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Finance problem recursive formula

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  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Daniel deposits $400 at the end of each month for 7 years in an account paying 1.6% compounded monthly. He then puts the total amount on deposit in another account paying 2.2% interest compounded semiannually for another 8 years. Find the final amount on the deposit after the entire 15 year period.



    2. Relevant equations

    How would this be plugged into a ti-83?
    For the first part i believe i use Payment to sinking fund formula: R=S*i/((1+i)^n -1)
    I think i need to use the recursive formula for annuities but i am just not sure, it goes as such Un=(1+i) u(n-1)+R

    3. The attempt at a solution

    400 = R(.016/12)/((1+.016/12)^12(7) -1)

    After that i am just totally lost...especially with how to use the recursive formula and how its plugged into the ti-83 to create a table to thus find the answer.


    Please anyone who knows about this help me!
    Thank you
     
  2. jcsd
  3. Oct 20, 2012 #2

    MarneMath

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    Just keep it simple. [itex] A = P(1+\frac{r}{n})^{nt} [/itex]. Your initial P = 400, r = rate as a decimal point, n = number of times interest compound in a year and t number of years. Putting that all together [itex] A = 400(1+\frac{.016}{12})^{12*7-1} [/itex]. You may be wondering why it's 12*7 - 1, but think about when the last 400 dollars is put into the account (end of the month). So that's the easy par. The hard part is now to realize that the first 400 dollars put into the account is the only one gets all that, the next 400 dollars you put in will get 1 month less, and the following 400 dollars will get 2 months less, until eventually the last 400 dollars gets nothing extra. So we end up with this formula that likes this [itex] A_t = 400(1+\frac{.016}{12})^{12*7-1} + 400(1+\frac{.016}{12})^{12*7-2} + ....+400 [/itex] Can you take it from here? Or do you need help reducing this sum into something easier?
     
    Last edited: Oct 20, 2012
  4. Oct 20, 2012 #3
    Thank you. I am still a bit confused on the -1 part, why does that exist exactly? as for the numbers decreasing that makes sense because of the recursive formula continues to decrease.
    so A_t = 400(1+\frac{.016}{12})^{12*7-1} is 0 and 400(1+\frac{.016}{12})^{12*7-2} is 446.20 then 400(1+\frac{.016}{12})^{12*7-3} is 445.60 would it go on like this for 7 years?

    Idk i still feel a bit confused here on how i actually get an answer.
     
  5. Oct 20, 2012 #4

    Ray Vickson

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    I guess the issue is about the very last deposit in the 7-year (84-month) period. Certainly, he/she deposits $400 at the end of months 1,2,3,...,83, but what about at the end of month 84? A *strict* interpretation of the actual words used in the problem (IF you have copied them correctly!) implies that $400 is also deposited at the end of month 84. But it would then be instantly withdrawn again, along with all the other money deposited and earned over the previous 83 months, and then put into the next investment. In other words, our investor would put in $400 and then instantly withdraw it again---but that is what the strict wording implies!

    On the other hand, perhaps there is no deposit made at the end of month 84, so what is withdrawn at the end of month 84 is the accumulation of everything accrued in months 1--83.

    If I were doing the problem I would point out that the wording seems ambiguous, and I would then solve BOTH versions of the problem and present both solutions, clearly labelled. Once I had figured out how to do one version (using either formulas or a spreadsheet, or whatever) doing the other version as well would not require much extra work. But that's just me.

    RGV
     
  6. Oct 21, 2012 #5
    So then i am looking for the 83rd month not the 84th?

    What I've figured out so far is how to plug the recursive sequence/system into the calculator. On the Ti-83 it goes like:
    U(n)=(1+.016/12)u(n-1)+400
    u(nmin)={0}

    Then i hit 2nd-tblset and i hit TblStart= 0
    Then i hit 2nd-TABLE and then i get a chart, yet at this point i am not sure what i am looking for is it number 83 or number 84 on the left hand side with the values on the right.

    And then after i have the correct amount then what do i need to plug that into to get the final amount on the deposit after the entire 15 year period?

    thank you
     
  7. Oct 21, 2012 #6

    Ray Vickson

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    To answer your first question: you tell me. Which version of the problem are you solving?

    RGV
     
  8. Oct 21, 2012 #7
    Well judging from the material i've been given it's going to be 83 due to the rounding of the value for the payment. I think the last payment becomes $.00. So then i get the table with 83 which is 35528.82669 after what should i do?
     
  9. Oct 21, 2012 #8
    Think i've figured this out once and for all. It goes like this: K=(1+.016/12) which is the accumulation for one months. Which equals: 1.001333333.
    Now i put the constant times the geometric formula: 400(k^84-1)/(k-1) = 35528.82669
    SO that's the answer for the first part
    Then for the last 8 years (96 months): (1+.022/12)^96 = 1.192245929

    Does this look correct to you all?

    The only part i am having confusion on is where i need to find the final amount on the deposit after the entire 15 year period? Is that to say add the two answers together or something more?

    Thank you
     
  10. Oct 21, 2012 #9

    Ray Vickson

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    I have not checked the arithmetic, but the formulas for the first part look OK. Now, just take that whole $35528.82669 and put it on deposit into the second account---that is what the question tells you. Are you sure about the factor 1 + 0.022/12 in the second part? Look again at what the question actually SAYS.

    RGV
     
  11. Oct 21, 2012 #10
    AH yes semiannually so it would be 1+.022/2 of course! thanks. How would i put it on deposit into the second account i am a bit confused by how to do that.

    Thank you
     
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