Author 
Topic 

effort
Senior Member
USA
38 Posts 
Posted  03/20/2013 : 13:10:27

If $2000.00 is invested at 3% interest compounded every four months, what will be the amount in the investment after 10 years?
I use the formula A = P(1 + (r/n))raised to the nt
A= 2000(1+(.03/3))raised to (310) So A = 2000(1.01)raised to 30 which equals $2696.70
My first question is why can't I solve it by 2000(1 + .03) raised to the 30? Why do I need the rate divided by 3 for the top answer of 2696.70? Working the problem as 2000(1 + .03)raised to the 30 yields $4854.5 which isn't the same amount as using the above formula which yielded $2696.70
My third question is why if I use 2000(1.03) and get the total and then do it again til I have done it 30 times.....why doesn't it equal to 2696.70 if I am just adding the amounts of up?
Since posting this problem, I have kept thinking about it. Does the difference in the answers have anything to do with the interest rate being compounded? Is the problem saying that 3% is the total year interest rate? If so, I am assuming this would mean that you would have to divided the interest rate by 4 to get the actual rate for each four months? I'm not sure about my reason so any help on this is appreciated. 
Edited by  effort on 03/20/2013 15:07:05 


Ultraglide
Advanced Member
Canada
299 Posts 
Posted  03/20/2013 : 16:48:09

When the interest rate is quoted, i.e. 3%, that is the annual rate so in this case you want the interest per period which is 1% (3%/3). Since it is calculated 3 times per year, there are 30 (3x10) interest conversion periods. Hence your original calculation is correct. 


the_hill1962
Advanced Member
USA
1468 Posts 
Posted  03/22/2013 : 11:06:38

You have a typo in your answer. It is $2695.70, not $2696.70



TchrWill
Advanced Member
USA
79 Posts 
Posted  03/25/2013 : 11:19:54

quote: Originally posted by effort
If $2000.00 is invested at 3% interest compounded every four months, what will be the amount in the investment after 10 years?
I use the formula A = P(1 + (r/n))raised to the nt
A= 2000(1+(.03/3))raised to (310) So A = 2000(1.01)raised to 30 which equals $2696.70
My first question is why can't I solve it by 2000(1 + .03) raised to the 30? Why do I need the rate divided by 3 for the top answer of 2696.70? Working the problem as 2000(1 + .03)raised to the 30 yields $4854.5 which isn't the same amount as using the above formula which yielded $2696.70
My third question is why if I use 2000(1.03) and get the total and then do it again til I have done it 30 times.....why doesn't it equal to 2696.70 if I am just adding the amounts of up?
Since posting this problem, I have kept thinking about it. Does the difference in the answers have anything to do with the interest rate being compounded? Is the problem saying that 3% is the total year interest rate? If so, I am assuming this would mean that you would have to divided the interest rate by 4 to get the actual rate for each four months? I'm not sure about my reason so any help on this is appreciated.
Simple Interest
Simple interest is defined as Principal x Interest Rate per unit time x Number of units of time or I = Pin where I is the interest on a principal P at an interest rate of i per period for n periods. The interest rate is expressed as a decimal. For example, the simple interest for one year on $100 at a 5% annual interest rate (per year) is $100 x .05 x 1 = $5. The interest on $100 at a 5% annual interest rate for two years is $100 x .05 x 2 = $10. The interest on $100 at a 6% annual interest rate for one year, but paid monthly, is $100 x .06/12 x 12 = 50 cents per month x 12 = $6 for the year.
Compound Interest
With compound interest, the interest due and paid at the end of the interest compounding period is added to the initial starting principal to form a new principal, and this new principal becomes the amount on which the interest for the next interest period is based. The original principal is said to be compounded, and the difference between the the final total, the compound amount, accumulated at the end of the specified interest periods, and the original amount, is called the compound interest.
In its most basic use, if P is an amount deposited into an account paying a periodic interest, then Sn is the final compounded amount accumulated where
..........................Sn = P(1+i)^n
where i is the periodic interest rate in decimal form = %Int./(100m), n is the number of interest bearing periods, and m is the number of interest paying periods per year.
For example, the compound amount and the compound interest on $5000.00 resulting from the accumulation of interest at 6% annual interest compounded monthly for 10 years is as follows:
Since m = 12, i = .06/12 = .005. Since we are dealing with a total of 10 years with 12 interest periods per year, n = 10 x 12 = 120. From this we get
......Sn = $5000(1+.005)^120 = $5000(1.8194) = $9097.
Consequently, the compound interest realized is $9097  $5000 = $4097. Of course the compound interest rate can be calculated directly from the simple expression
.........................I = P[(1+i)^n  1].
The interest associated with annuities, loans, mortgages, savings accounts, IRA's, etc. are all based on the compound interest principle. Some additional applications of the use of compound interest are offered below.
Annuities
An annuity is a series of payments made at equal intervals of time. There are many types of annuities. Contingent annuities depend on uncertain future events for their number of payment periods. Examples are pensions which typically stop at the death of the pensioner and life insurance policies which guarantee payments for the insureds lifetime. Annuities certain guarantee a number of payment periods and the length of these periods. Perpetuities continue forever, the number of payments being unlimited while the length of the payment periods are definite. Annuities certain come in two varieties, ordinary annuities and annuities due. Ordinary annuities make payments at the end of each payment period while annuities due make payments at the beginning of each payment period. Deferred annuities are annuities certain where the payments do not start for a prestipulated number of payment periods. While dererred annuities are typically ordinary annuities, they are sometimes annuities due. Simple annuities have the interest bearing periods the same as the payment periods. General annuities have the interest bearing periods different from the payment periods.
Ordinary Annuity
What will an amount of R dollars deposited in a savings bank at the end of each year for N years, and earning I% compounded annually, amount to in N years. Stated another way, what will R dollars deposited in a bank account at the end of each of n periods per year, and earning interest at I/n, compounded n times per year, amount to in N years?
Remember, an ordinary annuity consists of a definite number of deposits made at the ENDS of equal intervals of time. An annuity due consists of a definite number of deposits made at the BEGINNING of equal intervals of time.
For an ordinary annuity over n payment periods, n deposits are made at the end of each period but interest is paid only on (n  1) of the payments, the last deposit drawing no interest, obviously. In the annuity due, over the same n periods, interest accrues on all n payments but there is no payment made at the end of the nth period. Consider the deposit of $R at the end of each year for 5 years at an iterest rate of I% compounded annually.
The compound interest on the first deposit at the end of the first year is S1 = R(1 + i)^(n1). The compound interest on the second deposit at the end of the second year is S2 = R(1+i)^(n2). The compound interest on the third deposit at the end of the third year is S3 = R(1+i)^(n3) The compound interest on the fourth deposit at the end of the fourth year is S4 = R(1+i). There is no interest gained on the fifth deposit at the end of the fifth year.
The final accumulation is S = R(1+i)^(n1) + R(1+i)^(n2) + R(1+i)^(n3) + R(1+i) + R. Reversing this, S = R + R(1+i) + R(1+i)^(n3) + R(1+i)^(n2) + R(1+i)^(n1) Note that this is a geometric progression with first term a = R, common factor r = (1+i) and the number of terms n = n. As we know, the sum of the terms of such a progression derives from
………………..S = a(r^n  1)/((r  1).
Therefore, the formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span, an ordinary annuity, becomes
………………..S(n) = R[(1+i)^n  1]/(1+i1) = R[(1+i)^n  1]/i
where S(n) = the accumulation over the period of n intervals, R = the periodic deposit, n = the number of interest paying periods, and i = the annual interest % divided by 100 divided by the number of interest paying periods per year.
When an annuity is computed on the basis of the payments being made at the beginning of each period, an annuity due, the total accumulation is based on one more period minus the last payment. Thus, the total accumulation becomes
.....................S(n+1) = R[(1 + i)^(n+1)  1]/i  R
Example: $200 deposited annually for 5 years at 12% annual interest compounded annually. Therefore, R = 200, n = 5, and i = .12.
Ordinary Annuity ..................................Deposit.......Interest.........Balance Beginning of month 1........0................0...................0 End of month..........1.....200...............0.................200 Beg. of month.........2.......0.................0................200 End of month..........2.....200..............24................424 Beg. of month.........3.......0.................0................424 End of month..........3.....200............50.88..........674.88 Beg. of month.........4.......0.................0.............674.88 End of month..........4.....200............80.98..........955.86 Beg. of month.........5.......0.................0.............955.86 End of month..........5.....200...........114.70........1270.56
....S = R[(1 + i)^n  1]/i = 200[(1.12)^5  1]/.12 = $1270.56
Annuity Due ..................................Deposit.......Interest..........Balance Beginning of month 1......200..............0..................200 End of month..........1.......0...............24.................224 Beg. of month.........2.....200..............0..................424 End of month..........2.......0.............50.88...........474.88 Beg. of month.........3.....200..............0...............674.88 End of month..........3.......0.............80.98...........755.86 Beg. of month.........4.....200..............0...............955.86 End of month..........4.......0...........114.70..........1070.56 Beg. of month.........5.....200..............0..............1270.56 End of month..........5.......0...........152.47..........1423.03
Sn = [R[(1 + i)^(n +1)  1]/i  R]= = 200[(1.12)^6  1]/.12 200 = $1,423.03
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Present Value
The present value of an ordinary annuity is the sum of the present values of the future periodic payments at the point in time one period before the first payment.
What is the amount that must be paid (Present Value) for an annuity with a periodic payment of R dollars to be made at the end of each year for N years, at an interest rate of I% compounded annually? For this scenario, we must compute the present value of each future payment at compound interest, at a time one period before the initial payment, and adding all the present values of all the payments.
Remember that the compounded amount of $P is S = P(1 + i)^n. Here, P is the present value, S = the compounded amount (or in this case, the periodic payment). Considering a simple example, find the present value of of a series of $200 payments made at the end of each year for 3 years in a fund that pays 4% interest compounded annually.
With the first periodic payment being made at the end of the first year, its present value is $200/(1 + .04)^1 = $200/1.04. With the second payment being made at the end of the second year, its present value is $200/(1.04)^2. With the third payment being made at the end of the third year, its present value is $200/(1.04)^3. The total present value of the three payments is therefore, $200/(1.04) + $200/(1.04)^2 + $200/(1.04)^3.
In general terms, we can write this as P = R/(1+i)^1 + R/(1+i)^2 _ R/(1+i)^3.......+R/(1+i)^(n1) + R/(1+i)^n. It is clear that the series of present valuse form a geometric progression with first term a = R/(1+i), the common factor r = 1/(1+i) and n = the number of terms, or payments. WIth the sum of such a progression being S = a(r^n  1)/(r  1), substitution leads us to
.........P = R[1  (1 + i)^(n)]/i
where P = the Present Value, R = the periodic payment (or rent), n = the number of payment periods, and i = I/100n.
Example: What is the present value of an annuity that must pay out $12,000 per year for 20 years with an annual interest rate of 6%? Here, R = 12,000, n = 20, and i = .06 resulting in ....P = 12000[1  (1.06)^20]/.06 = $137,639
Therefore, the purchase of an annuity for $137,639 bearing an annual interest of 6%, will anable the $12,000 annual payment over a 20 year period, for a total payout of $240,000.
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Periodic Payment
The periodic payment, often referred to as the rent, is what must be paid, knowing only the present value, annual effective interest rate, compounding periods and the number of payment periods.
Example: What is the periodic payment required to retire a debt of P dollars in n periods (months or years) if payments start at the end of the first period and bear I% interest compounded periodically? For this typical loan payment calculation,
............R = Pi/[1  (1 +i)^(n)]
where R = the rent (periodic payment), P = the amount borrowed, n = the number of payment periods, and i = I/100.
Example: What is the annual payment required to retire a loan of $10,000 over a period of 5 years at an annual interest rate of 8%? Here, P = 10,000, n = 5, and i = .08 resulting in
........R = 10000(.08)/[1  (1.08)^5] = $2504.56 per year
The total amount paid back becomes 5(2504.56) = $12,522.82 meaning that the use of the money cost the borrower $2,522.82. It is worthy of note that most loans are paid on a monthly basis. The significance of this to the borrower is that he is paying the money back more often, thus reducing the outstanding balance more rapidly. The effect of this is to reduce the total amount paid for the use of the money. Here, P = 10,000, n = 60, and i = .006666 resulting in
....R = 10000(.006666)/[1.006666)^60] = $202.76 per month.
The total amount paid back becomes 202.76(60) = $12,165.60, a saving of $357.22 by paying monthly.
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The other example of finding Periodic Payments is where the interest rate per period, number of periods and final amount are known. In this scenario, the typical problem is what is the annual deposit required to accumulate a fund of S dollars over a period of n years with deposits starting at the end of the first year and bearing an interest rate of I% compounded annually? For this typical retirement fund application,
....R = S(n)(i)/[(1 + i)^n  1]
where R = the annual deposit, n = the number of years, and i = I/100.
Example: How much must be deposited into an account at the end of each year, over a 20 year period, bearing interest at the rate of 6% compounded annually, that will grow to a total amount of $100,000? Here, P = 100,000, n = 20, and i = .06 resulting in
.......R = 100,000(.06)/[(1.06)^20  1] = $2,718.45 per year
for a total deposit over the 20 year period of 20(2718.45) = $54,369.
To fully understand the significance of increasing the number of deposits and interest paying periods, observe what happens to the periodic deposits if the deposit periods and interest paying periods are increased to monthly. Here, S remains at 100,000, but n increases to 240 (12(20), and i changes to .005 resulting in
......R = 100,000(.06)/[(1.005)^240  1] = $216.43 per month
This reduces the total annual deposit to 12(216.43) = $2,597.17 for a total deposit over the 20 year period of 12(2597.17) = $51,943.45.
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Another scenario involves not knowing the number of periods. For instance, how long will be required to accumulate an amount of S(n) dollars with a quarterly payment of R dollars per period into an account paying an annual interest rate of I% compounded compounded quarterly?
For this situation,
...........n = log[S(n)i/R + 1]/log[1 + i].
Example: How long will it take to accumulate $100,000 with a quarterly deposit of $654.83 into an account paying an annual interest of 6% compounded quarterly? Here, S(n) = 100,000, R = $654.83, and i = .015 resulting in
.........................n = log[(100000(.015))/654.83 + 1]/log(1.015) = 80 quarters = 160 months = 20 years
Note that this is just another version of the previous example.
There is a special formula for continuous compounding. Amount = Principal (e^(rt) ) where r is the annual rate compounded continuously and t is the number of years.




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