Coaching Centre
Future value interest factor tables
Using a table of future value interest factors to find the future value of an annuity, and working backwards to find the regular contribution needed to reach a savings goal.
The recurrence relation in 3D works, but $120$ rows is a lot of button-pressing. A future value interest factor table has already done the work for a contribution of $\$1$. Multiply by your contribution and you are done. HSC exam papers supply these tables, so the skill is reading them correctly.
The future value table
Each entry is the future value of an annuity of $\$1$ per period. The columns are the interest rate per period and the rows are the number of periods.
| Periods | $0.5\%$ | $1\%$ | $2\%$ | $3\%$ | $4\%$ | $5\%$ | $6\%$ |
|---|---|---|---|---|---|---|---|
| $1$ | $1.0000$ | $1.0000$ | $1.0000$ | $1.0000$ | $1.0000$ | $1.0000$ | $1.0000$ |
| $2$ | $2.0050$ | $2.0100$ | $2.0200$ | $2.0300$ | $2.0400$ | $2.0500$ | $2.0600$ |
| $3$ | $3.0150$ | $3.0301$ | $3.0604$ | $3.0909$ | $3.1216$ | $3.1525$ | $3.1836$ |
| $4$ | $4.0301$ | $4.0604$ | $4.1216$ | $4.1836$ | $4.2465$ | $4.3101$ | $4.3746$ |
| $5$ | $5.0503$ | $5.1010$ | $5.2040$ | $5.3091$ | $5.4163$ | $5.5256$ | $5.6371$ |
| $6$ | $6.0755$ | $6.1520$ | $6.3081$ | $6.4684$ | $6.6330$ | $6.8019$ | $6.9753$ |
| $7$ | $7.1059$ | $7.2135$ | $7.4343$ | $7.6625$ | $7.8983$ | $8.1420$ | $8.3938$ |
| $8$ | $8.1414$ | $8.2857$ | $8.5830$ | $8.8923$ | $9.2142$ | $9.5491$ | $9.8975$ |
| $9$ | $9.1821$ | $9.3685$ | $9.7546$ | $10.1591$ | $10.5828$ | $11.0266$ | $11.4913$ |
| $10$ | $10.2280$ | $10.4622$ | $10.9497$ | $11.4639$ | $12.0061$ | $12.5779$ | $13.1808$ |
| $12$ | $12.3356$ | $12.6825$ | $13.4121$ | $14.1920$ | $15.0258$ | $15.9171$ | $16.8699$ |
| $15$ | $15.5365$ | $16.0969$ | $17.2934$ | $18.5989$ | $20.0236$ | $21.5786$ | $23.2760$ |
| $20$ | $20.9791$ | $22.0190$ | $24.2974$ | $26.8704$ | $29.7781$ | $33.0660$ | $36.7856$ |
| $24$ | $25.4320$ | $26.9735$ | $30.4219$ | $34.4265$ | $39.0826$ | $44.5020$ | $50.8156$ |
Why is the factor for $1$ period exactly $1$? The single $\$1$ goes in at the end of the period and has earned no interest yet. The factor for $2$ periods at $5\%$ is $2.05$: the first dollar has earned $5$ cents and the second has just gone in.
Using the table
Kiara deposits $\$3000$ at the end of each year into an account paying $5\%$ p.a. compounded yearly. Find the value after $10$ years.
Ben saves $\$250$ at the end of every month for $2$ years at $6\%$ p.a. compounded monthly. Find the future value.
Choose the rate p.a., how often contributions are made and the number of years. The reader converts them to $r$ per period and $n$, then highlights the cell to use. Check Examples 1 and 2, then switch to Find the contribution and check Example 3.
Finding the contribution for a goal
If you know the goal, divide by the factor to find the regular contribution.
Lucas wants $\$50\,000$ in $8$ years for a house deposit. His investment earns $6\%$ p.a. compounded yearly. How much must he deposit at the end of each year?
Your turn
Ava contributes $\$500$ at the end of every quarter for $5$ years. Her account pays $8\%$ p.a. compounded quarterly. Find the future value.
Show the working
$r = 8\% \div 4 = 2\%$ per quarter, $n = 5 \times 4 = 20$. Factor $= 24.2974$. $FV = 500 \times 24.2974 = \$12\,148.70$.
Use the table to find the future value interest factor for $3\%$ per period and $12$ periods.
Read the $3\%$ column, row $12$: $14.1920$.
$\$2000$ is deposited at the end of each year for $15$ years at $4\%$ p.a. compounded yearly. Find the future value.
Factor at $4\%$, $15$ periods $= 20.0236$. $FV = 2000 \times 20.0236 = \$40\,047.20$.
Contributions are made monthly for $2$ years at $6\%$ p.a. compounded monthly. Which column and row of the table should be used?
$r = 6\% \div 12 = 0.5\%$ per month and $n = 2 \times 12 = 24$ months.
$\$500$ is contributed at the end of every quarter for $5$ years at $8\%$ p.a. compounded quarterly. Find the future value.
$r = 2\%$, $n = 20$. Factor $= 24.2974$. $FV = 500 \times 24.2974 = \$12\,148.70$.
How much must be deposited at the end of each year for $10$ years at $4\%$ p.a. to reach $\$20\,000$? Answer to the nearest cent.
Factor $= 12.0061$. Contribution $= \dfrac{20\,000}{12.0061} \approx \$1665.82$.
$\$1000$ is deposited at the end of each year for $20$ years at $6\%$ p.a. Find the interest earned.
Factor $= 36.7856$. $FV = 1000 \times 36.7856 = \$36\,785.60$. Contributions $= 20 \times 1000 = \$20\,000$. Interest $= \$16\,785.60$.
Why is every factor in the row for $1$ period equal to $1.0000$?
Contributions are made at the end of each period, so after one period the account holds just the $\$1$ that has just gone in.
Lucas needs $\$50\,000$ in $8$ years. His investment earns $6\%$ p.a. compounded yearly. Find the yearly contribution, to the nearest cent.
Factor $= 9.8975$. Contribution $= \dfrac{50\,000}{9.8975} \approx \$5051.78$.
Exam style: Chloe wants $\$15\,000$ in $5$ years. She can make equal contributions at the end of each quarter into an account paying $8\%$ p.a. compounded quarterly. Find the quarterly contribution, to the nearest cent.
$r = 8\% \div 4 = 2\%$, $n = 20$. Factor $= 24.2974$. Contribution $= \dfrac{15\,000}{24.2974} \approx \$617.35$.
Exam style: Plan A: $\$2400$ at the end of each year for $10$ years at $5\%$ p.a. compounded yearly. Plan B: $\$1200$ at the end of every half-year for $10$ years at $4\%$ p.a. compounded half-yearly. Both put in $\$24\,000$. How much more is the better plan worth?
A: factor $(5\%, 10) = 12.5779$, $FV = 2400 \times 12.5779 = \$30\,186.96$. B: $r = 2\%$, $n = 20$, factor $= 24.2974$, $FV = 1200 \times 24.2974 = \$29\,156.88$. Plan A is worth $30\,186.96 - 29\,156.88 = \$1030.08$ more.