Making regular withdrawals
In a reverse situation to annuities where we make regular repayments, we may start with a single sum investment from which regular, equal amounts are withdrawn. A common example is of this type of annuity is superannuation.
Say you retire with a superannuation fund of $\$500000$$500000. The interest compounds monthly at $3%$3% per annum. You withdraw $\$5000$$5000 at the end of each month.
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After $1$1 month, the initial $\$500000$$500000 will have earned interest of $0.25%$0.25%, and then the first $\$5000$$5000 will be withdrawn, so there will be $\$\left(500000\times1.0025-5000\right)$$(500000×1.0025−5000) left in the account.
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This balance will then earn a month’s interest before another $\$5000$$5000 is withdrawn, and so on.
The funds in the account will decrease and eventually the balance will be $\$0$$0 because the regular withdrawals are more than the interest we make on the balance. So in this case, we’d be interested in how long our superannuation will last us. Unsurprisingly, this kind of annuity can also be modelled as a geometric series.
Worked example
Example 1
Murray retires with $\$320000$$320000 in his superannuation account. He wishes to make monthly withdrawals for the next $25$25 years, while the account will continue to earn interest at $6.48%$6.48% p.a. compounding monthly. How much can he withdraw each month?
Think: When modelling these annuities, we must consider the balance of the account at the end of each month. Let $A_n$An be the balance after $n$n months. $25$25 years is $300$300 months, and the monthly interest rate is $0.54%$0.54% per month.
Do: After one month, Murray's account will have earned interest before he withdraws his monthly amount, $M$M:
$A_1=320000(1.0054)-M$A1=320000(1.0054)−M
$A_1$A1 becomes the starting balance for the $A_2$A2 calculation. This accrues interest, then another withdrawal $M$M is made:
| $A_2$A2 | $=$= | $A_1\times1.0054-M$A1×1.0054−M |
| $=$= | $(320000(1.0054)-M)\times1.0054-M$(320000(1.0054)−M)×1.0054−M | |
| $=$= | $320000(1.0054)^2-1.0054M-M$320000(1.0054)2−1.0054M−M |
The pattern continues:
| $A_3$A3 | $=$= | $A_2\times1.0054-M$A2×1.0054−M |
| $=$= | $(320000(1.0054)^2-1.0054M-M)\times1.0054-M$(320000(1.0054)2−1.0054M−M)×1.0054−M | |
| $=$= | $320000(1.0054)^3-(1.0054)^2M-1.0054M-M$320000(1.0054)3−(1.0054)2M−1.0054M−M |
The next two expressions would be:
| $A_4$A4 | $=$= | $320000(1.0054)^4-(1.0054)^3M-(1.0054)^2M-1.0054M-M$320000(1.0054)4−(1.0054)3M−(1.0054)2M−1.0054M−M |
| $A_5$A5 | $=$= | $320000(1.0054)^5-(1.0054)^4M-(1.0054)^3M-(1.0054)^2M-1.0054M-M$320000(1.0054)5−(1.0054)4M−(1.0054)3M−(1.0054)2M−1.0054M−M |
We can see that the first term accrues interest with its power matching the number of months. The number of withdrawals also matches $n$n, with the powers stepping down from $(n-1)$(n−1). This gives us the general expression:
$A_n=320000(1.0054)^n-(1.0054)^{n-1}M-(1.0054)^{n-2}M-(1.0054)^{n-3}M-\ldots-1.0054M-M$An=320000(1.0054)n−(1.0054)n−1M−(1.0054)n−2M−(1.0054)n−3M−…−1.0054M−M
And in this case, where $n=300$n=300, we have
$A_{300}=320000(1.0054)^{300}-(1.0054)^{299}M-(1.0054)^{298}M-\ldots-M$A300=320000(1.0054)300−(1.0054)299M−(1.0054)298M−…−M
To find $M$M, we need to factorise and use the formula for the sum of a geometric series to evaluate the expression inside the brackets. Take care to factorise the negative sign:
$A_{300}=320000(1.0054)^{300}-M((1.0054)^{299}+(1.0054)^{298}+...+1)$A300=320000(1.0054)300−M((1.0054)299+(1.0054)298+...+1)
Our geometric series has $a=1$a=1, $r=1.0054$r=1.0054 and $n=300$n=300. Murray plans to use all of the money over the $25$25 years, so, $A_{300}=0$A300=0.
$0=320000(1.0054)^{300}-M(1(\frac{1.0054^{300}-1}{1.0054-1}))$0=320000(1.0054)300−M(1(1.0054300−11.0054−1))
Rearranging:
| $M$M | $=$= | $\frac{320000(1.0054)^{300}}{(\frac{1.0054^{300}-1}{1.0054-1})}$320000(1.0054)300(1.0054300−11.0054−1) |
| $M$M | $=$= | $\$2156.67$$2156.67 |
Loan repayments
If we make regular, equal deposits on our loan then the repayments form an annuity. The same process we used previously can be used to model the repayment of a loan. You may see these loans referred to as reducing balance loans, where the periodic repayments reduce the amount owing while compound interest continues to apply to the outstanding balance. Taking out a loan from a bank or a financial institution is a common way for people to buy expensive items such as cars or houses. If we borrow money, we will receive our desired amount (called the principal) from the lender (usually a bank), and we will sign a contract outlining specific details such as the interest rate and the length of the loan. The rate at which we repay the debt will determine the overall amount of interest we end up paying.
Worked example
Example 2
Mike borrowed $\$12000$$12000 at $6%$6% p.a. The interest is compounded monthly on the money owing and the loan is to be repaid in equal monthly repayments of $\$180$$180. After how many years will the loan be repaid? How much interest does Mike pay in total?
Think: Let $A_n$An be the balance of the loan after $n$n months. $6%$6% p.a is equivalent to $0.05%$0.05% per month. Do: After $1$1 month, the loan accrues interest and Mike makes one repayment:
$A_1=12000\left(1.005\right)-180$A1=12000(1.005)−180
Then $A_1$A1 becomes the new loan balance and the process repeats:
$A_2=A_1\left(1.005\right)-180=12000\left(1.005\right)^2-180\left(1.005\right)-180$A2=A1(1.005)−180=12000(1.005)2−180(1.005)−180
Continuing this pattern, we get the following:
$A_n=12000(1.005)^n-180(1.005)^{n-1}-180(1.005)^{n-2}-\ldots-180(1.005)-180$An=12000(1.005)n−180(1.005)n−1−180(1.005)n−2−…−180(1.005)−180
After factorising we get:
$A_n=12000(1.005)^n-180(1+1.005+\ldots+(1.005)^{n-2}+(1.005)^{n-1})$An=12000(1.005)n−180(1+1.005+…+(1.005)n−2+(1.005)n−1)
Evaluating the sum of the geometric series within the brackets using $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1 with $a=1$a=1, $r=1.005$r=1.005 and the unknown $n$n staying as $n$n gives:
$A_n=12000(1.005)^n-180(\frac{1.005^n-1}{1.005-1})$An=12000(1.005)n−180(1.005n−11.005−1)
We can remove the fraction from the expression by evaluating $\frac{180}{1.005-1}$1801.005−1 to get:
$A_n=12000(1.005)^n-36000(1.005^n-1)$An=12000(1.005)n−36000(1.005n−1)
The loan is to be repaid, so $A_n=0$An=0. We need to collect the like terms of the expression involving powers of $n$n:
| $0$0 | $=$= | $12000(1.005)^n-36000(1.005^n)+36000$12000(1.005)n−36000(1.005n)+36000 |
| $24000(1.005)^n$24000(1.005)n | $=$= | $36000$36000 |
| $(1.005)^n$(1.005)n | $=$= | $1.5$1.5 |
| $n$n | $=$= | $\frac{\log1.5}{\log1.005}=81.2955...$log1.5log1.005=81.2955... |
The loan, therefore, must be repaid after $82$82 repayments, which is $6$6 years and $10$10 months.
To work out how much interest is paid, we calculate the total repayments and subtract the loan amount. Mike paid $82\times180=\$14760$82×180=$14760. As the loan was $\$12000$$12000, this means the interest paid was $\$2760$$2760.
The outstanding balance of a reducing balance loan decreases in a non-linear fashion due to the fact that interest accrues on that balance: the more that you owe, the more interest that will be added at the end of every compounding period. Although your repayments are constant, the underlying principal is reduced by a greater amount as time passes. The image below illustrates this for a $30$30 year loan: the principal, coloured in dark blue, will be reduced to half its original size in approximately $20$20 years.
Analysing a reducing balance loan with a recurrence relation
Consider the example of using a short term loan service to purchase a new pair of shoes for $\$300$$300. Interest of $10%$10% per week is charged on the balance of the loan, and we are able to pay back $\$100$$100 each week.
The interest charged in the first week is $300\times0.1=30$300×0.1=30, so if $\$100$$100 is repaid, the amount due at the end of the first week is $300+30-100=\$230$300+30−100=$230.
The interest added in the second week is then $230\times0.1=23$230×0.1=23, so if another $\$100$$100 is repaid, the amount due at the end of the second week will be $230+23-100=\$153$230+23−100=$153.
The reducing balance for the first $4$4 weeks of the loan is shown in the table.
| Period (n) | Value at beginning of period | Interest added this period | Repayment | Value at end of period |
|---|---|---|---|---|
| 1 | $\$300$$300 | $\$30$$30 | $\$100$$100 | $\$230$$230 |
| 2 | $\$230$$230 | $\$23$$23 | $\$100$$100 | $\$153$$153 |
| 3 | $\$153$$153 | $\$15.30$$15.30 | $\$100$$100 | $\$68.30$$68.30 |
| 4 | $\$68.30$$68.30 | $\$6.83$$6.83 | $\$75.13$$75.13 | $\$0$$0 |
Notice a recurring pattern:
$\text{Balance at the end of the week }=\text{Balance at the beginning }\times1.1-100$Balance at the end of the week =Balance at the beginning ×1.1−100
The balance at the end of each week is carried forward to be the starting balance in the account. Hence, using the notation we previously encountered using recurrence relations, we can write this as
$A_n=A_{n-1}\times1.1-100,A_0=300$An=An−1×1.1−100,A0=300
Practice questions
Question 1
Sandro has a superannuation fund of $\$600000$$600000. He withdraws $\$40000$$40000 at the beginning of each year as an income stream in his retirement. The interest compounds monthly on the remaining balance at $3%$3% per annum.
Calculate the balance in the account at the end of the first year, correct to the nearest dollar.
By how much did the balance decrease in the first year? Is this the same amount as his withdrawal? Explain your answer.
Question 2
Rochelle invests $\$190000$$190000 at a rate of $7%$7% per annum compounded annually, and wants to work out how much she can withdraw each year to ensure the investment lasts $20$20 years.
We will use geometric sequences and series to determine what Rochelle's annual withdrawal amount should be if she wants the investment to last $20$20 years.
The amount in the account after $n$n years can be expressed as the $n$nth term of a geometric sequence minus the sum of a different geometric sequence.
Write an expression for the amount in the investment account after $n$n years.
Use $x$x to represent the amount to be withdrawn each year.
Hence determine Rochelle's annual withdrawal amount, correct to the nearest cent.
Question 3
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?