Finance (Grade 12 NSC Matric Mathematics): Revision Notes

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Present Value Annuities

What is a present value annuity?

A present value annuity involves making regular equal payments or instalments over a specific time period to repay a loan or bond. The remaining balance of the loan typically earns compound interest at a fixed rate. Understanding present value annuities helps you determine how much money needs to be invested today to support a series of future payments.

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Present value annuities are fundamental in financial mathematics because they help us understand the time value of money - a dollar today is worth more than a dollar tomorrow due to its earning potential through compound interest.

Understanding the concept through an example

Let's examine how present value annuities work using a practical scenario:

Kate wants to withdraw R1000 from her bank account each year for the next three years. Her account earns 10% compound interest annually. We need to calculate how much she must deposit initially to make these future withdrawals possible.

To solve this problem, we use the compound interest formula and make the principal (P) the subject:

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

Rearranging: P=A(1+i)−nP = A(1 + i)^{-n}

For each withdrawal, we calculate the required present value:

  • First withdrawal: P=1000(1+0.1)−1=R909.09P = 1000(1 + 0.1)^{-1} = \text{R}909.09
  • Second withdrawal: P=1000(1+0.1)−2=R826.45P = 1000(1 + 0.1)^{-2} = \text{R}826.45
  • Third withdrawal: P=1000(1+0.1)−3=R751.31P = 1000(1 + 0.1)^{-3} = \text{R}751.31

The total initial deposit required: R909.09 + R826.45 + R751.31 = R2486.85

This table verifies our calculation by showing how the account balance changes each year with compound interest and withdrawals.

Deriving the present value annuity formula

The calculation above shows a pattern that can be expressed as a geometric series:

R2486.85=1000(1+0.1)−1+1000(1+0.1)−2+1000(1+0.1)−3\text{R}2486.85 = 1000(1 + 0.1)^{-1} + 1000(1 + 0.1)^{-2} + 1000(1 + 0.1)^{-3}

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Understanding Geometric Series in Finance

This geometric series has:

  • First term: a=1000(1+0.1)−1a = 1000(1 + 0.1)^{-1}
  • Common ratio: r=(1+0.1)−1r = (1 + 0.1)^{-1}
  • Number of terms: n=3n = 3

Using the geometric series formula: Sn=a(1−rn)(1−r)S_n = \frac{a(1 - r^n)}{(1 - r)}

After substituting and simplifying the terms, we arrive at the general formula for present value annuities.

Key formulas

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Essential Present Value Annuity Formulas

Present value of a series of payments: P=x⋅[1−(1+i)−n]iP = x \cdot \frac{[1 - (1 + i)^{-n}]}{i}

Payment amount calculation: x=P×i[1−(1+i)−n]x = \frac{P \times i}{[1 - (1 + i)^{-n}]}

Where:

  • P = present value of the annuity
  • x = regular payment amount
  • i = interest rate per period
  • n = number of payment periods

The first formula helps determine what lump sum is needed today to fund a series of future payments. The second formula helps determine what regular payments are needed to repay a known loan amount.

Worked examples

Example 1: Student loan calculation

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Worked Example: Student Loan Calculation

Question: Andre takes out a student loan for his first year of civil engineering. The repayment period is 1.5 years with financial assistance, and the loan has an interest rate of 10.5% per annum compounded monthly. If Andre pays monthly instalments of R1446.91, calculate the original loan amount.

Step 1: Write down the given information and formula P=x⋅[1−(1+i)−n]iP = x \cdot \frac{[1 - (1 + i)^{-n}]}{i}

Given:

  • x=R1446.91x = \text{R}1446.91
  • i=0.10512i = \frac{0.105}{12} (monthly interest rate)
  • n=1.5×12=18n = 1.5 \times 12 = 18 months

Step 2: Substitute values and calculate

P=1446.91⋅[1−(1+0.10512)−18](0.10512)P = 1446.91 \cdot \frac{[1 - (1 + \frac{0.105}{12})^{-18}]}{(\frac{0.105}{12})}

P=R24 000.14P = \text{R}24\,000.14

Therefore, Andre's original student loan was R24 000.

Step 3: Calculate total interest paid

  • Total repayments: R1446.91 × 18 = R26 044.38
  • Total interest: R26 044.38 - R24 000 = R2044.38

Example 2: Monthly repayments calculation

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Worked Example: Monthly Repayments Calculation

Question: Hristo wants to buy a wine farm worth R8 500 000. He plans to use his current home sale proceeds of R3 400 000 as a deposit. He secures a loan with a 10-year repayment period at 9.5% interest compounded monthly. Calculate his monthly repayments.

Step 1: Identify the loan amount and use the payment formula x=P×i[1−(1+i)−n]x = \frac{P \times i}{[1 - (1 + i)^{-n}]}

  • P=R8 500 000−R3 400 000=R5 100 000P = \text{R}8\,500\,000 - \text{R}3\,400\,000 = \text{R}5\,100\,000
  • i=0.09512i = \frac{0.095}{12} (monthly rate)
  • n=10×12=120n = 10 \times 12 = 120 months

Step 2: Calculate the monthly payment

x=5 100 000×0.09512[1−(1+0.09512)−120]x = \frac{5\,100\,000 \times \frac{0.095}{12}}{[1 - (1 + \frac{0.095}{12})^{-120}]}

x=R65 992.75x = \text{R}65\,992.75

Hristo's monthly repayments will be R65 992.75.

Step 3: Calculate total interest

  • Total repayments: R65 992.75 × 120 = R7 919 130
  • Total interest: R7 919 130 - R5 100 000 = R2 819 130

Example 3: Outstanding loan balance

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Worked Example: Outstanding Loan Balance

Question: A school takes out a loan for a new bus costing R330 000 after a 15% deposit. The loan has prime + 1% interest compounded monthly over 3 years. Calculate the monthly repayments and the outstanding balance after the first year.

Step 1: Calculate loan details

  • P=R330 000−(15100×R330 000)=R280 500P = \text{R}330\,000 - (\frac{15}{100} \times \text{R}330\,000) = \text{R}280\,500
  • i=0.09512i = \frac{0.095}{12} (assuming prime rate of 8.5% + 1%)
  • n=3×12=36n = 3 \times 12 = 36 months

Step 2: Calculate monthly payments

x=280 500×0.09512[1−(1+0.09512)−36]x = \frac{280\,500 \times \frac{0.095}{12}}{[1 - (1 + \frac{0.095}{12})^{-36}]}

x=R8 985.24x = \text{R}8\,985.24

Monthly repayments are R8985.24.

Step 3: Calculate outstanding balance after 12 payments

The balance equals the present value of the remaining 24 payments: P=8 985.24⋅[1−(1+0.09512)−24](0.09512)P = 8\,985.24 \cdot \frac{[1 - (1 + \frac{0.095}{12})^{-24}]}{(\frac{0.095}{12})}

P=R195 695.07P = \text{R}195\,695.07

The outstanding balance after one year is R195 695.07.

Prime lending rate

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Understanding Prime Lending Rate

The prime lending rate serves as a benchmark interest rate that private banks use when lending money to the public. Banks use this rate as a reference point for determining interest rates on various loan types, including small business loans, home loans, and personal loans.

Interest rates may be expressed as a percentage above or below the prime rate. For calculations in this topic, we typically assume the prime lending rate is 8.5% per annum.

Remember!

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Key Points to Remember:

  • Present value annuities help determine today's value of a series of future equal payments
  • The key formula is P=x⋅[1−(1+i)−n]iP = x \cdot \frac{[1 - (1 + i)^{-n}]}{i} for finding present value
  • For payment calculations, rearrange the formula to x=P×i[1−(1+i)−n]x = \frac{P \times i}{[1 - (1 + i)^{-n}]}
  • Monthly compounding requires dividing annual rates by 12 and multiplying years by 12
  • Outstanding loan balances can be calculated using the present value of remaining payments

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