1.7 Amortisation and annuities

Loans and how they are repaid

Borrowing with interest

A bank that lends money expects it back, with interest as payment for the use of the money over time. The amount borrowed is the loan's principal, and the interest charged each period is a percentage of the amount still owed at that time, which falls as the loan is repaid.

The conditions of a loan state three things: the interest rate, how often interest is charged (the compounding period), and how long the borrower has to repay. The length of time is fixed when the loan is agreed.

Interest rates on loans are quoted as a nominal annual interest rate, for example 6% a year compounded monthly. The rate for one month is then $\frac{6\%}{12} = 0.5\%$, and each month the lender adds 0.5% of the amount still owed.

What happens in one payment period

The borrower pays the same amount at the end of every period. Part of that payment covers the interest charged in the period; the rest reduces the debt.

What is still owed after a payment is the balance that the next period's interest is charged on.

If $B_k$ is the outstanding balance after $k$ payments, $i$ is the interest rate per period as a decimal and $PMT$ is the payment, then each period follows the same two steps: add the interest, then subtract the payment.

The multiplier $1 + i$ is the same one that appears in compound interest; the subtraction of $PMT$ is what makes the balance fall instead of grow. The part of the payment that reduces the debt is $PMT - iB_k$, the payment minus the interest.

Amortization

Repaying a loan by a fixed number of equal payments, each covering that period's interest and part of the principal, is called amortization. The payment is chosen so that the last payment brings the balance exactly to zero.

Because the balance falls with every payment, the interest charged falls too, and so a growing share of each fixed payment goes towards the principal. On a long loan such as a mortgage, the early payments are mostly interest and the late payments mostly principal. A table that tracks this period by period is the clearest way to see it.

The schedule shows the pattern of every amortized loan. The interest column falls, the principal column rises by the same amount, and their sum stays fixed at the payment. The principal repaid also grows by about 1% a month, because each month's saving in interest is added to the next month's repayment.

The TVM solver with regular payments

Building a schedule by hand works for six payments, but a mortgage has 300 or more. Finding the payment that brings the balance to exactly zero is done with the GDC's TVM solver, introduced in subtopic 1.4 for compound interest with $PMT = 0$. Loans and annuities use the same solver with $PMT \neq 0$.

The seven entries

The solver links seven quantities, including the present value and the future value from subtopic 1.4. Any one of the first five can be found when the other six are entered.

With $P/Y$ set, $N$ counts payments, not years. A 20-year loan repaid monthly has $N = 20 \times 12 = 240$. The compounding entry $C/Y$ is usually the same as $P/Y$, but not always: a loan can charge interest compounded annually while payments are monthly, and then $P/Y = 12$ and $C/Y = 1$.

On a TI-84 Plus CE the solver is under APPS → Finance → TVM Solver; on a Casio fx-CG50 it is MENU → Financial → COMPND. Both show the seven entries, and the unknown is found by placing the cursor on it and pressing ALPHA ENTER (TI-84) or choosing the entry to solve for from the function keys (Casio).

The sign convention for loans and annuities

The solver needs to know which way money flows. It uses the sign convention from subtopic 1.4: from the point of view of one person, money received is positive and money paid out is negative. For a loan, the borrower receives the loan at the start ($PV$ positive) and pays out every instalment ($PMT$ negative). For a savings plan, the saver pays in every instalment ($PMT$ negative) and receives the fund at the end ($FV$ positive).

The cash-flow diagrams for a loan and a savings plan show why the signs must differ. If $PV$ and $PMT$ are both entered as positive for a loan, the solver treats the borrower as receiving the loan and also receiving every payment. No interest rate makes that balance out, so the solver gives an error or a meaningless answer. The lender's point of view, with every sign reversed, gives the same numbers, so either view works as long as it is used consistently.

Payments at the end of each period

Payments in a loan or annuity of this kind are made at the end of each period: the first monthly payment is made one month after the loan is received, and the first deposit into a savings plan earns no interest until the following month. A series of equal payments like this has its own name.

When the payments come at the end of each period, the annuity is called ordinary. Calculators can also handle payments at the start of each period, which some leases and rents use; that arrangement is called an annuity due.

Finding the regular payment

To find the payment on a loan, enter everything that is known, with $FV = 0$ because the debt is cleared at the end, and solve for $PMT$.

The Revision Village lesson on loans sets out the same entries for several loans and shows the solver screen at each step. It also shows how the outstanding balance part-way through a loan is found by changing $N$.

When payments and compounding differ

Sometimes the interest is compounded less often than payments are made. Then $P/Y$ and $C/Y$ differ, and the solver converts the rate for each payment period itself. Entering the wrong $C/Y$ changes the answer, though often only slightly, which makes the error hard to spot.

Mortgages, deposits and the total cost of a loan

Mortgages and deposits

The largest loan most people take out is used to buy a home. The home itself is security for the loan: if the borrower stops paying, the lender can take and sell it.

Lenders rarely lend the whole price. The buyer pays part of the price at the start from their own money, and borrows the rest. The word deposit is also used for money paid into a bank account; for a loan it means the up-front part of a price.

A deposit is often given as a percentage of the price, and the loan is the price minus the deposit:

where $d\%$ is the deposit as a percentage of the price. A UK lender quoting a rate "at 75% loan-to-value" lends up to 75% of the price, so the buyer pays a deposit of at least 25%.

Total amount paid and total interest

Once the payment is known, the total cost follows from two multiplications and a subtraction. Every payment is the same (ignoring a small adjustment to the last one), so the total of the payments is the number of payments times the payment.

The two forms in the definition agree because the deposit is included on both sides: $(\text{deposit} + N \times PMT) - \text{price} = N \times PMT - (\text{price} - \text{deposit})$.

Source: Bank of England, quoted household interest rates, series IUMBV34 (2-year fixed, 75% LTV) and IUMBV42 (5-year fixed, 75% LTV), September 2026.

Where the money goes over the term

The schedule for the £250,000 mortgage has 300 rows, but the pattern of the six-month loan repeats on a larger scale. In the first month the interest is $250\,000 \times \frac{0.0513}{12} = £1068.75$, so only £411.72 of the £1480.47 payment reduces the debt. The bar chart of each year's payments shows how slowly that changes at first.

The yearly payments of the mortgage split into a shrinking interest part and a growing principal part. More than two-thirds of the first year's payments are interest, and the two parts only become equal in year 12. This is why borrowers who sell or repay early in a mortgage find that they still owe most of what they borrowed.

The outstanding balance part-way through a loan

Using the solver for the balance

Borrowers often need to know what they still owe: to repay a loan early, to sell a house, or to move to a new rate. The outstanding balance after $m$ payments is the future value of the loan after $m$ periods with the known payment. So the solver is used with $N = m$ and the unknown is $FV$.

The entries are the same as for the whole loan except two: $N$ is the number of payments made so far, and $PMT$ is now known while $FV$ is the unknown. The $FV$ found is negative from the borrower's point of view, because it is an amount still to be paid out.

The graph of the outstanding balance against time is a curve that falls slowly at first, while most of each payment is interest, and steeply near the end. The straight line drawn with it shows how the balance would fall if the same amount of principal, £10,000, were repaid every year.

The balance curve of the mortgage stays above the straight line throughout. Solving $FV = -125\,000$ for $N$ gives $N \approx 195$ payments, so half the debt is still owed after about 16.25 years, almost two-thirds of the way through the term.

Interest paid over part of a loan

The interest paid in the first $m$ payments is the part of those payments that did not reduce the debt. It is found in two steps: first the fall in the balance, then the payments minus that fall.

For the mortgage, the first 5 years' payments total $60 \times 1480.47 = £88\,828.20$, and the principal repaid is $250\,000 - 221\,907.46 = £28\,092.54$. The interest paid in the first 5 years is therefore $88\,828.20 - 28\,092.54 = £60\,735.66$: more than two-thirds of the money paid.

Changing the conditions of a loan

A change of interest rate

Many loans fix the rate for only part of the term. When the fixed period ends, the outstanding balance becomes, in effect, a new loan: its $PV$ is the balance, its $N$ is the number of payments left, and its $I\%$ is the new rate. The method always has two stages: find the balance with the old conditions, then find the new payment with the new ones.

Source: Bank of England, quoted household interest rates, series IUMBV34, December 2021 and December 2023.

Many households whose low fixed rates ended in 2023 faced payment rises of a similar size. Because fixed rates last two or five years, a rise in interest rates takes that long to reach many borrowers' budgets. The model assumes the new rate is fixed for the remaining 23 years, which in reality it will not be.

Repaying a loan early

A borrower who repays the whole balance early pays the outstanding balance instead of all the remaining payments. The interest saved is the difference between the payments that would have been made and the balance paid instead.

For the £250,000 mortgage repaid in full after 5 years, the remaining payments would total $240 \times 1480.47 = £355\,312.80$, while the balance is £221,907.46, so $355\,312.80 - 221\,907.46 = £133\,405.34$ of interest is avoided. Real lenders may charge a fee for early repayment during a fixed-rate period, which reduces the saving.

Finding the number of payments or the interest rate

Solving for N

When the payment is fixed in advance ("pay $300 a month until the loan is repaid"), the unknown is the number of payments. The solver gives $N$ as a decimal, and the decimal must be interpreted in context.

A non-integer $N$ means that the last regular payment would more than clear the debt. In practice the borrower makes the whole number of full payments and then one smaller final payment, so the number of payments is $N$ rounded up.

Solving for the interest rate

When the loan amount, the payment and the term are known, the solver finds the nominal annual rate. This is how a borrower can work out the true cost of an offer that is described only by its payments ("pay $185 a month for two years").

Comparing quoted rates

Lenders in many countries must state the cost of a loan as a single yearly percentage so that offers can be compared.

Because an APR includes fees, two loans with the same interest rate can have different APRs. The TVM solver works with the nominal annual interest rate entered as $I\%$, with the compounding entered separately as $C/Y$; fees, if there are any, have to be handled as separate amounts.

Comparing two options

Borrowers often have two ways of paying for the same thing. There is rarely a single right answer: one option may cost less in total while the other needs no deposit or has a lower monthly payment. A full answer states a choice and gives a reason based on a correct comparison.

Annuities: saving with regular payments

The future value of a savings plan

An annuity can run the other way: instead of repaying a debt, a saver pays a fixed amount into an account at the end of each period, and the account earns compound interest. The value of the fund at the end is the future value. The first payment earns interest for $N - 1$ periods, the second for $N - 2$ periods and the last for none, so the fund is the sum of a geometric series.

In the solver, the saver starts with nothing ($PV = 0$), pays in every period ($PMT$ negative) and receives the fund at the end ($FV$ positive, the unknown).

The interest earned by a savings annuity is smaller than it would be if the whole €10,800 were invested at the start, because most of the money is in the account for only part of the 6 years. The graph of the balance would curve slightly upwards above the straight line of total payments.

A lump sum and regular payments together

The solver can handle a starting amount and regular payments at the same time. Both are paid in by the saver, so both $PV$ and $PMT$ are negative, and $FV$ is positive.

The Revision Village annuities lesson works through both kinds of annuity with the solver. The change of sign between saving and drawing an income is the point most worth watching.

The real value of a savings plan

Over several years, prices rise, so a fund of €12,035 buys less than €12,035 does today. Subtopic 1.4 shows how to express a future amount in today's money, its real value, using the real interest rate. The same idea applies to a savings annuity: the approximate real rate is the nominal rate minus the inflation rate, and that rate is entered as $I\%$ to give the fund's value in today's prices.

The exact real rate is $\frac{1 + \text{nominal}}{1 + \text{inflation}} - 1$ (rates as decimals). For rates of a few per cent the two methods differ by only a few hundredths of a percentage point.

The real value (€11,192) is still above the €10,800 paid in, so the saver's purchasing power has grown, but by much less than the nominal interest suggests.

Annuities that pay an income

The fund needed for a regular income

A retirement annuity reverses the savings plan. A fund is built up while working; at retirement the fund pays a fixed income at the end of each month while the remaining money continues to earn interest. The fund is designed to run out exactly when the payments stop, so $FV = 0$.

From the retiree's point of view, the fund is paid in at the start ($PV$ negative) and the income is received ($PMT$ positive). The present value needed is the amount that, with interest, pays every instalment of the income.

This two-stage structure (find the fund that pays the income, then the saving that builds the fund) is the standard pattern for retirement questions. The $FV$ of the first stage is the $PV$ of the second, with its sign reversed for the new point of view.

Amortization schedules with a spreadsheet

Spreadsheets are the other standard tool for amortization, and banks build schedules this way. A spreadsheet applies the two-step rule for one payment period to every row: the interest is the previous balance times the rate per period, the principal repaid is the payment minus the interest, and the new balance is the previous balance minus the principal repaid. Copying one row down 300 times gives the whole schedule.

First and last rows of the schedule for £250,000 over 25 years at 5.13% compounded monthly, with interest rounded to the penny each month. Rate: Bank of England IUMBV34, September 2026.

The last payment is £1.60 larger than the others, because the payment of £1480.47 is slightly less than the exact value £1480.4728…. The interest column of the full schedule sums to £194,142.60, against £194,141.00 from $300 \times 1480.47 - 250\,000$. Real lenders adjust the final payment in this way; the simpler $N \times PMT$ is the usual estimate of the total paid.

Some GDCs can also read values from a schedule directly. On the TI-84 Plus CE the finance menu has bal( for the balance after a given payment, and ΣInt( and ΣPrn( for the interest and principal paid between two payments. These are useful checks on the TVM working.

The mathematics behind the solver

The formulas behind the solver come from geometric series. They show why the solver gives the values it does, and they explain when a payment can and cannot clear a debt.

The future value of a savings annuity

Let $i$ be the interest rate per period and $PMT$ the payment at the end of each of $N$ periods. The payment made at the end of period $k$ earns interest for the remaining $N - k$ periods, so by the final payment its value is $PMT(1+i)^{N-k}$. Adding all $N$ payments, from the last (no interest) to the first ($N - 1$ periods of interest):

The second line is the sum of a geometric series with first term $PMT$ and common ratio $1 + i$, from subtopic 1.3. For the student saving €150 a month at 3.6%, $i = 0.003$ and $N = 72$: $150 \times \frac{1.003^{72} - 1}{0.003} = 12\,035.06$, matching the solver.

The payment on a loan

For a loan of $PV$, the balance after $N$ payments is the loan grown for $N$ periods minus the grown value of all the payments, and it must equal zero:

Rearranged for $PV$, this is the annuity formula $PV = PMT \times \frac{1 - (1+i)^{-N}}{i}$, which gives the fund needed for an income. For the six-month loan of $3000 at 1% a month, $PMT = \frac{3000 \times 0.01}{1 - 1.01^{-6}} = 517.645\ldots$, the value the solver gave.

The balance as an exponential model

Replacing $N$ by $k$ in the balance gives the outstanding balance after $k$ payments, which can be rearranged into a form from subtopic 2.5:

This is an exponential function of $k$ plus a constant. The constant $\frac{PMT}{i}$ is the balance at which the interest charged, $i \times B$, exactly equals the payment. If the loan is larger than this, the coefficient $PV - \frac{PMT}{i}$ is positive and the debt grows for ever; if the loan is smaller, the coefficient is negative and the balance falls ever faster until it reaches zero. A payment can only clear a debt if it is larger than the first period's interest, $PMT > i \times PV$.

Case study: paying off a credit card

Credit cards are one of the most widely used forms of high-interest borrowing. In September 2026 the representative rate on UK credit-card lending to households was 24.71% a year, up from 21.42% in December 2021. A card balance that is not cleared each month is a loan with no fixed term: the holder chooses how much to repay, subject to a minimum.

Source: Bank of England, quoted household interest rates, series IUMCCTL (sterling credit-card lending to households).

Treating the rate as a nominal annual rate compounded monthly, the monthly rate is $\frac{24.71\%}{12} \approx 2.06\%$. On a balance of £2000 the first month's interest is $2000 \times \frac{0.2471}{12} = £41.18$. By the condition from the balance formula, any fixed payment of £41.18 or less never clears the debt.

The solver gives the number of payments for three fixed monthly payments on a £2000 balance ($I\% = 24.71,\ PV = 2000,\ FV = 0,\ P/Y = C/Y = 12$), with no new spending on the card.

Totals include a smaller final payment. Rate: Bank of England IUMCCTL, September 2026.

The balance curves for £40, £50 and £100 a month show three very different outcomes from small differences in the payment.

The case shows the effect of a high rate on a small payment. At £50 a month the holder repays more than twice the amount borrowed, and the interest (£2257) is larger than the original debt. Doubling the payment to £100 cuts the interest by about 73%, from £2257 to £604. At the December 2021 rate of 21.42%, the same £50 a month would have cleared the debt in 71 payments, so the rise in rates added more than a year of payments.

The model has limitations. Real card minimum payments are usually a percentage of the balance (with a floor), not a fixed amount, so they shrink as the debt falls and take even longer to clear it. The representative rate is an average, and many cards charge more. Interest on cards is often worked out daily, and new purchases add to the balance. Even so, the calculation raises an ethical question. A borrower who can amortize a debt can see when repayments are mostly paying interest, while a lender earns the most interest from customers who pay only a little more than the interest each month.

Exam focus

Amortization and annuity questions appear on Paper 1 as short items of 3–6 marks and on Paper 2 as finance questions of 10–16 marks that often combine loans and annuities with compound interest, depreciation or inflation from subtopic 1.4. The command terms are find, calculate, write down (the loan amount after a deposit), hence (a new payment from a balance found earlier), justify and give a reason (choosing between options). Questions often begin with an accuracy instruction, such as "give all answers correct to 2 decimal places" or "to the nearest dollar", which applies to every part.

The recurring shapes are: the payment on a loan; the total paid and the total interest; the balance after some years; the interest paid in the first few years from the payment and the balance; a new payment after a rate change; the number of payments, rounded up; the interest rate for given payments; the fund needed for an income, then the saving to build it; and a savings plan's real value with inflation. Markschemes give M1 for an attempt to use the financial app (or the annuity formula) and A1 for all entries correct, so every entry must be written down, including $P/Y$ and $C/Y$. A substituted compound interest formula earns no method mark when payments are involved.

Part (e) is where marks are most often lost on questions like this. The interest is the amount paid minus the fall in the debt, so the balance from part (d) is only one step in the working. In part (f) the new loan runs for the 20 years that remain, so $N = 240$ rather than 300.

Three further habits earn marks on every finance question. Write $N$ as a calculation ($25 \times 12 = 300$) so that the number of periods is visible. Give money answers as positive amounts with the currency, even though the solver shows a negative payment. When $N$ comes out as a decimal, state the number of payments rounded up and say why.

Practice questions

1. A loan is repaid with monthly payments over 15 years. Write down the values of $N$ and $P/Y$ to enter in the TVM solver. [2]

2. A loan of $12,000 is taken at 6.5% a year, compounded monthly, and repaid with monthly payments over 5 years. Find the monthly payment and the total interest paid, to the nearest cent. [4]

3. A saver pays £80 into an account at the end of every month for 4 years. The account pays 2.9% a year, compounded monthly. Calculate the value of the account after 4 years, to the nearest penny. [3]

4. A mortgage of $180,000 is repaid over 20 years at 5.5% a year, compounded monthly, with monthly payments. Find the monthly payment and the outstanding balance after 8 years, to the nearest cent. [5]

5. A credit card balance of $3500 is charged 18% a year, compounded monthly. The holder pays $150 at the end of each month and makes no new purchases. Find the number of payments needed to clear the balance. [3]

6. A phone costing $6000 is paid for with 24 monthly payments of $290 and no deposit. Find the nominal annual interest rate, compounded monthly, correct to 3 significant figures. [3]

7. A retirement fund must pay $1500 at the end of every month for 20 years, earning 3.8% a year compounded annually. (a) Find the amount needed in the fund, correct to 3 significant figures. [3] (b) Find the monthly saving needed for 30 years, at the same rate, to build this fund. [3]

8. An investor pays $250 at the end of every month for 8 years into an account paying 4.2% a year, compounded monthly. The inflation rate is expected to be 2.7% a year. (a) Find the value of the account after 8 years. [3] (b) Find the real value of the account after 8 years, using an approximate real interest rate. [3]

9. Explain why, for a loan repaid by equal monthly payments, the part of each payment that reduces the debt increases over time. [2]

10. A laptop costs $1500. Option A: pay $1500 now. Option B: a 10% deposit and 12 monthly payments of $120. Find the total cost of Option B and justify which option a buyer with $1500 in savings should choose. [4]

Summary

• Amortization repays a loan by equal payments at the end of each period; each payment covers that period's interest and repays part of the principal, so the interest share falls and the principal share rises.

• Each period: interest $= i \times$ balance, and new balance $=$ balance $+$ interest $-$ $PMT$; an amortization schedule records this row by row.

• The TVM solver uses $N$ (number of payments), $I\%$, $PV$, $PMT$, $FV$, $P/Y$ and $C/Y$; money received is positive and money paid out negative, so for a loan $PV$ and $PMT$ have opposite signs and $FV = 0$.

• Total interest $= N \times PMT -$ amount borrowed; total cost $=$ deposit $+ N \times PMT$.

• The outstanding balance after $m$ payments is $FV$ with $N = m$; interest paid so far $= m \times PMT - (\text{amount borrowed} - \text{balance})$

• A rate change makes the balance a new loan: its $PV$ is the balance and its $N$ the payments remaining.

• When solving for $N$, round up: a decimal means one smaller final payment.

• A savings annuity has $PV = 0$ (or a lump sum), $PMT$ negative and $FV$ positive; an income annuity needs a fund $PV$ with $FV = 0$

• Real values with inflation use the real rate $\approx$ nominal rate $-$ inflation rate as $I\%$.

• Behind the solver, every annuity is a geometric series; a payment clears a debt only if it is larger than the first period's interest.

Self-check

• To do:

  I can explain why the interest part of a loan payment falls over time.

• To do:

  I can build an amortization schedule for a short loan, including an adjusted final payment.

• To do:

  I can enter a loan in the TVM solver with the correct signs, $N$, $P/Y$ and $C/Y$.

• To do:

  I can find a loan's regular payment and write down every TVM entry.

• To do:

  I can find the amount borrowed after a deposit given as a percentage.

• To do:

  I can calculate the total amount paid, the total interest and the total cost of an item.

• To do:

  I can find the outstanding balance after a given number of payments.

• To do:

  I can calculate the interest paid over the first few years from the payment and the balance.

• To do:

  I can find a new payment after an interest-rate change, using the balance as the new $PV$.

• To do:

  I can find the number of payments and round it up correctly, with the reason.

• To do:

  I can find the interest rate implied by a loan's payments.

• To do:

  I can find the future value of a savings annuity, with or without a starting lump sum.

• To do:

  I can find the fund needed for a regular income and the saving needed to build it.

• To do:

  I can find the real value of a savings plan using an approximate real interest rate.

• To do:

  I can justify a choice between two financial options with a numerical comparison.

29 038.20 (USD) A1

Note: Accept 29 038.46 from the unrounded payment. Follow through from part (a).

[2 marks]