Borrowing money from the bank to buy a car or a house usually means getting a reducing balance loan. The most common reducible balance loan type is a mortgage. This is when we borrow money to buy a property.
This is the sort of loan where we are charged compound interest as the fee for borrowing money, and at regular time periods, usually months, we are required to make repayments to the financial institution to slowly pay off the loan. The repayments are usually all the same amount, except for the last repayment which may be less than the full amount - since payments are generally rounded up we see the cumulative affect of small additional payments.
When we make each repayment the amount owed (the balance) reduces, hence the name for this type of loan.
As well as calculating the length of time it takes to pay off a loan, we are often interested in the total amount paid on the loan and the total amount of interest paid to the bank.
Analysing a reducing balance loan with a table
Worked example
example 1
Mr and Mrs Roberts take out a mortgage to purchase a house. They borrow $\$550000$$550000 from a bank that charges them $5.8%$5.8% interest, compounded monthly. At the end of each month the Roberts make a repayment of $\$3500$$3500.
We can represent this situation using a table of values or a spreadsheet. The first and last months of the loan are shown below.
Note the calculation used to create this spreadsheet
- In cell B3, opening balance = closing balance for previous month
- In cell C2, monthly interest = annual rate/12 * opening balance
- In cell D2, repayment = fixed amount (except for last month)
- In cell E2, closing balance = opening balance + interest - repayment
(a) Calculate the value of $X$X in the table.
Think: $X$X is situated in the interest column, so we need to find the interest on the amount at the start of month $2$2. We must first divide the interest rate by $12$12 as interest is calculated monthly.
Do:
| $X$X | $=$= | $\frac{0.058}{12}\times549158.33$0.05812×549158.33 |
| $=$= | $2654.26$2654.26 |
Reflect: Notice the interest to be paid slowly reduces as the balance is reduced.
(b) Calculate the value of $Y$Y in the table.
Think: Usually finding the opening balance is easy because it's the same as the closing balance from the previous time period. But we can't see it in this case. So we'll need to work backwards by writing an equation and solving it.
Do: Using the equation for the closing balance we have:
| opening balance + interest - payment | $=$= | closing balance |
| $Y+56.59-3500$Y+56.59−3500 | $=$= | $8908.81$8908.81 |
| $Y$Y | $=$= | $12349.12$12349.12 |
(c) Calculate the value of $Z$Z in the table.
Think: This is the month where the Roberts will crack open a bottle of champagne since they've finally paid off their loan! Their final repayment will be less than their usual amount. It will be equal to the amount owing at the end of the second last month plus the interest on the amount owing.
Do:
| $Z$Z | $=$= | $1978.22+\frac{0.058}{12}\times1978.22$1978.22+0.05812×1978.22 |
| $=$= | $1987.78$1987.78 |
(d) Calculate the total repayments made by the Roberts.
Think: The Roberts have made $295$295 repayments each of $\$3500$$3500 plus the final repayment of $\$1987.78$$1987.78
Do:
| Total repayments | $=$= | $295\times\$3500+\$1987.78$295×$3500+$1987.78 |
| $=$= | $\$1034487.78$$1034487.78 |
(e) Hence, determine the total interest the Roberts paid for the loan.
Think: The total interest paid on the loan is the total repayments minus the amount borrowed.
Do:
| Interest paid | $=$= | Total repayments - Loan |
|
| $=$= | $\$1034487.78-\$550000$$1034487.78−$550000 |
|
|
| $=$= | $\$484487.78$$484487.78 |
|
So you can see that they nearly paid double the purchase price over the $296$296 months, or $25$25 years.
Reflect: How could they pay off their house sooner?
They could try to do one or both of the following:
- Halve the monthly payment and make this new payment fortnightly. This saves a significant amount as they will effectively be making an extra month's payment per year and will pay the loan off sooner. They would end up making an extra repayment each year since there are $26$26 fortnights in a year and$\frac{26}{2}=13$262=13 so they would have paid for $13$13 months instead of $12$12.
- Increase the amount they repay. Intuitively, if you make larger repayments each month or fortnight, you'll be able to pay off your loan sooner, saving you a lot of interest and thus helping you pay less overall.
Total amount paid = total of full repayments + final adjusted repayment
Total interest paid = total amount paid - initial amount borrowed
Practice question
Question 1
Ivan takes out a car loan for $\$24000$$24000. He is charged $8.1%$8.1% per annum interest, compounded monthly. Ivan makes repayments of $\$450$$450 at the end of each month.
Complete the values in the empty cells in the table below. Give your answers correct to the nearest cent.
Month Opening Balance Interest Repayment Closing Balance 1 $24000$24000 $162.00$162.00 $450$450 $23712.00$23712.00 2 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 3 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
Modelling a reducing balance loan with a recurrence relation
A reducing balance loan uses compound interest as well as making regular repayments, this mirrors the situations of an investment with regular withdrawals (present value annuity). Instead of having an initial investment and then regular withdrawals, we have an initial loan with regular repayments. The sequence forming the balance of a reducing balance loan is the same as the one we encountered for present value annuities.
For a principal loan, $P$P, at the compound interest rate of $i$i per period and a payment of $d$d per period, the sequence of the value of the loan over time forms a first order linear recurrence.
The sequence which generates the value, $A_n$An, of the loan at the end of each instalment period is:
$A_{n+1}=rA_{n-1}-d$An+1=rAn−1−d, where $A_0=P$A0=P and $r=1+i$r=1+i
Worked example
Example 2
Tim is starting up his own small business. He has saved $\$15\ 000$$15 000 to buy equipment and borrows another $\$50\ 000$$50 000 from the bank. He is charged interest at a rate of $4.5%$4.5% per annum, compounded monthly, and he makes monthly repayments of $\$400$$400.
(a) How much does Tim owe at the end of the first month?
Think: Starting from the amount Tim has borrowed, add the interest on and then take away his repayment. Remember the annual rate of $4.5%$4.5% must be divided by $12$12 to get a monthly rate of $0.375%$0.375%, or $0.00375$0.00375.
Do:
| Amount owing | $=$= | Opening balance + interest - payment |
| $=$= | $50\ 000+50\ 000\times0.00375-400$50 000+50 000×0.00375−400 | |
| $=$= | $49\ 787.50$49 787.50 |
(b) Write a recurrence relation which gives the balance $B_{n+1}$Bn+1 in terms of $B_n$Bn, and an initial condition $B_0$B0.
Think: Each month there is an increase due to interest (which can be done with a multiplication) and then a decrease due to the repayment. Using the sequence which generates the balance, $B_n$Bn, of the loan at the end of each instalment $B_{n+1}=rB_n-d$Bn+1=rBn−d, where $B_0=P$B0=P. We have an initial loan $B_0=50\ 000$B0=50 000, a loan rate of $i=0.375%$i=0.375% per month and a payment of $d=400$d=400.
Do:
$B_n=1.00375\times B_{n-1}-400$Bn=1.00375×Bn−1−400 ; $B_0=50\ 000$B0=50 000
(c) Determine how many months it will take Tim to pay off the loan.
Think: To do this we will need to use our recursive rule to create a table of values and then scroll through our table of values to find the first month in which the balance becomes negative.
Do: Create a column for the month number and another for the balance of the account, enter the following formulas and drag down to generate the sequence:
| A | B | |
|---|---|---|
| 1 | $\text{Month}$Month | $B_n$Bn |
| 2 | $0$0 | $\$50\ 000$$50 000 |
| 3 | =A2+1 | =1.00375*B2-400 |
| 4 |
Scrolling through your created table you should see the following:
| $n$n | $B_n$Bn |
|---|---|
| $166$166 | $\$1186.61$$1186.61 |
| $167$167 | $\$791.06$$791.06 |
| $168$168 | $\$394.03$$394.03 |
| $169$169 | $-\$4.49$−$4.49 |
So we can see that Tim pays off the loan in month $169$169.
(d) Calculate the amount of his final repayment.
Think: The amount of the final repayment will be equal to the remaining amount owing in month $168$168, plus the interest generated that month.
Do:
| Final repayment | $=$= | $394.03+0.00375\times394.03$394.03+0.00375×394.03 |
| $=$= | $395.51$395.51 |
So the value of the final repayment is $\$395.51$$395.51.
(e) Hence or otherwise, determine the total amount Tim paid for the equipment.
Think: The equipment cost Tim an initial $\$15000$$15000 plus the cost of the loan. The cost of the loan was $168$168 repayments of $\$400$$400 each, plus one final repayment of $\$395.51$$395.51.
Do:
| Total amount paid | $=$= | $15000+168\times400+395.51$15000+168×400+395.51 |
| $=$= | $82595.51$82595.51 |
So the total amount Tim paid is $\$82595.51$$82595.51.
Practice question
Question 2
Bart borrows $\$61000$$61000 from a banking institution. He is charged $6.6%$6.6% per annum interest, compounded monthly. At the beginning of each month, before interest is charged, he makes a repayment of $\$400$$400.
Fill in the missing values in the table. Give all values correct to the nearest cent and use your rounded answers for all subsequent calculations in the table.
Month Opening Balance Repayment Interest Closing Balance 1 $61000$61000 $400$400 $333.30$333.30 $[60933.29\le60933.3\le60933.31]$[60933.29≤60933.3≤60933.31] 2 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 3 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 4 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Write a recursive rule that gives the closing balance, $B_n$Bn, at the end of month $n$n.
Write both parts of the rule (including for $B_0$B0) on the same line, separated by a comma.
Use your calculator to determine how much is owing on the loan after $4$4 years.
Give your answer to the nearest cent.
At the end of which year and month will the loan have been repaid?
$\editable{}$ year(s) $\editable{}$ month(s)
Formulas for finding the balance and payments
As the sequence of a reduce balance loan mirrors that of a present value annuity we can also apply the formulas from that section to find the balance of a loan after a given number of repayments or the minimum repayments required to pay off a loan in a given time frame.
The balance of a reducing balance loan:
$A=P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=P(1+i)n−M((1+i)n−1i)
The minimum repayment required for a loan of $P$P:
$M=\frac{Pi}{1-\left(1+i\right)^{-n}}$M=Pi1−(1+i)−n
Where:
- $A$A is the balance of the loan after $n$n repayments
- $P$P is the principal amount of the loan
- $M$M is the repayment per period
- $i$i is the interest rate per period
- $n$n is the total number of payment periods
For this formula we require payments per year and compounds per year to be equal.
Worked example
Example 3
Royston borrows $\$350\ 000$$350 000 to purchase a home at a rate of $5.4%$5.4% per annum compounded monthly over $20$20 years.
(a) Calculate the minimum monthly repayments.
Think: Use the formula to find the minimum repayment with $P=\$350\ 000$P=$350 000, $i=\frac{0.054}{12}=0.0045$i=0.05412=0.0045 and $n=12\times20=240$n=12×20=240.
Do:
| $M$M | $=$= | $\frac{Pi}{1-\left(1+i\right)^{-n}}$Pi1−(1+i)−n |
| $=$= | $\frac{350\ 000\times0.0045}{1-\left(1+0.0045\right)^{-240}}$350 000×0.00451−(1+0.0045)−240 | |
| $=$= | $\$2387.89$$2387.89 |
Hence, Royston needs to repay a minimum of $\$2387.89$$2387.89 each month.
Reflect: The payment has been rounded up to to the nearest cent ensure the loan is paid off.
(b) Find the balance of the loan after $4$4 months.
Think: Using our recursive rule we can see each step we multiply the previous value by $1.0045$1.0045 and subtract $\$2387.89$$2387.89. As this is not a large number of repeated calculations we can find this value using the answer key on the calculator to find the balance at the end of each month for four months.
Do:
- Enter the principal loan amount $350\ 000$350 000 and press enter (
or equivalent). This is the starting balance. - Type in the next step using the answer key, in this case $1.0045\times$1.0045×
$-2387.89$−2387.89. This is the balance after $1$1 month. - Continue to press to obtain the next term in the sequence. One press: balance after $2$2 months, Two presses: balance after $3$3 months, $\ldots$….
The balance after $4$4 months is $\$346\ 726.43$$346 726.43
(c) Find the balance of the loan after $5$5 years.
Think: $5$5 years is $60$60 payment periods. As this is a significant number of payment periods and keeping track using the answer key becomes difficult and time consuming, let's use the formula to find a future value.
Do:
| $A$A | $=$= | $P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{i}\right)$P(1+i)n−M((1+i)n−1i) |
| $=$= | $350\ 000\left(1+0.0045\right)^{60}-2387.89\left(\frac{\left(1+0.0045\right)^{60}-1}{0.0045}\right)$350 000(1+0.0045)60−2387.89((1+0.0045)60−10.0045) | |
| $=$= | $\$294\ 150.62$$294 150.62 |
Royston would still owe $\$294\ 150.62$$294 150.62 on his loan after five years.
(d) Find the interest charged in over the term of the loan. (Ignoring final payment adjustment.)
Think: If we ignore the adjustment on the final payment we assume we have $240$240 equal repayments of $\$2387.89$$2387.89. The interest is the difference between the total amount paid and the initial loan.
Do:
| $\text{Interest}$Interest | $=$= | $\text{Total paid}-\text{Initial loan}$Total paid−Initial loan |
| $=$= | $\$2387.89\times240-\$350\ 000$$2387.89×240−$350 000 | |
| $=$= | $\$223\ 093.60$$223 093.60 |
A total of approximately $\$223\ 093.60$$223 093.60 was paid in interest.
Reflect: The payment was rounded up so the interest paid is a slight over estimate as the final payment would have been adjusted to account for the accumulated effect of the slightly larger repayment.
Practice question
Question 3
A young couple take out a personal loan for $\$28500$$28500 with reducible interest of $9.6%$9.6% p.a. calculated monthly. Repayments of $\$750$$750 per month are made. The table below charts the progress of the loan for the first four months.
| Month | Amount owing at beginning of month ($\$$$) | Interest ($\$$$) | Repayment ($\$$$) | Amount owing at end of month ($\$$$) |
|---|---|---|---|---|
| $1$1 | $28500$28500 | $228$228 | $750$750 | $27978$27978 |
| $2$2 | $27978$27978 | $223.82$223.82 | $750$750 | $27451.82$27451.82 |
| $3$3 | $27451.82$27451.82 | $219.61$219.61 | $750$750 | $26921.44$26921.44 |
| $4$4 | $26921.44$26921.44 | $X$X | $750$750 | $Y$Y |
Complete the recursive rule that describes this situation, where $A_n$An is the amount owing after the $n$nth month.
$A_{n+1}=$An+1=$\editable{}$$\times A_n-$×An−$\editable{}$, $A_0=$A0=$\editable{}$
Calculate the values of $X$X and $Y$Y, to the nearest cent.
$X=$X=$\editable{}$
$Y=$Y=$\editable{}$
Calculate how many whole months it takes to pay off the loan.
Calculate the total amount of interest paid during the loan.
Round your answer to the nearest cent.
If the couple increase their payments to $\$1000$$1000 per month, how many whole months will it take until the loan is paid off?
If the couple wishes to pay off the loan in $4$4 years exactly, what monthly payment should they make?
Round your answer to the nearest cent.
Reducing balance loans using a financial solver
Often it's more convenient to analyse various situations for a reducible balance loan using an online financial solver. Remember that the present value will be the value of the loan and is entered as a positive value, as the lender is receiving money from the bank. The payments will be negative as these are paid to the bank.
Worked example
example 4
Audrey takes out a car loan for $\$24000$$24000. The finance company charge her $8.5%$8.5% interest compounded monthly.
How much should Audrey repay each month if she wants to repay this loan in $5$5 years?
Using a financial solver we enter the details as follows:
| Compound Interest | ||
|---|---|---|
| N | $60$60 | There are $12$12 months per year, and $5$5 years. |
| I% | $8.5$8.5 | The annual interest rate |
| PV | $24000$24000 | PV is positive as the bank is giving Audrey money |
| PMT | This is the value we are trying to find. | |
| FV | $0$0 | The future balance will be zero. |
| P/Y | $12$12 | $12$12 payments per year |
| C/Y | $12$12 | Compounding monthly |
Solving we get: PMT = $-492.40$−492.40
So Audrey should pay $\$492.40$$492.40 each month.
Note: PMT displays as a negative number since the payments will be made from Audrey to the financial institution.
Final payment using a financial solver
Worked example
Example 5
Audrey takes out a car loan for $\$24000$$24000. The finance company charge her $8.5%$8.5% interest compounded monthly. Audrey decides she can only afford to pay $\$480.00$$480.00 each month which means it will take her a little more than $5$5 years to pay off the loan. Calculate the amount of Audrey's final payment on the loan.
There are two methods in which a final payment can be calculated.
Method 1:
Think: Calculate the amount owing for the final month and increase this by the monthly interest rate.
Do: First find the number of whole payments it will take.
| Compound Interest | ||
|---|---|---|
| N | Number of payments is unknown |
|
| I% | $8.5$8.5 | The annual interest rate |
| PV | $24000$24000 | PV is positive as the bank is giving Audrey money |
| PMT | $-480$−480 | PMT is negative as Audrey is giving money to the bank |
| FV | $0$0 | The future balance will be zero. |
| P/Y | $12$12 | $12$12 payments per year |
| C/Y | $12$12 | Compounding monthly |
Solving for N we find the number of months to pay the loan is $61.94$61.94. therefore it will take $61$61 full payments of $\$480.00$$480.00 plus a smaller final part payment.
Now find the balance remaining after $61$61 whole payments. Put N = $61$61 and solve for FV. We find that FV = $-449.38$−449.38.
Add the monthly interest to this final balance:
| Final payment | $=$= | $449.38+449.38\times\frac{0.085}{12}$449.38+449.38×0.08512 |
| $=$= | $452.56$452.56 |
The final payment is $\$452.56$$452.56.
Method 2:
Think: Calculate the FV for the month in which the loan will be paid. This will be an overpayment. Subtract the overpayment amount from the usual monthly payment.
Do: Calculate the FV balance for N = $62$62 months. The FV is $27.44$27.44.
Subtracting $27.44$27.44 from the usual payment gives:
| Final payment | $=$= | $480-27.44$480−27.44 |
| $=$= | $452.56$452.56 |
The final payment is $\$452.56$$452.56.
Practice question
Question 4
Mr and Mrs Derek held a mortgage for $25$25 years. Over that time they made monthly repayments of $\$5500$$5500 and were charge a fixed interest rate of $4.4%$4.4% per annum, compounded monthly.
We will use a financial solver to determine how much they initially borrowed.
Which variable on the solver do we want to solve for?
PmtAFVBNCPVDI%EFill in the value for each of the following:
N:$\editable{}$I%:$\editable{}$Pmt:$\editable{}$FV:$\editable{}$PpY:$\editable{}$CpY:$\editable{}$Hence, state how much Mr. and Mrs. Derek initially borrowed, correct to the nearest dollar.
What if number of payments $\ne$≠ number of compounding periods?
In real life, banks usually calculate interest on loan accounts monthly but people can choose to make fortnightly or even weekly repayments.
When the number of payments is not equal to the number of compounding periods the financial solvers are great tool.
N is the total number of payments $=\text{payments per year}\times\text{number of years}$=payments per year×number of years
P/Y is number of payments per year
C/Y is the number of times interest is calculated per year
We can also use the online TVM solver to explore these problems.
Practice question
Question 5
Joanne borrows $\$345000$$345000 to buy an apartment. The bank offers a reducing balance loan with an interest rate of $2.35%$2.35% p.a. compounded monthly. Joanne opts to make fortnightly payments of $\$1250$$1250 in order to pay off the loan. Answer the following questions.
Assume there are $26$26 fortnights in a year.
What is the balance, in dollars, after $100$100 weeks?
Round your answer to the nearest cent.
Approximate how long it takes her to pay off the loan in years.
Round your answer to two decimal places.