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January 21st, 2012, 05:20 AM  #11  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help) Quote:
I think you used .10/12 as rate: slightly too high: rate compounds semiannually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments.  
January 21st, 2012, 10:49 AM  #12  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help) Quote:
Code: j = 10% i = 12% Year Deposit Interest Balance 0 .00 .00 .00 1 1000.00 .00 1000.00 2 1100.00 120.00 2220.00 3 1210.00 266.40 3696.40 F = Future value (?) D = initial deposit (1000) j = deposit increase (.10) i = interest rate (.12) n = number of years (3) F = D*[(1 + i)^n  (1 + j)^n] / (i  j) F = 1000(1.12^3  1.10^3) / (.12  .10) = 3696.40 Basically all that simple...google will give ya lotsa sites, like: http://www.financeformulas.net/Growing ... Value.html  
January 21st, 2012, 07:33 PM  #13  
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Very difficult question (need help) Quote:
 
January 21st, 2012, 07:49 PM  #14  
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Very difficult question (need help) Quote:
This is a present value relative to the withdrawals, payments growing at 5% annually, but earning (we assume) 10% semiannually. That equation is growing to a future value, not shrinking from a present value.  
January 22nd, 2012, 06:00 AM  #15  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help) Quote:
This was the rate per the problem: "in an account that earns an average of 10% compounded semiannually". 10% cpd semiannually means 10.25% annually ; 1.05^2  1 = .1025 So the problem statement could as well be: ....10.25% compounded annually (same thing). It is standard practice to convert the effective ANNUAL rate such that it matches the payments frequency; on this case, to 12 periods annually; so: (1 + i)^12 = 1.1025 ; i = .0081648... So the "semiannual equivalent deposit" is ~1852.14, not ~1853.24. Another example:If given rate is 12% annual cpd. quarterly and payment frequency is 6 (every 2 months), then equivalent rate becomes: (1 + i)^6 = (1 + .12/4)^4 ; i = .019901...slightly lower than 2%, as expected.  
January 22nd, 2012, 06:19 AM  #16  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help) Quote:
I used it that way because (being lazy) it was easier to "do the typing"! Take the ending balance of 3696.40 and get its Present Value: 3696.40 / 1.12^3 = ~2631.02. So this means that if a payment of $1000 increasing by 10% yearly is made for 3 years, then the amount that can be "borrowed" is $2631.02 Code: j = 10% i = 12% Year Payment Interest Balance 0 .00 .00 2631.02 1 1000.00 315.72 1946.74 2 1100.00 233.61 1080.35 3 1210.00 129.65 .00 of $132,691.91 yearly increasing by 5% yearly would basically be obtained the same way. So you don't scold me again(!), here's the direct formula: P = Payment (?) A = Amount borrowed (2631.02) n = number of periods (3) i = interest rate (.12) j = payment increase (.10) P = A(i  j) / [1  (1 + j)^n / (1 + i)^n] P = 2631.02(.12  .10) / (1  1.10^3 / 1.12^3) = 1000  
January 22nd, 2012, 09:31 AM  #17 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help)
More on "the formula"! Code: Payments accumulation Loan, no payments 0 .00 2631.02 1 1000.00 .00 1000.00 315.72 2946.74 2 1100.00 120.00 2220.00 353.61 3300.35 3 1210.00 266.40 3696.40 396.05 3696.40 the loan in a separate account; both accounts end up the same, of course. The "payments account" is looked at this way: 1000(1.12)^2 = 1254.40 1100(1.12)^1 = 1232.00 1210(1.12)^0 = 1210.00 ==================== total.......... = 3696.40 So the requirement is made up of a SUM; this is shown as such in Mark's formula. The formula I've shown produces same results, in a direct way. 
January 22nd, 2012, 10:28 AM  #18 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,415 Thanks: 1025  Re: Very difficult question (need help) Code: # DATE DEP/WD INTEREST BALANCE 1 Jan.1/1 300.00 .00 300.00 2 Feb.1/1 300.00 2.45 602.45 3 Mar.1/1 300.00 4.92 907.37 .... 419 Nov.1/35 300.00 8,680.68 1,072,157.77 420 Dec.1/35 300.00 8,754.00 1,081,211.77 1 Jan.1/36 133,978.13 8,827.93 956,061.57 2 Jan.1/37 140,677.04 97,996.31 913,380.84 .... 9 Jan.1/44 197,946.72 36,694.91 196,747.71 10 Jan.1/45 207,844.05 20,166.63 9,070.29 KB Jan.1/46 929.71 10,000.00 : KB = Kicks Bucket! there is a 1 year period between last withdrawal and $10,000 donation, the 1st withdrawal will be $133,978.13. Quite close to Mark's $133,930.49, where $10,000 donation is not 1 year later. 
January 23rd, 2012, 02:42 AM  #19  
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Very difficult question (need help) Quote:
Oh, actually, you know what. We're both right. .8333% per month is still correct, as long as you treat it like simple interest within the six month period, which is actually what I learned and should have done. (i.e. 300*.8333*6 + 300*.8333*5, etc) I just looked back at my work and realized I compounded the .8333% per month! I didn't mean to do that. (Probably sounds like the hard way, but most of my classes were using charts, not equations.)  

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