document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=535717>Question 535717</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You deposit $3000 in an account that pays 3.5% interest compounded once a year.  Your friend deposits $2500 in an account that pays 4.8% interest compounded monthly. Use this information to answer the following questions.\r<BR>\n" );
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document.write( "Who will have more money in their account after one year?  How much more? <BR>\n" );
document.write( "Who will have more money in their account after five years?  How much more? <BR>\n" );
document.write( "Who will have more money in their account after 20 years?  How much more?  <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #351987 by </B><A HREF=/tutors/aboutme.mpl?userid=lmeeks54><B>lmeeks54(111)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lmeeks54><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=lmeeks54><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=351987\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There are a couple different ways to solve this problem.  I'll talk about two: the one you're probably asked for, that is how to apply compound interest formulas to two separate accounts to compare their performance, or setting it up on a spreadsheet.  <BR>\n" );
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document.write( "This is worth noting because very few financial calculations for transactions occuring over time (e.g., regular car or mortgage loan payments, annuity disbursments, principal growth through compound interest, etc.) are done one at a time by hand using formulas.  They are almost all done in spreadsheets or other financial software.  The reason for this--contrary to this problem, which asks you for comparison between the accounts at only two or three specified time periods--is that analysts and even normal everyday people want to see data about their accounts at more than just one or two discrete time periods.<BR>\n" );
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document.write( "Okay, the long-winded intro into, 'learn how to do spreadsheets someday, they will make your life a LOT easier,' is over.  Let's solve this problem for you...<BR>\n" );
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document.write( "The data we care about for problems such as these are:<BR>\n" );
document.write( "Principal - the money being borrowed, lent, or saved<BR>\n" );
document.write( "Interest rate - usually expressed as an annual percentage<BR>\n" );
document.write( "Term - the length of time (usually years or months) of the loan, security, or annuity.  This often gets turned in the # of compounding periods<BR>\n" );
document.write( "Frequency - for compound interest problems, you care about monthly, quarterly, or annual compounding (it makes a difference...sometimes a BIG difference)<BR>\n" );
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document.write( "Let P = deposit principal<BR>\n" );
document.write( "Let F = the future value of the money on deposit <BR>\n" );
document.write( "Let r = annual interest rate<BR>\n" );
document.write( "Let n = # of compounding periods<BR>\n" );
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document.write( "The basic formula for compound interest is:<BR>\n" );
document.write( "F = P*(1 + r)^n<BR>\n" );
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document.write( "Note: we have an interesting situation that is not hard to deal with, but which can be confusing and must be kept track of.  Namely, one account compounds annually and one account compounds monthly.  But they both have their interest rates expressed in annual terms.  For account 1, which has an annual rate and annual compounding, we don't do anything special.  For account 2, which has an annual rate and monthly compounding, we have to convert the annual rate to a monthly rate, which is simple: we just divide by 12.  Don't forget this step in problems like this or you'll get really squirrelly answers (i.e., the earned (called \"accrued\") interest will get too big really fast).<BR>\n" );
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document.write( "Let's go back to our variables and formula and change them slightly.  Let's add:<BR>\n" );
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document.write( "Let y = year # for calculation purposes  (remember, this will be 1, 5, or 20)<BR>\n" );
document.write( "Let n = # periods per year for compounding (this will be 1 or 12 in our example, but could be 4 for quarterly compounding, which is common in some places)<BR>\n" );
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document.write( "Let's also amend the basic compounding formula to account for y and the newly defined n:<BR>\n" );
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document.write( "F = P*(1 + r/n)^(y*n)<BR>\n" );
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document.write( "That's all the background so we can set up this problem.  Now, with the givens for Acct1 and Acct2, we need to calculate F at y = 1, 5, and 20 years for both accounts:<BR>\n" );
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document.write( "P1 = $3,000<BR>\n" );
document.write( "r1 = 3.5%<BR>\n" );
document.write( "n1 = 1  (for annual compounding)<BR>\n" );
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document.write( "P2 = $2,500<BR>\n" );
document.write( "r2 = 4.8% (but we will have to divide it by n2 to use it)<BR>\n" );
document.write( "n2 = 12   (for monthly compounding)<BR>\n" );
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document.write( "For both accounts, we will use y = 1, 5, 20...so we have to do 6 calculations.<BR>\n" );
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document.write( "Let's do the first calculation for Acct1 and Acct2 for the end of year 1:<BR>\n" );
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document.write( "F1 = $3,000*(1 + .035/1)^(1*1)<BR>\n" );
document.write( "F1 = $3,105.00<BR>\n" );
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document.write( "Note: this represents the $3,000 starting principal + yr 1's one annual interest payment.  Giving us a new deposit principal going into year 2<BR>\n" );
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document.write( "F2 = $2,500*(1 + .048/12)^(1*12)<BR>\n" );
document.write( "F2 = $2,622.68<BR>\n" );
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document.write( "Note: this also represents the $2,500 starting principal + yr 1's 12 monthly compound interest payments.  Also giving us a new principal going into year 2.  Of course, with monthly compounding, the principal grows every month.<BR>\n" );
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document.write( "The 1st question was at the end of year 1, which account had more money, and how much more?  The answer is account 1 has more money and the difference is $482.32.  Note: acct1 started out year 0 with $500 more, so acct2 is already starting to catch up, though it will take a few years to pass acct1.<BR>\n" );
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document.write( "That takes care of y = 1 for year 1.  Now all that is left is y = 5 for F1 and F2 again and y = 20 for F1 and F2 again.<BR>\n" );
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document.write( "Note: everything is exactly the same except for the y term.  I'll leave you to set that up for yourself, however, I'll give you the correct answers so you can check your work:<BR>\n" );
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document.write( "y = 5<BR>\n" );
document.write( "F1 = $3,563.06<BR>\n" );
document.write( "F2 = $3,176.60<BR>\n" );
document.write( "Acct1 still winning, but now by a lot less: $386.46<BR>\n" );
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document.write( "y = 20<BR>\n" );
document.write( "F1 = $5,969.37<BR>\n" );
document.write( "F2 = $6,516.75<BR>\n" );
document.write( "Acct2 is NOW winning, with a delta of: $547.38<BR>\n" );
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document.write( "Another interesting fact to know is when Acct2 passes Acct1. It turns out because they are compounding differently, between the end of the 11th year and the beginning of the 13th year, the lead for who has more money swaps back and forth.  However, after the beginning of the 13th year, Acct2 is in the lead forever, and pulling away more and more as time goes on.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Lee<BR>\n" );
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document.write( "Ps.  Email me at: w.lee.meeks@gmail.com if you have any more questions about this problem. </SPAN>\n" );
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