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aama100

  • 3 years ago

Jason is a new college student who has saved $10000 when he was a teenager and is invested at 12 percent in a financial assent that pays interest monthly. He expects to graduate 4 years from today. a. How much can Jason withdraw every month while he is in college if the first withdrawal occurs today ? (Answer 260.73) b. How much can Jason withdraw every month while he is in college if he waits until the end of this month to make the first withdrawal ? ( Answer 263.34)

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  1. amistre64
    • 3 years ago
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    I cant tell if the information is saying: 12% annually is divided up into monthly installments (1+(.12/12)); or if it means that the account pays out 12% each month. (1+ .12). But if I see it right the equations go along the lines of: a. B{n} = [B{n-1} - withdrawal] * (1 + interest) and b. B{n} = [B{n-1} * (1 + interest)] - withdrawal

  2. amistre64
    • 3 years ago
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    i got (b) worked out :) \begin{align} B_n&=B_{n-1}(1.01)-w\\ &=B_{n-2}(1.01)^2-w(1.01)-w\\ &=B_{n-3}(1.01)^3-w(1.01)^2-w(1.01)-w\\ &\ or \ simply\\ B_{n}&=B_{n-3}(1.01)^3-w(1.01^2+1.01+1)\\ &=B_{n-4}(1.01)^4-w(1.01^3+1.01^2+1.01+1)\\ &=B_{n-r}(1.01)^r-w(1.01^{r-1}+1.01^{r-2}+...+1.01^2+1.01+1)\\ \end{align} \(B_{n-r}=B_{0}\); when r=n \begin{align} B_{n} = B_{0}(1.01)^n - w(1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1)\\ \end{align} We can clean this up by suming up those \((1.01^{n-1}...+1)\) parts like this. \begin{align} A&=\{\hspace{3.5em}1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1\}\\ (1.01)A&=\{1.01^{n}+1.01^{n-1}+1.01^{n-2+}...+1.01^2+1.01\hspace{2em}\}\\ \end{align} \begin{align} A&=\{\hspace{3.5em}1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01+1\}\\ -(1.01)A&=\{1.01^{n}+1.01^{n-1}+1.01^{n-2}+...+1.01^2+1.01\hspace{2em}\}\\ ----&=-----------------------\\ -.01A&=1 - 1.01^n\\ ----&-------\\ A &= \frac{1-1.01^n}{-.01} \end{align} \begin{align} B_{n} &= B_{0}(1.01)^n -w(A)\\ B_n-B_0(1.01)^n&=-w(A)\\ \end{align} \[w=\frac{B_n-B_0(1.01)^n}{-A}\] \[w=\frac{B_{48}-10000(1.01)^{48}}{-A}\] Thats one part of it :)

  3. amistre64
    • 3 years ago
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    We want B{48} to equal 0 which makes this equation simply, and I use that term loosely :) \[w = \frac{-10000(1.01)^{48}(.01)}{1-1.01^{48}}=263.338...\]

  4. amistre64
    • 3 years ago
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    And the other one I finally worked out to be: \[w = \frac{-10000(1.01)^{48}(.01)}{1.01^{48}(.01)+1.01^{48}-1.01}=-260.73104....\] ended up with a negative in there somehow, butI think its pretty close :)

  5. aama100
    • 3 years ago
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    Marvelous work! Thank you!

  6. aama100
    • 3 years ago
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    so from where did you get 1.01 and .01 ?

  7. amistre64
    • 3 years ago
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    the 1.01 is just the interest rate of 12% compounded monthly; 1 + .12/12 = 1.01 the ".01" comes in as a result of summing up the interest over any given amount of time. The equations start out as recursions and can be manipulated into functions with a great deal of effort :)

  8. amistre64
    • 3 years ago
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    the equations are by hand so they dont follow any sort of given formula as is, but they should be close since I pretty much had to reinvent the wheel to solve them :)

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