Step 1: What really is going on with compound interest?
Each year (because the compounding is annual), the bank takes Mary's
account balance, multiplies it by the interest rate, i, and
adds that amount of money to her account. i is usually given
as a percent, but percents are just fractions over one hundred. So
we shall assume i is a fraction. What the bank does when it
compounds Mary's account is equivalent to simply taking her old balance,
multiplying it by
x = i + 1 eq. 5.3-2-1Then each time the bank compounds her account they are just multiplying by x
Now suppose Mary had just deposited $1000 back in 1992 and left it there without making any further deposits. What would have happened? On January 1 of 1993, the bank would have multiplied her balance by x. On January 1 of 1994, the bank would have multiplied her balance by x again. And again on January 1 of 1995 and again on January 1 of 1996 and yet again on January 1 of 1997. Altogether the bank would have multplied her account by x five times. So on January 1 of 1997, her account balance would stand at $1000 times x5.
But Mary did make further deposits. And when you think about it, the $1000 from 1992 was exposed to five compoundings, the $1500 from 1993 was exposed to four compoundings, the second $1500 from 1994 was exposed to three compoundings, the $2500 from 1995 was exposed to two compoundings, and the $3500 from 1996 was exposed to only one compounding. Do you see where this is leading? Can you make the polynomial interpretation of this problem now? And remember that her balance in 1997 was $11953.84. On January 1 of 1997, what would Mary have to do to her account to make the balance be zero? Clearly she would have to withdraw $11953.84 from it. So can you use that to determine the constant term of the polynomial function you are going to apply Newton-Raphson to? I really want you to think about this before going on. Try to write the polynomial without peeking at the next step.
Step 2: The polynomial function you should have gotten is
f(x) = 1000x5 + 1500x4 + 1500x3 + 2500x2 + 3500x - 11953.84
eq. 5.3-2-2
Do you see why? Make sure you understand that. Reread the last step if
you need to.
We are now looking at finding the x that makes f(x) be zero. And Newton-Raphson is the tool to do it. I will choose the first guess, x0, wishfully. If banks were paying 10% interest, then x would be 1.10. And so that is what I'll choose for x0. You can choose any reasonable interest rate to construct an x0 from and still get the same answer at the end of this problem. We also need to know f'(x). See if you can find it yourself before proceeding to the next step.
Step 3: When you took the derivative of f(x) you should have gotten
f'(x) = 5000x4 + 6000x3 + 4500x2 + 5000x + 3500 eq. 5.3-2-3Now we go on to substitute equations 5.3-2-2 and 5.3-2-3 into 5.3-13, which is repeated here.
f(xn)
xn+1 = xn - eq. 5.3-13
f'(xn)
This I know you can do for yourself.
You have f(x) and f'(x).
Go ahead and make the substitution.
Step 4: Here is what you should have gotten for the Newton-Raphson iteration formula:
1000xn5 + 1500xn4 + 1500xn3 + 2500xn2 + 3500xn - 11953.84
xn+1 = xn -
5000xn4 + 6000xn3 + 4500xn2 + 5000xn + 3500
eq. 5.3-2-4
If you got this far, then consider yourself to have done all the thinking
part of the problem. The rest is just grinding it out on a calculator
by applying 5.3-2-4 over and over. Here is the table I get starting with
the x0 of 1.10 I selected:
x0 = 1.10000000000000 x1 = 1.07565433675613 x2 = 1.07500042749496 x3 = 1.07499996987911 x4 = 1.07499996987889That gets f(x) to within a billionth of a dollar of zero. Good enough. To within the accuracy any bank would publish its interest rate, the rate here is 7.5% per annum. And the reason that on the second to last iteration the accuracy did not double was due to roundoff errors in the calculator.
email me at hahn@netsrq.com