ANNUITY: EXPLAINED WITH 11 EXAMPLES

An annuity is conceptualized as a series of equal or varying payments/receipts made or received at a given interval, such as annually, monthly, quarterly, daily and so on.

It is a fixed sum of money paid or received at regular intervals.

There are two main types of annuity, namely: Ordinary annuity and annuity due.

Furthermore, there are some special cases of an annuity such as deferred annuity and perpetual annuity.

Ordinary annuity refers to a series of equal payments/receipts at the end of a specified interval.

Monthly salary payments, annual salary payments, bond repayment in equal payments at the end of an interval (monthly, semi-annually, etc.) are all examples of ordinary annuities

Annuity due refers to a series of equal or varying payments/receipts at the beginning of an interval e.g beginning of the month's rental payments.

Now, for the purpose of the time value of money concept, we might be interested in the future or present value of each type of annuity

Future value of an ordinary annuity

The Future Value of an ordinary annuity with interest compounded annually is computed thus:

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

or

$\sum^n_{x=1}A(1+i)^{n-x}$

Where A is the periodic cash flows

      i is the annual interest rate expressed in percentage

      n is the number of years

      x is the starting period of the annuity

Example 1

What is the accumulated value of a series of the annual income of N15,000 compounded at a rate of 14% annually for 5 years?

Solution:

Recalled that $FV=\frac{A\left[(1+i)^n-1\right]}{i}$, Hence

$FV=\frac{15,000\left[(1+0.14)^5-1\right]}{0.14}$

$FV=\frac{15,000\left[(1.14)^5-1\right]}{0.14}=99,151.56$

You can also solve it using sigma notation

$\sum^5_{x=1}15000(1.14)^{5-x}$

$$15000(1.15)^{5-1}+15000(1.15)^{5-2}+15000(1.15)^{5-3}+15000(1.15)^{5-4}+15000(1.15)^{5-5}$$

$15000(1.15)^{4}+15000(1.15)^{3}+15000(1.15)^{2}+15000(1.15)^{1}+15000=99,151.56$

Note: when the accumulated/future value is given but the annuity is unknown, the first formula is preferable

Example 2

Dr Johnson predicted that his company's equipment would need to be changed in four years, costing N1.6 million.

How much will be put in the account per year for the next four years if money can be invested at a rate of 15% compounded annually, so that the fund may be used to replace the equipment?

Solution:

Fv=1.6m, n=4, i=0.15, A=?

$Fv=\frac{A\left[(1+i)^n-1\right]}{i}$

$1,600,000=\frac{A\left[(1.15)^4-1\right]}{0.15}$

$1,600,000 \times 0.15=A(1.74900625-1)$

$240,000=A(0.74900625)$

$\frac{240,000}{0.74900625}=A$

$A=N320,424.56$

Hence N320,424.56 is required series of equal payment of equal annual payment for 4 years

So far, we have assumed that interest compounds annually, but interest do compound intra-yearly such as l semi-annually, quarterly, monthly, weekly, or even daily.

To solve this type of annuity, we use the following formula:

$Fv=\frac{A\left[(1+\frac{i}{m})^{nm}-1\right]}{\frac{i}{m}}$

or 

$\sum^{nm}_{x=1}A\left(1+\frac{i}{m}\right)^{nm-x}$

Example 3

Given that Obed was paid a monthly salary of N65,000, Calculate the future value of his pay assuming he receives it for 3 years and the interest rate is 21% compounded monthly.

Solution:

$Fv=\frac{65,000\left[\left(1+\frac{0.21}{12}\right)^{12\times 3}-1\right]}{\frac{0.21}{12}}$

$Fv=\frac{65,000\left[1.0175)^{36}-1\right]}{0.0175}=N3,221,798.42$

 or

$Fv=\sum^{36}_{x=1}65,000\left(1+\frac{0.21}{12}\right)^{36-x}$

$Fv=65,000\left(1+\frac{0.21}{12}\right)^{36-1}+65,000\left(1+\frac{0.21}{12}\right)^{36-2}............65,000\left(1+\frac{0.21}{12}\right)^{36-36}$

$FV=65,000\left(1+\frac{0.21}{12}\right)^{35}+65,000\left(1+\frac{0.21}{12}\right)^{34}............65,000\left(1+\frac{0.21}{12}\right)^{0}=N3,221,798.42$$

The first formula is preferable if we need to compute a variable other than the future value.

Example 4

How long will it take James to accumulate N1,000,000 if he invests N15,000 per month in a 16% compounded monthly investment?

Solution:

Fv=1,000,000, A=15,000, n=? i=0.16, m=12

$1,000,000=\frac{15,000\left[\left(1+\frac{0.16}{12}\right)^{12n}-1\right]}{\frac{0.16}{12}}$

$1,000,000=\frac{15,000\left[(1.01333333)^{12n}-1\right]}{0.01333333}$

Now, let's cross multiply

$13333.33=15,000\left[(1.01333333)^{12n}-1\right]$

$\frac{13333.33}{15,000}=(1.01333333)^{12n}-1$

$0.8889=(1.01333333)^{12n}-1$

$1.8889=(1.01333333)^{12n}$

$1.8889=(1.01333333)^{12n}$

Adding logarithm to both sides

$\log 1.8889=\log 1.01333333^{12n}$

Recalled that $\log b^a=a\log b$, hence

$\log 1.8889=12n \log 1.01333333$

$\frac{\log 1.8889}{\log 1.01333333}=12n$

$48.01701=12n$

$n=\frac{48.01701}{12}=4$

Therefore, it will take roughly 4 years

Future value of an increasing or growing ordinary annuity

For varying annuity or increasing annuity, the nth term of an A.P is added to the formula:

If the interest compound annually, we use the following formula:

$Fv=\sum^n_{x=1}A+d(x-1)(1+i)^{n-x}$

For intra-yearly compounding, use the formula below.

$Fv=\sum^{nm}_{x=1}A+d(x-1)\left(1+\frac{i}{m}\right)^{nm-x}$

Example 5

Consider a man with a monthly salary of N100, which increases by N15 for each subsequent month. 

Find the future value of the whole accumulated salary provided it earn an interest of 15% compounded monthly for two years.

Solution:

Using sigma notation:

$Fv=\sum^{2 \times 12 }_{x=1}100+15(x-1)\left(1+\frac{0.15}{12}\right)^{24-x}=6945.7$

The upper limit of the sigma notation is 24 and the lower limit is one.

$Fv=100+15(x-1)\left(1+\frac{0.15}{12}\right)^{24-1}........100+15(x-1)\left(1+\frac{0.15}{12}\right)^{24-24}=6945.7$

Hence, the accumulated value is equal to 6945.7

Future value of an annuity due

To calculate the future value of an annuity due that compounds annually, it is as follows:

$Fv=\frac{A\left[(1+i)^{n+1}-(1+i)\right]}{i}$$ or $$\sum^n_{x=1}A(1+i)^{(n+1)-x}$

The following formula is used to calculate the future value of an annuity due that compounds on a quarterly, weekly, or monthly basis

$Fv=\frac{A\left[\left(1+\frac{i}{m}\right)^{nm+1}-(1+\frac{i}{m})\right]}{\frac{i}{m}}$ or $\sum^{nm}_{x=1}A\left(1+\frac{i}{m}\right)^{\left((nm+1)-x\right)}$$l

Example 6

Given a 9.5 per cent compounded yearly interest rate, what is the future value of an N100,000 rental payment made at the start of each year for ten years?

Solution

$Fv=\frac{100,000\left[(1.095)^{10+1}-(1.095)\right]}{0.095}=1703851.83$

Or

$Fv=\sum^{10}_{x=1}100,000(1.095)^{\left(10+1)-x\right)}$

Expressing the notation

$100,000(1.095)^{(10+1)-1}..........100,000(1.095)^{(10+1)-11}=1703851.83$

The first formula is preferred when the future value of the annuity due is known.

Example 7

Mr Joshua, who currently earns N280,000 per year, will need N800,000 in five years to cover the cost of buying a home.

In order to meet this objective, Mr joshua will contribute an equal amount monthly out of his salary towards the investment plan.

Any cash contributed will earn a monthly interest rate of 36% compounded. How much should Mr Joshua put into the investment plan starting now in order to achieve his objective?

Solution:

Here, $Fv=800,000$, $n=5$, $m=12$. $i=0.36$

$800,000=\frac{A\left[\left(1+\frac{0.36}{12}\right)^{\left(12 \times 5)+1\right)}-\left(1+\frac{0.36}{12}\right)\right]}{\frac{0.36}{12}}$

$800,000=\frac{A\left[(1.03)^{61}-(1.03)\right]}{0.03}$

$24,000=A\left[(1.03)^{61}-(1.03)\right]$

$A=\frac{24,000}{(1.03)^{61}-(1.03)}=4763.46$

Approximately N4763.46 is required to be contributed into achieving this objective.

Future value of an increasing or growing annuity due

For varying Annuity due/ or an annuity increasing in arithmetic progression with a common difference. We use the following formulas

If the interest is compounded annually, the formula is as follows:

$Fv=\sum^n_{x=1}A+d(x-1)(1+i)^{(n+1)-x}$

For an annuity that compounds intra-yearly

$Fv=\sum^{nm}_{x=1}A+d(x-1)(1+\frac{i}{m})^{((nm+1)-x)}$

Example 8

What is the future value of a series of N45,000 that's put into an account with a quarterly increment of 3000, earning 13% interest compounded quarterly for four years?

Solution:

$Fv=\sum^{4 \times 4}_{x=1} 45,000+3000(x-1)\left(1+\frac{0.13}{4}\right)^{(16+1)-x}=1154522.52$

Present value of the ordinary annuity

The present value of an ordinary annuity For an interest that compounds annually.

$Pv=\frac{A\left[1-(1+i)^{-n}\right]}{i}$ or $\sum^n_{x=1}A(1+i)^{-n}$

for an interest that compounds intra-yearly

$Pv=\frac{A\left[1-(1+\frac{i}{m})^{-nm}\right]}{\frac{i}{m}}$ 

or

$\sum^{nm}_{x=1}A(1+\frac{i}{m})^{-nm}$

Example 9

Rather than paying you N250 a month for the next twenty years, the driver who hit you with his vehicle is willing to pay a lump sum to settle the claim now.

Calculate the lump sum if the interest rate is 12% compounded annually

Solution

Using intra-yearly formula

$Pv=\frac{250\left[1-\left(1+\frac{0.12}{12}\right)^{-240}\right]}{\frac{0.12}{12}}=22,704.85408$ 

or

$Pv=\sum^{240}_{x=1}250\left(1+\frac{0.12}{12}\right)^{-x}=22,704.85408$

Present value of an increasing or growing ordinary annuity

For varying annuity/increasing in arithmetic progression with a common difference of

For annually

$pv=\sum^n_{x=1}A+d(x-1)(1+i)^{-x}$

For intra-yearly

$Pv=\sum^{nm}_{x=1}A+d(x-1)\left(1+\frac{i}{m}\right)^{-x}$

Example 10

Baba Ijebu intends to start a business for his wife.

To accomplish this, he plans to deposit N10,000 at the end of each quarter, with the sum increasing by N300 for each consecutive quarter, yielding a 16 per cent interest rate compounded quarterly

To accomplish so, he plans to deposit N10,000 at the end of each quarter, with the sum increasing by N300 for each consecutive quarter, yielding a per cent annual interest rate compounded quarterly.

How much will his wife be indifferent supposing the deposits starting from now, last for five years

Solution:

$Pv=\sum^{4 \times 5}_{x=1}10,000+300(x-1)\left(1+\frac{0.16}{4}\right)^{-x}=300255.44$

Present value of an annuity due

Where interest compounds annually 

$Pv=\frac{A\left[(1+i)-(1+i)^(-n+1)\right]}{i}$ or $\sum^n_{x=1}A(1+i)^{-x+1}$

Where interest compounds Intra-yearly

$Pv=\frac{A\left[(1+\frac{}{m})-(1+\frac{}{m})^{(-n+1)}\right]}{\frac{i}{m}}$ or $\sum^n_{x=1}A\left(1+\frac{i}{m}\right)^{-x+1}$

Example 11

Shola must pay off N500,000 in liabilities over six years by amortizing the loan monthly.

What is the monthly payment if the interest rate is 18 per cent per year

Solution

Here $n=6$, $m=12$, $i=0.18$ and $Pv=500,000$

$500,000=\frac{A\left[\left(1+\frac{0.18}{12}\right)-\left(1+\frac{0.18}{12}\right)^{-(12 \times 6)+1}\right]}{\frac{0.18}{12}}$

$500,000=\frac{A\left[(1.015)-(1.015)^{-71}\right]}{0.15}$

$7500=A\left[(1.015)-(1.015)^{-71}\right]$

$A=\frac{7500}{(1.015)-(1.015)^{-71}}$

A=N11,235.8651 at the required monthly payment to cover the obligation.

Present value of an increasing or growing annuity due

For varying annuity in AP, 

We use the formula for an interest that compound Annually 

$\sum^n_{x=1}A+d(x-1) \times(1+i)^{-x-1}$

We use the formula for an interest that compound intrayearly

$\sum^{nm}_{x=1}A+d(x-1) \times\left(1+\frac{i}{m}\right)^{-x-1}$

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