An annuity at due is a series of periodic payments made at the beginning of each period

The value of a stream of equal payments made at the start of the period at a future date is the **future value of an annuity at due.**

It is calculated thus:

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]\left(1+\frac{i}{m}\right)$$

Where,

C is periodic cash flows or payments

i is the interest rate

m is several times compounding occurs in a year.

n is the number of years

**Example 1**

Supposed N800 is deposited at the start of each year for the next 4 years in an account paying 10% per year compounded annually. How much will be in the account at the end of the term of the annuity?

**Solution:**

As previously noted, the future value of an ordinary annuity is calculated thus:

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]\left(1+\frac{i}{m}\right)$$

Here C=800, i=0.1, m=1 and n=4

$FV=800\left[\frac{\left(1+\frac{0.1}{1}\right)^{4(1)}-1}{\frac{0.1}{1}}\right]\left(1+\frac{0.1}{1}\right)$

$FV=800\left[\frac{\left(1.1\right)^{4}-1}{0.1}\right]\left(1 .1\right)$

$FV=800\left[\frac{1.4641-1}{0.1}\right]\left(1.1\right)$

$FV=800\left[\frac{0.4641}{0.1}\right]\left(1.1\right)=N4084.08$

__Example 2__

Supposed N2000 is deposited at the beginning of each month for the next 3 years in an account paying 5% per annum compounded monthly. How much will be in the account at the end of the term of the annuity?

**Solution:**

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]\left(1+\frac{i}{m}\right)$$

$FV=2000\left[\frac{\left(1+\frac{0.05}{12}\right)^{3(12)}-1}{\frac{0.05}{12}}\right]\left(1+\frac{0.05}{12}\right)$

* Note:* m is 12 because there are 12 months in a year.

$FV=2000\left[\frac{\left(1+0.0041667\right)^{36}-1}{0.0041667}\right]\left(1.0041667\right)$

$FV=2000\left[\frac{\left(1.0041667\right)^{36}-1}{0.0041667}\right]\left(1.0041667\right)$

$FV=2000\left[\frac{\left(1.16147\right)-1}{0.0041667}\right]\left(1.0041667\right)$

$FV=2000\left[\frac{0.1614736}{0.0041667}\right]\left(1.0041667\right)=N77830$

__Example 3__

Kenny plans to invest N1200 at the beginning of each quarter for the next 4 years. He expects to earn an interest rate of 10% that will be compounded quarterly. What would be the future value of these payments at the end of the 4 years?

**Solution:**

$FV=1200\left[\frac{\left(1+\frac{0.1}{4}\right)^{4(4)}-1}{\frac{0.1}{4}}\right]\left(1+\frac{0.1}{4}\right)$

** Note:** m is 4 because a year is equivalent to 4 quarters.

$FV=1200\left[\frac{\left(1+0.025\right)^{16}-1}{0.025}\right]\left(1+0.025\right)$

$FV=1200\left[\frac{\left(1.025\right)^{16}-1}{0.025}\right]\left(1.025\right)$

$FV=1200\left[\frac{\left(1.484505\right)-1}{0.025}\right]\left(1.025\right)$

$FV=1200\left[\frac{0.484505}{0.025}\right]\left(1.025\right)=N23838$

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