Year 12 · Financial mathematics
Quick tips — memory joggers
What an annuity is
Table of interest factors
Recurrence relation
Matching the rate & period
Level 1 · Fluency
An annuity is built from equal regular payments earning compound interest. True or false: its future value equals (payment) × (number of payments).
False.
False. Each payment also earns compound interest, so the future value is greater than simply payment × number of payments.
True or false: for a set payment and interest rate, increasing the number of regular payments increases the future value of an annuity.
True.
True. Each extra payment is another deposit that also earns compound interest, so more payments give a larger future value.
True or false: in a compound-interest annuity, an earlier payment earns more interest by the end than a later payment of the same size.
True.
True. An earlier payment is invested for more compounding periods, so it grows more by the end of the annuity.
Level 2 · Application
$300 is deposited at the end of each month into an account paying 4.5% p.a. compounded monthly. The future-value interest factor for 60 monthly payments at 0.375% per month is 67.1456. Calculate the value of the account after 5 years.
$20,143.67
FV = payment × factor = 300 × 67.1456 = $20,143.67.
$500 is deposited at the end of each month into an account. The future-value interest factor for 48 monthly payments is 53.0143. Calculate the value of the account after 4 years.
$26,507.15
FV = payment × factor = 500 × 53.0143 = $26,507.15.
$1000 is invested at the end of each quarter for 5 years at 6% p.a. compounded quarterly. The future-value interest factor for 20 payments at 1.5% per quarter is 23.1237. Calculate the future value of the investment.
$23,123.70
FV = payment × factor = 1000 × 23.1237 = $23,123.70.
Level 3 · Further Application
A $400 000 home loan is charged 5% p.a. compounded monthly and repaid over 25 years.
(a) It is an annuity because it is repaid by equal regular (monthly) payments earning/charging compound interest; here the borrower makes the regular payments to the lender.
(b) Repayment = 400000 ÷ 171.0600 = $2,338.36.
A $35 000 car loan is charged 9% p.a. compounded monthly and repaid over 5 years.
(a) Equal monthly repayments with compound interest. (b) $726.54
(a) It is repaid by equal regular monthly payments while compound interest is charged on the balance, so it fits the annuity model.
(b) Repayment = present value ÷ factor = 35000 ÷ 48.1734 = $726.54.
A savings account pays 3.6% p.a. compounded monthly. Nadia deposits $250 at the end of each month for 6 years.
(a) 0.3% per month. (b) $20,153.88
(a) Monthly rate = 3.6% ÷ 12 = 0.3% per month.
(b) FV = payment × factor = 250 × 80.6155 = $20,153.88.
Level 1 · Fluency
A superannuation account receives a fixed deposit every month. Is this an annuity of contributions or of withdrawals?
in
Contributions — money is regularly paid in and grows.
A retiree draws a fixed pension payment out of an invested lump sum each month. Is this an annuity of contributions or of withdrawals?
withdrawals
Withdrawals — money is regularly drawn out of the lump sum, so the balance falls.
A student pays $50 into a savings account at the end of every week. Is this an annuity of contributions or of withdrawals?
contributions
Contributions — equal amounts are paid in each week and the balance grows.
Level 2 · Application
A retiree has $500 000 earning 4% p.a. and withdraws $30 000 at the end of each year. Calculate the balance at the end of the first year, immediately after the withdrawal.
Balance = 500000 × 1.04 − 30000 = $490,000.00.
A worker starts with $0 and contributes $2000 at the end of each year to a fund earning 5% p.a. Calculate the balance at the end of the second year, immediately after the deposit.
$4,100.00
End of year 1: 0 × 1.05 + 2000 = $2,000.00.
End of year 2: 2000 × 1.05 + 2000 = 2100 + 2000 = $4,100.00.
A fund of $80 000 earns 3% p.a. and $10 000 is withdrawn at the end of each year. Calculate the balance immediately after the second withdrawal.
$64,572.00
End of year 1: 80000 × 1.03 − 10000 = 82400 − 10000 = $72,400.00.
End of year 2: 72400 × 1.03 − 10000 = 74572 − 10000 = $64,572.00.
Level 3 · Further Application
Explain the difference between the two uses of an annuity, using one specific example of each:
(a) Regular contributions: equal amounts are paid in over time and grow with compound interest — e.g. paying $500/month into superannuation to build a retirement balance.
(b) Regular withdrawals: a lump sum earns interest while equal amounts are drawn out — e.g. a retiree drawing $30 000/year from $500 000, with the balance falling over time.
A couple pays $800 per month into an investment account for 20 years, then at retirement draws $2000 per month from the accumulated balance.
(a) The 20-year paying-in phase is an annuity of regular contributions; the retirement paying-out phase is an annuity of regular withdrawals.
(b) During the contributions phase the balance rises (deposits plus interest); during the withdrawals phase the balance falls, because each withdrawal is larger than the interest earned.
Classify each of the following as behaving like an annuity of contributions or of withdrawals, giving a reason:
(a) Withdrawals (a reducing annuity): the loan is a lump-sum debt that is drawn down by equal repayments, so the balance owed falls each month.
(b) Contributions: equal amounts are paid in each month, so the balance grows over time.
Level 1 · Fluency
For a recurrence relation An = An−1(1.006) − 500 with A0 = 20 000, what does the “− 500” represent?
A regular withdrawal (or repayment) of $500 taken out each period, after interest is added.
In the recurrence relation An = An−1(1.005) + 300 with A0 = 5000, what does multiplying by the factor “1.005” represent?
Adding 0.5% compound interest to the balance each period (a monthly interest rate of 0.5%).
The recurrence relation An = An−1(1.004) + 600 with A0 = 0 models an annuity. Does the “+ 600” add money to, or remove money from, the account each period?
It adds money — a regular contribution of $600 is deposited each period, after interest, so the balance grows.
Level 2 · Application
A savings account earns 0.4% interest per month. At the end of each month, after interest, $400 is deposited. The balance follows An = An−1(1.004) + 400, with A0 = 20 000. Calculate the balance after 3 months.
$21,445.77
A₁ = 20000(1.004)+400 = $20,480.00.
A₂ = $20,480.00(1.004)+400 = $20,961.92.
A₃ = $20,961.92(1.004)+400 = $21,445.77.
A loan of $8000 is charged 1% interest per month. At the end of each month, after interest, a repayment of $700 is made. The balance follows An = An−1(1.01) − 700, with A0 = 8000. Calculate the balance after 3 repayments.
$6,121.34
A₁ = 8000(1.01)−700 = 8080 − 700 = $7,380.00.
A₂ = 7380(1.01)−700 = 7453.80 − 700 = $6,753.80.
A₃ = 6753.80(1.01)−700 = 6821.34 − 700 = $6,121.34.
A savings plan follows An = An−1(1.005) + 250, with A0 = 10 000. Calculate the balance after 4 months.
$11,209.03
A₁ = 10000(1.005)+250 = $10,300.00.
A₂ = 10300(1.005)+250 = $10,601.50.
A₃ = 10601.50(1.005)+250 = $10,904.51.
A₄ = 10904.51(1.005)+250 = $11,209.03.
Level 3 · Further Application
Tina invests $60 000 at 0.5% per month. Immediately after the interest is added each month, she withdraws $800. The balance follows An = An−1(1.005) − 800, with A0 = 60 000.
$58,492.49; $892.49
(a) A₁ = 60000(1.005)−800 = $59,500.00; A₂ = $58,997.50; A₃ = $58,492.49.
(b) Interest each month = balance × 0.005: 60000(0.005) + $59,500.00(0.005) + $58,997.50(0.005) = $300.00 + $297.50 + $294.99 = $892.49.
A car loan of $24 000 is charged 0.6% interest per month. Repayments of $500 are made at the end of each month, after interest, so the balance follows An = An−1(1.006) − 500, with A0 = 24 000.
$22,925.58; $425.58
(a) A₁ = 24000(1.006)−500 = 24144 − 500 = $23,644.00; A₂ = 23644(1.006)−500 = $23,285.86; A₃ = 23285.86(1.006)−500 = $22,925.58.
(b) Interest each month = balance × 0.006: 24000(0.006) + 23644(0.006) + 23285.86(0.006) = $144.00 + $141.86 + $139.72 = $425.58.
An investment follows An = An−1(1 + r) + 400, with A0 = 5000.
(a) $5,852.13. (b) Higher.
(a) A₁ = 5000(1.005)+400 = 5025 + 400 = $5,425.00; A₂ = 5425(1.005)+400 = 5452.13 + 400 = $5,852.13.
(b) Higher — a larger interest rate adds more interest to the balance each month, so the account grows faster.
Level 1 · Fluency
A future-value annuity factor tells you the future value of an annuity of $1 per period. If the factor for 8 payments is 8.583, what is the future value of $1000 paid each period for 8 periods?
$8583
1000 × 8.583 = $8583.
A present-value annuity factor tells you the present value of an annuity of $1 per period. If the factor for 5 payments is 4.3295, what is the present value of $2000 paid each period for 5 periods?
$8659
PV = 2000 × 4.3295 = $8659.
A future-value annuity factor for 10 payments is 12.006. What is the future value of $500 paid each period for 10 periods?
$6003
FV = 500 × 12.006 = $6003.
Level 2 · Application
$250 is invested at the end of each year for 7 years at 5% p.a. compounded annually. The future-value interest factor for 7 payments at 5% is 8.1420. Calculate the future value of the annuity.
$2,035.50
FV = 250 × 8.1420 = $2,035.50.
$600 is invested at the end of each year for 10 years at 6% p.a. compounded annually. The future-value interest factor for 10 payments at 6% is 13.1808. Calculate the future value of the annuity.
$7,908.48
FV = 600 × 13.1808 = $7,908.48.
A loan is repaid by $1500 at the end of each year for 6 years at 7% p.a. The present-value interest factor for 6 payments at 7% is 4.7665. Calculate how much was borrowed (the present value).
$7,149.75
PV = 1500 × 4.7665 = $7,149.75.
Level 3 · Further Application
Sam saves for a $3200 goal in 3 years by making equal payments at the end of every six months into an account earning 4% p.a. compounded six-monthly (2% per half-year).
| Periods (n) | Factor at r = 2% |
|---|---|
| 6 | 6.3081 |
$510
(a) Interest is applied every six months, so over 3 years there are 6 periods; the six-monthly rate is 4% ÷ 2 = 2%.
(b) Payment = 3200 ÷ 6.3081 = $507.28. To reach the goal, round up to the nearest $10: $510.
Priya wants $10 000 in 4 years by paying equal amounts at the end of each year into an account earning 5% p.a. compounded annually.
| Periods (n) | Factor at r = 5% |
|---|---|
| 4 | 4.3101 |
$2330
(a) Interest is applied once a year for 4 years, so n = 4; the annual rate is 5%, so r = 5%.
(b) Payment = 10000 ÷ 4.3101 = $2320.13. To reach the goal, round up to the nearest $10: $2330.
Leo invests $450 at the end of every quarter for 2 years at 8% p.a. compounded quarterly (2% per quarter).
| Periods (n) | Factor at r = 2% |
|---|---|
| 8 | 8.5830 |
$3,862.35
(a) Interest is applied every quarter for 2 years, so there are 8 periods; the quarterly rate is 8% ÷ 4 = 2%.
(b) FV = payment × factor = 450 × 8.5830 = $3,862.35.