# quick interest question ...

• Jan 8th 2007, 05:14 PM
asiankatt
quick interest question ...
To provide an annual scholarship for 25 years, a donation of $50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship? I would greatly appreciate it if I could be given some help. Thanks so much in advance. • Jan 9th 2007, 04:41 AM ticbol Quote: Originally Posted by asiankatt To provide an annual scholarship for 25 years, a donation of$50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship?

I would greatly appreciate it if I could be given some help.

I don't know if there is a ready formula for this. Let us derive one.

Let X = amount of scholarship to be deducted every year
A = amount of investment after any year
P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)] S(24) = (1.055)[(1 -(1.055)^24) / (1 -(1.055)] S(24) = (1.055)[(-2.61459)/(-0.055)] S(24) = 50.15259 ----------------------*** Substituting that and the givens into (i), A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) which is zero, so, 0 = (50,000)(1.055)^25 -X(50.15259) -X 0 = 190,670 -X(51.15259) X = (190,670)/(51.15259) X =$3727.47 -----------------answer.
• Jan 9th 2007, 07:07 AM
Soroban
Hello, ticbol!

There is a "Sinking Fund" formula . . . which you derived from scratch.

. . Lovely work!

One of its forms looks like this: .$\displaystyle A \;=\;P\frac{i(1 + i)^n}{(1+i)^n - 1}$

where: .$\displaystyle \begin{array}{cccc} P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ A & = & \text{periodic withdrawl} \end{array}$

• Jan 9th 2007, 10:02 AM
ticbol
Quote:

Originally Posted by Soroban
Hello, ticbol!

There is a "Sinking Fund" formula . . . which you derived from scratch.

. . Lovely work!

One of its forms looks like this: .$\displaystyle A \;=\;P\frac{i(1 + i)^n}{(1+i)^n - 1}$

where: .$\displaystyle \begin{array}{cccc} P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ A & = & \text{periodic withdrawl} \end{array}$

Good morning, Soroban.

So that is a Sinking Fund. Umm.
I just had time last night to play with it. I thought the question was interesting.
• Jan 9th 2007, 10:30 AM
Soroban
Hello again, ticbol!

I can appreciate the reasoning and the algebra
. . that went into your derivation.

Every year or so, I can't my list of favorite formulas,
. . and I've been forced to derive the Amortization Formula.