Advanced Financial Modelling and Applications
Real world financial applications often involve more complexity than basic interest and depreciation methods. Many include compounding elements with regular payments that are made throughout the loan or investment period. These payments may also vary based on personal changing circumstances which adds further layers of complexity. Examples of financial structures with fixed payment terms established at the start and maintained throughout will be analysed using recurrence relations. These include reducing balance loans, annuities, interest-only loans, perpetuities, and annuity investments, where the interest accrued occurs before the payments are made.
For simplicity, it will be assumed that the payment periods align with the compounding periods in all cases. As compounding interest is common across these financial structures, the growth factor remains the same for all cases:
\[R = 1 + \frac{r}{100}\]
Where:
- \(R\) is the growth factor
- \(r\) is the interest rate (in percentage)
Use this page to revise the following concepts of advanced financial modelling:
- Reducing Balance Loans
- Annuities
- Interest Only Loans
- Perpetuities
- Annuity Investments
- Amortisation Tables
- Other calculations
Reducing Balance Loans
Reducing balance loans are a type of compound interest loan where regular payments are made to reduce the value of the loan and interest is accrued on the remaining amount each time period. This reduces the amount of interest accrued each time period and allows the loan to be paid off over a fixed period of time.
This can be represented using the recurrence relation:
\[V_0 = a,\quad\quad V_{n + 1} = RV_n - D\]
Where:
- \(V_0\) is the principal amount
- \(a\) is the constant (representing the principal value)
- \(V_{n + 1}\) is the future value after \(n + 1\) time periods
- \(R\) is the growth factor
- \(V_n\) is the value at the nth time period
- \(n\) is the number of payment and compounding periods
- \(D\) is the payment amount
Worked Example
A car is purchased for \(\$27\ 000\) with a reducing balance loan, which accrues interest at a rate of \(17\%\text{ p.a.}\) compounding quarterly. Quarterly payments of \($1\ 500\) are made. Calculate the value of the loan at the end of each quarter over a period of a year.
Determine the given variables
\[\begin{align}&V_0 = 27000 \\ &r = \frac{17}{4} = 4.25 \\ &n = 4 \\ &D = 1500\end{align}\]
Calculate the growth factor
\[\begin{align}R &= 1 + \frac{4.25}{100} \\ &= 1.0425\end{align}\]
Construct the recurrence relation
\[V_0 = 27000,\quad\quad V_{n+ 1} = 1.0425V_n - 1500\]
Calculate the values each period
\[\begin{align}V_1 &= 1.0425(27000) - 1500 = 26647.5 \\ V_2 &= 1.0425\left(26647.5\right) - 1500 = 26280.01875 = 26280.02 \\ V_3 &= 1.0425\left(26280.01875\right) - 1500 = 25896.919\ldots = 25896.92 \\ V_4 &= 1.0425\left(25896.919\ldots\right) - 1500 = 25497.538\ldots = 25497.54\end{align}\]
The value of the loan at the end of the first quarter is \($26\ 647.50\), at the end of the second quarter is \($26\ 280.02\), at the end of the third quarter is \($25\ 896.92\), and at the end of the fourth quarter (first year) is \($25\ 497.54\).
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Use these tabs to view the worked solutions for the question above.
Calculations must be taken at each time period. First, determine \(V_0\) and \(r\) from the question.
\[\begin{align}&V_0 = \$42\ 000 \\ &r = \frac{6.2}{12}\end{align}\]
Next, determine the growth factor \(R\)
\[R = 1 + \frac{r}{100} = 1 + \frac{6.2/12}{100}\]
Now applying this to the recurrence relation \(V_{n + 1} = RV_n - D\):
\[\begin{align} &V_1 = \left(1 + \frac{6.2/12}{100}\right)\left(42000\right) - 600 = \$41\ 617 \\ &V_2 = \left(1 + \frac{6.2/12}{100}\right)\left(41617\right) - 600 = \$41\ 232.02 \\ &V_3 = \left(1 + \frac{6.2/12}{100}\right)\left(41232.02\right) - 600 = \$40\ 845.0 \end{align}\]
Annuities
Annuities are a type of compound interest investment where regular payments are taken out from it to provide income over a fixed period. Interest is accrued on the remaining amount in the annuity each time period. Annuities are calculated exactly the same way as a reducing balance loan. However, the key difference is that in an annuity, the investor is effectively lending money to the institution, which returns it with interest over time, whereas in a loan, the borrower is repaying money borrowed from a lender.
Worked Example
\(\$12\ 000\) is invested into an account that earns an interest rate of \(5\%\text{ p.a.}\) compounding yearly. \(\$1\ 000\) is withdrawn at the end of each year. Calculate the value of the investment at the end of \(4\) years.
Determine the given variables
\[\begin{align}&V_0 = 12000 \\ &r = 5 \\ &n = 4 \\ &D = 1000\end{align}\]
Calculate the growth factor
\[\begin{align}R &= 1 + \frac{5}{100} \\ &=1.05\end{align}\]
Construct the recurrence relation
\[V_0 = 12000,\quad\quad V_{n + 1} = 1.05V_n - 1000\]
Calculate the values each period
\[\begin{align}&V_1 = 1.05(12000) - 1000 = 11600 \\ &V_2 = 1.05\left(11600\right) - 1000 = 11180 \\ &V_3 = 1.05(11180) - 1000 = 10739 \\ &V_4 = 1.05(10739) - 1000 = 10275.95\end{align}\]
The value of the investment at the end of the fourth year is \(\$10\ 275.95\).
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Interest Only Loans
Interest only loans are a type of compound interest loan where regular payments are made equal to the interest accrued on the loan. This keeps the value of the loan the same as the principal amount for the entire period of the loan. Hence the name “interest only” as only the interest is being paid. The principal amount is often paid out in full at a later date. This is calculated the same way as reducing balance loans and annuities, except that the payments are equal to the interest accrued:
\[D = I = \frac{r}{100}\ \times\ V_0\]
Where:
- \(V_0\) is the principal amount
- \(R\) is the growth factor
- \(D\) is the payment amount
Note\(V_0\) is used as \(V_n = V_0\). The payment amount remains the same each period as the principal amount remains the same. |
Worked Example
A \(\$7\ 000\) interest only loan is taken that accrues interest at a rate of \(12.6\%\text{ p.a.}\) compounding quarterly. Calculate the amount of the quarterly repayments and determine how much interest would be paid over a period of a year.
Determine the given variables
\[\begin{align}&V_0 = 7000 \\ &r = \frac{12.6}{4} = 3.15\end{align}\]
Calculate the payment amount
\[\begin{align}D &= \frac{3.15}{100}\ \times\ 7000 \\ &= 220.5\end{align}\]
The quarterly repayment for the interest only loan is \(\$220.50\).
Calculate the yearly total interest amount – there are four quarters in a year
\[\text{total interest} = 4\ \times\ 220.50 = $882\]
This is the same as calculating the yearly interest amount as compounding does not have any effect on interest only loans.
\[\text{total interest} = \frac{12.6}{100}\ \times\ 7000 = $882\]
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Perpetuities
Perpetuities are a type of compound interest investment where regular payments are taken out equal to the interest accrued on the investment. This keeps the value of the investment the same as the principal amount for the entire period of the investment. The principal amount is not usually paid out, hence the name “perpetuity ” as this investment provides perpetual payments. Perpetuities are calculated exactly the same way as interest only loans.
Worked Example
\(\$40\ 000\) is invested in a perpetuity that accrues interest at a rate of \(2.85\%\text{ p.a.}\) compounding yearly. Calculate the amount of the yearly payments.
Determine the given variables
\[\begin{align}&V_0 = 40000 \\ &r = 2.85\end{align}\]
Calculate the payment amount
\[\begin{align} D&= \frac{2.85}{100}\ \times\ 40000 \\ &=1140\end{align}\]
The amount for the yearly payments from the perpetuity is \(\$1140\).
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Annuity Investments
Annuity investments are a type of compound interest investment where regular payments are made to the account. This further increases the amount of interest accrued as it is also calculated on the additional payments made.
This can be represented using the recurrence relation:
\[V_0 = a,\quad\quad V_{n + 1} = RV_n + D\]
Worked Example
\(\$23\ 000\) is made into an annuity account that earns an interest rate of \(2.6\%\text{ p.a.}\) compounding yearly. Additional payments of \(\$1500\) are made to the investment at the end of each year. Calculate the value of the investment at the end of \(4\) years.
Determine the given variables
\[\begin{align}&V_0 = 23000 \\ &r = 2.6 \\ &n = 4 \\ &D = 1500\end{align}\]
Calculate the growth factor
\[\begin{align}R &= 1 + \frac{2.6}{100} \\ &=1.026\end{align}\]
Construct the recurrence relation
\[V_0 = 23000,\quad\quad V_{n + 1} = 1.026V_n + 1500\]
Calculate the values each period
\[\begin{align}&V_1 = 1.026(23000) + 1500 = 25098 \\ &V_2 = 1.026\left(25098\right) + 1500 = 27250.548 = 27250.55 \\ &V_3 = 1.026\left(27250.548\right) + 1500 = 29459.062\ldots = 29459.06 \\ &V_4 = 1.026\left(29459.062\ldots\right) + 1500 = 31724.997\ldots = 31725.00\end{align}\]
The value of the investment at the end of the fourth year is \(\$31\,725.00\).
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Amortisation Tables
Amortisation tables detail the payments and interest over the life of a loan or investment. They break down each component of a payment, including the total amount, interest portion, principal portion, and remaining balance. The headings of the table vary depending on the type of loan or investment to match their respective financial structures.
| Payment number \((n + 1)\) | Payments made or received \((D)\) | Interest accrued \(\left(I = \frac{r}{100}\ \times\ V_n\right)\) | Principal reduction or increase \(\left(P = D \pm I\right)\) | Balance of loan or investment \((V_{n + 1} = V_n \pm P)\) |
|---|---|---|---|---|
| 0 | - | - | - | \(V_0\) |
| 1 | ||||
| 2 | ||||
| \(\ldots\) |
Each row corresponds to a single step in the recurrence relation and can be filled out according to the type of financial structure.
Worked Example
\(\$23\ 000\) is made into an annuity account that earns an interest rate of \(2.6\%\text{ p.a.}\) compounding yearly. Additional payments of \(\$1500\) are made to the investment at the end of each year. Create an amortisation table of the annuity over the first \(4\) years. Round all values to the two decimal places.
| Payment number \((n + 1)\) | Payments made | Interest accrued \(\left(I = \frac{r}{100}\ \times\ V_n\right)\) | Principal | Balance of \((V_{n + 1} = V_n + P)\) |
|---|---|---|---|---|
| 0 | - | - | - | \(23000\) |
| 1 | 1500 | \(\frac{2.6}{100}\ \times\ 2300 = 598\) | \(1500 + 598 = 2098\) | \(23000 + 2098 = 25098\) |
| 2 | 1500 | \(\frac{2.6}{100}\ \times\ 25098 = 652.55\) | \(1500 + 652.55 = 2152.55\) | \(25098 + 2152.55 = 27250.55\) |
| 3 | 1500 | \(\frac{2.6}{100}\ \times\ 27250.55 = 708.51\) | \(1500 + 708.51 = 2208.51\) | \(27250.55 + 2208.51 = 29459.06\) |
| 4 | 1500 | \(\frac{2.6}{100}\ \times\ 29459.06 = 765.94\) | \(1500 + 765.94 = 2265.94\) | \(29459.06 + 2265.94 = 31725.00\) |
Other calculations
Other values for these applications may also be determined such as calculating the:
- Future value of an investment \((FV)\)
- Initial principal needed \((PV)\)
- Interest rate required \((I\%)\)
- Payment amount required \((PMT)\)
- Number of time periods (\(n)\)
- Total payments made\( \left(PMT_{\text{total}} = PMT\ \times\ n\right)\)
The total amount of payments made is the payments multiplied by the number of payment periods. This is also often adjusted for the final payment – in which case the total payments can be calculated using \(PMT\ \times\ \left(n - 1\right) + PMT_{\text{final}}\) - Total interest paid \(\left(I_{total} = PMT_{\text{total}} - PV\right)\)
The principal value subtracted from the total payments will give you the additional amount that was paid which is the total interest amount
Some of these calculations are more complex (1 – 5) and require the use of a calculator or financial solver.
Additionally, final payments are generally not equal to exactly the regular payment amount. This is due to the context of the problem or rounding that is required. This additional amount owed or in excess is usually made as an adjustment to the final amount.