finance, growth and decay

GRade 12

Curriculum Statement

  • Solve problems involving present value and
    future value annuities.

 

  • Make use of logarithms to calculate the value of n, the time period, in the equations:

                                                     and

 

  • Critically analyse investment and loan options and make informed decisions as to best option(s) (including pyramid).
A=P(1+i)^n
A=P(1+i)nA=P(1+i)^n
A=P(1-i)^n
A=P(1−i)nA=P(1-i)^n

Revision

Grade 11 - Financial mathematics

simple

Interest or depreciation is calculated using the initial amount.

compound

Interest or depreciation is calculated using the current value of the loan or investment.

A=P(1+in)
A=P(1+in)A=P(1+in)
A=P(1+i)^n
A=P(1+i)nA=P(1+i)^n
A=P(1-in)
A=P(1−in)A=P(1-in)
A=P(1-i)^n
A=P(1−i)nA=P(1-i)^n

Simple interest

Straight-line depreciation

Compound interest

Reducing-balance depreciatation

Nominal interest

An interest rate which is compounded more than once a year

Effective interest

The interest rate which would affect the same change per year as a nominal interest rate

i_{eff} = \left( 1 + \frac{i_{\text{nom}}}{n} \right)^n - 1
ieff=(1+inomn)n−1i_{eff} = \left( 1 + \frac{i_{\text{nom}}}{n} \right)^n - 1

...for converting the nominal rate to the annual effective rate.

compounded

daily, monthly...

only compounded

once every year

Part one

Calculating the time period of a loan or investment

concept

Using knowledge of logarithms, one is now able to solve for a particular time period of a loan or investment as follows:

Example 1

R4000 is deposited in a savings account paying 10% per annum compounded annually. How long will it take for the savings account to double?

Example 2

The interest rate for a loan is 7% compounded half-yearly. How many years will it take the loan amount to double?

Example 3

A motor car depreciates at 12% per annum using the reducing-balance method. After what length of time will the value of the car be 80% of the original selling price?

Part TWo

Investments or loans involving annuities

DEfinition:

An annuity is a series of fixed, equal payments to a loan or investment at regular intervals subject to a rate of interest over the period of the annuity

Future Value annuities

Money is invested at regular intervals such that all the payments - as well as the compound interest earned - contribute to the total value of the investment when it matures.

investments

Type Example Effective Rate
Short term 5-year savings policy ~9% p.a
Long term Retirement Annuity ~8% p.a

If R500 is invested every month, with the account having an interest rate of 12% per annum compounded monthly, calculate the savings after 3 months.

concept

Time Interest Payment Balance
0 months R0 R500 R500
1 months R5 R500 R1005
2 months R10.05 R500 R1515.05
3 months R15.15 R500 R2030.20

Example

R1200 is deposited into a savings plan - while the monthly payments continue at the end of every month for the next 20 years. If the interest remains fixed at 8% per annum compounded monthly:

 

  • How much money will be in the account at the end of the 20th year?
F = \frac{x\left[(1 + i)^{n}-1\right]}{i}
F=x[(1+i)n−1]iF = \frac{x\left[(1 + i)^{n}-1\right]}{i}

Example - continued...

  • At the end of the 20 years, the money is left in the account for a further year. How much money has accumulated in the account now?

Present Value annuities

Instalments are made at regular intervals such that all the repayments contribute to the total repayment of the loan - as well as the interest charged during the term.

Loans

Type Example Effective Rate
Short term Unsecured Loan 30-60% p.a
Long Term Home Loan ~10.5% p.a

Example 1

A loan is taken out to the value of R200 000. A 15% deposit is paid, and the balance is settled through fixed monthly instalments for the next 10 years. The interest rate is 12% per annum compounded monthly.

 

  • Calculate the value of the monthly payments
P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}

Example 2

A business takes out a 20 year loan of R500 000. The loan is repaid by means of equal monthly repayments, starting 6 months after granting the loan. The interest rate is 24% per annum compounded monthly.

 

  • Calculate the value of the monthly payments
P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}

Part THree

Analysing Loan Options

Example

A customer is searching for the most affordable home loan for R950 000. He receives the following offers from 3 different banks shown below:

Comparing Investments and loans

Bank Term Interest
Option A 20 years 15%
Option B 15 years 18%
Option C 30 years 12%

(Interest is calculated per annum, compounded monthly)

  • Decide which option would be the best financially? Motivate a response using the appropriate calculations.

option A

A customer is searching for the most affordable home loan for R950 000. 

Term Interest
20 years 15%

(Interest is calculated per annum, compounded monthly)

P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}

option B

A customer is searching for the most affordable home loan for R950 000. 

Term Interest
15 years 18%

(Interest is calculated per annum, compounded monthly)

P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}

option C

A customer is searching for the most affordable home loan for R950 000. 

Term Interest
30 years 12%

(Interest is calculated per annum, compounded monthly)

P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}

A customer is searching for the most affordable home loan for R950 000. He receives the following offers from 3 different banks shown below:

Summary

Bank Term Interest Monthly Repayments Total Amount
Option A 20 yrs 15% R12 509.50 R3 002 208.25
Option B 15 yrs 18% R15 299.00 R2 753 819.98
Option C 30 yrs 12% R9 771.82 R3 517 855.08

(Interest is calculated per annum, compounded monthly)

NOTE: Option C has the lowest monthly repayment amount, so while it may be easier from a month-to-month basis, it also generates the largest total amount paid by the end of the term.

Option B has the lowest total amount required to be paid, and therefore should be the best option - as long as the monthly repayments are feasible according to the person's monthly budget.

Part FOUR

BALANCE OUTSTANDING ON A LOAN

Balance Outstanding on a Loan at a given time

It is useful to be able to calculate the balance of a loan, and to know it at a specific time during the course of the loan.

Example 1

A R25 000 loan for 2 years is paid back through monthly payments of R1 176.84 starting one month after the granting of the loan. If interest is calculated at 12% per annum compounded monthly:

  • Calculate the balance outstanding at the end of one year.

Example 1 - continued...

Example 2

A R120 000 loan is granted over four years at an interest rate of 18% per annum compounded monthly.

  • Calculate the outstanding balance after the 20th payment has been made.

Part FIVE

sinking funds

Example

A R20 000 car is bought which depreciates in value at 15% per annum on a reducing-balance. A new car in 5 years time is estimated to have increased in value by 10% every year. The car will be sold at the end of the 5 years, so a sinking fund is set up in order to buy the new car. The sinking fund will grow at 12% per annum compounded monthly, with a first payment made immediately, and the final payment at the end of 5 years.

Sinking Funds

A company may purchase assets which will be used for a certain period of time and then sold or disposed of when new assets are purchased. A savings plan is often set up in order to prepare for this future, and this account is often known as a SINKING FUND.

Example - Continued...

A R20 000 car is bought which depreciates in value at 15% per annum on a reducing-balance. A new car in 5 years time is estimated to have increased in value by 10% every year. The car will be sold at the end of the 5 years, so a sinking fund is set up in order to buy the new car. The sinking fund will grow at 12% per annum compounded monthly, with a first payment made immediately, and the final payment at the end of 5 years.

Calculate the scrap value of the car

Example - Continued...

A R20 000 car is bought which depreciates in value at 15% per annum on a reducing-balance. A new car in 5 years time is estimated to have increased in value by 10% every year. The car will be sold at the end of the 5 years, so a sinking fund is set up in order to buy the new car. The sinking fund will grow at 12% per annum compounded monthly, with a first payment made immediately, and the final payment at the end of 5 years.

Calculate the COST OF a car in 5 years time

Example - Continued...

A R20 000 car is bought which depreciates in value at 15% per annum on a reducing-balance. A new car in 5 years time is estimated to have increased in value by 10% every year. The car will be sold at the end of the 5 years, so a sinking fund is set up in order to buy the new car. The sinking fund will grow at 12% per annum compounded monthly, with a first payment made immediately, and the final payment at the end of 5 years.

FIND the amount needed in the sinking fund

Example - Continued...

The sinking fund will grow at 12% per annum compounded monthly, with a first payment made immediately, and the final payment at the end of 5 years.

FIND the equal monthly payments required to reach this needed R23 336.09 in 5 years time

Example - EXTENSION...

If the business decides to withdraw R1 000 from the account at the end of each year to pay for a car service...

calculate the reduced value of the fund

Example - EXTENSION...

Calculate the adjusted monthly repayments to allow for these withdrawals...

summary

Revision of key concepts and formulae

Simple interest

A = P(1 + in)
A=P(1+in)A = P(1 + in)

Compound interest

A = P(1 + i)^n
A=P(1+i)nA = P(1 + i)^n

Simple Depreciation

A = P(1 - in)
A=P(1−in)A = P(1 - in)

Compound Depreciation

A = P(1 - i)^n
A=P(1−i)nA = P(1 - i)^n

Future value

Present value

P = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
P=x[1−(1+i)−n]iP = \frac{x \left[1 - (1 + i)^{-n} \right]}{i}
x = \frac{P \times i}{\left[1 - (1 + i)^{-n}\right]}
x=P×i[1−(1+i)−n]x = \frac{P \times i}{\left[1 - (1 + i)^{-n}\right]}

or

F = \frac{x\left[(1 + i)^{n}-1\right]}{i}
F=x[(1+i)n−1]iF = \frac{x\left[(1 + i)^{n}-1\right]}{i}
x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}
x=F×i[(1+i)n−1]x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}

or

Grade 12 - Finance, Growth and Decay

By dviljoen

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Grade 12 - Finance, Growth and Decay

A summary of the key concepts for students following the South African CAPS (FET) Mathematics curriculum.

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