Looking ahead to retirement, you sign up for automatic savings in a fixed income 401K plan that pays 5% per year compounded annually. You plan to invest $2500 at the end of the year for the next 25 years. How much will your account have in it at the end of 25 years?
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Automatice savings
- Thread starter kayesal
- Start date
Looking ahead to retirement, you sign up for automatic savings in a fixed income 401K plan that pays 5% per year compounded annually. You plan to invest $2500 at the end of the year for the next 25 years. How much will your account have in it at the end of 25 years?
The following should enable you to come up with your answer.
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
The following should enable you to come up with your answer.
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