Annuities are investments which involve series of payments made at regular intervals and come in two main forms:
- Future value annuities: This is where the investor deposits regular payments to a to save for a future requirement or goal. Here the investor is saving to a future value.
- Present value annuities: Is where an investor withdraws regular payments often used as part of an income such as after retirement. Here the investor is starting with a present value and drawing down from it.
This lesson will look at future value annuities - investments with regular payments and our next lesson will look at present value annuities - investments with regular withdrawals.
Future value annuities with no initial deposit
When saving for a future goal it is not generally realistic to invest a lump sum and leave it to accumulate interest. Making regular payments into an account where interest is regularly compounded is a common approach to reach a savings goal.
Focusing on investments consisting of regular payments without an initial lump sum deposit, we have the following models and formulas:
For an annuity with no initial deposit, regular deposits of $d$d per period and earning compound interest at a rate of $i$i per payment period, the recursive sequence which generates the value, $A_n$An, of the investment at the end of each instalment period is:
$A_{n+1}=rA_n+d$An+1=rAn+d, where $A_0=0$A0=0 and $r=\left(1+i\right)$r=(1+i)
For an investment with regular payments and no initial deposit the future value of the investment, $A$A, can be found using the following formula:
$A=M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=M((1+i)n−1i)
Where $M$M is the regular payment, $i$i is the interest rate per period, and $n$n is the total number of payment periods.
This can be rearranged to find the payment, $M$M, required to reach a given goal, $A$A:
$M=\frac{Ai}{\left(1+i\right)^n-1}$M=Ai(1+i)n−1
Worked examples
Example 1
Laura wants to save $\$5800$$5800 for a holiday in three years time. She plans to make regular monthly payments into an account paying $2.7%$2.7% p.a. compounded monthly.
(a) What is the least amount Laura would need to deposit each month so that she reaches her goal in time for her trip?
Think: We have a goal of $A=\$5800$A=$5800, an interest rate of $i=\frac{0.027}{12}$i=0.02712 per month and a time frame of $n=36$n=36 months. Substituting these values into the formula for the payment we have:
Do:
| $M$M | $=$= | $\frac{Ai}{\left(1+i\right)^n-1}$Ai(1+i)n−1 |
| $=$= | $\frac{5800\times0.00225}{\left(1+0.00225\right)^{36}-1}$5800×0.00225(1+0.00225)36−1 | |
| $=$= | $\$154.86$$154.86 |
Hence, Laura will need to deposit $\$154.86$$154.86 per month to reach her goal.
(b) Laura can afford to set aside $\$200$$200 each month. How much sooner will she reach her goal?
Think: Using the formula for the future value of the investment with $M=200$M=200 and $i=0.00225$i=0.00225, we can use guess and check for values of $n$n to find when $A>5800$A>5800. Since $\$154.86$$154.86 took $36$36 months, trying values around $30$30 months is a good starting point.
Do:
$n=28$n=28,
| $A$A | $=$= | $M\left(\frac{\left(1+i\right)^n-1}{i}\right)$M((1+i)n−1i) |
| $=$= | $200\left(\frac{\left(1+0.00225\right)^{28}-1}{0.00225}\right)$200((1+0.00225)28−10.00225) | |
| $=$= | $\$5773.46$$5773.46 |
$n=29$n=29,
| $A$A | $=$= | $M\left(\frac{\left(1+i\right)^n-1}{i}\right)$M((1+i)n−1i) |
| $=$= | $200\left(\frac{\left(1+0.00225\right)^{29}-1}{0.00225}\right)$200((1+0.00225)29−10.00225) | |
| $=$= | $\$5986.45$$5986.45 |
Hence it will take Laura $29$29 months to reach her goal which is $7$7 months earlier.
(c) If she could not go on the trip early but continued to put aside $\$200$$200 each month, how much additional money will she have to spend on the holiday over the original budget?
Think: Use the formula for the future value of the investment with $M=200$M=200, $i=0.00225$i=0.00225 and $n=36$n=36.
Do:
| $A$A | $=$= | $M\left(\frac{\left(1+i\right)^n-1}{i}\right)$M((1+i)n−1i) |
| $=$= | $200\left(\frac{\left(1+0.00225\right)^{36}-1}{0.0025}\right)$200((1+0.00225)36−10.0025) | |
| $=$= | $\$7490.87$$7490.87 |
Hence, Laura will have $\$7490.87-\$5800=\$1690.87$$7490.87−$5800=$1690.87 additional to take on her trip.
Practice questions
Question 1
Uther invests $\$2300$$2300 each year for $30$30 years into an account with an interest rate of $2.5%$2.5%p.a., compounded annually. Determine the future value $A$A of this annuity at the end of the investment period, rounding your answer to the nearest cent.
Question 2
To save money for a master’s degree, you deposit $\$2880$$2880 at the end of each year in an annuity with a rate of $3%$3% compounded annually.
How much money $A$A will you have saved at the end of $4$4 years?
Give your answer to the nearest cent.
Find the total interest earned.
Give your answer to the nearest cent.
A different bank offers you an investment plan where the annual interest rate is the same, but interest is compounded monthly and you make monthly contributions of $\$240$$240.
Under this plan, how much money $A$A would you have saved after $4$4 years?
Tables and sequences for investments with a regular payment (including initial deposit)
Often an investor will make an initial deposit as well as a regular payment to contribute to the growth of the investment. Tables and recursive rules are useful for more complex finance problems, such as a compound interest investment with regular deposits, or a reducing balance loan with regular payments. Investments involving compound interest are often displayed in a table of values so that the growth in the value of the investment can be clearly seen.
The sequence to generate the value of investment is the same as given for future value annuities in the box above with the exception of the initial value now set as the principal investment, that is $A_0=P$A0=P.
Worked example
Example 2
Fiona invests $\$14000$$14000 in an investment account that compounds annually. To further grow the investment she also makes a regular contribution of $\$2000$$2000 per year. The progression of the investment is shown in the table below.
| Year | Account balance at start of year ($) |
Interest ($) | Payment ($) | Account balance at end of year ($) |
|---|---|---|---|---|
| 1 | $14000$14000 | $350$350 | $2000$2000 | $16350$16350 |
| 2 | $16350$16350 | $408.75$408.75 | $2000$2000 | $18758.75$18758.75 |
| 3 | $18758.75$18758.75 | $468.97$468.97 | $2000$2000 | $21227.72$21227.72 |
(a) Determine the interest rate for the investment.
Think: We can use any of the table rows to do this. Use the formula
$\text{interest rate }=\frac{\text{amount of interest in year n}}{\text{account balance start of year n }}\times100$interest rate =amount of interest in year naccount balance start of year n ×100%
Do: Using the figures from Year 1 we find:
| Rate | $=$= | $\frac{350}{14000}$35014000 |
| $=$= | $0.025$0.025 | |
| $=$= | $2.5%$2.5% |
Reflect: We can check we obtain the same result using the values from Year 2 or Year 3.
(b) Write a recursive rule for $A_{n+1}$An+1 in terms of $A_n$An that gives the value of the account after $n$n years and an initial condition $A_0$A0.
Think: Using the sequence which generates the value, $A_n$An, of the investment at the end of each instalment, $A_{n+1}=rA_n+d$An+1=rAn+d, where $A_0=P$A0=P and $r=\left(1+i\right)$r=(1+i). We have an initial investment of $A_0=14000$A0=14000, a regular deposit of $d=2000$d=2000 and using the rate we found in part (a) we have $r=(1+0.025)$r=(1+0.025).
Do:
$A_{n+1}=(1+0.025)\times A_n+2000$An+1=(1+0.025)×An+2000, where $A_0=14000$A0=14000.
Example 3
Patricia invests $\$1500$$1500 at $4.2%$4.2% p.a. compounded monthly and makes an additional monthly payment of $\$100$$100 at the end of each month.
(a) Write a recurrence relation for this situation, where $A_n$An is the balance at the end of the $n$nth month and $A_0$A0 is the initial investment.
Think: Using the form $A_{n+1}=rA_n+d$An+1=rAn+d , where $A_0=$A0= the initial investment and $r=\left(1+\frac{0.042}{12}\right)$r=(1+0.04212). Substitute in the values for $r$r and $A_0$A0.
Do: $A_{n+1}=\left(1.0035\right)\times A_n+100$An+1=(1.0035)×An+100, where $A_0=\$1500$A0=$1500
(b) Find the value of the investment after $4$4 months.
Think: Using our recursive rule we can see each step we multiply the previous value by $1.0035$1.0035 and add $100$100. We can use the answer key on the calculator to find the balance at the end of each month for four months.
Do:
- Enter the principal amount $1500$1500 and press enter (
or equivalent). This is the starting balance - Type in the next step using the answer key, in this case $1.0035\times$1.0035×
+$100$100. This is the balance after $1$1 month. - Continue to press to obtain the next term in the sequence. One press: balance after $2$2 months, Two presses: balance after $3$3 months, $\ldots$….
The balance after $4$4 months is $\$1923.22$$1923.22
(c) Determine how many whole months it takes for the balance to exceed twice the original investment.
Think: Continue using the sequence as above in the calculator and count until the balance first exceeds $\$3000$$3000.
Do: At $4$4 months the balance was $\$1923.22$$1923.22, pressing ten more times we see the balance first exceed $\$3000$$3000, so it will take $14$14 months.
Practice question
Question 3
Jenny is saving for a European holiday which will cost $\$15000$$15000. She puts the $\$5000$$5000 she has already saved in a savings account which offers interest compounded quarterly. She also makes a quarterly contribution of $\$350$$350, at the end of each quarter.
The progression of the investment is shown in the table below.
| Quarter | Account balance at start of quarter ($\$\quad$$ ) | Interest ($\$\quad$$ ) | Payment ($\$\quad$$ ) | Account balance at end of quarter ($\$\quad$$ ) |
|---|---|---|---|---|
| $1$1 | $5000$5000 | $46.88$46.88 | $350$350 | $5396.88$5396.88 |
| $2$2 | $5396.88$5396.88 | $50.60$50.60 | $350$350 | $5797.48$5797.48 |
| $3$3 | $5797.48$5797.48 | $A$A | $350$350 | $B$B |
Find the interest rate per quarter for this account, as a percentage to four decimal places.
Calculate the value of $A$A in dollars.
Round your answer to the nearest cent.
Calculate the value of $B$B in dollars.
Round your answer to the nearest cent.
Write a recursive rule, $T_{n+1}$Tn+1, in terms of $T_n$Tn that gives the balance in the account at the beginning of the $\left(n+1\right)$(n+1)th quarter.
$T_{n+1}=$Tn+1=$\editable{}$$\times T_n+$×Tn+$\editable{}$
$T_1=$T1=$\editable{}$
Using your calculator, how many whole quarters will it take to until she has enough money to go on the holiday?
If she changes the payment to $\$400$$400 per quarter, how much sooner, in whole quarters, will she be able to go on holidays?
Formulas for finding the future value
To find the future value of an investment after many payment periods a formula can be useful. For example if we wanted to find the future value of in investment compounded monthly with monthly repayments, it would be inconvenient to use $60$60 repeated calculations in a scientific calculator. We could use a spreadsheet or the following formulas.
Recall the future value of an investment without regular payments can be found using the compound interest formula:
$A=P\left(1+i\right)^n$A=P(1+i)n
Where $A$A is the future value of the investment, $P$P is the principal amount invested, $i$i is the interest rate per period, and $n$n is the total number of compound periods.
And from the start of the lesson we have future value of an investment with regular payments but no initial deposit the can be found using the following formula:
$A=M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=M((1+i)n−1i)
Where $A$A is the future value of the investment, $M$M is the regular payment, $i$i is the interest rate per period, and $n$n is the total number of payment periods.
Combing these we can find the future value of an investment with regular payments and an initial deposit:
$A=P\left(1+i\right)^n+M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=P(1+i)n+M((1+i)n−1i)
For this formula we require payments per year and compounds per year to be equal.
Worked example
Example 4
Michael makes regular deposits of $\$250$$250 at the end of each month into an account earning $4.5%$4.5% p.a. compounded monthly.
(a) If Michael makes no initial deposit, write a recurrence relation for this situation, where $A_n$An is the balance at the end of the $n$nth month and $A_0$A0 is the initial investment.
Think: Using the form $A_{n+1}=rA_n+d$An+1=rAn+d , where $A_0=$A0= the initial investment and $r=\left(1+\frac{0.045}{12}\right)$r=(1+0.04512). Substitute in the values for $r$r , $d$d and $A_0$A0.
Do: $A_{n+1}=\left(1.00375\right)\times A_n+250$An+1=(1.00375)×An+250, where $A_0=\$0$A0=$0
(b) Find the value of the investment after $4$4 months.
Think: Using our recursive rule we can see each step we multiply the previous value by $1.00375$1.00375 and add $250$250. As this is not a large number of repeated calculations we can find this value as we did in example 3 by using the answer key on the calculator to find the balance at the end of each month for four months.
Do: Following the steps from example 3 we find the value after $4$4 months is $\$1005.64$$1005.64.
(c) Find the value of the investment after $3$3 years.
Think: $3$3 years is $36$36 payment periods, as this is a significant number of payment periods and keeping track using the answer key becomes more difficult. Let's use the formula to find a future value with no initial deposit.
Do:
| $A$A | $=$= | $M\left(\frac{\left(1+i\right)^n-1}{i}\right)$M((1+i)n−1i) |
| $=$= | $250\left(\frac{\left(1+0.00375\right)^{36}-1}{0.00375}\right)$250((1+0.00375)36−10.00375) | |
| $=$= | $\$9616.52$$9616.52 |
Michael would have $\$9616.52$$9616.52 in his account after three years.
(d) If Michael had made an initial deposit of $\$2000$$2000 at the beginning of the first year, what would be the value of the investment after three years?
Think: We need to include this initial deposit and interest earned on it, notice this is what the final formula does by adding the previous calculation to the compound interest formula with the initial deposit.
Do:
| $A$A | $=$= | $P\left(1+i\right)^n+M\left(\frac{\left(1+i\right)^n-1}{i}\right)$P(1+i)n+M((1+i)n−1i) |
| $=$= | $2000\left(1+0.00375\right)^{36}+250\left(\frac{\left(1+0.00375\right)^{36}-1}{0.00375}\right)$2000(1+0.00375)36+250((1+0.00375)36−10.00375) | |
| $=$= | $\$11\ 905.02$$11 905.02 |
Michael would have $\$11\ 905.02$$11 905.02 in his account after three years with the initial deposit included.
Practice question
Question 4
Mr Jones opened a bank account for his granddaughter Susana on the day she was born: January 5, 2005. He deposited $\$2000$$2000.
Mrs Jones, Susana’s grandmother, also deposited money into this account on that day, and continues to do so by depositing $\$400$$400 every $3$3 months.
The balance at the end of each quarter for this investment, where interest is compounded quarterly, is given by $A_n=1.03A_{n-1}+400$An=1.03An−1+400, $A_0=2400$A0=2400.
State the quarterly interest rate.
State the nominal annual interest rate.
Determine the balance on the day after Susana's first birthday.
Round your answer to the nearest cent.
Determine the balance on the day after Susana's $13$13th birthday.
Round your answer to the nearest cent.
Using spreadsheets to model compound interest with payments
We previously introduced modelling compound interest with spreadsheets. Spreadsheets can also include payment details and are a useful tool for solving financial problems as the progression of the investment can be clearly seen as well as the effect of changing interest rates and payments.
Worked example
Example 5
Thomas needs $\$15\ 000$$15 000 for a house deposit. He invests $\$10\ 000$$10 000 in the bank with an interest rate of $3%$3% p.a. compounded monthly and also makes a payment of $\$150$$150 per month. He creates the following spreadsheet to help him do "what if analysis" to examine the problem.
| A | B | C | D | E | |
|---|---|---|---|---|---|
| 1 | Principal | $\$10\ 000$$10 000 | |||
| 2 | Annual interest rate | $3%$3% | |||
| 3 | Compounds per year | $12$12 | |||
| 4 | Payment per period | $\$150$$150 | |||
| 5 | |||||
| 6 | Month | Balance start of month | Interest | Payment | Balance end of month |
| 7 | 1 | $\$10\ 000$$10 000 | $\$25$$25 | $\$150$$150 | $\$10\ 175$$10 175 |
| 8 | 2 | $\$10\ 175$$10 175 | $\$25.44$$25.44 | $\$150$$150 | $\$10\ 350.44$$10 350.44 |
Some of the formulae Thomas used to create this spreadsheet are shown in the table below.
| A | B | C | D | E | |
|---|---|---|---|---|---|
| 6 | Month | Balance start | Interest | Payment | Balance end |
| 7 | 1 | =B1 | =$B$2/$B$3*B7 | =$B$4 | =B7+C7+D7 |
| 8 | =A7+1 | =E8 | |||
| 9 |
Note:
- The spreadsheet is designed so that if the values in cells B1, B2, B3 and B4 are changed, the whole page instantly updates. This makes it quick and easy to investigate different investment options.
- The $ signs in the cell references make the reference absolute. That means the cell name will not change as the formula is copied down the column.
- The month numbers here have been calculated using a formula in cell A8.
(a) What is the purpose of the formula in cell C7?
Think: Ignore the $ signs. The formula is =B2/B3*B7, which calculates $3%\div12\times\$10\ 000$3%÷12×$10 000
Do: It calculates the amount of monthly interest by taking the yearly interest rate (cell B2) and dividing by the number of compounds per year (cell B3) then multiplying by the amount at the start of the month (cell B7).
(b) What is the purpose of the formula in cell E7?
Think: The formula is =B7+C7+D7 so it is added three values together.
Do: It calculates the balance at the end of the month by calculating start balance + interest + payment.
(c) Write the formula that will be in cell E8.
Think: This will be the start amount + interest + payment for row 8.
Do: The formula is =B8+C8+D8
(d) Use a spreadsheet program to recreate this spreadsheet. How long does it take Thomas to save the $\$15\ 000$$15 000?
Reflect: The spreadsheet should show that it takes him $28$28 months to save the money.
Practice question
question 5
The spreadsheet below shows the first year of an investment with regular deposits.
| $A$A | $B$B | $C$C | $D$D | $E$E | |
| $1$1 | Year | Beginning Balance | Interest | Deposit | End Balance |
| $2$2 | $1$1 | $6000$6000 | $660$660 | $500$500 | $7160$7160 |
| $3$3 | |||||
| $4$4 | |||||
| $5$5 |
Calculate the annual interest rate for this investment.
Write a formula for cell $B$B$3$3.
Enter only one letter or number per box.
$B$B$3$3 $=$= $\editable{}$$\editable{}$
Write a formula for cell $C$C$6$6 in terms of $B$B$6$6.
Enter one letter or a number per box.
$C$C$6$6 $=$= $\editable{}$$\ast$∗$\editable{}$$\editable{}$
Which of the following is the correct formula for cell $E$E$5$5?
$B$B $5$5 $\ast$∗ $C$C $5$5 $+$+ $D$D $5$5
A$B$B $5$5 $+$+ $C$C $5$5 $-$− $D$D $5$5
B$B$B $5$5 $+$+ $C$C $5$5 $+$+ $D$D $5$5
C$B$B $5$5 $+$+ $C$C $5$5 $\ast$∗ $D$D $5$5
D$B$B $5$5 $\ast$∗ $C$C $5$5 $\ast$∗ $D$D $5$5
E$B$B $5$5 $-$− $C$C $5$5 $-$− $D$D $5$5
FUse technology to reproduce this spreadsheet and determine the end balance for the $4$4th year.
Calculate the total interest earned over these $4$4 years.
Financial solvers for investments with a regular payment
An online TVM solver can be a powerful tool for financial problems when used correctly. The examples above can also be answered using a financial solver.
- If you are investing money then PV is negative and FV is positive. Hint: think of investing as 'giving' your money to the bank so from your point of view the money is negative.
- If you are borrowing money PV is positive and FV is negative. Hint: think of borrowing as 'receiving' money from the bank so from your point of view the money is positive.
- Payments (PMT) made to the bank for either investments or loans are negative, again we can think of this as 'giving' your money to the bank.
- N is the total number of payments.
We can also use the online TVM solver to explore these problems.
Practice questions
question 6
Valerie is saving for a European holiday which will cost $\$20000$$20000. She puts the $\$8000$$8000 she has already saved in a savings account which offers $3.75%$3.75% p.a compounded quarterly. She also makes a quarterly contribution of $\$400$$400. Valerie is interested in finding how long it will take her to save up for her holiday.
If $N$N is the number of payments, complete the table of values showing the variables required to use a financial solver to solve this problem.
Variable Value $N$N - $I\left(%\right)$I(%) $\editable{}$$%$% $PV$PV $\editable{}$ $Pmt$Pmt $\editable{}$ $FV$FV $\editable{}$ $P/Y$P/Y $\editable{}$ $C/Y$C/Y $\editable{}$ Determine the number of whole quarters it will take for Valerie to save the $\$20000$$20000 required for the holiday.
question 7
An investor deposits $\$20000$$20000 into a high earning account with interest of $4.5%$4.5% p.a. compounded weekly and makes $\$150$$150 weekly deposits into the account.
If $N$N is the number of payments, complete the table of values showing the variables required to use a financial solver to determine how long it takes for the original investment to double in value.
Assume there are $52$52 weeks in a year.
Variable Value $N$N - $I\left(%\right)$I(%) $\editable{}$$%$% $PV$PV $\editable{}$ $Pmt$Pmt $\editable{}$ $FV$FV $\editable{}$ $P/Y$P/Y $\editable{}$ $C/Y$C/Y $\editable{}$ Calculate the number of whole weeks it will take for the investment to double.
How much should the investor deposit each week in dollars if they want the original investment to double at the end of three years?
Again, assume there are $52$52 weeks in a year.
Round your answer to the nearest cent.