My Math Forum Finance Problem

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 January 30th, 2014, 04:03 PM #1 Newbie   Joined: Jan 2014 Posts: 4 Thanks: 0 Finance Problem Hi there I have difficulty solving this question. Any help appreciated Question: Once you retire you will be withdrawing $48,000 at the beginning of every year for another 30 years. To make your retirement plans and for the coming 30 years, you will start depositing at the end of every year over 24 years a certain amount of money in your bank account. Since you are expecting that your salary will grow by 2% every year for the coming 24 years then you expect your yearly savings to grow by the same rate. If your bank is paying you 3.25% Annual Percentage Rate compounded weekly. What should be your first deposit so that you can achieve your retirement plan?  January 31st, 2014, 07:49 PM #2 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 Re: Finance Problem Hello. Have a look at the thread by "Noritaka" which is almost identical to yours. If you need more help, come back and ask.  February 2nd, 2014, 01:14 PM #3 Newbie Joined: Jan 2014 Posts: 4 Thanks: 0 Re: Finance Problem Ok so i get that its three step process going backwards. First I need to know effective annual interest rate k = (1 + 0.0325/52)^52 - 1 = 0.03302 Then we need to know how much money I have to have to allow to take 48,000 per year for 30 years. PV1 = 48000 / 0.03302 * ( 1 - 1/1.03302^30 ) = 905,132.061 ( so after 30 year I have to have this amount in my bank to be able to take 48,000 dollars a year for 30 years right? ) Then because I was depositing money only for 24 years for 6 years money was growing on interest PV2 = 905,132.061 / (1 + 0.003302)^6 = 744,834.131 ( SO after depositing for 24 I need to have this amount in my banck if I want to retire after 6 years and take 48000 per year for 30 years) And now Lastly is the part I do not get. What formula do I use here? Growing annuity? ? ? is it this one? But this one is present value but i need future value right? PV0 = PMT/(k-g) * ( 1 - ((1+g)/(1+k))^n Lets solve for PMT as I need to know PMT PMT = PV0 * (k - g)/ (1 - ((1+g)/(1+k))^n) = 744834.131 * (0.03302 - 0.02) / (1-((1-0.02)/(1+0.003302))^24) = 36,951.5616 ( SO this means that I have to start by depositing 36,951 first time to achieve it. But the answer is not correct as there are a) 18,495.93 b) 20,590.39 c) 21,270.29 d) 442,895.39 e) non of above So please help me if you can I am pretty sure I am wrong with last formula as I dont need PV I need FV right? Or am I totally lost in finance? February 2nd, 2014, 07:08 PM #4 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 Re: Finance Problem Quote:  Originally Posted by Marcipanas Then because I was depositing money only for 24 years for 6 years money was growing on interest PV2 = 905,132.061 / (1 + 0.003302)^6 = 744,834.131 ( SO after depositing for 24 I need to have this amount in my banck if I want to retire after 6 years and take 48000 per year for 30 years) That's wrong. There is no 6 years involved. Deposits are made for 24 years, then the 48,000 starts right away (end of 24th year) or 1 year later (end of 25th year). Here's the calculations: GIVENS: r = .0325, j = .02, p = 48000, n = 24, m = 30 Step 1; convert rate: i = (1 + r/52)^52 - 1 : .0330234.... Step 2; get required value v at end of n years: v = p[1 - 1/(1 + i)^m] / i : 905092.957848.... Step 3; get 1st deposit d (which increases by j annually) which will accumulate to v : d = v(i - j) / [(1 + i)^n - (1 + j)^n] : 20590.39 .....which is choice b) This assumes that the 1st 48000 starts at end of 25th year. IF 1st 48000 starts at end of 24th year, (as your problem states "$48,000 at the beginning of every year"; but does not say which year)
then the formula at Step 2 needs to be slightly revised to:

Step 2; get required value v at end of n years:
v = {p[1 - 1/(1 + i)^m] / i} * (1 + i) : which gives 934982.209896....
Step 3 remains same:
d = v(i - j) / [(1 + i)^n - (1 + j)^n] : 21270.29 .....which is choice c)

Looks like your teacher wants you to decide if there is a 1 year "waiting" period or not!

 February 2nd, 2014, 10:04 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 Re: Finance Problem In case you insist that there is a 6 years delay: GIVENS: r = .0325, j = .02, p = 48000, n = 24, m = 30, w = 6 (delay) Step 1; convert rate: i = (1 + r/52)^52 - 1 : .0330234.... Step 2; get required value v at end of n years: v = p[1 - 1/(1 + i)^m] / i : 905092.957848.... (same as before) Step 3; get 1st deposit d (which increases by j annually for n years, then interest only for w years) which will accumulate to v (notice change to formula): d = v(i - j) / {[(1 + i)^n - (1 + j)^n](1 + i)^w} : 16942.5428... This assumes that the 1st 48000 starts at end of 31st year. IF 1st 48000 starts at end of 30th year, (as your problem states "$48,000 at the beginning of every year"; but does not say which year) then the formula at Step 2 needs to be slightly revised to: Step 2; get required value v at end of n years: v = {p[1 - 1/(1 + i)^m] / i} * (1 + i) : which gives 934982.209896....(same as before) Step 3; get 1st deposit d (which increases by j annually for n years, then interest only for w years) which will accumulate to v: d = v(i - j) / {[(1 + i)^n - (1 + j)^n](1 + i)^w} : 17502.0432... So we have 16942.54 or 17502.04: both NOT in your answer choices...convinced?! February 3rd, 2014, 08:15 AM #6 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Finance Problem Quote:  Originally Posted by Marcipanas a) 18,495.93 b) 20,590.39 c) 21,270.29 d) 442,895.39 e) non of above I get a number rather close to c). I used a spreadsheet with =B2*(1+B$1)+C3
in B3 (dragged to B26),
=(B26-48000)*(1+B$1) in B27 (dragged to B56), =C3*1.02 in C4 (dragged to C26), = (1 + 0.0325/52)^52 - 1 in B1, and the answer to the problem in C3. This should leave B56 at 0.  February 3rd, 2014, 11:32 AM #7 Newbie Joined: Jan 2014 Posts: 4 Thanks: 0 Re: Finance Problem Thank you guys. I see my mistakes as english is not my mother tongue I got messed up with extra 6 years and I forgot to use Annuity due formula and took simple annuity. February 3rd, 2014, 11:34 AM #8 Newbie Joined: Jan 2014 Posts: 4 Thanks: 0 Re: Finance Problem Quote: Originally Posted by CRGreathouse Quote:  Originally Posted by Marcipanas a) 18,495.93 b) 20,590.39 c) 21,270.29 d) 442,895.39 e) non of above I get a number rather close to c). I used a spreadsheet with =B2*(1+B$1)+C3
in B3 (dragged to B26),
=(B26-48000)*(1+B$1) in B27 (dragged to B56), =C3*1.02 in C4 (dragged to C26), = (1 + 0.0325/52)^52 - 1 in B1, and the answer to the problem in C3. This should leave B56 at 0. Our teacher said to round intermediate calculations to 5 digits after point ( y.xxxxx) . And excel takes every number after point. February 3rd, 2014, 01:52 PM #9 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Finance Problem Quote:  Originally Posted by Marcipanas Our teacher said to round intermediate calculations to 5 digits after point ( y.xxxxx) . And excel takes every number after point. You could force it to round however you like, of course. February 3rd, 2014, 08:07 PM #10 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 Re: Finance Problem Quote:  Originally Posted by Marcipanas I forgot to use Annuity due formula and took simple annuity. Would have saved lots of time if that had been specified... The problem can be "SIMPLY" worded this way: An annuity due (or annuity immediate) allowing$48,000 annual for 30 years, starting in 24 years
is to be set up with 24 annual deposits starting in 1 year, the deposits increasing by 2% annually.
The annual savings rate for the full period is 3.25% compounded weekly.

That's it! No need for word-happy teachers to build a story confusing the issue

Solved as following:

GIVENS: r = .0325, j = .02, p = 48000, n = 24, m = 30

Step 1; convert rate:
i = (1 + r/52)^52 - 1 : .033023405816

Step 2; get required value v at end of n years:
v = {p[1 - 1/(1 + i)^m] / i} * (1 + i) : which gives 934982.209896983414

Step 3; get 1st deposit d (which increases by j annually for n years) that accumulates to v:
d = v(i - j) / [(1 + i)^n - (1 + j)^n] : 21269.129877079599
The "choice" given is 21,270.29 ; I'm using greater accuracy as you can see...

The whole thing will look like this (bank statement format):
Code:
YEAR    TRANSACTION       INTEREST         BALANCE
1       21,269.13            .00       21,269.13
2       21,694.51         702.38       43,666.02
...
24       33,539.28      28,817.07      934,982.21
24      -48,000.00            .00      886,982.21
25      -48,000.00      29,291.17      868,273.38
...
52      -48,000.00       3,019.84       46,465.55
53      -48,000.00       1,534.45             .00
Notice that 1st deposit of 21269.13 * 1.02^23 = 33539.28, the last deposit.

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