First of all, 25 (per week) times 52 (weeks) = 1300...not 1200 Secondly, the interest rate...is it really compounded on a yearly basis (extremely rare) or is it compounded at 2.3% on a monthly basis? It makes a big difference. 2.3% compounded on a monthly basis would be 0.001916666 each month and with money going in each week it would increase exponentially based on the element of time. Also the exact moment the deposits are made in relation to when the monthly rate is calculated in the long run will certainly have an impact on your bottom line. Example: deposits made on Friday in that first month will have a different amount to calculate at the moment of calculation than perhaps deposits made on a Monday. Since 30 days is your monthly average based on 360 days, we all know there´s more days than that in a given year. <img src=smilies/icon_smile.gif width=15 height=15 border=0 align=middle> So a not so easy solution to your problem persists...unless of course you´re just looking for a ballpark estimate then is becomes relativly easy to compute your money on hand each year if you differentiate exactly how the interest rate is really calculated. Damn that was a mouthful! <img src=smilies/icon_smile.gif width=15 height=15 border=0 align=middle> Rob "Stinger" Lordier You like poems? --> <A href=´http://www.voicesnet.com´ Target=_Blank> Voicesnet.com </a> Creator of the original <A href=´http://solutions.h1.ru/faq/privater.shtml´ Target=_Blank> Privateer FAQ </a> <b> old </b> Favorite saying - No good deed goes unpunished Edited by - Stinger on 3/29/2005 11:40:20 PM
A question for the math-guru´s
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If you´re going with an annual interest rate then it becomes fairly easy to work out, especially with fixed-size contributions. The formula you´re looking for is as follows: y(n) = P(1 + r)n + c[((1 + r)^(n + 1) - (1 + r))/r Where: <b>y(n) </b> is the balance after n years <b>P </b> is the initial balance <b>r </b> is the interest rate, expressed as a decimal (0.023 in your case) <b>c </b> is the value of the annual contributions. For example: y(0) = 0*(1+0.023)^0 + 1200*(((1+0.023)^(0+1)-(1+0.023))/0.023) <b>= 0 </b> y(1) = 0*(1+0.023)^1 + 1200*(((1+0.023)^(1+1)-(1+0.023))/0.023) <b>= 1227.6 </b> y(2) = 0*(1+0.023)^2 + 1200*(((1+0.023)^(2+1)-(1+0.023))/0.023) <b>= 2483.44 </b> etc. As long as the initial balance is 0 you can ignore the first part of the equation completely, since it equates to 0^n which is always 0. You are then left with: c[((1 + r)^(n + 1) - (1 + r))/r which is a (simple) geometric series. Class dismissed <img src=smilies/icon_smile_tongue.gif width=15 height=15 border=0 align=middle> Edited by - Accushot on 3/30/2005 2:51:16 AM
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@T E T I think you´ve missed the annual contributions out of the formula. That formula is correct for a fixed amount of money gaining compound interest every year, but you also have to take into account the 1300 euros going in every year as well. Crude
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Thank you Accushot, that was exactly what I was looking for! Cheers mate! Edit: As I told you guys, my bank sets 30 days in a month, and 360 in a year. I however, would set 28 days in a month, just to make it easier. Therefore, I´d contribute 4 times a month, times twelve months, which sets my annual contribution at 1200 instead of 1300. Then I can always take the remaining days per month, mulitply them by twelve (2*12=24) and make an extra contribution (let´s say an additional 100) right before the end of the financiel year (or something like that). But thanks again guys <img src=smilies/icon_smile.gif width=15 height=15 border=0 align=middle> Edited by - Navillia on 3/30/2005 3:11:17 AM