whitetoastisgood
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Would anyone be able to help me find out if interest problems (time-value-of-money) expressed as exponential formulas?
thanks a mil,
thanks a mil,
Interest Problems
- Thread starter whitetoastisgood
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Interest on bank accounts or investments comes in two flavors, simple interest and compound interest.
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 S is the final compounded amount accumulated where S = P(1+i)^n, 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 S = $5000(1+.005)^120 = $5000(1.8194) = $9097. Consequently, the compound interest realized is $9097 - $5000 = $4097. Of course the compound interest 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.
1--What will R dollars deposited in a bank account at the end of each of n periods, and earning interest at I%, compounded n times per year, amount to in N years?
This is called an ordinary annuity, differeing from an annuity due. 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 and there is no payment made at the end of the nth period.
The formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span is
S(n) = R[(1 + i)^n - 1]/i
where S(n) = the accumulation over the period of n inter, P = 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. This is known as an ordinary annuity.
When an annuity is cumputed 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 = R[{(1 + i)^(n + 1) - 1}/i - 1]
Simple 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
S = [R[(1 + i)^(n +1) - 1]/i - R] = 200[(1.12)^6 - 1]/.12 - 200 = $1,423.03
2--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, P = R[1 - (1 + i)^(-n)]/i where P = the Present Value, R = the periodic payment, n = the number of payment periods, and i = I/100.
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 bearing an annual interest of 6% for $137, 639, will anable the $12,000 annual payment over a 20 year period, for a total payout of $240,000.
3--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 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.
4--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 deposit, and interest paying, periods, obseve what happens to the periodic deposits if the depost 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.
5--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 in case 4.
There are other variations of these basic applications which may be found in any good book on Business Mathematics or Accounting.
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 S is the final compounded amount accumulated where S = P(1+i)^n, 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 S = $5000(1+.005)^120 = $5000(1.8194) = $9097. Consequently, the compound interest realized is $9097 - $5000 = $4097. Of course the compound interest 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.
1--What will R dollars deposited in a bank account at the end of each of n periods, and earning interest at I%, compounded n times per year, amount to in N years?
This is called an ordinary annuity, differeing from an annuity due. 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 and there is no payment made at the end of the nth period.
The formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span is
S(n) = R[(1 + i)^n - 1]/i
where S(n) = the accumulation over the period of n inter, P = 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. This is known as an ordinary annuity.
When an annuity is cumputed 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 = R[{(1 + i)^(n + 1) - 1}/i - 1]
Simple 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
S = [R[(1 + i)^(n +1) - 1]/i - R] = 200[(1.12)^6 - 1]/.12 - 200 = $1,423.03
2--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, P = R[1 - (1 + i)^(-n)]/i where P = the Present Value, R = the periodic payment, n = the number of payment periods, and i = I/100.
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 bearing an annual interest of 6% for $137, 639, will anable the $12,000 annual payment over a 20 year period, for a total payout of $240,000.
3--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 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.
4--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 deposit, and interest paying, periods, obseve what happens to the periodic deposits if the depost 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.
5--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 in case 4.
There are other variations of these basic applications which may be found in any good book on Business Mathematics or Accounting.
whitetoastisgood
New member
- Joined
- Feb 27, 2006
- Messages
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Thank you
Wow, Thank you so much! That is perfect. You put it in terms which i can understand. I will be able to answer my question.
Thanks again!
Wow, Thank you so much! That is perfect. You put it in terms which i can understand. I will be able to answer my question.
Thanks again!
whitetoastisgood
New member
- Joined
- Feb 27, 2006
- Messages
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Confusion
I have math theory questions. Am i not able to request help on those?
Thanks a mil,
I have math theory questions. Am i not able to request help on those?
Thanks a mil,
- Joined
- Apr 12, 2005
- Messages
- 11,339
If you actually have a math theory question, post it in the "Other Math Help" section. So far, you have not asked such a question. Your response to the textbook section TchrWill wrote indicates you really are not asking "Math Theory" questions. There are many text books available and many review websites that you can find equally useful.