Loan Amortization

When you take out a car loan or a mortgage, you pay it back with equal regular payments. Each payment covers that period’s interest first, and whatever is left over pays down the amount you owe. Paying off a loan this way is called amortization. In IB Mathematics: Applications and Interpretation you work these problems with the finance solver on your GDC (or a spreadsheet), so the skill is setting the problem up correctly and reading the results sensibly.

Key ideas

The finance (TVM) solver

Your GDC’s finance app has a time value of money (TVM) solver. You fill in all but one field, then solve for the missing one.

Field	Meaning
$N$	total number of payments
$I\%$	annual interest rate, as a percentage (type $5.4$, not $0.054$)
$PV$	present value: the amount at the start (the loan, or a starting deposit)
$PMT$	the regular payment
$FV$	future value: the amount at the end ($0$ when a loan is fully paid off)
$P/Y$	payments per year ($12$ for monthly)
$C/Y$	compounding periods per year ($12$ for monthly, $2$ for semi-annually)

In IB exams, payments are made at the end of each period, so make sure your GDC is set to END (not BEGIN).

Sign convention

The solver tracks money from your point of view: money you receive is positive, money you pay out is negative.

• Loan: the bank gives you the money, so $PV$ is positive. Your payments go out, so $PMT$ is negative. At the end you owe nothing: $FV = 0$.
• Savings plan: your deposits go out, so $PMT$ is negative (and $PV$ is $0$ or negative). The money you get back at the end is positive $FV$.

If the solver gives an error or a strange answer, the signs are the first thing to check.

Amortization tables

An amortization table (repayment schedule) shows what happens to each payment. For each period:

1. Interest $=$ opening balance $\times$ periodic rate (for monthly compounding, $\dfrac{I\%}{12}$ as a decimal).
2. Principal repaid $=$ payment $-$ interest.
3. Closing balance $=$ opening balance $-$ principal repaid.

Early on, the balance is large, so most of each payment is interest. As the balance falls, less goes to interest and more to the principal. That’s why the balance in the graph below falls slowly at first and faster near the end.

The balance of the mortgage in Example 3. After $5$ of the $25$ years, only about $35 000 of the $320 000 has been paid off.

Balances and total interest

• Balance after $k$ payments: enter $N = k$ (keep $I\%$, $PV$, $PMT$, $P/Y$, $C/Y$ the same) and solve for $FV$. A negative $FV$ means you still owe that amount. Many GDCs also have built-in amortization functions for the balance, and for the interest and principal paid over a range of payments.
• Total interest $=$ total of all payments $-$ amount borrowed $= N \times PMT - PV$ (using the payment as a positive number, rounded to the cent).
• Interest paid in the first $k$ payments $= k \times PMT - (\text{amount borrowed} - \text{balance after } k)$.

The formula behind the solver

The solver uses the present value formula from annuities: present value (and future value for savings). For a loan with periodic rate $i$ and $n$ payments,

$$R = \frac{PV \times i}{1 - (1 + i)^{-n}}$$

The guide says knowing the annuity formula helps understanding, but it isn’t examined: in IB exams you can always use technology. It’s still a good way to check a GDC answer.

Worked examples

Example 1: A car loan

Mira borrows $24 000 to buy a car. The interest rate is $6.6\%$ per year, compounded monthly, and she repays the loan with monthly payments over $5$ years.

• (a) Find her monthly payment.
• (b) Find the total interest she pays.

Solution.

(a) Enter:

$N$	$I\%$	$PV$	$PMT$	$FV$	$P/Y$	$C/Y$
$60$	$6.6$	$24\,000$	?	$0$	$12$	$12$

Solving gives $PMT = -470.712\ldots$. The negative sign means it’s a payment, so her monthly payment is $470.71.

Check with the formula: $i = \dfrac{0.066}{12} = 0.0055$ and

$$R = \frac{24\,000 \times 0.0055}{1 - 1.0055^{-60}} = 470.71$$

(b) Total paid $= 60 \times 470.71 = 28\,242.60$. Total interest $= 28\,242.60 - 24\,000$, which is $4242.60.

Example 2: The start of an amortization table

A loan of $5000 at $9\%$ per year, compounded monthly, is repaid with monthly payments over $2$ years. Find the monthly payment and complete the first three rows of the amortization table.

Solution. TVM: $N = 24$, $I\% = 9$, $PV = 5000$, $FV = 0$, $P/Y = C/Y = 12$. This gives $PMT = -228.42$, so the payment is $228.42.

The monthly rate is $\dfrac{9\%}{12} = 0.75\% = 0.0075$.

Month	Opening balance ($)	Interest ($)	Principal repaid ($)	Closing balance ($)
1	$5000.00$	$37.50$	$190.92$	$4809.08$
2	$4809.08$	$36.07$	$192.35$	$4616.73$
3	$4616.73$	$34.63$	$193.79$	$4422.94$

For month 1: interest $= 5000 \times 0.0075 = 37.50$, principal $= 228.42 - 37.50 = 190.92$, closing balance $= 5000 - 190.92 = 4809.08$. Each later row starts from the previous closing balance. Notice the interest shrinking and the principal part growing each month.

Check: setting $N = 3$ in the TVM solver and solving for $FV$ gives $FV = -4422.933\ldots$, matching the table to within a cent. (The table rounds each month’s interest to the cent, so tiny differences like this are normal.)

Example 3: A Canadian mortgage

In Canada, fixed-rate mortgage rates are compounded semi-annually, even though payments are monthly. A family takes a $320 000 mortgage at $5.4\%$ per year, compounded semi-annually, with monthly payments over $25$ years.

• (a) Find the monthly payment.
• (b) Find the balance after $5$ years.
• (c) Find the interest paid in those first $5$ years.

Solution.

(a) $N = 25 \times 12 = 300$, $I\% = 5.4$, $PV = 320\,000$, $FV = 0$, $P/Y = 12$, $C/Y = 2$. Solving gives $PMT = -1934.665\ldots$, so the payment is $1934.67.

(b) Change $N$ to $60$, enter $PMT = -1934.67$, and solve for $FV$: $FV = -284\,972.44$. The family still owes $284 972.44.

(c) In $5$ years they paid $60 \times 1934.67 = 116\,080.20$ dollars. Of this, the principal repaid was $320\,000 - 284\,972.44 = 35\,027.56$ dollars. So the interest paid was

$$116\,080.20 - 35\,027.56 = 81\,052.64$$

which is $81 052.64. About $70\%$ of what they paid in the first five years was interest.

Example 4: Comparing payment sizes

Jonah has a student loan of $18 000 at $7.2\%$ per year, compounded monthly. Compare paying $300 per month with paying $400 per month: how many payments does each take, and roughly how much interest does each cost?

Solution. Solve for $N$ with $I\% = 7.2$, $PV = 18\,000$, $FV = 0$, $P/Y = C/Y = 12$.

• $PMT = -300$: $N = 74.60\ldots$. That’s $74$ full payments and a smaller $75$th payment, so just over $6$ years. Interest $\approx 74.60\ldots \times 300 - 18\,000 \approx 4381$, about $4381.
• $PMT = -400$: $N = 52.60\ldots$. That’s $53$ payments, about $4.4$ years. Interest $\approx 52.60\ldots \times 400 - 18\,000 \approx 3044$, about $3044.

Paying $100 more each month clears the loan about $22$ months sooner and saves about $1340 in interest. Bigger payments cut the balance faster, so less interest builds up.

Common mistakes

Getting the signs wrong. For a loan, $PV$ is positive and $PMT$ is negative (or the other way round, as long as they’re opposite). If both have the same sign, the solver either fails or gives nonsense, like a loan that is never paid off.

Typing the rate as a decimal. $I\%$ wants $6.6$ for $6.6\%$, not $0.066$. Entering $0.066$ means a rate of $0.066\%$, and the payment comes out far too small.

Forgetting to change P/Y and C/Y. Monthly payments need $P/Y = 12$, and the compounding needs its own $C/Y$. A Canadian mortgage has $P/Y = 12$ but $C/Y = 2$. Using $12$ for both gives a slightly different (wrong) payment.

Mixing up N with the number of years. $N$ counts payments. A $25$-year mortgage with monthly payments has $N = 300$, not $25$.

Using the annual rate in an amortization table. Each row’s interest uses the rate per period: for $9\%$ compounded monthly that’s $0.75\%$, not $9\%$.

Treating every payment as interest-free principal. The amount you owe doesn’t drop by the full payment each month. Only the principal part reduces the balance, and early on that’s the smaller part.

Practice

1. (Warm-up) Sam borrows $12 000 at $5.5\%$ per year, compounded monthly, to be repaid monthly over $4$ years. Write down the values to enter in the TVM solver, and find the monthly payment.

Solution

$N = 48$, $I\% = 5.5$, $PV = 12\,000$, $PMT = ?$, $FV = 0$, $P/Y = 12$, $C/Y = 12$.

Solving gives $PMT = -279.077\ldots$, so the monthly payment is $279.08.

2. (Warm-up) Lea deposits $150 at the end of every month into an account paying $4.2\%$ per year, compounded monthly. How much is in the account after $10$ years?

Solution

$N = 120$, $I\% = 4.2$, $PV = 0$, $PMT = -150$, $P/Y = C/Y = 12$. Solve for $FV$: $FV = 22\,321.968\ldots$

She has $22 321.97 after $10$ years. (She deposited $120 \times 150 = 18\,000$ dollars, so about $4322 is interest.)

Check with the future value formula: $i = 0.0035$, and $\dfrac{150\,(1.0035^{120} - 1)}{0.0035} = 22\,321.97$.

3. (Warm-up) A loan of $9000 is repaid with $36$ monthly payments of $276.50. Find the total interest paid.

Solution

Total paid $= 36 \times 276.50 = 9954$ dollars. Total interest $= 9954 - 9000$, which is $954.

4. (Core) A loan of $2400 at $12\%$ per year, compounded monthly, is repaid with $6$ monthly payments.

• (a) Find the monthly payment.
• (b) Complete the full amortization table.
• (c) Explain why the final payment is adjusted slightly.

Solution

(a) $N = 6$, $I\% = 12$, $PV = 2400$, $FV = 0$, $P/Y = C/Y = 12$ gives $PMT = -414.116\ldots$, so the payment is $414.12.

(b) The monthly rate is $1\% = 0.01$.

Month	Opening ($)	Interest ($)	Principal ($)	Closing ($)
1	$2400.00$	$24.00$	$390.12$	$2009.88$
2	$2009.88$	$20.10$	$394.02$	$1615.86$
3	$1615.86$	$16.16$	$397.96$	$1217.90$
4	$1217.90$	$12.18$	$401.94$	$815.96$
5	$815.96$	$8.16$	$405.96$	$410.00$
6	$410.00$	$4.10$	$410.02$	$-0.02$

(c) The payment was rounded up from $414.116\ldots$ to $414.12$, so the six payments overpay by $2$ cents. In practice the last payment is reduced to $410.00 + 4.10 = 414.10$ dollars so the balance ends at exactly $0$.

5. (Core) A store sells a $1500 TV on a plan of $24$ monthly payments of $70. Find the annual interest rate, compounded monthly, that the store is charging, to 3 s.f. How much interest does the plan cost?

Solution

$N = 24$, $PV = 1500$, $PMT = -70$, $FV = 0$, $P/Y = C/Y = 12$. Solve for $I\%$: $I\% = 11.126\ldots$

The rate is about $11.1\%$ per year (3 s.f.). Total interest $= 24 \times 70 - 1500$, which is $180.

6. (Core) A $30 000 car can be bought in two ways:

• Option A: $0\%$ dealer financing, with $60$ equal monthly payments.
• Option B: a $3000 cash discount, and the remaining $27 000 borrowed from a bank at $4.5\%$ per year, compounded monthly, repaid monthly over $5$ years.

Which option costs less in total, and by how much?

Solution

Option A: $\dfrac{30\,000}{60} = 500$ dollars per month, so the total is $30 000.

Option B: $N = 60$, $I\% = 4.5$, $PV = 27\,000$, $FV = 0$, $P/Y = C/Y = 12$ gives $PMT = -503.361\ldots$, so $503.36 per month. Total $= 60 \times 503.36 = 30\,201.60$ dollars.

Option A costs less, by $30\,201.60 - 30\,000 = 201.60$, that is, by $201.60.

7. (Core) A $450 000 mortgage has a rate of $4.6\%$ per year, compounded semi-annually, with monthly payments over $25$ years.

• (a) Find the monthly payment.
• (b) Find the balance after $5$ years.
• (c) Find the interest paid in the first $5$ years.

Solution

(a) $N = 300$, $I\% = 4.6$, $PV = 450\,000$, $FV = 0$, $P/Y = 12$, $C/Y = 2$ gives $PMT = -2515.706\ldots$, so the payment is $2515.71.

(b) $N = 60$, $PMT = -2515.71$, solve for $FV$: $FV = -395\,734.75$. The balance is $395 734.75.

(c) Paid: $60 \times 2515.71 = 150\,942.60$ dollars. Principal repaid: $450\,000 - 395\,734.75 = 54\,265.25$ dollars. Interest: $150\,942.60 - 54\,265.25 = 96\,677.35$, which is $96 677.35.

8. (Challenge) Priya saves $400 at the end of every month for $30$ years in a fund earning $6\%$ per year, compounded monthly. She then moves the whole amount into an account earning $4\%$ per year, compounded monthly, and withdraws an equal amount at the end of every month for $25$ years, until the account is empty. Find:

• (a) the amount she has saved after $30$ years
• (b) her monthly withdrawal

Solution

(a) $N = 360$, $I\% = 6$, $PV = 0$, $PMT = -400$, $P/Y = C/Y = 12$. Solve for $FV$: $FV = 401\,806.016\ldots$, about $401 806.02.

(b) Now she deposits this amount (it goes out of her hands into the account) and receives the withdrawals: $N = 300$, $I\% = 4$, $PV = -401\,806.016\ldots$ (use the stored value), $FV = 0$, $P/Y = C/Y = 12$. Solve for $PMT$: $PMT = 2120.880\ldots$

She can withdraw $2120.88 per month. (She deposited $360 \times 400 = 144\,000$ dollars in total, but receives $300 \times 2120.88 \approx 636\,000$ dollars: that’s the power of compound interest over a long time.)

9. (Challenge) Dev borrows $20 000 at $6\%$ per year, compounded monthly, repaid with monthly payments over $5$ years. Right after his 24th payment, he pays an extra lump sum of $5000. He keeps making the same monthly payment.

• (a) Find the monthly payment and the balance just before the lump sum.
• (b) How many more payments does he need after the lump sum?
• (c) Estimate how much interest he saves compared with the original plan.

Solution

(a) $N = 60$, $I\% = 6$, $PV = 20\,000$, $FV = 0$, $P/Y = C/Y = 12$ gives $PMT = -386.656\ldots$, so $386.66. Then $N = 24$, $PMT = -386.66$, solve for $FV$: $FV = -12\,709.68$. He owes $12 709.68.

(b) After the lump sum he owes $12\,709.68 - 5000 = 7709.68$ dollars. Solve for $N$ with $PV = 7709.68$, $PMT = -386.66$, $I\% = 6$, $FV = 0$: $N = 21.05\ldots$

He needs $22$ more payments ($21$ full payments and a small final one), instead of the $36$ left on the original plan.

(c) Original total interest: $60 \times 386.66 - 20\,000 = 3199.60$ dollars.

New total paid $\approx 24 \times 386.66 + 5000 + 21.05\ldots \times 386.66 \approx 22\,421.73$ dollars, so new interest $\approx 22\,421.73 - 20\,000 = 2421.73$ dollars.

He saves about $3199.60 - 2421.73 \approx 778$, that is, roughly $778 in interest.