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A fund of $30,000 is used to award scholarships of amount 3000, one per year, at the end of each year for as long as possible. If i=0.09, find the number of scholarships which can be awarded and the amount left over in the fun one year after the last scholarship has been awarded.

Can someone please explain to me how I can do this?

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Please do add all the necessary information: what is "i"? Is this a problem on compound interest or what? What have you done...? – DonAntonio Feb 22 '13 at 3:57
the effective interest rate – Yenny Chen Feb 22 '13 at 4:32

This is a straight amortization problem. It is exactly the same as calculating the number of \$3000 payments on a \$30,000 mortgage, where the interest is 0.09 per payment period. This has a nice closed-form solution, which we can derive.

If you have a principal amount $N$ at the start of some period, then at the end of that period, it grows to $N(1 + i)$. Furthermore, $M$ is subtracted from it, leaving $N(1 + i) - M$. This gives us a recurrence relation:

$$N_{j+1} = N_j(1 + i) - M$$

which is of the form

$$N_{j+1} = A N_j + B$$

This is an order 1, linear, inhomogeneous recurrence relation. It is linear because the previous value is only multiplied by a constant. It is order 1, because only one previous value is considered. For instance Fibonacci is order two because a subsequent value is formed by combining two prior values. And it is inhomogenous because of the nonzero constant term.

Here is one way to solve this sucker. Let us first shift over to $j+2$:

$$N_{j+2} = A N_{j+1} + B$$

Now, we have two equations, and we can subtract them from each other. This cancels out the pesky B term common to both:

$$N_{j+2} - N_{j+1} = A N_{j+1} - A N_j + B - B$$

$$N_{j+2} - N_{j+1} = A N_{j+1} - A N_j$$

Now, yo, check this. We rearrange the equation, and we obtain something new: a linear recurrence which is order 2, but homogeneous:

$$N_{j+2} = N_{j+1} + A N_{j+1} - A N_j$$

$$N_{j+2} = (A + 1)N_{j+1} - A N_j$$

As if by magic, B has vanished. This means that no matter what the monthly payment is or the scholarship award, as the case may be, the recurrence relation is the same. What kicks off different amortizations is the first two values. Analogy: the Fibonacci recurrence relation $F_{i+2} = F_{i+1} + F_i$ does not give rise only to the Fibonacci sequence, but to others: the Fibonacci sequence is the one which begins with $1, 1$. IF you begin with, say, $3, 41$, then you get something else.

Anyhoo, back to solving:

$$N_{j+2} = (A + 1)N_{j+1} - A N_j$$

One way to attack this is through some matrix algebra. First we rewrite the recurrence as a 2x2 matrix multiplying a vector:

$$\begin{bmatrix}N_{j+2}\\N_{j+1}\end{bmatrix} = \begin{bmatrix}A + 1 & - A\\1 & 0\end{bmatrix}\begin{bmatrix}N_{j+1}\\N_j\end{bmatrix}$$

Basically, we are now visualizing the recurrence as a Markov chain process whereby we start with some pair of values which form an initial vector $I$, and we multiply that by the matrix to produce a new vector. We can do this as many times as we want.

$$\begin{bmatrix}N_{j+2}\\N_{j+1}\end{bmatrix} = {\begin{bmatrix}A + 1 & - A\\1 & 0\end{bmatrix}}^jI$$

So now all we need to do is raise the matrix to a power. And secondly, to establish the initial vector $I$ (where we will finally resurrece the $B$ coefficient).

When all that is done, we can substitute the quantities from the original problem: $A = i + 1 = 1.09$ (interest rate plus one) and $B = -M = -3000$ (award amount).

Determining I is very easy. It consists of the principal amount, and then the next period's value: the principal amount inflated by the interest and then robbed of the award figure:

$$I = \begin{bmatrix}29700\\30000\end{bmatrix}$$

(to be continued ... until then, that is your homework)

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One approach is just to use a spreadsheet. You start with 30,000. Each year you add 9% of the balance, then subtract 3000. See when the balance goes below zero. If you lay it out reasonably, you can just copy down once you have made the second year of formulas.

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This is a recursive function.

$x[k+1] = (x[k]\times1.09)-3000$


$x[0] = 30000$

You basically will have to simulate it.

fund = 30000;
years = 0
while (fund>=3000)
  years = years+1;

disp fund;
disp years;


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