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8.2: Compound Interest


Learning Objectives

In this section, you will learn to:

  • Find the future value of a lump-sum.
  • Find the present value of a lump-sum.
  • Find the effective interest rate.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Simplify each expression.

a. (100(3+2^2))

b. (100(3+2)^2)

Click here to check your answer

a. (700)

b. (2500)

If you missed this problem, review here. (Note that this will open a different textbook in a new window.)

2. If an amount of $2,000 is borrowed at a simple interest rate of 10% for 3 years, how much is the interest?

Click here to check your answer

($600)

If you missed this problem, review Section 8.1. (Note that this will open in a new window.)

3. You borrow $4,500 for six months at a simple interest rate of 8%. How much is the interest?

Click here to check your answer

($180)

If you missed this problem, review Section 8.1. (Note that this will open in a new window.)

4. John borrows $2400 for 3 years at 9% simple interest. How much will he owe at the end of 3 years?

Click here to check your answer

($7848)

If you missed this problem, review Section 8.1. (Note that this will open in a new window.)

Compound Interest

In the last section, we examined problems involving simple interest. Simple interest is generally charged when the lending period is short and often less than a year. When the money is loaned or borrowed for a longer time period, if the interest is paid (or charged) not only on the principal, but also on the past interest, then we say the interest is compounded.

Suppose we deposit $200 in an account that pays 8% interest each year. At the end of one year, we will have $200 + $200(.08) = $200(1 + .08) = $216.

Now suppose we put this amount, $216, in the same account. After another year, we will have $216 + $216(.08) = $216(1 + .08) = $233.28.

So an initial deposit of $200 has accumulated to $233.28 in two years. Further note that had it been simple interest, this amount would have accumulated to only $232. The reason the amount is slightly higher is because the interest ($16) we earned the first year, was put back into the account. And this $16 amount itself earned for one year an interest of $16(.08) = $1.28, thus resulting in the increase. So we have earned interest on the principal as well as on the past interest, and that is why we call it compound interest.

Now suppose we leave this amount, $233.28, in the bank for another year, the final amount will be $233.28 + $233.28(.08) = $233.28(1 + .08) = $251.94.

Now let us look at the mathematical part of this problem so that we can devise an easier way to solve these problems.

After one year, we had $200(1 + .08) = $216

After two years, we had $216(1 + .08)

But $216 = $200(1 + .08), therefore, the above expression becomes

[$ 200(1+.08)(1+.08) = $ 200(1+.08)^2=$ 233. 28 onumber]

After three years, we get

[$ 233.28(1+.08)=$ 200(1+.08)(1+.08)(1+.08) onumber]

which can be written as

[$ 200(1+.08)^{3}=$ 251.94 onumber]

Suppose we are asked to find the total amount at the end of 5 years, we will get

[200(1+.08)^{5}=$ 293.87 onumber]

We summarize as follows:

The original amount

$200

= $200

The amount after one year

$200(1 + .08)

= $216

The amount after two years

$200(1 + .08)2

= $233.28

The amount after three years

$200(1 + .08)3

= $251.94

The amount after five years

$200(1 + .08)5

= $293.87

The amount after t years

$200(1 + .08)t

COMPOUNDING PERIODS

Banks often compound interest more than one time a year. Consider a bank that pays 8% interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay an interest equal to one-fourth of 8%, or 2%.

Now if we deposit $200 in the bank, after one quarter we will have ($ 200left(1+frac{.08}{4} ight)) or $204.

After two quarters, we will have ($ 200left(1+frac{.08}{4} ight)^{2}) or $208.08.

After one year, we will have ($ 200left(1+frac{.08}{4} ight)^{4}) or $216.49.

After three years, we will have ($ 200left(1+frac{.08}{4} ight)^{12}) or $253.65, etc.

The original amount

$200

= $200

The amount after one quarter

($ 200left(1+frac{.08}{4} ight))

= $204

The amount after two quarters

($ 200left(1+frac{.08}{4} ight)^{2})

= $208.08

The amount after one year

($ 200left(1+frac{.08}{4} ight)^{4})

= $216.49

The amount after two years

($ 200left(1+frac{.08}{4} ight)^{8})

= $234.31

The amount after three years

($ 200left(1+frac{.08}{4} ight)^{12})

= $253.65

The amount after five years

($ 200left(1+frac{.08}{4} ight)^{20})

= $297.19

The amount after t years

($ 200left(1+frac{.08}{4} ight)^{4t})

We can see the formula for compound interest emerge.

Definition: Compound Interest, (n) times per year

If a lump-sum amount of (P) dollars is invested at an interest rate (r), compounded (n) times a year, then after (t) years the final amount is given by

[A=Pleft(1+frac{r}{n} ight)^{n t} ]

(mathbf{P}) is called the principal and is also called the present value.

Example (PageIndex{1})

If $3500 is invested at 9% compounded monthly, what will the future value be in four years?

Solution

Clearly an interest of .09/12 is paid every month for four years. The interest is compounded (4 imes 12 = 48) times over the four-year period. We get

[mathrm{A}=$ 3500left(1+frac{.09}{12} ight)^{48}=$ 3500(1.0075)^{48}=$ 5009.92 onumber]

$3500 invested at 9% compounded monthly will accumulate to $5009.92 in four years.

Example (PageIndex{2})

How much should be invested in an account paying 9% compounded daily for it to accumulate to $5,000 in five years?

Solution

We know the future value, but need to find the principal.

[egin{array}{l}
$ 5000=Pleft(1+frac{.09}{365} ight)^{365 imes 5}
$ 5000=P(1.568225)
$ 3188.32=P
end{array} onumber]

$3188,32 invested into an account paying 9% compounded daily will accumulate to $5,000 in five years.

Example (PageIndex{3})

If $4,000 is invested at 4% compounded annually, how long will it take to accumulate to $6,000?

Solution

(n = 1) because annual compounding means compounding only once per year. The formula simplifies to (A=(1+r)^{t}) when (n = 1).

[egin{aligned}
$ 6000 &=4000(1+.04)^{t}
frac{6000}{4000} &=1.04^{t}
1.5 &=1.04^{t}
end{aligned} onumber]

We use logarithms to solve for the value of (t) because the variable (t) is in the exponent.

[t=log _{1.04}(1.5) onumber]

Using the change of base formula we can solve for (t):

[t=frac{ln (1.5)}{ln (1.04)}=10.33 ext { years } onumber]

It takes 10.33 years for $4000 to accumulate to $6000 if invested at 4% interest, compounded annually

Example (PageIndex{4})

If $5,000 is invested now for 6 years what interest rate compounded quarterly is needed to obtain an accumulated value of $8000.

Solution

We have (n = 4) for quarterly compounding.

[egin{aligned}
$ 8000 &=$ 5000left(1+frac{r}{4} ight)^{4 imes 6}
frac{$ 8000}{$ 5000} &=left(1+frac{r}{4} ight)^{24}
1.6 &=left(1+frac{r}{4} ight)^{24}
end{aligned} onumber]

We use roots to solve for (t) because the variable (r) is in the base, whereas the exponent is a known number.

[sqrt[24]{1.6}=1+frac{mathrm{r}}{4} onumber]

Many calculators have a built in “nth root” key or function. In the TI-84 calculator, this is found in the Math menu. Roots can also be calculated as fractional exponents; if necessary, the previous step can be rewritten as

[1.6^{1 / 24}=1+frac{mathrm{r}}{4} onumber]

Evaluating the left side of the equation gives

[egin{array}{l}
1.0197765=1+frac{mathrm{r}}{4}
0.0197765=frac{mathrm{r}}{4}
mathrm{r}=4(0.0197765)=0.0791
end{array} onumber]

An interest rate of 7.91% is needed in order for $5000 invested now to accumulate to $8000 at the end of 6 years, with interest compounded quarterly.

Effective Interest Rate

Banks are required to state their interest rate in terms of an “effective yield” or “effective interest rate”, for comparison purposes. The effective rate is also called the Annual Percentage Yield (APY) or Annual Percentage Rate (APR).

The effective rate is the interest rate compounded annually would be equivalent to the stated rate and compounding periods. We often suppose we invest $1 over the course of one year to determine the effective rate, as is shown in the next example.

To examine several investments to see which has the best rate, we find and compare the effective rate for each investment.

Example (PageIndex{5})

If Bank A pays 7.2% interest compounded monthly, what is the effective interest rate?
If Bank B pays 7.2% interest compounded semiannually, what is the effective interest rate?

Which bank pays more interest?

Solution

Bank A: Suppose we deposit $1 in this bank and leave it for a year, we will get

[egin{array}{l}
1left(1+frac{0.072}{12} ight)^{12}=1.0744
mathrm{r}_{mathrm{EFF}}=1.0744-1=0.0744
end{array} onumber]

We earned interest of $1.0744 - $1.00 = $.0744 on an investment of $1.

The effective interest rate is 7.44%, often referred to as the APY.

Bank B: The effective rate is calculated as

[mathbf{r}_{mathrm{EFF}}=1left(1+frac{0.072}{2} ight)^{2}-1=.0733 onumber]

The effective interest rate is 7.33%.

Bank A pays slightly higher interest, with an effective rate of 7.44%, compared to Bank B with effective rate 7.33%.

Definition: Effective Interest Rate, compounded (n) times per year

If a bank pays an interest rate (r) per year, compounded (n) times a year, then the effective interest rate is given by [mathbf{r}_{mathrm{EFF}}=left(1+frac{r}{n} ight)^{n}-1]

This is also reffered to as the annual percentage yield, or APY.

Continuous Compounding

Interest can be compounded yearly, semiannually, quarterly, monthly, and daily. Using the same calculation methods, we could compound every hour, every minute, and even every second. As the compounding period gets shorter and shorter, we move toward the concept of continuous compounding.

But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts? When interest is compounded "infinitely many times", we say that the interest is compounded continuously. Our next objective is to derive a formula to model continuous compounding.

Suppose we put $1 in an account that pays 100% interest. If the interest is compounded once a year, the total amount after one year will be ($ 1(1+1)=$ 2).

  • If the interest is compounded semiannually, in one year we will have ($ 1(1+1 / 2)^{2}=$ 2.25)
  • If the interest is compounded quarterly, in one year we will have ($ 1(1+1 / 4)^{4}=$ 2.44)
  • If the interest is compounded monthly, in one year we will have ($ 1(1+1 / 12)^{12}=$ 2.61)
  • If the interest is compounded daily, in one year we will have ($ 1(1+1 / 365)^{365}=$ 2.71)

We show the results as follows:

Frequency of compounding

Formula

Total amount

Annually

($ 1(1 + 1))

$2

Semiannually

($ 1(1+1 / 2)^{2})

$2.25

Quarterly

($ 1(1+1 / 4)^{4}=$ 2.44)

$2.44140625

Monthly

($ 1(1+1 / 12)^{12})

$2.61303529

Daily

($ 1(1+1 / 365)^{365})

$2.71456748

Hourly

($ 1(1+1 / 8760)^{8760})

$2.71812699

Every minute

($1(1+1 / 525600)^{525600})

$2.71827922

Every Second

($ 1(1+1 / 31536000)^{31536000} )

$2.71828247

Continuously

($ 1(2.718281828 ldots))

$2.718281828...

We have noticed that the $1 we invested does not grow without bound. It starts to stabilize to an irrational number 2.718281828... given the name "e" after the great mathematician Euler.

In mathematics, we say that as (n) becomes infinitely large the expression equals (left(1+frac{1}{n} ight)^{n}) = e.

Therefore, it is natural that the number e play a part in continuous compounding.
It can be shown that as (n) becomes infinitely large the expression (left(1+frac{r}{n} ight)^{n t}=e^{r t})

Therefore, it follows that if we invest $(P) at an interest rate (r) per year, compounded continuously, after (t) years the final amount will be given by

[ A = P cdot e^{rt} onumber ]

Definition: Continuously Compounded Interest

If an amount (mathrm{P}) is invested for (t) years at an interest rate (r) per year, compounded continuously, then the future value is given by [mathrm{A} = mathrm{P}e^{rt}]

Example (PageIndex{6})

$3500 is invested at 9% compounded continuously. Find the future value in 4 years.

Solution

Using the formula for the continuous compounding, we get (A=Pe^{rt}).

egin{aligned}
A &=$ 3500 e^{0.09 imes 4}
A &=$ 3500 e^{0.36}
A &=$ 5016.65
end{aligned}

Example (PageIndex{7})

If an amount is invested at 7% compounded continuously, what is the effective interest rate?

Solution

If we deposit $1 in the bank at 7% compounded continuously for one year, and subtract that $1 from the final amount, we get the effective interest rate in decimals.

[egin{array}{l}
mathrm{r}_{mathrm{EFF}}=1 mathrm{e}^{0.07}-1
mathrm{r}_{mathrm{EFF}}=1.0725-1
mathrm{r}_{mathrm{EFF}}=.0725 ext { or } 7.25 \%
end{array} onumber]

Definition: Effective Interest Rate, compounded continously

If a bank pays an interest rate (r) per year, compounded continuously, then the effective interest rate is given by [mathrm{r}_{mathrm{EFF}}=e^{mathbf{r}}-1]

Example (PageIndex{8})

If an amount is invested at 7% compounded continuously, how long will it take to double?

Solution

We don’t know the initial value of the principal but we do know that the accumulated value is double (twice) the principal.

[mathrm{P} cdot {e}^{0.07t}=2 mathrm{P} onumber]

We divide both sides by (mathrm{P})

[e^{.07 t}=2 onumber]

Using natural logarithm:

[egin{array}{l}
.07 mathrm{t}=ln (2)
mathrm{t}=ln (2) / .07=9.9 : mathrm{years}
end{array} onumber]

It takes 9.9 years for money to double if invested at 7% continuous interest.

Example (PageIndex{9})

a. At the peak growth rate in the 1960’s the world's population had a doubling time of 35 years. At that time, approximately what was the growth rate?

b. As of 2015, the world population’s annual growth rate was approximately 1.14%. Based on that rate, find the approximate doubling time.

Solution

We expect the world's population to grow continuously, not in discrete intervals such as years or months. Therefore, we will use the formula (A = Pe^{rt}).

a. Substituting (2 mathrm{P}) for (A) and 35 for (t) gives us the equation

[2 mathrm{P}=mathrm{P} cdot {e}^{r(35)} onumber]

We divide both sides by (mathrm{P}):

[2 ={e}^{r(35)} onumber]

Using natural logarithm:

( ln (2) = r(35) onumber)

Dividing both sides by 35:

( dfrac{ln(2)}{35}=r onumber)

( 0.0198=r onumber)

The growth rate was approximately (1.98\%).

b. Substituting (2 mathrm{P}) for (A) and (0.0114) for (r) gives us the equation

[2 mathrm{P}=mathrm{P} cdot {e}^{0.0114t} onumber]

We divide both sides by (mathrm{P}):

[2 ={e}^{0.0114t} onumber]

Using natural logarithm:

( ln (2) = 0.0114t onumber)

Dividing both sides by 0.014:

( dfrac{ln(2)}{0.0114}=t onumber)

( 60.8=t onumber)

If the world population were to continue to grow at the annual growth rate of 1.14% , it would take approximately 60.8 years for the population to double.

SECTION 8.2 SUMMARY

Below is a summary of the formulas we developed for calculations involving compound interest:

COMPOUND INTEREST (n) times per year

  1. If an amount (mathrm{P}) is invested for (t) years at an interest rate (r) per year, compounded (n) times a year, then the future value is given by [A=Pleft(1+frac{r}{n} ight)^{n t} onumber] (mathbf{P}) is called the principal and is also called the present value.
  2. If a bank pays an interest rate (r) per year, compounded (n) times a year, then the effective interest rate is given by [mathbf{r}_{mathrm{EFF}}=left(1+frac{r}{n} ight)^{n}-1 onumber]

CONTINUOUSLY COMPOUNDED INTEREST

  1. If an amount (mathrm{P}) is invested for (t) years at an interest rate (r) per year, compounded continuously, then the future value is given by [mathrm{A} = mathrm{P}e^{rt} onumber]
  2. If a bank pays an interest rate (r) per year, compounded continuously, then the effective interest rate is given by [mathrm{r}_{mathrm{EFF}}=e^{mathbf{r}}-1 onumber]

Compound Interest Formula With Examples

Compound interest, or 'interest on interest', is calculated with the compound interest formula.

The formula for compound interest is P (1 + r/n)^(nt), where P is the initial principal balance, r is the interest rate, n is the number of times interest is compounded per time period and t is the number of time periods.

Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. Subtract the principal if you want just the compound interest.

The above assumes interest is compounded once per period (yearly). When incorporating multiple compounds per period (monthly compounding or quarterly compounding, etc), the formula changes. It looks like this:

The concept of compound interest is that interest is added back to the principal sum so that interest is gained on that already-accumulated interest during the next compounding period. How important is it? Just ask Warren Buffett, one of the world's most successful investors:

"My wealth has come from a combination of living in America, some lucky genes, and compound interest."

In this article, we'll take a look at the compound interest formula in more depth. We'll also go through an example and discuss other variations of the formula that can help you to calculate the interest rate and time factor or to incorporate regular contributions. Should you wish to try some calculations using your own figures, you can use our popular compound interest calculator.


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Compound Interest (CI) is the addition of Interest to the Initial principal value and also the accumulated interest of previous periods of a loan or any deposit. Use this online compound interest calculator to calculate C.I compounded for annually, half-yearly, quarterly.

Formula:

For compounding interest calculation, select an option (annually or half-yearly or quarterly) from the drop-down menu of 'Interest Compounded' box and enter the inputs, the compound interest calculator will update you the CI with ease. Compounding interest is the interest calculated on the initial principal and also on the accumulated interest of previous periods of a deposit or loan. It is also referred as 'interest on interest'. Higher the number of compounding periods, higher is the CI. Apart from CI, this compound interest calculator helps you to calculate principal and rate of interest too.

Example

If a person deposit Rs. 5000 paying 6% interest for 5years.What will be the compound interest on the same sum at the same rate for the same period, compounded annually?

Compound Interest= P(1+(R/100))^n
=5000(1+(6/100))^5
=Rs. 1691.13
Total Amount = Principal + Compound Interest
= Rs. 6691.13


Compound Interest Formulas III

The fifth group in Table 1-5 covers a set of problems that uniform series of equal investments, A, occurred at the end of each time period for n number of periods at the compound interest rate of i. In this case, the cumulated present value of all investments, P, needs to be calculated. In summary, P is unknown and A, i, and n are given parameters. And the problem can be noted as P / A i , n and displayed as:

Figure 1-6: Uniform Series Present-Worth Factor, P / A i , n

If we replace substitute F in Equation 1-3 from Equation 1-2, we will have the present value as:

Equation 1-3: F = A [ ( 1 + i ) n − 1 ] / i Equation 1-2: F = P ( 1 + i ) n P ( 1 + i ) n = A [ ( 1 + i ) n − 1 ] / i

Equation 1-5 gives the cumulated present value, P, of all uniform series of equal investments, A, as P = A [ ( 1 + i ) n − 1 ] / [ i ( 1 + i ) n ] . And also can be noted as: P = A * P / A i , n .The factor [ ( 1 + i ) n − 1 ] / [ i ( 1 + i ) n ] is called the “uniform series present-worth factor” and is designated by P / A i , n . This factor is used to calculate the present sum, P that is equivalent to a uniform of equal end of period payments, A. Then P / A i , n = A [ ( 1 + i ) n − 1 ] / [ i ( 1 + i ) n ]

Note that n is the number of time periods that equal series of payments occur.

Please review the following video, Uniform Series Present Worth Factor (Time 3:35).

PRESENTER: The fifth group covers the set of problems that P is a known parameter, A, i and n are given variables. In these problems, we have uniform series of equal investments, A, in the end of each time period, for n number of periods, at the compound interest rate of I.

And the problem asks you to calculate the accumulated present value of all investments, P. We can summarize these questions using the factor notation. P is the unknown variable, and should be on the left side. And A is the given, and should be written on the right side.

As explained before, Equation 1-3 returns the future value, F, from A, i and n. And Equation 1-2 calculates the future value, F, from present value, P, interest rates, i and n number of periods. So if we substitute F in Equation 1-3 from Equation 1-2, we will have this new equation-- 1-5. This equation gives us the accumulated present value of equal series payments, A, paid for n period, at interest rate of i.

Equation 1-5 can also be written according to factor notation. P equals A times the factor P over A. This factor is called Uniform Series Present-Worth Factor, which is used to calculate the presence on P that is equivalent to a uniform series of equal payments, end of the period payments, A.

For example, what would be the present value of 10 uniform investments of $2,000, invested at the end of each year, for interest rate of 12%, compounded annually? First, we draw the time line. Left hand side is a present time, time zero payment, which needs to be calculated. N equals 10, because there are 10 uniform investments.

So we have 10 years. And above each year, we have $2,000, starting from year one to year 10. So A equals $2,000, n is 10, and interest rate is 12%. Using the factorization, P equals A, multiply the factor-- i is 12%, and n is 10. And the result.

So if you save $2,000 per year, at the end of each year for 10 years, starting from year one to year 10, the accumulated money is equal to $11,300 at present time. It has the same value as $11,300 at the present time.

Example 1-5:

Calculate the present value of 10 uniform investments of 2000 dollars to be invested at the end of each year for interest rate 12% per year compound annually.

P=? A=$2000 A=$2000 A=$2000 A=$2000 0
0 1 2 . 9 10

Using Equation 1-5, we will have:
P = A * P / A i , n = A [ ( 1 + i ) n − 1 ] / [ i ( 1 + i ) n ] P = A * P / A 12 % , 10 = 2000 * [ ( 1 + 0.12 ) 10 − 1 ] / [ 0.12 ( 1 + 0.12 ) 10 ] P = 2000 * 5.650223 = $ 11 , 300.45

Note that we use the factor P / A i , n when we have equal series of payments. i is the interest rate and n is the number of equal payments. There is an important assumption here, the first payment has to start from year 1. In that case P / A i , n will return the equivalent present value of the equal payments.

Now let's consider the case that we have equal series of payments and the first payment doesn't start from year 1. In that case the factor P / A i , n will give us the equivalent single value of equal series of payments in the year before the first payment. However, we want the present value of them (at year 0). So, we need to multiply that with the factor P / F i , n and discount it to the present time (year 0).

P=? A=$2000 A=$2000 A=$2000 0
0 1 2 . 10 11

Note that there are 10 equal series of $2,000 payments. But the first payment is not in year 1. The factor P / A 12 % , 10 returns the equivalent value of these 10 payments to the year before the first payment, which is year 1.

P=? $2000(P/A12%,10) 0
0 1 2 . 10 11

However, we want the present value. So, we need to discount the value by one year to have the present value of 10 equal payments.

P=? $2000(P/A12%,10)(P/F12%,1) 0
0 1 2 . 10 11
Present value = 2 , 000 ( P / A 12 % , 10 ) ( P / F 12 % , 1 )

Example: Now consider the the following case that the first payment starts at year 3:

P=? A=$2000 A=$2000 A=$2000 0
0 1 2 3 . 10 12

Present value = 2 , 000 ( P / A 12 % , 10 ) ( P / F 12 % , 2 )

Table 1-10: Uniform Series Present-Worth Factor, P/Ai,n
Factor Name Formula Requested variable Given variables
P / A i , n Uniform Series Present-Worth Factor [ ( 1 + i ) n − 1 ] / [ i ( 1 + i ) n ] P: Present value of uniform series of equal investments A: uniform series of equal investments
n: number of time periods
i: interest rate

6.Capital-Recovery Factor

The sixth group in Table 1-5 belongs to set of problems that A is unknown and P, i, and n are given parameters. In this category, uniform series of an equal sum, A, is invested at the end of each time period for n periods at the compound interest rate of i. In this case, the cumulated present value of all investments, P, is given and A needs to be calculated. It can be noted as A / P i , n .

Figure 1-7: Capital-Recovery Factor, A / P i , n

Equation 1-5 can be rewritten for A (as unknown) to solve these problems:

Equation 1-6 determines the uniform series of equal investments, A, from cumulated present value, P, as A = P [ i ( 1 + i ) n ] / [ ( 1 + i ) n − 1 ] . The factor [ i ( 1 + i ) n ] / [ ( 1 + i ) n − 1 ] is called the “capital-recovery factor” and is designated by A/Pi,n. This factor is used to calculate a uniform series of end of period payment, A that are equivalent to present single sum of money P.

Note that n is the number of time periods that equal series of payments occur.

Please watch the following video, Capital Recovery Factor (Time 3:37).

PRESENTER: The sixth group belongs to the set of problems that A is unknown and P, i, and n are given parameters. This category is similar to the fifth group, but P is given and A needs to be calculated. In this category of problems, we know the present value P, or accumulated present value of all payments. And we want to calculate the uniform series of equal sum A that are invested in the end of each time period for n periods at the compound interest rate of i.

So we have present value P, and we want to calculate equivalent A, given interest rate of i and number of periods n. The proper factor to summarize these questions is A over P, or A/P. A is the unknown variable, is on the left side, and P, given variable, on the right side.

Equation to calculate A is straightforward. We just need to rewrite the equation in 1-5 for A as unknown, and we will have equation 1-6 that calculates A from P, i, and n. If we write the equation 1-6 according to the factor notation, we will have factor A over P. The factor is called capital recovery factor and is used to calculate uniform sales of end of period payments A that are equivalent to present single sum of money P.

Let's work on this example. We want to know the uniform series of equal investment for five years at interest rate of 4% which are equivalent to $25,000 today. Let's say you want to buy a car today for $25,000, and you can finance the car for five years and 4% of interest rate per year, compounded annually. And you want to know how much you have to pay each year.

First, we draw the timeline. Left side is the present time, which we have $25,000. n equals 5, and above each year, starting from year one to year five, we have A that has to be calculated. For the factor, we have i equal 4% and n is five and the result, which tells us $25,000 at present time is equivalent to five uniform payments of $5,616 starting from year one to year five with 4% annual interest rate. Or $25,000 at present time has the same value of five uniform payments of $5,616 starting from year one to year five with 4% annual interest rate.

Example 1-6:

Calculate uniform series of equal investment for 5 years from present at an interest rate of 4% per year compound annually which are equivalent to 25,000 dollars today. (Assume you want to buy a car today for 25000 dollars and you can finance the car for 5 years with 4% of interest rate per year compound annually, how much you have to pay each year?)

P=$25,000 A=? A=? A=? A=? A=? 0
0 1 2 3 4 5

Using Equation 1-6, we will have:
A = P * A / P i , n = P [ i ( 1 + i ) n ] / [ ( 1 + i ) n − 1 ] A = P * A / P 4 % , 5 = 25 , 000 * [ 0.04 ( 1 + 0.04 ) 5 / [ ( 1 + 0.04 ) 5 − 1 ] ] A = 25 , 000 * 0.224627 = 5615.68

So, having $25,000 at the present time is equivalent to investing $5,615.68 each year (at the end of the year) for 5 years at annual compound interest rate of 4%.

Table 1-11: Capital-Recovery Factor, A/Pi,n
Factor Name Formula Requested variable Given variables
A / P i , n Capital-Recovery Factor [ i ( 1 + i ) n ] / [ ( 1 + i ) n − 1 ] A: uniform series of equal investments P: Present value of uniform series of equal investments
n: number of time periods
i: interest rate

A / P i , n = A / F i , n * F / P i , n = P [ i ( 1 + i ) n ] / [ ( 1 + i ) n − 1 ]

Using these six techniques, we can solve more complicated questions.

Example 1-7:

Assume a person invests 1000 dollars in the first year, 1500 dollars in the second year, 1800 dollars in the third year, 1200 dollars in the fourth year and 2000 dollars in the fifth year. At an interest rate of 8%:
1) Calculate time zero lump sum settlement “P”.
2) Calculate end of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments.
3) Calculate five uniform series of equal payments "A", starting at year one, that is equivalent to above values.

P=? 1000 1500 1800 1200 2000 F=?
0 1 2 3 4 5

1) Time zero lump sum settlement “P” equals the summation of present values:

P = 1000 * ( P / F 8 % , 1 ) + 1500 * ( P / F 8 % , 2 ) + 1800 * ( P / F 8 % , 3 ) + 1200 * ( P / F 8 % , 4 ) + 2000 * ( P / F 8 % , 5 ) P = 1000 * 0.92593 + 1500 * 0.85734 + 1800 * 0.79383 + 1200 * 0.73503 + 2000 * 0.68058 P = 5884.03

2) End of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments equals the summation of future values:

F = 1000 * ( F / P 8 % , ( 5 − 1 ) ) + 1500 * ( F / P 8 % , ( 5 − 2 ) ) + 1800 * ( F / P 8 % , ( 5 − 3 ) ) + 1200 * ( F / P 8 % , ( 5 − 4 ) ) + 2000 F = 1000 * ( F / P 8 % , 4 ) + 1500 * ( F / P 8 % , 3 ) + 1800 * ( F / P 8 % , 2 ) + 1200 * ( F / P 8 % , 1 ) + 2000 F = 1000 * 1.36049 + 1500 * 1.25971 + 1800 * 1.1664 + 1200 * 1.08 + 2000 F = 8645.58

Please note that in the factor subscript, n is the number of time period difference between F (the time that future value has to be calculated) and P(the time that the payment occurred). For example, 1800 payment occurs in year 3 but we need its future value in year 5 (2 year after) and time difference is 2 years. So, the proper factor would be: ( F / P 8 % , ( 5 − 3 ) ) or ( F / P 8 % , 2 ) .

3) Uniform series of equal payments "A" can be calculated from either P or F :
A = 5884.03 * A / P 8 % , 5 = 5884.03 * 0.25046 = 1473.7 or
A = 8645.58 * A / F 8 % , 5 = 8800.71 * 0.17046 = 1473.7

Example 1-8: repeat your calculations for the following payments:

P=? 800 1000 1000 1600 1400 F=?
0 1 2 3 4 5

1) Time zero lump sum settlement “P” equals the summation of present values: P = 800 + 1000 * ( P / F 8 % , 1 ) + 1000 * ( P / F 8 % , 2 ) + 1600 * ( P / F 8 % , 3 ) + 1400 * ( P / F 8 % , 4 ) P = 800 + 1000 * 0.92593 + 1000 * 0.85734 + 1600 * 0.79383 + 1400 * 0.73503 P = 4882.44

2) End of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments equals the summation of future values: F = 800 * ( F / P 8 % , 5 ) + 1000 * ( F / P 8 % , 4 ) + 1000 * ( F / P 8 % , 3 ) + 1600 * ( F / P 8 % , 2 ) + 1400 * ( F / P 8 % , 1 ) F = 800 * 1.46933 + 1000 * 1.36049 + 1000 * 1.25971 + 1600 * 1.1664 + 1400 * 1.08 F = 7173.9

3) Uniform series of equal payments "A" can be calculated from either P or F:
A = 4882.44 * A / P 8 % , 5 = 4882.44 * 0.25046 = 1222.84 or
A = 7173.9 * A / F 8 % , 5 = 7173.9 * 0.17046 = 1222.84


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Compound Interest

What's compound interest and what's the formula for compound interest in Excel? This example gives you the answers to these questions.

1. Assume you put $100 into a bank. How much will your investment be worth after 1 year at an annual interest rate of 8%? The answer is $108.

2. Now this interest ($8) will also earn interest (compound interest) next year. How much will your investment be worth after 2 years at an annual interest rate of 8%? The answer is $116.64.

3. How much will your investment be worth after 5 years? Simply drag the formula down to cell A6.

4. All we did was multiplying 100 by 1.08, 5 times. So we can also directly calculate the value of the investment after 5 years.

Note: there is no special function for compound interest in Excel. However, you can easily create a compound interest calculator to compare different rates and different durations.

5. Assume you put $100 into a bank. How much will your investment be worth after 5 years at an annual interest rate of 8%? You already know the answer.

Note: the compound interest formula reduces to =100*(1+0.08/1)^(1*5), =100*(1.08)^5

6. Assume you put $10,000 into a bank. How much will your investment be worth after 15 years at an annual interest rate of 4% compounded quarterly? The answer is $18,167.

Note: the compound interest formula reduces to =10000*(1+0.04/4)^(4*15), =10000*(1.01)^60

7. Assume you put $10,000 into a bank. How much will your investment be worth after 10 years at an annual interest rate of 5% compounded monthly? The answer is $16,470.

Note: the compound interest formula always works. If you're interested, download the Excel file and try it yourself!


ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.2

Question 1.
Calculate the compound interest on ₹6000 at 10% per annum for two years.
Solution:

Question 2.
Salma borrowed from Mahila Samiti a sum of ₹ 1875 to purchase a sewing machine. If the rate of interest is 4% per annum, what is the compound interest that she has to pay after 2 years?
Solution:

Question 3.
Jacob invests ₹12000 for 3 years at 10% per annum. Calculate the amount and the compound interest that Jacob will get after 3 years.
Solution:

Question 4.
A man invests ₹46875 at 4% per annum compound interest for 3 years.
Calculate:
(i) the interest for the first year.
(ii) the amount standing to his credit at the end of second year.
(iii) the interest for the third year.
Solution:

Question 5.
Calculate the compound interest for the second year on ₹6000 invested for 3 years at 10% p.a. Also find the sum due at the end of third year.
Solution:

Question 6.
Calculate the amount and the compound interest on ₹5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.
Solution:

Question 7.
Calculate the difference between the compound interest and the simple interest on ₹20000 in 2 years at 8% per annum.
Solution:


What Happens To An Account With Compounded Interest And No Withdrawals?


Consider now an account in which P0 is invested at the beginning of a compounding period, with a nominal interest rate r and compounding K times per year (so each compounding period is (1/K) th of one year). How much will be in the account after n compounding periods? Let P j denote the balance in the account after j compounding periods, including the interest earned in the last of these j periods. NOTE THAT WE HAVE JUST DEFINED A SEQUENCE OF REAL NUMBERS. To review what these sequences are, in general, see sequences of real numbers. Note that we have a recursive definition of this sequence:

Pj+1=P j + the interest earned by Pj in one compounding period.

In words, the balance at the end of a new compounding period is the balance at the end of the preceding period plus the interest that older balance earned during the compounding period. The interest earned is r * (1/K) * Pj,, as described above in the interest calculation for one period. Thus, at the end of the (j+1) th period,
Pj+1 = Pj + the interest earned by Pj in one compounding period
= Pj + (nominal rate)*(compounding period as a fraction of a year)*Pj
= Pj + r * (1/K) * Pj
= Pj + (r/K) * Pj
= Pj * (1 + r/K)

Importance of compounding intervals

The frequency of compounding and wealth accumulation are directly related. The higher the frequency of compounding, more the accumulation of wealth. Let&rsquos look at the growth of Rs 10,000 at 10% interest compounded at different frequencies.

Time Annual Quarterly Monthly
1 11,000.00 11,038.13 11,047.13
5 16,105.10 16,386.16 16,453.09
10 25,937.42 26,850.64 27,070.41

It is very clear from the above example that the higher the compounding interval, higher is the wealth accumulated. Also, longer the investment tenure higher is the wealth accumulated.

Let the magic of compounding work for you by investing regularly and staying invested for long horizons and increasing the frequency of loan payments. By familiarizing yourself with such concepts you can make better financial decisions and earn higher returns.


Watch the video: : Investments Using Compound Interest Pt. 1 (October 2021).