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January 21st, 2012, 05:20 AM   #11
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
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Re: Very difficult question (need help)

Quote:
 Originally Posted by Erimess Because it's not compounding, I just figured out what six $300 payments would come out to at the end of six months, starting at the beginning of the month. (The problem doesn't say beginning or end, which is a serious flaw in the problem, but since the retirement is at the beginning, I'm going to just go with that throughout. I haven't yet re-worked it for end of the month.)$300 a month, not compounding but earning, at the end of six months is $1853.24. (i.e.$300 for 6 months, $300 for 5 months, etc.) So that's how much there is at the end of the first six months. That amount will now just continue to compound at 5% per six months. The next six months the same thing is going to happen. So there's another 1853.24 plus the amount compounded on the first 1853.24. This seems to me is the equivalent of making a 1853.24 payment at the end of every six months for 70 periods at 5%. No? That gave me 1,090,681.80. Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, 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: 14,597 Thanks: 1038 Re: Very difficult question (need help) Quote:  Originally Posted by Erimess I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that. 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 Li'l "picture",$1000 deposit increasing by 10% annually, rate 12% annually.

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
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From: USA

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Re: Very difficult question (need help)

Quote:
 Originally Posted by Denis Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments.
I didn't miss the compounding. Looks like you're doing this as though the 10% were a yield and I did it as though it's a rate.

January 21st, 2012, 07:49 PM   #14
Senior Member

Joined: Apr 2011
From: USA

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Re: Very difficult question (need help)

Quote:
Originally Posted by Denis
Quote:
 Originally Posted by Erimess I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that.
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
Li'l "picture", $1000 deposit increasing by 10% annually, rate 12% annually. 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 This isn't the same thing. I could've done that in Excel easily enough. (Though it never hurts to add an equation to the repertoire.) This is a present value relative to the withdrawals, payments growing at 5% annually, but earning (we assume) 10% semi-annually. 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: 14,597 Thanks: 1038 Re: Very difficult question (need help) Quote: Originally Posted by Erimess Quote:  Originally Posted by Denis Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments. I didn't miss the compounding. Looks like you're doing this as though the 10% were a yield and I did it as though it's a rate. Huh? I simply can't follow that... This was the rate per the problem: "in an account that earns an average of 10% compounded semi-annually". 10% cpd semi-annually 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: 14,597 Thanks: 1038 Re: Very difficult question (need help) Quote: Originally Posted by Erimess Quote: Originally Posted by Denis Quote:  Originally Posted by Erimess I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that. 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 Li'l "picture",$1000 deposit increasing by 10% annually, rate 12% annually.

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
This isn't the same thing. I could've done that in Excel easily enough. (Though it never hurts to add an equation to the repertoire.)

This is a present value relative to the withdrawals, payments growing at 5% annually, but earning (we assume) 10% semi-annually. That equation is growing to a future value, not shrinking from a present value.
YES, it's the same thing...though hard to see, I'll admit.
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
In the actual problem, the balance required in the account after 35 years, to accomodate a withdrawal
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: 14,597 Thanks: 1038 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 As shown, we can look at the payments as accumulating in a separate account, 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: 14,597 Thanks: 1038 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! Ok, since we're in agreement that 1st deposit is at beginning, plus 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:  Originally Posted by Denis This was the rate per the problem: "in an account that earns an average of 10% compounded semi-annually". 10% cpd semi-annually means 10.25% annually ; 1.05^2 - 1 = .1025 Yeah. But that would also mean 5% at the semi-annual mark, which would turn into 10.25% annually once compounded. But I first figured out what all the$300 payments would have earned at the semi-annual point, which would be at 5%, or .8333% per month. It doesn't compound til you tack the interest on. So the second half of the year, you're earning 5% on the original plus the interest, equating to 10.25% as a yield. But it's still 5% per semi-annual period. Way I learned it - way I've always done it.

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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