Compound interest to a specific amount
Date: 07/24/97 at 13:24:59 From: Patrick Burke Subject: Compound interest to a specific amount Hi, I've looked at your FAQ, and found dozens of compound interest formulae and explanations, but I'm sort of math-impaired and haven't been able to use them to find what I want. I'm not interested in the final amount, I'm interested in the time it would take for a regular deposit account to equal one million dollars, assuming monthly payments of $100.00 at, say, 8, 10 and 12 percent interest. The problem for me is t, time, is an exponential function in the compound interest formulae, so I have been ending up with the mt-1 root of $100,00, and things like that, and I don't know how to take the n-1 root of some number, so I get stuck. Here is my question: If I have a brokerage account, and I deposit $100 into it every month, how long will it take that account to reach $1,000,000 at: 8 percent 10 percent 12 percent assuming interest compounded monthly? Thanks, Patrick Burke Training Specialist The University of Chicago
Date: 07/24/97 at 14:20:33 From: Doctor Rob Subject: Re: Compound interest to a specific amount Patrick, Let i be the monthly interest rate. This is 1/12 the yearly interest rate. The main information you need to work out the right formula is that with a given principal 100, the interest to be added is i*100, so that the amount accumulated after one month has increased from 100 to (1+i)*100. After 1 month, the amount accumulated is 100*(1+i). If, then, a payment is made of 100, the amount accumulated is 100*(1+i)+100. If we repeat this process after 2 months, the amount accumulated then would be (1+i)*[100*(1+i)+100]+100 = 100*(1+i)^2 + 100*([1+i] + 1) After 3 months, the amount accumulated would be 100*(1+i)^3 + 100*([1+i]^2 + [1+i] + 1) and after 4 months, 100*(1+i)^4 + 100*([1+i]^3 + [1+i]^2 + [1+i] + 1) = 100*(1+i)^4 + 100*([1+i]^4 - 1)/([1+i] - 1) From this you can see (and prove) that the pattern is that after k months, the amount accumulated is 100*(1+i)^k + 100*([1+i]^k - 1)/i Now after time k, the amount accumulated is 1000000. Then 1000000 = 100*(1+i)^k + 100*([1+i]^k - 1)/i) This gives an equation relating i and k. For a given interest rate, you solve this equation for k: 1000000*i/100 = (1+i)^(k+1) - 1, (1+i)^(k+1) = 10000*i + 1, (k+1)*log(1+i) = log(10000*i+1), k = -1 + log(10000*i+1)/log(1+i). This trick, to take the logarithms of both sides, is often useful when the unknown appears in an exponent. Remember that the annual interest rate is 12*i, so your 8 percent annual rate translates to i = 2/300 = 0.0066666..., which we use in the above formula. Now you substitute in your interest rate and get the number of months until the $1,000,000.00 has accumulated. Of course you will have to round upwards, since the answer will probably not be an integer. Then the last payment will probably take you over $1,000,000.00, as opposed to hitting it on the nose! For your 8 percent rate, I get 634 months, or nearly 53 years. You can calculate the others yourself. If this is not clear, or not what you wanted, write back and we can try again. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 07/24/97 at 16:43:13 From: Patrick Burke Subject: Re: Compound interest to a specific amount Thank you. Quite helpful. Exactly what I was looking for.
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