, 02.

Example : 10. A student invests R500 for 5 years at 12,5 % per year simple interest. Calculate how much money will he have at the end of 5 years.

Soln. Amount after n years is A = P (1 + nr/100). Here P = R500; n = 5 and r = 12, 5. Thus A = 500(1 + 5(0, 125)) = 500(1, 625) = R812, 50.

2. A student invests R500 for 5 years at 12, 5 % compound interest p.a. calculated quarterly. Find the amount after 5 years.

Soln. Here we need to convert the yearly rate to the quarterly rate. There are 4

quarters in a year , thus the quarterly rate is 12, 5÷4 % . Also 5 years becomes 5 × 4 = 20 quarters. So the amount is A = P (1 + r/100)n _{= 500(1 +} 12,5

400) 20₌

R925, 23

3. R1000 amounts to R1500 after 2,5 years compound interest. The interest was calculated monthly. Find the interest rate per year.

Soln. A = P (1 + r/100)n _{implies (}A P)

1/n_{= 1 + r/100. Thus r = 100[(}A P)

1/n_{− 1]} per month. so n = 2, 5 × 12 = 30 months. Therefore r = 100[(1500₁₀₀₀)1/30− 1] = 100[(1.5)1/30_{− 1] = 1.360725 per month. Hence per year (12 months) the rate} is r = 1.360725 × 12 = 16.3287.

Calculating the period n:

Let A = P (1 + r/100)n. Then (A/P ) = (1 + r/100)n implies log(A/P ) =

n log(1 + r/100), thus n = _log(1+r/100)log(A/P ) .

4. Calculate how long it will take an investor to make a gain of R1000 on an initial investment of R3000 if the compound interest rate is 11,5 % p.a. calculated monthly.

Soln. Here the total interest to be made is R1000. So the final amount must be R4000. r = 11, 5 ÷ 12 per month. Thus the period in months is

n = _log(1+r/100)log(A/P ) = _{log(1+(11.5)/1200)}log(4/3) = 301626 months, translating to about 2, 51 years.

Depreciation

Depreciation is the reverse of compound interest. An item loses value the mo-ment it leaves a store, and we say it depreciates. Suppose its initial value is P rands and depreciates at the rate of r % p.a. What is its value after n years? After 1 year: value V1 = P − P (r/100) = P (1 − r/100)

After 2 years: value V2 = P (1 − r/100) − P (1 − r/100)(r/100) = P (1 − r/100)2_, and so forth. Thus its value after n years is Vn = P (1 − r/100)n.

Example 1. A car costs R180 000 and depreciates at a rate of 8 % p.a. calcu-lated on the diminishing balance. Calculate the value of the car after 5 years.

Soln. Its value is V5 = P (1 − r/100)n = 180000(1 − 8/100)5 = R118634, 67. Calculating the period n:

Let Vn = P (1 − r/100)n. Then (Vn/P ) = (1 − r/100)n implies log(Vn/P ) =

n log(1 − r/100), thus n = log(Vn/P )

Annuities. An annuity is a series of fixed amounts paid at regular intervals,

and accumulating compound interest in the process. A value of the annuity after a given interval is the amount available at the end of the interval with due interest also added. If I pay fixed amounts of R2000 quarterly into a savings account for 10 years, then the series of those fixed amounts is an annuity. The value of the annuity is the amount available after the specified period.

Future value annuity: This is the value of an annuity after the specified period, for example in investing for retirement, the amount that should be available is the future value. Alternatively the amout to which the annuity grows is the future value of the annuity.

Present value annuity: the single sum that needs to be invested now to get a precalculated future value after a certain period of time, is the present value of the annuity.

Sinking Fund: This is an investment made to replace valuable and expensive items at a later stage. E.g. a car, tractor or furniture.

Example 1. Calculate the value of an annuity in which R1000 is invested

at the end of each year at 10 % per annum for 5 years.

soln. Annuities involve compound interest, see the following table:

Amount/Value end year 1 1000 end year 2 1000 + 1000(1 + 10/100) end year 3 1000 + 1000(1 + 10/100) + 1000(1 + 10/100)2 end year 4 1000 + 1000(1 + 10/100) + 1000(1 + 10/100)2 _{+ 1000(1 + 10/100)}3 end year 5 1000 + 1000(1 + 10/100) + 1000(1 + 10/100)2 _{+ 1000(1 + 10/100)}3₊ 1000(1 + 10/100)4 = R6105, 10

So the required value of the annuity after 5 years is R6105, 10 (future value). Now let P = fixed deposit of an annuity, A = future value of the annu-ity, n = period of the annuannu-ity, and r = interest rate per period. If the de-posits are made at the end of the year, then the above example shows that

A =Pn−1_k=0P (1 + r/100)k_.

Now from the summation of a geometric series, A = P100_r [(1 +₁₀₀r )n− 1].

NOTE: When money is invested at the beginning of the year, the future value

is A = P100_r (1 + r/100)[(1 + ₁₀₀r )n_{− 1] because interest is added at the end of}

Example 10_{. Calculate the value of an annuity in which R1000 is invested} at the beginning of each year at 10 % per annum for 5 years.

Soln. Future value is FV = A = P100_r (1 + r/100)[(1 +₁₀₀r )n− 1] = 1000100₁₀(1 + 10/100)[(1 + ₁₀₀10)5_{− 1] = R6715.61}

Example 2. Themba invests R500 every 6 months at 9% p.a. interest,

compounded half-yearly for 10 years. Calculate the value of the annuity after 10 years.

soln. Convert to half-years: P = R500; 10 years = 20 half years; rate = 9/2 per

half-year. Using A = P100_r [(1+₁₀₀r )n_{− 1], we have A = 500}200 9 [(1+

9 200)

20_{− 1] =}

R15685.71.

Now consider A = P100_r [(1 + ₁₀₀r )n _{− 1] and find a formula for the fixed}

amount payable periodically in an annuity:

Soln. P = _100[(1+rAr 100)n−1]

Example 3. John wants to have an amount of R10000 at the end of 5 years. The best compound interest rate that he can have is 8 % p.a.. Calculate his yearly fixed deposits (deposited at the end of the year) in order to achieve his goal.

Soln. Fixed deposit = P = _100[(1+rAr 100)n−1]

= 8(10000)

100[(1+₁₀₀8 )5_−1] = R1704.56

Present value: This is a single sum of money invested under the same terms as an annuity. Instead of paying fixed deposits at regular intervals, only one large sum is invested. The financial institution then works out fixed deposits as in the future value.

Let PV = present value to be invested; FV = future value; n = period of investement, r = interest rate. Assume the lump sum is deposited at the be-ginning of the year. After 1 year, F V = P V (1 + r/100)1. After 2 years,

F V = P V (1 + r/100)2. It is clear that after n years F V = P V (1 + r/100)n. Example 4. Tshidi has an annuity to which she contributes R1000 p.a. (at the end of the year) at 6% annual compound interest. The annuity will mature in 25 years. Calculate the present value of the annuity.

Soln. FV = P100_r [(1 + ₁₀₀r )n_{− 1] = 1000}100 6 [(1 +

6 100)

F V = P V (1 + r/100)n _{implies P V =} F V

(1+r/100)n =

54864.51

(1+6/100)25 = R12783.36.

Example 40. Tshidi has an annuity to which she contributes R1000 p.a. (at the beginning of the year) at 6% p.a compound interest annually. The annuity will mature in 25 years. Calculate the present value of the annuity.

Soln. FV = P100_r (1+r/100)[(1+₁₀₀r )n−1] = 1000100₆ (1+6/100)[(1+₁₀₀6 )25−1] = (54864.51((1.06) = 58156.38. Now F V = P V (1 + r/100)n _{implies P V =} F V (1+r/100)n = 58156.38 (1+6/100)25 = R13550.36.

Formula involving present value and fixed deposits:

Recall: F V = P100_r [(1 + ₁₀₀r )n_{− 1] where P = fixed deposit carried out}

peri-odically at the end of the year.

F V = P V (1 + r/100)n_{. Thus P V (1 + r/100)}n _{= P}100 r [(1 + r 100) n_{− 1]. Solving} for PV, we have P V = P 100 r [(1+ r 100)n−1] (1+r/100)n = 100P [(1+₁₀₀r )n_−1] r(1+r/100)n .

Similarly if deposits are made at the beginning of the year.

Example 5. Which of the following will give greater financial return? (a) An annuity of R100 deposited per month (at the end) for 20 years at 12 % p.a. interest compounded half-yearly.

(b) A single deposit of R10000 (at the beginning) invested for 20 years at 12 % p.a. interest compounded half-yearly.

Soln. (a) Converting to half year period, 1 month = 1/6 half year, 20

years = 40 half years; R100/month = R600/half year. Therefore FV =

P100_r [(1 + ₁₀₀r )n_{− 1] = 600}200 12[(1 + 12 200) 40_{− 1] = R92857.18.} (b) F V = P V (1 + r/100)n _{= 10000(1 + 12/200)}40 _{= R102857.18.} Therefore (b) yields a bigger return.

Loan Repayments.

Loan repayment works just like annuity (paid at the end of an interval). The difference is that you pay the financial institution interest.

Example 6. Calculate the total cost of repaying a car loan of R100 000 at 9 % p.a. in equal monthly repayments over a 25-year period.

Soln. The total cost of repayment is just like a future value in an

annu-ity with present value PV = 100 000. Now P V = P

100 r [(1+ r 100) n_−1] (1+r/100)n implies P = P V ((1+r/100)100 n) r [(1+ r 100)n−1] = rP V ((1+r/100)_100[(1+ r n) 100)n−1] = (9/12)100000((1+9/1200)300) 100[(1+₁₂₀₀9 )300_−1] = R839.20 =

monthly repayments. Thus the total payable = R839.20 × 300 = R251758.91 . (Note that 25 years = 300 months).

Sinking Fund. This is an investment made to replace expensive items in a few years’ time. Scrap value is the value of the depreciated item after a period of time.

Example. Machinery is purchased at a cost of R550 000 and is expected to rise in cost at 15 % p.a. compound interest and depreciate in value at 8 % per annum compounded annually.

A sinking fund is started to make provision for replacing the old machine. The sinking fund pays 16 % interest p.a. compounded monthly, and you make (fixed) monthly payments into this account for 10 years, starting immediately and ending one month before the purchase of the new machine. Determine: (a) the replacement cost of a new machine 10 years from now;

(b) the scrap value of the machine in 10 years’ time;

(c) the monthly payments into the sinking fund that will make provision for the replacement of the new machine.

Soln. (a) Let P V = R550000. Then the future value of the machine is F V = P V (1 + r/100)n _{= 550000(1 + 15/100)}10 _{= R2225056.76 =} Replace-ment Cost

(b) Scrap value is SV = P V (1 − r/100)n _{= 550000(1 − 8/100)}10 _{= R238913.65} (c) Now FV - SV = R2 225 056.76 - R238 913.65 = 1986143.11 is the re-quired value (SF) of the sinking fund. Consider SF = P100_r (1 + r/100)[(1 +

r

100)

n

− 1] where P = fixed instalment. Thus P = 100 SF r [(1+ r 100) n_{−1](1+r/100)} = 1986143.11 1200[(1+ 16 1200)120−1](1+16/1200)