My Math Forum Very difficult question (need help)

 Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance

 January 19th, 2012, 09:03 PM #2 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,556 Thanks: 101 Math Focus: The calculus Re: Very difficult question (need help) Every six months $1800 is deposited and interest compounded, so after 35 years, or 70 compounding periods, the account has a value V in dollars given by: $V=1800\sum_{k=1}^{70}(1.05)^k\approx1112318.89$ Now subtract the$10,000.00 going to the university and we are left with: $1102318.89 To find the first year's withdrawal W, we set: $1.05^{18}V-W\sum_{k=9}^{18}1.05^k=0$ $W=\frac{1.05^{18}V}{\sum_{k=9}^{18}1.05^k}\approx\ frac{2652861.8308853353}{19.5124395879424565}\appr ox135957.47$ Close, but no cigar. I post my incorrect response so perhaps someone who knows how to actually solve this can show me where I went wrong. January 20th, 2012, 01:27 AM #3 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 Re: Very difficult question (need help) Quote:  Originally Posted by MarkFL Every six months$1800 is deposited and interest compounded, so after 35 years, or 70 compounding periods, the account has a value V in dollars given by: $V=1800\sum_{k=1}^{70}(1.05)^k\approx1112318.89$
You have to obtain a MONTHLY equivalent rate (since deposits are monthly):
(1 + r)^12 = 1.05^2
Then use 300 (not 1800) and 420 (not 70).

 January 20th, 2012, 09:06 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 Re: Very difficult question (need help) Code: Time Dep/-Wd Interest Balance 0 .00 1 300.00 .00 300.00 2 300.00 2.45 602.45 3 300.00 4.92 907.37 ..... 419 300.00 8,680.68 1,072,157.76 420 300.00 8,754.01 1,081,211.77 1 -132,841.67 948,370.10 (same time as 420th deposit) 2 -139,483.75 97,207.94 906,094.29 ..... 9 -196,267.65 36,468.57 195,991.85 10 -206,081.03 20,089.18 10,000.00 I make the 1st withdrawal $132,841.67 (instead of 132,691.91), as shown above. I am interpreting your "unclear problem" this way (make sure you tell your teacher to quit making up stories that are not clear math-wise!) : 420 monthly deposits of$300 are made in an account earning 10% compounded semi-annually; the deposits are made at month-end. On the same day as the last deposit is made, $W is withdrawn; withdrawals continue annually for the next 9 years, increasing by 5% annually. The rate remains at 10% compounded semi-annually (note that this was not stated in problem). "EDIT" After the last withdrawal,$10,000.00 remains in the account. Calculate W If instead I started the monthly deposits 1 month sooner (beginning of month), it would make the situation worse, since the account would be higher after 420 months. Anyway, 132,891.67 - 132,691.91 = 199.76: such a difference is really negligible when you consider a span of 45 years. If the "scenario" is different from what I assumed, let me know.
 January 20th, 2012, 01:31 PM #5 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,556 Thanks: 101 Math Focus: The calculus Re: Very difficult question (need help) To obtain the equivalent monthly rate: $$$1+\frac{r}{12}$$^12=$$1+\frac{0.10}{2}$$^2$ $$$1+\frac{r}{12}$$=$$1.05$$^{\frac{1}{6}}$ $V=300\sum_{k=1}^{420}$$1.05$$^{\frac{k}{6}}\approx 1090039.70$ To find the first year's withdrawal W, we set: $1.05^{18}V-W\sum_{k=9}^{18}1.05^k=10000$ $W=\frac{1.05^{18}V-10000}{\sum_{k=9}^{18}1.05^k}\approx\frac{2613310. 5045594033}{19.5124395879424565}\approx133930.49$
January 20th, 2012, 02:26 PM   #6
Math Team

Joined: Oct 2011

Posts: 3,073
Thanks: 32

Re: Very difficult question (need help)

Quote:
 Originally Posted by MarkFL $W=\frac{1.05^{18}V-10000}{\sum_{k=9}^{18}1.05^k}\approx\frac{2613310. 5045594033}{19.5124395879424565}\approx133930.49$
Agree, IF 1st $300 deposit is at START of month. Using actual year, last (420th)$300 deposit is made Dec.1(year 35).
Interest is earned/credited on Dec.31(year 35).
The 1st withdrawal is theoretically on same date...even if Jan.1(year 36).

So really we're at the 44 year point when last withdrawal is made.
Death is 1 year later: so I think we have to "re-visit" this, and calculate
to end up with ~9,070.29 after the last withdrawal:
so at death (1 year later or end of 45th year), we end up with
9070.20 * 1.1025 = 10,000.00

What you say to that, Mark? Am I clear?

 January 20th, 2012, 02:41 PM #7 Joined: Jan 2012 Posts: 2 Thanks: 0 Re: Very difficult question (need help) The answer is what is stated in the problem, I have asked the professor. I am pulling my hair out because of this problem. There is some trick with that 5% inflation rate. I am getting an answer very similar to you Dennis, but apparently, that is the wrong answer.
January 20th, 2012, 07:17 PM   #8
Math Team

Joined: Oct 2011

Posts: 3,073
Thanks: 32

Re: Very difficult question (need help)

Quote:
 Originally Posted by xtifer The answer is what is stated in the problem, I have asked the professor. I am pulling my hair out because of this problem. There is some trick with that 5% inflation rate. I am getting an answer very similar to you Dennis, but apparently, that is the wrong answer.
Did you ask Mr Prof about the final $300 deposit and the first annual withdrawal: are they together, or a month apart, or something else? And did you ask about the$10,000 donation: was it made 1 year after the last annual withdrawal?
Problem states: "You also plan to bequeath, upon your death at 75 years of age, ....";
but the last withdrawal is at 74 years old (beginning of year).

Ask him/her if this is the intended cash flow (starting with year 1 for simplicity):
Jan.1/01 300.00 (deposit#1)
Feb,1/01 300.00 (deposit#2)
...
Nov.1/35 300.00 (deposit#419)
Dec.1/35 300.00 (deposit#420)

Jan.1/36 - W (wd#1)
Jan.1/37 - (1.05)W (wd#2)
Jan.1/38 - (1.05^2)W (wd#3)
...
Jan.1/45 - (1.05^9)W (wd#10)

Jan.1/46 -10,000.00 .....leaves account at exactly ZERO

Surely they'll tell you if that's the correct interpretation of their confusing "wording"!

January 21st, 2012, 03:00 AM   #9

Joined: Apr 2011
From: USA

Posts: 782
Thanks: 1

Re: Very difficult question (need help)

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.

So I will merely toss in my thoughts on what's been posted and you two can grab it from there.

Quote:
 Originally Posted by Denis You have to obtain a MONTHLY equivalent rate (since deposits are monthly): (1 + r)^12 = 1.05^2 Then use 300 (not 1800) and 420 (not 70).
I did this an entirely different way, just on my own, and came up with something about $600 off Mark's second shot. 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.

Quote:
 The rate remains at 10% compounded annually (note that this was not stated in problem).
Why annually when it was semi-annual the rest of the time? Again, the problem lacks this information, but I also see no reason it would suddenly stop compounding. But I see even less reason it would suddenly change to annual.

Quote:
 And did you ask about the $10,000 donation: was it made 1 year after the last annual withdrawal? Problem states: "You also plan to bequeath, upon your death at 75 years of age, ...."; but the last withdrawal is at 74 years old (beginning of year). I think this is clear. Last withdrawal is at the beginning of age 74, and the$10,000 is one year later. I don't see the two as being connected or any reason to interpret it differently.

Quote:
 Originally Posted by MarkFL Now subtract the $10,000.00 going to the university and we are left with:$1102318.89
I totally disagree with this one. Assuming that we continue the semi-annual compounding, that $10,000 would come from compounded money. We're not pulling out$10,000 at age 65 and just letting it sit doing nothing. It can be separated out, but we need the present value of it, which would be 3768.89. And I think it's easier just to separate that out.

But once that's subtracted off, I have no clue how to do the rest of the retirement part of it.

January 21st, 2012, 06:04 AM   #10
Math Team

Joined: Oct 2011

Posts: 3,073
Thanks: 32

Re: Very difficult question (need help)

Quote:
Originally Posted by Erimess
I
Quote:
 The rate remains at 10% compounded annually (note that this was not stated in problem).
Why annually when it was semi-annual the rest of the time? Again, the problem lacks this information, but I also see no reason it would suddenly stop compounding. But I see even less reason it would suddenly change to annual.
Thanks...typo (changed it); but at least I did say "remains"

January 21st, 2012, 06:20 AM   #11
Math Team

Joined: Oct 2011

Posts: 3,073
Thanks: 32

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, 11:49 AM #12 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 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, 08:33 PM   #13

Joined: Apr 2011
From: USA

Posts: 782
Thanks: 1

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, 08:49 PM   #14

Joined: Apr 2011
From: USA

Posts: 782
Thanks: 1

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, 07:00 AM   #15
Math Team

Joined: Oct 2011

Posts: 3,073
Thanks: 32

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.

 Tags difficult, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Iamthatdude Math Events 1 August 28th, 2013 12:13 PM frankpupu Linear Algebra 0 February 21st, 2012 05:44 PM ahmed-ar Calculus 3 December 9th, 2011 01:44 PM DragonScion Advanced Statistics 0 October 29th, 2011 02:54 PM grahamlee Complex Analysis 0 July 27th, 2009 09:56 PM

 Contact - Home - Top