Present value annuities
A present value annuity is a style of investment from which individuals usually withdraw a regular amount of funds. The withdrawal in many cases is repeated until the funds run out, that is the value of the annuity reaches $\$0$$0. In our next lesson we will look at the special case of perpetuities where rather than the funds dwindling the investment value remains constant.
We can model present value annuities with a recurrence relation, this is similar to modelling future value annuities with the payment in this case being withdrawn and hence, negative in the form shown below.
For an annuity with initial investment $P$P, regular withdrawals of $d$d per period and earning compound interest at a rate of $i$i per payment period, the recursive sequence which generates the value, $A_n$An, of the investment at the end of each instalment period is:
$A_{n+1}=rA_n-d$An+1=rAn−d, where $A_0=P$A0=P and $r=\left(1+i\right)$r=(1+i)
The present value, $A$A, which is the amount of money that must be invested in order to sustain the payments for the given duration is:
$A=M\left(\frac{1-\left(1+i\right)^{-n}}{i}\right)$A=M(1−(1+i)−ni)
Where $M$M is the regular payment, $i$i is the interest rate per period, and $n$n is the total number of payment periods.
This can be rearranged to find the maximum payment that can be sustained by an initial deposit of $A$A:
$M=\frac{Ai}{1-\left(1+i\right)^{-n}}$M=Ai1−(1+i)−n
Worked example
Example 1
James wants to take two years off and would like an income of $\$3000$$3000 per month to live off.
(a) What amount does he need to invest into an account to sustain the payments over $2$2 years if the account earns $4.5%$4.5% p.a. compounded monthly?
Think: Use the formula for the present value of the investment with $M=3000$M=3000, $i=\frac{0.045}{12}=0.00375$i=0.04512=0.00375 and $n=24$n=24.
Do:
| $A$A | $=$= | $M\left(\frac{1-\left(1+i\right)^{-n}}{i}\right)$M(1−(1+i)−ni) |
| $=$= | $3000\left(\frac{1-\left(1+0.00375\right)^{-24}}{0.00375}\right)$3000(1−(1+0.00375)−240.00375) | |
| $=$= | $\$68\ 731.97$$68 731.97 |
James would require an investment of $\$68\ 731.97$$68 731.97 to receive the income he wanted.
(b) If he only invests $\$50\ 000$$50 000, what is the maximum he can withdraw each month if the money is to last two years?
Think: Use the formula to find the payment with $A=50\ 000$A=50 000, $i=0.00375$i=0.00375 and $n=24$n=24.
Do:
| $M$M | $=$= | $\frac{Ai}{1-\left(1+i\right)^{-n}}$Ai1−(1+i)−n |
| $=$= | $\frac{50\ 000\times0.00375}{1-\left(1+0.00375\right)^{-24}}$50 000×0.003751−(1+0.00375)−24 | |
| $=$= | $\$2182.39$$2182.39 |
James could afford to withdraw payments of up to $\$2182.39$$2182.39 to make the money last $2$2 years.
example 2
Tahlia receives an inheritance of $\$250\ 000$$250 000 and decides to invest the entire amount in an annuity earning $7.2%$7.2% per annum compounded monthly. At the end of each month, once the interest has been paid into her account, Tahlia withdraws $\$2000$$2000 to help pay for living expenses.
(a) Calculate the value of her annuity at the end of the first month.
Think: To calculate the value at the end of the first month, we will first need to add the interest owed to the investment and then subtract the withdrawal. Remember the annual rate of $7.2%$7.2% must be divided by $12$12 to get a monthly rate of $0.6%$0.6%, or $0.006$0.006.
Do:
| $\text{Value }$Value | $=$= | $250\ 000\left(1+0.006\right)-2000$250 000(1+0.006)−2000 |
| $=$= | $\$249500$$249500 |
(b) Write a recursive rule that gives the value of the annuity, $A_n$An, at the end of $n$n months.
Think: The recursive rule for the value at the end of the month has the form $A_{n+1}=rA_n-d$An+1=rAn−d, where $A_0$A0 is the initial investment. We have an initial investment of $A_0=250\ 000$A0=250 000, a monthly rate of $i=0.6%$i=0.6% and a monthly withdrawal of $d=2000$d=2000.
Do:
$A_{n+1}=1.006A_n-2000$An+1=1.006An−2000, where $A_0=250000$A0=250000
(c) Use a spreadsheet to determine during which year and month Tahlia's annuity will end.
Think: To do this we will need to use our recursive rule to create a table of values and then scroll through our table of values until we find the first time the value of the annuity becomes a negative value.
Do: Create a column for the month number and another for the balance of the account, enter the following formulas and drag down to generate the sequence:
| A | B | |
|---|---|---|
| 1 | $\text{Month}$Month | $A_n$An |
| 2 | $0$0 | $\$250\ 000$$250 000 |
| 3 | =A2+1 | =1.006*B2-2000 |
| 4 |
Scrolling through your created table you should see the following:
| $\text{Month}$Month | $A_n$An |
|---|---|
| $229$229 | $\$5422.07$$5422.07 |
| $230$230 | $\$3454.60$$3454.60 |
| $231$231 | $\$1475.33$$1475.33 |
| $232$232 | $-\$515.82$−$515.82 |
We can see the annuity first holds a negative value during month $232$232. Therefore, it will take $19.33$19.33 years. So this occurs after $19$19 years and $4$4 months.
(d) How much should Tahlia withdraw each month if she wishes her annuity to last indefinitely (become a perpetuity)?
Think: If Tahlia never wants her annuity to run out, then she should only withdraw the amount of interest earned in the first month, so that the value of her investment forever remains at $\$250000$$250000.
Do:
| $\text{Interest }$Interest | $=$= | $0.006\times250000$0.006×250000 |
| $=$= | $\$1500$$1500 |
(e) If Tahlia could find a higher interest rate for her investment, but kept the withdrawal amount the same, would the annuity end sooner or later?
Think: What does it mean if the interest rate was higher? It means she would earn more from her annuity each month.
Do: Therefore, if Tahlia earns more but withdraws the same amount, her annuity will last longer.
Practice questions
QUESTION 1
Lachlan received an inheritance of $\$100000$$100000. He invests the money at $8%$8% per annum with interest compounded annually at the end of the year. After the interest is paid, Lachlan withdraws $\$9000$$9000 and the amount remaining in the account is invested for another year.
How much is in the account at the end of the first year?
Write a recursive rule for $A_n$An in terms of $A_{n-1}$An−1 that gives the value of the account after $n$n years and an initial condition $A_0$A0.
Write both parts on the same line separated by a comma.
What is the value of the investment at the end of year $10$10?
Round your answer to the nearest cent.
By the end of which year will the annuity have run out?
Question 2
Bill has won $\$260000$$260000 and sets up an annuity earning $4.8%$4.8% interest per annum, compounded annually.
At the end of each year Bill withdraws $\$18000$$18000.
Complete the table below, rounding each answer to the nearest cent, and using the rounded answer to calculate the amounts for the following year.
Year
Balance at beginning of year ($) Interest ($) Withdrawal ($) Balance at end of year ($) 1 $260000$260000 $12480$12480 $18000$18000 $254480$254480 2 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 3 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 4 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ 5 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ If Bill had chosen to withdraw $\$18000$$18000 in twelve monthly instalments of $\$1500$$1500 instead of in one lump sum, what would be the result compared to the original situation? Choose the most accurate statement.
There would be no difference in the balance of his annuity at the end of 5 years.
AHe would have less money in his annuity at the end of 5 years.
BHe would have more money in his annuity at the end of 5 years.
C
Formulas for finding the future value
To find the future value of an investment after many payment periods we could use a spreadsheet or the following formula. This is similar to the formula for the future value of an investment with an initial deposit and regular payments but here we subtract the accumulated payments and their interest.
The future value of an investment with regular withdrawals is:
$A=P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=P(1+i)n−M((1+i)n−1i)
Where:
- $A$A is the future value of the investment
- $P$P is the principal investment
- $M$M is the payment per period
- $i$i is the interest rate per period
- $n$n is the total number of payment periods
For this formula we require payments per year and compounds per year to be equal.
If you are interested in how we get this formula, the extension section of this lesson explains it step-by-step.
Worked example
Example 3
Nadira invests $\$150\ 000$$150 000 at a rate of $9%$9% per annum compounded monthly. At the end of each month she withdraws $\$2000$$2000 from the investment after the interest is paid and the balance is reinvested in the account.
(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 $r=\left(1+\frac{0.09}{12}\right)$r=(1+0.0912). Substitute in the values for $r$r, $d$d and $A_0$A0.
Do: $A_{n+1}=\left(1.0075\right)\times A_n-2000$An+1=(1.0075)×An−2000, where $A_0=\$150\ 000$A0=$150 000
(b) Find the value of the investment after $4$4 months.
Think: Using our recursive rule we can see that in each step we multiply the previous value by $1.0075$1.0075 and subtract $2000$2000. As this is not a large number of repeated calculations we can find this value using the answer key on the calculator to find the balance at the end of each month for four months.
Do:
- Enter the principal amount $150\ 000$150 000 and press enter (
or equivalent). This is the starting balance. - Type in the next step using the answer key, in this case $1.0075\times$1.0075×
$-2000$−2000. 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 $\$146\ 460.43$$146 460.43
(c) Find the value of the investment after $5$5 years.
Think: $5$5 years is $60$60 payment periods. As this is a significant number of payment periods and keeping track using the answer key becomes difficult and time consuming, let's use the formula to find a future value.
Do:
| $A$A | $=$= | $P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{i}\right)$P(1+i)n−M((1+i)n−1i) |
| $=$= | $150\ 000\left(1+0.0075\right)^{60}-2000\left(\frac{\left(1+0.0075\right)^{60}-1}{0.0075}\right)$150 000(1+0.0075)60−2000((1+0.0075)60−10.0075) | |
| $=$= | $\$84\ 003.88$$84 003.88 |
Nadira would have $\$84\ 003.88$$84 003.88 in her account after five years.
Practice question
Question 3
Eileen invests an inheritance in an annuity from which she makes monthly withdrawals. This annuity pays and calculates her interest annually.
The balance of her annuity (in dollars) at the end of the $n$nth year can be defined recursively as:
$V_n=\left(1+0.064\right)\times V_{n-1}-30000$Vn=(1+0.064)×Vn−1−30000; where $V_0=280000$V0=280000.
How much did Eileen inherit?
What annual interest rate does she earn on her annuity?
What is the monthly amount she withdraws?
What is the balance of her annuity after four years? Give your answer correct to the nearest cent.
During which year will her annuity close?
Modelling a present value annuity with financial solvers
We can solve problems involving present value annuities using an online financial solver by setting the present value (PV) as the initial value invested and future value (FV) as the remaining funds at the end of the investment (often $\$0$$0). The present value should be entered as a negative to indicate depositing the money for investment and the payment (PMT) will be positive as this is returned to the investor.
We can also use the online TVM solver to explore these problems.
Practice question
QUESTION 4
Kathleen invests $\$110000$$110000 at a rate of $12%$12% per annum compounded monthly.
We will use a financial solver to determine what Kathleen's equal monthly withdrawal should be if she wants the investment to last $30$30 years.
Fill in the value for each of the following. Type an $X$X next to the variable we wish to solve for.
$N$N $\editable{}$ $I$I$%$% $\left(\editable{}\right)%$()% $PV$PV $\editable{}$ $PMT$PMT $\editable{}$ $FV$FV $\editable{}$ $P$P$/$/$Y$Y $\editable{}$ $C$C$/$/$Y$Y $\editable{}$ Hence determine the amount of the monthly withdrawal.
Give your answer to the nearest cent.
Deriving the future value of investment formula
We know that we start with a principal investment $P$P which gains interest each compounding period according to some rate $i$i, and after this we then withdraw a payment of $M$M.
Let's start by representing this information in an equation for the first few compounding periods:
| $V_0$V0 | $=$= | $P$P |
| $V_1$V1 | $=$= | $P\left(1+i\right)-M$P(1+i)−M |
| $V_2$V2 | $=$= | $\left(P\left(1+i\right)-M\right)\times\left(1+i\right)-M$(P(1+i)−M)×(1+i)−M |
| $=$= | $P\left(1+i\right)^2-M\left(\left(1+i\right)+1\right)$P(1+i)2−M((1+i)+1) | |
| $V_3$V3 | $=$= | $\left(P\left(1+i\right)^2-M\left(\left(1+i\right)+1\right)\right)\times\left(1+i\right)-M$(P(1+i)2−M((1+i)+1))×(1+i)−M |
| $=$= | $P\left(1+i\right)^3-M\left(\left(1+i\right)^2+\left(1+i\right)+1\right)$P(1+i)3−M((1+i)2+(1+i)+1) |
Because the sequence is recursive, we begin to see a pattern emerge after expanding and simplifying the result of each compounding period. In fact, we can generalise these equations to get:
$V_n=P\left(1+i\right)^n+M\left((1+i)^{n-1}+(1+i)^{n-2}+\ldots+(1+i)^2+(1+i)+1\right)$Vn=P(1+i)n+M((1+i)n−1+(1+i)n−2+…+(1+i)2+(1+i)+1)
However, this looks quite messy, so we also want to simplify this equation by applying a useful equivalence.
For any sequence of powers we have the equivalence:
$x^{n-1}+x^{n-2}+\ldots+x^2+x+1=\frac{x^n-1}{x-1}$xn−1+xn−2+…+x2+x+1=xn−1x−1
Test this equivalence for yourself by checking that:
$\left(x^{n-1}+x^{n-2}+\ldots+x^2+x+1\right)(x-1)=x^n-1$(xn−1+xn−2+…+x2+x+1)(x−1)=xn−1
Using this equivalence, and letting $A=V_n$A=Vn be the future value after $n$n compounding periods, we can rewrite the general equation as:
$A=P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{1+i-1}\right)$A=P(1+i)n−M((1+i)n−11+i−1)
which simplifies to:
$A=P\left(1+i\right)^n-M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=P(1+i)n−M((1+i)n−1i)
where $A$A is the future value after $n$n compounding periods, $P$P is the principal investment, $i$i is the interest rate per compounding period, and $M$M is the amount being withdrawn at the end of each compounding period.
and if we instead make regular deposits, we add $M$M instead of subtracting $M$M each compounding periods which gives us:
$A=P\left(1+i\right)^n+M\left(\frac{\left(1+i\right)^n-1}{i}\right)$A=P(1+i)n+M((1+i)n−1i)
where $A$A is the future value after $n$n compounding periods, $P$P is the principal investment, $i$i is the interest rate per compounding period, and $M$M is the amount being deposited at the end of each compounding period.