Present Value Annuity
In this article, we will be looking at present value annuties. We use present value annuities in loan situtaions. We also use it to determine investment payments (where you invest money and want to be paid back every month/quarter/year, etc).
In your grade 12 mathematics exam, you are expected to calculate the loan amount, the monthly repayment, the duration of the loan, etc. This article will you through the topic by covering the following:
- Introductory scenario.
- Deriving the Present Value Annuity Formula
- Finding loan balance
- Delayed payments
- Introductory scenario
In the opening paragraphs, we mentioned that we use present value annuities when we calculate loans or want to know how much our investments can pay pay us per month/quarter/semester, etc.
As an example, let us consider a situation where Lesego wants to withdraw R1000 at the end of each year from her savings acount for four years. How much money must she deposit into this account in order to be able to do this if the bank offers interest at 10% p.a.?
To answer the question, we must first calculate how much she must deposit for the first withdrawal, the second, the third and the fourth withdrawal.
Let's use the compound interest formula:
$${A=P(1+i)^n}$$Rearranging to solve for the principal amount
$${ P=A(1+i)^{-n}}$$| Year | How much Lesego must invest |
|---|---|
| 1 | $${1000(1+,1)^{-1}=R909,09}$$ |
| 2 | $${1000(1+0.1)^{-2}=R826,45}$$ |
| 3 | $${1000(1+0.1)^{-3}=R751,31}$$ |
| 4 | $${1000(1+0.1)^{-4}=R683,01}$$ |
| Total | R3 169,86 |
We can see that Lesego has to invest \(R3 169,86\). Now let us see if this money will allow her to make R1000 withdrawals at the end of each year for 4 years.
| Year | Accumulated | Balance |
|---|---|---|
| 1 | \begin{gather*} \small 3169,86(1+0.1)\\ \small =R3486,85 \end{gather*} | \begin{gather*} \small 3486,85-1000\\ \small =R2486,85 \end{gather*} |
| 2 | \begin{gather*} \small 2486,85(1+0.1)\\ \small =R2735,54 \end{gather*} | \begin{gather*} \small 2735,54-1000\\ \small =R1735,54 \end{gather*} |
| 3 | \begin{gather*} \small 1735,54(1+0.1)\\ \small =R1909,09 \end{gather*} | \begin{gather*} \small 1909,09-1000\\ \small =R909,09 \end{gather*} |
| 4 | \begin{gather*} \small 909,09(1+0.1)\\ \small =R1000 \end{gather*} | \begin{gather*} \small 1000-1000\\ \small =R0 \end{gather*} |
So we have seen that the money we invest grows at an interest. On the other hand, the R1000 is withdrawn at regular intervals. This makes it an annuity (future value annuity).
Generally, we can say that accumulated amount of the loan money is equal to the future value of the regular payments.
- Deriving the Present Value Annuity Formula
We just found that the accumulated amount of the loan is equal to the future value of the regular payments. Mathematically:
\begin{gather*} \small A=F\\ \small P(1+i)^{n}=\frac{x[(1+i)^{n}-1]}{i}\\ \end{gather*}Making P subject of the formula.
\begin{gather*} \small P=\frac{x[(1+i)^{n}-1]}{(1+i)^{n}(i)}\\ \small P=\frac{x[(1+i)^{n}-1](1+i)^{-n}}{i}\\ \small P=\frac{x[(1+i)^{n-n}-(1+i)^{-n}]}{i}\\ \small P=\frac{x[1-(1+i)^{-n}]}{i} \end{gather*}where,
P = loan amount
\(x\) = installment
\(i\) = interest
\(n\) = number of installments
Below is a timeline that represents the formula above. Normally, loan repayments begin the following month/year/quarter, etc., depending on how your interest is quoted.
When you have drawn an accurate timeline, you will be able to find an accurate number of installments. This is very important for large number of installments. The following equation will help you get correct number of installments.
$${\small n=T_n-T_i+1}$$Where \(T_n\) is the time of the last payment and \(T_i\) is the time of the first payment.
Example: Colin takes a loan at a bank that charges him interest of 10% pa. If he pays R8760 annually for four yours, how much loan did he take?
Answer
\(i\) = 10% pa = 0,1
\(x\)= R8760
\(n\) = 4
P = ?
- Finding loan balance
You will also be asked to find the loan balance. This is maybe along the way after you have made some payments and you want to know how much you need to pay to settle the debt. This can also be used in the case where payments are delayed or deferred.
To calculate loan balance, we make use of two formulas:
\begin{align} \small Balance =\frac{x[1-(1+i)^{-n}]}{i}\label{eq:eq1}\tag{1}\\ \end{align}Or
\begin{align} \small Balance = A -F\label{eq:eq2}\tag{2} \end{align}When using equation (1), the \(n\) represents number of remaining installments. To find the \(n\), use the following formula:
$${\small n= T_n - T_{nn}}$$The \(T_n\) is the expected final time of the loan payment and \(T_{nn}\) is the actual time at which the last payment is/was made.
The above equation is not standard. We use it to help you get the correct \(n\).
NB: Your exam can also give you a time at which you must calculate balance. This time will be your \(T_{nn}\).
Equation (2) can be expanded:
\begin{gather*} \begin{split} \small Balance &= A -F\\ \small &=P(1+i)^n-\frac{x[(1+i)^n-1]}{i} \end{split} \end{gather*}The \(n\) used here is the time specified in the question.
Example : Colin pays R8760 at the end of every year towards the loan he took. If it takes him four yours to setle the loan, what is his balance after two years? The bank charges him interest of 10% pa.
Answer
Method 1:
\(i\) = 10% pa = 0,1
\(x\)= R8760
\(n\) = 4-2=2
∴ Colin will be owing R15 203,31.
Method 2:
\(i\) = 10% pa = 0,1
\(x\)= R8760
\(n\) =2 (Given on the question)
P=R27768,02
We will calculate the A and F separately.
\begin{gather*} \begin{split} \small A&=P(1+i)^n\\ \small &=27768,02(1+0,1)^2\\ \small &=\textrm{R33 599,30}\\ \end{split} \end{gather*} \begin{gather*} \begin{split} \small F&=\frac{x[(1+i)^n-1]}{i}\\ \small &=\frac{8760[(1+0,1)^2-1]}{0,1}\\ \small &=\textrm{R18 396} \end{split} \end{gather*}Now we can find the balance.
\begin{gather*} \begin{split} \small Balance &= A -F\\ \small &=R33599,30-18396\\ \small &=\textrm{R15 203,30} \end{split} \end{gather*}∴ Colin will be owing R15 203,30.
The differences in the cents are due to the rounding errors. Do not worry about this difference, it is expected.
- Delayed Payments
Now let us look at a case where one misses payment as quoted by their interest rate. In such cases, you would be requested to find balance at a time or maybe find a new installment.
So what is actually happening here is that if you miss payment(s), your debt is not reducing. As a result, you might need to change your installment or the loan duration might change too if you will not increase your installment.
Let us try typical exam question.
Example: Delayed payments
Bank A offers Colin a loan of R27 768,02 at 10% pa for six years. If his first payment is on year three, calculate his installment.
Answer
P = R27 768,02
\(i\) = 10% = 0,1
\(n=6 - 3 + 1=4\). Refer to Fig 2 below.
\(x\) = ?
The timeline below represents the case.
First thing we need to do is to drag the loan amount to a time before the first installment, i.e \(T_2\). This means you need to find the accumulated amount of the loan at \(T_2\).
\begin{gather*} \begin{split} \small A&=P(1+i)^n\\ \small &=27768,02(1+0,01)^2\\ \small &=\textrm{R33 599,30} \end{split} \end{gather*}Now we can find the installment. The value above is the new loan amount before you start paying. We can now use the present value formula.
\begin{gather*} \begin{split} \small P &=\frac{x[1-(1+i)^{-n}]}{i}\\ \small 33599,30&=\frac{x[1-(1+0,1)^{-4}]}{0,1}\\ \small 33599,30&=3,169865446x\\ \small x&= \textrm{R10 599,60} \end{split} \end{gather*}Lastly, let us look at another example where one misses payments in the middle of a loan period.
Example: Missed payments along the way during the loan period.
Colin took a R120 000 loan at Bank A. The bank charges interest at 10% pa for 10 years. Colin makes annual paymnets of R19 529 at the end of each year. He couldn't make payments at the end of the fourth, fifth and sixth year, so he decided to increase his installments so he can settle his loan at the end of 10 years. Calculate his new installmment.
Answer
The timeline below represents the case.
To answer this question, we need to realise that there are three missed payments from \(T_4\) to \(T_6\).
Now, because there are payments made between \(T_1\) and \(T_3\), we must find the balance at \(T_3\). We will use equation (1).
\(x\) = R19 529
\(i\) = 0,1
\(n\) = \(T_n\) - \(T_{nn}\) = 10 - 3 = 7
This is the amount owed at \(T_3\). Now since there are no payments for the next three years, this money will gain interest. The new balance at \(T_6\) is:
\begin{gather*} \begin{split} \small Balance &= \textrm{Balance at }T_3(1+i)^n\\ \small &= 95075,35(1+0,1)^3\\ \small &= \textrm{R126 545,29} \end{split} \end{gather*}NB: We have used \(n =3\). This number is the period during which there were no paymentss made, i.e., \(T_4\) to \(T_6\).
Now we can find the new installment. The \(n\) for the last phase of payments is \(n=T_n-T_{nn}=10-7+1=4\)
\begin{gather*} \begin{split} \small P &=\frac{x[1-(1+i)^{-n}]}{i}\\ \small x &=\frac{Pi}{1-(1+i)^{-n}}\\ \small x &=\frac{126545,29 \times 0,1}{1-(1+0,1)^{-4}}\\ \small &= \textrm{R39 921,34} \end{split} \end{gather*}