# Present Value

The present value of a given sum of money which is due at the end of a certain period is that sum which if invested now at the given rate of interest accumulates to the said sum at the end of the period. In this article, we will look at the calculation of the present value of money along with some examples.

## Present Value Formula

Let’s say that P is the Principal [Present Value] and the rate of interest is r% per period. Therefore, after using compounding the period-wise interest, the amount ‘A’ due after ‘n’ periods is:

A = P(1 + $$\frac {r}{100}$$)n

⇒ P = $$\frac {A}{(1 + \frac {r}{100})^n}$$

This is the present value of ‘A’ due at the end of ‘n’ years. Therefore, the present value of the amount ‘A’ which is due at the end of period ‘n’ and at the rate of r% per annum = $$\frac {A}{(1 + \frac {r}{100})^n}$$. Further, the present value or capital value of an annuity is the sum of the present values of all payments.

Also, the present value of Rs. A paid at the beginning of the second period = $$\frac {A}{(1 + \frac {r}{100})^1}$$ = $$\frac {A}{1 + i}$$.

Further, the present value of Rs. A paid at the beginning of the nth period = $$\frac {A}{(1 + \frac {r}{100})^{n-1}}$$.

## Present Value of Annuity

This is the sum of the present values of all the payments received in an annuity. It relies on the concept of the time value of money.

The time value of money states that a Rupee today is worth more than the same Rupee at a future date. Therefore, the present value of annuity is

= A + $$\frac {A}{(1 + \frac {r}{100})}$$ + $$\frac {A}{(1 + \frac {r}{100})^2}$$ + … + $$\frac {A}{(1 + \frac {r}{100})^{n-1}}$$

= A$$\left\{1 + \frac {1}{(1 + \frac {r}{100})} + \frac {1}{(1 + \frac {r}{100})^2} + … + \frac {1}{(1 + \frac {r}{100})^{n-1}}\right\}$$

= A$$\left\{ \frac {1 – \frac {1}{(1 + \frac {r}{100})^n}}{1 – \frac {1}{(1 + \frac {r}{100})}} \right\}$$ = A$$\left\{ \frac {1 – (1 + \frac {r}{100})^{-n}}{1 – (1 + \frac {r}{100})^{-1}} \right\}$$

Further, when an annuity becomes a perpetuity, that is n→∞, then

(1 + $$\frac {r}{100})^{-n}$$ = 0

Therefore,

Present Value of Perpetuity = $$\left\{ \frac {A}{1 – (1 + \frac {r}{100})^{-1}} \right\}$$

If the rate of interest per rupee, per period i = $$\frac {r}{100}$$, then

The present-value of annuity due for ‘n’ periods = A$$\left\{ \frac {1 – (1 + i)^{-n}}{1 – (1 + i)^{-1}} \right\}$$

Also, the present-value for perpetuity = $$\frac {A}{1 – (1 + i)^{-1}}$$

### Example of Present Value

Find the amount of annuity of Rs. 4,000 per annum for 10 years reckoning compound interest at 10% per annum.

Considering immediate annuity, the required amount

= 4000 (1 + $$\frac {10}{100}$$)9 + 4000 (1 + $$\frac {10}{100}$$)8 + … + 4000 (1 + $$\frac {10}{100}$$)1 + 4000

For the sake of simplicity, let’s write this in the reverse order. Therefore, we have

4000 + 4000 (1 + $$\frac {10}{100}$$) + 4000 (1 + $$\frac {10}{100}$$)2 + … + 4000 (1 + $$\frac {10}{100}$$)9

= 4000 [1 + 1.1 + (1.1)2 + … + (1.1)9] … because (1 + $$\frac {10}{100}$$) = 1.1

= 4000 $$\left\{ \frac {(1.1)^{10} – 1}{1.1 – 1} \right\}$$ = 4000$$\left\{ \frac {(2.594 – 1)}{0.1} \right\}$$ = $$\frac {4000 × 1.594}{0.1}$$ = Rs. 63,760 (Sixty three thousand even hundred and sixty).

## Solved Question

Q1. Find the amount of annuity of Rs. 6,000 per annum for 10 years reckoning compound interest at 10% per annum.

Solution:

As we have seen in the example above, if we consider an immediate annuity, then the required amount is

= 6000 (1 + $$\frac {10}{100}$$)9 + 6000 (1 + $$\frac {10}{100}$$)8 + … + 6000 (1 + $$\frac {10}{100}$$)1 + 6000

For the sake of simplicity, let’s write this in the reverse order. Therefore, we have

6000 + 6000 (1 + $$\frac {10}{100}$$) + 6000 (1 + $$\frac {10}{100}$$)2 + … + 6000 (1 + $$\frac {10}{100}$$)9

= 6000 [1 + 1.1 + (1.1)2 + … + (1.1)9] … because (1 + $$\frac {10}{100}$$) = 1.1

= 6000 $$\left\{ \frac {(1.1)^{10} – 1}{1.1 – 1} \right\}$$ = 6000$$\left\{ \frac {(2.594 – 1)}{0.1} \right\}$$ = $$\frac {6000 × 1.594}{0.1}$$ = Rs. 95,640 (Ninety five thousand six hundred and forty).

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