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6. Reducing Balance Loans and Annuities
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VCE 12 General 2023

6.01 Reducing balance loans

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Introduction

A reducing balance loan is the type of loan where compound interest is charged as a fee for borrowing money, and repayments are made at regular time periods to the financial institution to slowly pay off the loan.

As repayments are made each time period, the amount that is owed (the balance) reduces, leading to the name "reducing balance loan". The most common reducing balance loan type is a mortgage. This is when money is borrowed to buy a property.

Reducing balance loan with an amortisation table

An amortisation table displays how a loan or investment changes over time on a step by step basis. It shows the payment number, the interest, the principal reduction and the balance of the loan after a payment has been made. The following steps are made with each payment:

  1. Interest needs to be calculated: compound interest rate per payment times the previous balance

  2. Reduction in principal: payment made minus the interest

  3. Balance of the loan: balance owing minus the reduction in principal

Examples

Example 1

Mr. and Mrs Roberts take out a mortgage to purchase a house. They borrow \$550\,000 from a bank that charges them 5.8\% interest, compounded monthly. At the end of each month the Roberts' make a repayment of \$3500. This is represented in the amortisation table below. All values are given in dollars.

PeriodPaymentInterest PaidReduction in PrincipalBalance of Loan
0000550\,000
135000.004\,833 \times 550\,000=2658.333500-2658.33=841.67550\,000−841.67=549\,158.33
23500 845.73548\,312.60
335002650.18849.82547\,462.78
\ldots\ldots\ldots\ldots\ldots
\ldots\ldots\ldots\ldots
293350059.693440.318908.81
294350043.063456.945451.87
295350026.353473.651978.22
296 1978.220

The row for period 1 shows how each value has been calculated.

a

Calculate the interest for period 2 in the table.

Worked Solution
Create a strategy

Multiply the previous balance by the compound interest rate per payment.

Apply the idea

We can find it using a similar calculation to period 1:

\displaystyle \text{Interest}\displaystyle =\displaystyle 549\,158.33 \times \dfrac{0.058}{12}Use the previous balance, \$549\,158.33
\displaystyle =\displaystyle \$2654.26Evaluate using a calculator
b

Calculate the balance of the loan immediately before period 293 in the table.

Worked Solution
Create a strategy

With the same period, 293, add the loan balance and reduction in principal.

Apply the idea

We can see that the balance at the end of period 293 is \$8908.81, and the total reduction in principal during this period is \$3440.31. So we have that:

\displaystyle \text{Previous loan balance}\displaystyle =\displaystyle 8908.81+3440.31Substitute the values
\displaystyle =\displaystyle \$12\,349.12Evaluate

So the balance at the end of period 292 is \$12\,349.12.

c

Calculate the repayment made in the final period in the table.

Worked Solution
Create a strategy

Add the previous loan balance to the interest generated that month.

Apply the idea

The amount left to pay at the end of the second last period (period 295) is \$1978.22.

\displaystyle \text{Final repayment}\displaystyle =\displaystyle 1978.22+\dfrac{0.058}{12}\times 1978.22
\displaystyle =\displaystyle \$1987.78
d

Calculate the total repayments made by the Roberts'.

Worked Solution
Create a strategy

Multiply the loan repayment by the number of repayments, then add the final repayment.

Apply the idea

The Roberts' have made 295 repayments each of \$3500 plus the final repayment of \$1978.78.

\displaystyle \text{Total repayment}\displaystyle =\displaystyle 295\times 3500+1987.78
\displaystyle =\displaystyle \$1\,034\,487.78
e

Determine the total interest that the Roberts' paid over the course of the loan.

Worked Solution
Create a strategy

Subtract the amount borrowed from the total amount repaid.

Apply the idea

We know from part (d) that the total repayment is \$1\,034\,487.78, and the amount he borrowed is \$550\,000. This means that:

\displaystyle \text{Total interest paid}\displaystyle =\displaystyle 1\,034\,487.78-550\,000
\displaystyle =\displaystyle \$484\,487.78

So the Roberts' paid a total of \$484\,487.78 in interest. This is almost equal to the initial value of the loan - the Roberts' paid nearly double the purchase price in total over the 296 months (25 years).

Reflect and check

Unless their house doubles in value over the 25 years, the Roberts' have paid more for their house than what it's worth.

To help mitigate this, they could try to do a combination of the following:

  • Make fortnightly repayments. By making more regular repayments, they would give the loan less time to accumulate interest and so they would end up paying back less interest in total.

  • Increase the amount they repay. Intuitively, by making larger repayments each time period the Roberts' will be able to pay off the loan sooner, which also leads to less interest being accumulated (as well as finishing paying off the loan sooner).

Idea summary

An amortisation table displays how a loan or investment changes over time on a step by step basis.

The following steps are made with each payment:

  1. Interest needs to be calculated: compound interest rate per payment times the previous balance

  2. Reduction in principal: payment made minus the interest

  3. Balance of the loan: balance owing minus the reduction in principal

Reducing balance loan with a recurrence relation

Since a reducing balance loan uses compound interest as well as making regular repayments, this type of recurrence relation involves a combination of an arithmetic sequence and a geometric sequence.

Interest only loans

In some situations, banks will offer interest only loans. In this type of loan, the amount of the repayment exactly matches the amount of interest charged on the loan. This means that the balance of the loan will never decrease. If using the finance solver on a CAS calculator for an interest only loan, the FV and PV will be the same.

Interest on an reducing balance loan can be modelled using the following recurrence relation: V_{n+1}=RV_n-d,\,V_0=P where V_{n+1} is the value of the investment after n+1 time periods, R equals 1+\dfrac{r\%}{100}, usually expressed as a decimal, where r\% is the interest rate, d is the amount subtracted per time period, and P is the initial value of the investment or loan (the principal value).

Examples

Example 2

Tim is starting up his own small business. He has saved \$15\,000 to buy equipment and borrows another \$50\,000 from the bank. He is charged interest at a rate of 4.5\% per annum, compounded monthly, and he makes monthly repayments of \$400.

a

How much does Tim owe at the end of the first month?

Worked Solution
Create a strategy

Subtract the repayment from the opening balance, then add the interest.

Apply the idea

To find the interest, multiply the initial amount borrowed by the monthly interest rate. To find the rate per month, we divide the annual rate of 4.5\%=0.045 by 12.

\displaystyle \text{Interest}\displaystyle =\displaystyle 50\,000\times \left(\dfrac{0.045}{12}\right)
\displaystyle =\displaystyle \$187.50
\displaystyle \text{Amount owing}\displaystyle =\displaystyle 50\,000-400+187.50Substitute the values
\displaystyle =\displaystyle \$49\,787.50Evaluate

So at the end of the first month, Tim owes \$49\,787.50.

b

Write a recurrence relation which gives the balance B_{n+1} in terms of B_n, and an initial condition B_0.

Worked Solution
Create a strategy

Use the recursive rule V_{n+1}=R \times V_{n}-d, where V_{0} = P, and find R,\,d,\, and P.

Apply the idea

Each month the repayment is d=\$400. The initial amount borrowed was P=\$50\,000.

The balance is increased by 4.5\% =0.045 each year. To find the rate per month, we divide by 12 to get: R=1+\dfrac{0.045}{12}=1.003\,75. So the recursive rule becomes:

\displaystyle B_{n+1}\displaystyle =\displaystyle R\times B_{n} - d,\,B_{0}=PWrite the recursive rule
\displaystyle =\displaystyle 1.003\,75 \times B_{n} - 400,\,B_{0}=50\,000Substitute R, \,d, \,P
\displaystyle B_{n+1}\displaystyle =\displaystyle 1.003\,75B_{n} - 400,\,B_{0}=50\,000Simplify
c

Determine how many months it will take Tim to pay off the loan.

Worked Solution
Create a strategy

Enter the recursive rule and closing balance we found in part (b) into your calculator and find when the balance of the loan first drops below \$0.

Apply the idea

Using your calculator you should get that the value drops below 0 at N=169.

So Tim pays off the loan in month 169.

d

Calculate the amount of his final repayment.

Worked Solution
Create a strategy

Add the remaining amount owing to the interest generated that month.

Apply the idea

We know from part (c) that the remaining amount owing in month 168 is 394.03.\text{Final repayment}=394.03+\dfrac{0.045}{12}\times 394.03=\$395.51

e

Determine the total amount Tim paid for the equipment.

Worked Solution
Create a strategy

Multiply the loan repayment by the number of repayments, then add the final repayment and the starting money he saved.

Apply the idea

Tim has made 168 repayments of \$400 each, and one final repayment of \$395.51 on part (d).

\displaystyle \text{Total amount paid}\displaystyle =\displaystyle 168\times 400+395.51+15\,000Substitute the values
\displaystyle =\displaystyle \$82\,595.51Evaluate using a calculator
f

Graph the balance of the loan, B, against the number of months, n, for 0<n<168.

Worked Solution
Create a strategy

Enter the recursive rule and initial amount borrowed we found in part (b) into your calculator, then use the graph function on the calculator to graph the recurrence relation.

Apply the idea
A graphing calculator showing a decreasing concave down curve. Ask your teacher for more information.

We can see that the curve hits the horizontal axis shortly after month 160 which agrees with the finding that the loan was paid off in the 168 month. The curve hits the vertical axis at 50\,000, which agrees with the initial loan being for \$50\,000. And the values on the curve are decreasing over time which agrees with the balance of the loan decreasing over time as the loan gets paid off.

Example 3

Dylan takes out a loan to purchase a property. He makes equal monthly loan repayments of \$4600 over 27 years to pay it off. The interest of 8\% is compounded annually.

What is the total loan amount?

Worked Solution
Create a strategy

Use the formula: \text{Total loan amount}=\text{Number of repayments}\times \text{Loan repayments}

Apply the idea

The repayments occurred monthly, so 12 repayments will occur for every year that has passed, wherein the given states 27 years.

\displaystyle \text{Total loan amount}\displaystyle =\displaystyle 12\times 27\times 4600Subsitute the values
\displaystyle =\displaystyle \$1\,490\,400Evaluate

Example 4

Vincent takes out a loan for \$68\,000. He is charged 12\% per annum interest, compounded monthly. At the end of each month, he makes a repayment of \$750.

a

Fill in the missing values in the table. Give all values correct to the nearest cent and use your rounded answers for all subsequent calculations in the table.

MonthOpening balanceInterestRepaymentClosing balance
168\,00068075067\,930
2
3
4
Worked Solution
Create a strategy

We can complete the table by first finding the opening balance, the interest, then the repayment, and lastly the closing balance.

Apply the idea

Data for Month 2:

To find the opening balance, take the closing balance from the previous month:

\text{Opening balance} = \$67\,930

To find the interest, multiply the opening balance by the monthly interest rate. To find the rate per month, we divide the annual rate of 12\%=0.12 by 12.

\displaystyle \text{Interest}\displaystyle =\displaystyle 67\,930\times \left(\dfrac{0.12}{12}\right)
\displaystyle =\displaystyle \$679.30

The repayment is still the same every month.

\text{Repayment}=\$750

To find the closing balance, subtract the repayment from the opening balance, then add the interest.

\displaystyle \text{Closing Balance}\displaystyle =\displaystyle 67\,930-750+679.30
\displaystyle =\displaystyle \$67\,859.30

Follow the same process for months 3 and 4 to get the following table:

MonthOpening balanceInterestRepaymentClosing balance
168\,00068075067\,930
267\,930679.3075067\,859.30
367\,859.30 678.5975067\,787.89
467\,787.89 677.8875067\,715.77
b

Write a recursive rule that gives the closing balance, V_{n+1}, at the end of month n+1. Write both parts of the rule (including for V_0) on the same line, separated by a comma.

Worked Solution
Create a strategy

Use the recursive rule V_{n+1}=R \times V_{n}-d, where V_{0} = P, and find R,\,d,\, and P.

Apply the idea

Each month the repayment is d=\$750. The initial amount borrowed was P=\$68\,000.

The balance is increased by 12\% =0.12 each year. To find the rate per month, we divide by 12 to get: R=1+\dfrac{0.12}{12}=1.01. So the recursive rule becomes:

\displaystyle V_{n+1}\displaystyle =\displaystyle R \times V_{n} - d,\,V_{0}=PWrite the recursive rule
\displaystyle =\displaystyle 1.01 \times V_{n} - 750,\,V_{0}=68\,000Substitute R, \,d, \,P
\displaystyle V_{n+1}\displaystyle =\displaystyle 1.01V_{n} - 750,\,V_{0}=68\,000Simplify
c

Use the sequence facility on your calculator to determine how much is owing on the loan after 3 years. Give your answer to the nearest cent.

Worked Solution
Create a strategy

Enter the recursive rule and closing balance we found in part (b) into your calculator and look for the loan balance owed after 3 years.

Apply the idea

3 years is equal to 36 months. So when generating the values for the recursive rule in your calculator, look for the value when N=36.

Using your calculator you should get that the value for the 36th month is \$64\,984.62.

d

At the end of which year and month will the loan have been repaid?

Worked Solution
Create a strategy

Enter the recursive rule and closing balance we found in part (b) into your calculator and find when the balance of the loan first drops below \$0.

Apply the idea

Using your calculator you should get that the value drops below 0 at N=239. This means after 239 months. To find the number of years, we divide by 12.

\displaystyle 239\div 12\displaystyle \approx\displaystyle 19.91667\ldotsDivide the number of months by 12
\displaystyle =\displaystyle 19\dfrac{11}{12}Write as a mixed numeral

This tells us that 239 months is equal to 19 years and 11 months.

So at the end of 19 years and 11 months the loan will be repaid.

Idea summary

Interest on a reducing balance loan can be modelled using the following recurrence relation:

\displaystyle V_{n+1}=R V_n -d,\,V_0=P
\bm{V_{n+1}}
is the value of the investment or loan after n+1 time periods
\bm{R}
is 1+\dfrac{r\%}{100 }, usually expressed as a decimal, where r\% is the interest rate
\bm{d}
is the amount subtracted per time period
\bm{P}
is the initial value of the loan (the principal value)

Outcomes

U3.AoS2.3

the concepts of financial mathematics including simple and compound interest, nominal and effective interest rates, the present and future value of an investment, loan or asset, amortisation of a reducing balance loan or annuity and amortisation tables

U3.AoS2.8

use a table to investigate and analyse on a step–by-step basis the amortisation of a reducing balance loan or an annuity, and interpret amortisation tables

U3.AoS2.4

the use of first-order linear recurrence relations to model compound interest investments and loans, and the flat rate, unit cost and reducing balance methods for depreciating assets, reducing balance loans, annuities, perpetuities and annuity investments

U3.AoS2.9

use technology with financial mathematics capabilities, to solve practical problems associated with compound interest investments and loans, reducing balance loans, annuities and perpetuities, and annuity investments

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