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December 17, 2014

Homework Help: math,help

Posted by student on Saturday, June 30, 2007 at 11:28am.

what formula do i use for the following problem:

which of the following investments is larger after 19years?
a) $7500 is deposited annually and earns 4.5% interest compounded annually.
b)$600 is deposited monthly and earns 4.5% interest compounded monthly.

Does he deposit the money at the beginning of the month/year (future annutity due) or the end of the month/year (ordinary annuity)?

(Broken Link Removed)

The answer depends on when the annuity is deposited

it doesn't say , so what do you think of the formula i should use.

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


what is the name of the formula you are using?

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