Accumulating InterestDate: 07/10/2003 at 18:25:25 From: Joey Subject: Accumulating interest I'm taking a course on problem-solving, in which we were given the following problem to solve: Michael has $355.00 left over each month after paying all his monthly bills. Instead of spending his leftover money each month, Michael chooses to open a savings account offering 3% interest, accrued monthly. Calculate the amount of money Michael will have in his savings account at the end of each of the following periods: a) 3 months b) 6 months c) 12 months Here is what I tried so far: Mon Amount --- -------------------------------------------------- 1 (.03) * 355.00 = 365.65 2 365.65 + 355.00 = 720.65 *.03 = 21.62 + 720.65 = $742.27 3 742.27 + 355.00 = 1,097.27 * .03 = 32.92 + 1097.27 = 1130.19 4 1130.19 + 355 = 1485.19 * .03 = 44.55 + 1485.19 = 1529.74 Am I on the right track? - Joey Date: 07/10/2003 at 19:33:08 From: Doctor Ian Subject: Re: Accumulating interest Hi Joey, One way to think about it is to imagine how you'd solve it using a spreadsheet. Each row of the spreadsheet would represent what happens at the end of a month. You have a current balance, which at first would be a single deposit: +--------------+ | Balance | | during month | +--------------+ | D | +--------------+ You compute some percentage of that, and add it to the balance to get the new balance. +--------------+---------------+ | Balance | After | | during month | interest | +--------------+---------------+ | D | iD | +--------------+---------------+ Note that I'm using i to mean (1 plus the interest rate), e.g., if the interest rate is 3%, then i is 1.03. Why? Because that's how you _add_ interest: x + 0.03*x = (1 + 0.03)*x = 1.03*x After interest is paid, you add another deposit: +--------------+---------------+---------------+ | Balance | After | After | | during month | interest | deposit | +--------------+---------------+---------------+ | D | iD | iD + D | +--------------+---------------+---------------+ and what you have at the end becomes the new balance during the next month: +--------------+---------------+---------------+ | Balance | After | After | | during month | interest | deposit | +--------------+---------------+---------------+ | D | iD | iD + D | +--------------+---------------+---------------+ | iD + D | +--------------+ Now we need to pay interest on the balance during the month. Just as before, we do this by multiplying i by whatever is in the first column: +--------------+---------------+---------------+ | Balance | After | After | | during month | interest | deposit | +--------------+---------------+---------------+ | D | iD | iD + D | +--------------+---------------+---------------+ | iD + D | i(iD + D) | +--------------+---------------+ And then we add the new deposit: +--------------+---------------+---------------+ | Balance | After | After | | during month | interest | deposit | +--------------+---------------+---------------+ | D | iD | iD + D | +--------------+---------------+---------------+ | iD + D | i(iD + D) | i(iD + D) + D | +--------------+---------------+---------------+ Now we can simplify that last item to look like this: +--------------+---------------+---------------+ | Balance | After | After | | during month | interest | deposit | +--------------+---------------+---------------+ | D | iD | iD + D | +--------------+---------------+---------------+ | iD + D | i(iD + D) | i^2D + iD + D | +--------------+---------------+---------------+ Does this make sense so far? If you actually _have_ a spreadsheet, you can go ahead and set it up this way. Assuming it makes sense, can you see why the third column will look like this? iD + D i^2 + iD + D i^3 + i^2D + iD + D i^4D + i^3D + i^2D + iD + D . . i^nD + i^(n-1)D + ... + i^2D + iD + D At each step, you're multiplying everything you already had by i (which is what you do to pay interest on it), and then adding D. We can use the distributive property to factor out a D, leaving us with (i^n + i^(n-1) + ... + i^2 + i + 1) * D Now, on the one hand, this still looks kind of complicated, although nowhere near as complicated as using actual numbers. But the good news is that a sequence like a, ar, ar^2, ar^3, ar^4, ... is a pretty well-known object, called a 'geometric sequence', and there is a formula for finding the sum of one in a single step: a + ar + ar^2 + ... + ar^n = (a - ar^(n+1))/(1-r) Since you're in a problem-solving class, there's a good chance that the point of assigning the problem was to get you to see why solving it in symbolic form is a better idea than just cranking out the calculations. One of the reasons I didn't actually check your results is that I couldn't really do that without going through all the steps that you went through. And here's the thing: mathematics is a kind of applied laziness. A mathematician is someone who has realized that in many cases, it's harder to solve a single _particular_ problem than it is to solve _all_ the related problems at once, in one shot, which is what I did here. What if I want to change the interest rate? No problem - I just change the value of the associated variable in my formula. What if I want to change the amount of each deposit? The same. And it's just a little bit more work to solve the problem in a way that lets me adjust for how often interest is compounded, e.g., weekly or daily. (And as a bonus, I didn't even have to write as much, because I was using single letters instead of strings of digits to represent my numbers.) But if you just solve the specific problem, using numbers, then to solve a slightly different problem is just as much work as the first time around. By the way, 3% accrued monthly works out to an annual interest rate of nearly 43 percent. Where is this bank? I'd like to open an account there! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 07/11/2003 at 07:45:57 From: Joey Subject: Accumulating interest The way I see doing this is: you start out with the amount which he has left over monthly which in my initial problems case is $355 and multiply it by the interest rate of .03. Then for the second month, take your balance from the previous month and add the deposit for the current month giving whatever and then multiply that by 1.03 and you have your amount for that month. Then just keep following these steps for the rest of the months. How exactly is that wrong is what I do not understand. I am calculating interest on his account balance after every month. Do you see what I am saying? Date: 07/11/2003 at 09:48:45 From: Doctor Ian Subject: Re: Accumulating interest Hi Joey, The problem with your method is that if you make the deposit _before_ computing the interest, you're paying interest immediately on the deposit. You don't get interest on money until after the bank has had a chance to use it. That's what they're paying interest for. So on January 1, you put $355 into the bank. Then on February 1, you put another $355 in, and the bank pays you interest... but only on the first $355. You don't get interest on the second $355 until March, at which time you put in another deposit. What makes these calculations a pain (and the reason people have developed formulas to avoid having to do them) is that you can't just compute the interest based on the current day's balance. You only get interest on the money that's been there for a while. (Otherwise I could put $10,000 into an account the day before interest gets paid, collect the interest, and withdraw all the money the next day. No bank would be able to stay in business if this were the case.) Here is another way to look at it. Suppose you deposit $100 on the first day of each month during the course of a year, but each time you start a new account instead of combining it all in one account. The bank pays you 5% interest per month on each account. At the end of the year, the account you opened on December 1 will have accrued interest for one month: Dec. acct: $100 * 1.05 What about the November account? At the end of November, you would get 5% interest; and you'd get it again at the end of December: Dec. acct: $100 * 1.05 Nov. acct: ($100 * 1.05) * 1.05 And the October account? You'll collect interest on that three times: Dec. acct: $100 * 1.05 Nov. acct: ($100 * 1.05) * 1.05 Oct. acct: (($100 * 1.05) * 1.05) * 1.05 And so on. Using exponents to tidy things up, Dec acct: $100 * 1.05^1 Nov acct: $100 * 1.05^2 Oct acct: $100 * 1.05^3 Sep acct: $100 * 1.05^4 Aug acct: $100 * 1.05^5 Jul acct: $100 * 1.05^6 Jun acct: $100 * 1.05^7 May acct: $100 * 1.05^8 Apr acct: $100 * 1.05^9 Mar acct: $100 * 1.05^10 Feb acct: $100 * 1.05^11 Jan acct: $100 * 1.05^12 Factoring out the original $100, that all adds up to $100 * (1.05^1 + 1.05^2 + ... + 1.05^12) which is our old friend, the geometric series. Now, this has to be the same amount of money as we'd get the other way. All we've done is change the situation around a little so that we're not mixing deposits and interests. The cost of that is having more accounts. But that's not _really_ a cost, because the accounts are imaginary. They just exist long enough to find a general solution to the class of problems, and then they disappear. However, even though there is no cost, there is a big benefit, which is that we get to solve 12 extremely simple problems rather than one complicated one. Does it make more sense now? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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