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Linear Gradient Series Derivation

  1. Aug 7, 2012 #1
    Howdy, I guess I need to explain this situation a little bit.. Im doing a project about a guy buying a house, but using a gradient series approach to do the payments, for example, say the original monthly payment for a $225,000 15 year loan is ~$1600. The guy will pay that the first month, and then add increasing amounts of $50 for each month to pay off the note faster. I was hoping someone could help me with this? I talked with my professor and he VERY briefly just mentioned that you can get the formula for it by adding the two series together.. Like the original month to month payment of $1600 and then add the series of increasing $50 amounts. I was hoping someone could help me with the derivation of this? I found this formula, but I don't want to just use something I don't know anything about.. This project is kind of an extra assignment, and therefore it's really out of the scope of my class.. so I don't have any kind of notes or anything to help.

    15 year note, 3.6% APR

    Formula I found - [itex] P = G \frac{(1+i)^{N}-iN-1}{i^{2}(1+i)^{N}} [/itex]

    Formula I have from class - [itex] P = A \frac{(1+i)^{N}-1}{i(1+i)^{N}} [/itex]

    When looking at the two, I can see similarities, but I'm not sure what the straight formula for just G is.. Anywho, hopefully this was clear..

    My end goal was to calculate how much faster he could pay the loan off, and how much money he would save.


    Thanks for any help!
     
  2. jcsd
  3. Aug 7, 2012 #2
    How would you derive the formula that you already know?
     
  4. Aug 7, 2012 #3
    Well that's the problem. I found this formula online, but I was hoping someone could show me the actual steps to get to this formula so I could understand it?
     
  5. Aug 7, 2012 #4
    You have one formula "from the class" - can you explain how it is obtained?
     
  6. Aug 7, 2012 #5
    (1) [itex] F=P(1+i)^{N} [/itex]
    (2) [itex] F=A \frac{(1+i)^{N}-1}{i} [/itex]

    (2) is derived from [itex] F = A(1+i)^{0} + A(1+i)^{1} + A(1+i)^{2} A(1+i)^{3} + ... + A(1+i)^{N-1} = A \frac{(1+i)^{N}-1}{1+i-1} [/itex]

    using (1) and (2) to solve for P gets the formula.
     
  7. Aug 8, 2012 #6

    Ray Vickson

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    The present-value of a stream of steady payments A, A, A,..., A at times t=0, 1, 2,..., N is
    [tex]\text{PV1} = A + A\rho + a \rho^2 + \cdots + A \rho^N = A \frac{1-\rho^{N+1}}{1-\rho},[/tex] where
    [tex] \rho = \frac{1}{1+i}.[/tex]
    However, the situation you describe is different: you have a string of increasing payments A, A+a, A+2a, ..., A+Na at times t = 0, 1, 2, ..., N, and the present value of that stream is
    [tex] \text{PV2} = A + (A+a)\rho + (A+2a) \rho^2 + \cdots + (A + Na) \rho^N
    = \frac{A(1-\rho^{N+1}) - aN \rho^{N+1}}{1-\rho} + \frac{a(\rho - \rho^{N+1})}{(1-\rho)^2}. [/tex]

    You already know A = 1600, a = 50 and you can calculate [itex]\rho[/itex]. Solve (numerically) the equation PV2 = 225000 to find the new N (months).

    RGV
     
    Last edited: Aug 8, 2012
  8. Aug 8, 2012 #7
    This may sound dumb, but how do I solve for N when it is a normal variable AND an exponent? .. I'm totally baffled.
     
  9. Aug 8, 2012 #8

    Ray Vickson

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    You need to use a numerical method. If you have access to Maple, Mathematica or Matlab, you can ask that package to solve the equation. If you have access to a spreadsheet, such as EXCEL, you can use the built-in Solver tool to get a solution. Alternatively, you can get a rough solution by plotting PV2 as a function of N, then get a more accurate solution by an iterative improvement such as Newton's method or any one of a dozen available methods (bisecting search, secant method, regula falsi, etc) Do a Google search on these names to get more details.

    You might also try the free on-line package "Wolfram Alpha".

    RGV
     
  10. Aug 8, 2012 #9
    Hmm.. ok! So if I don't have any of those fancy things, just 'guess and check' is as viable as anything?

    Thanks for your help Ray.
     
  11. Aug 9, 2012 #10

    Ray Vickson

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    If you have internet access you have free access to Wolfram Alpha. I tried it: in the input panel, type
    sum (1600+50*k)/(1 + 0.036/12)^k,k=0..N
    After a while you get a formula in another panel, with the unknown N in it. You can outline the formula using the mouse and copy it to another panel, then insert the word 'solve' (no apostrophes) in front of it and '= 225000' (no apostrophes) after it. In a few seconds you get a graph, a numerical solution, etc.

    Try it out: http://www.wolframalpha.com/

    RGV
     
    Last edited: Aug 9, 2012
  12. Aug 9, 2012 #11
    Ray, Thank you very much! I have used Wolfram Alpha before, but never to such an extent. Very cool.

    I did have a quick question though, on your previous formula, where you have :


    [itex] PV1 = A + Aρ + Aρ^{2} + ... + Aρ^{N} = A\frac{1-ρ^{N+1}}{1-ρ} [/itex]

    Where [itex] ρ=\frac{1}{1+i} [/itex]


    How did you get that formula? I am getting [itex] P = A\frac{(1+i)^{N}-1}{i(1+i)^{N}} [/itex]

    When I try to set your formula and mine equal to eachother, it never works out. Is my formula incorrect?
     
  13. Aug 9, 2012 #12

    Ray Vickson

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    [tex] \frac{1-\frac{1}{(1+i)^{N+1}}{1-\frac{1}{1+i}} = \frac{(1+i)^{N+1}-1}{i (1+i)^N}.[/tex]
    The two formulas are exactly the same.

    RGV
     
  14. Aug 9, 2012 #13
    The formula I was given in class was derived like this:

    You have a set of N payments of amount A. The sum of these payments is symmetrical to a geometric progression sum. [itex] a + ax + ax^{2} + ... + ax^{N-1} = \frac{a(x^{n}-1)}{x-1} [/itex]

    if you let [itex] x=(1+i) [/itex]

    Then [itex] F=A\frac{(1+i)^{N}-1}{i} [/itex]

    Using the simple formula [itex] F=P(1+i)^{N} [/itex] and and the one I just derived and solving for P, I get [itex] P=A\frac{(1+i)^{N}-1}{i(1+i)^{N}}[/itex]

    Rearranging that equation to solve for A gets: [itex] A=P\frac{i(1+i)^{N}}{(1+i)^{N}-1} [/itex]


    That's how I got my formulas. Are there mistakes?

    For your formula, when I try to make it equate to mine, I get:
    [itex] P=A\frac{(1+i)^{N}-1}{i(1+i)^{N}} = A\frac{1-(\frac{1}{1+i})^{N+1}}{1-(\frac{1}{1+i})} [/itex]

    which doesn't equate. I can alter your equation by simplifying the denominator and multiplying everything by [itex] \frac{(1+i)^{N+1}}{(1+i)^{N+1}}[/itex] to get [itex] P=A\frac{(1+i)^{N+1}-1}{i(1+i)^{N}} [/itex] which is so close except for the N+1 in the exponential in the numerator. Again, I can't see any problems, but I can't really take a step back from it. Any help, again, is much appreciated!
     
  15. Aug 9, 2012 #14

    Ray Vickson

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    The LaTeX in my previous response failed, so here is the argument in plain text. Look at the expression for P involving ρ. Its numerator = 1-[1/(1+i)]^(N+1) and its denominator = 1-1/(1+i) = i/(1+i). Multiply the numerator and denominator by (1+i)^(N+1), to get: new numerator = (1+i)^(N+1) - 1, and new denominator = (1+i)^(N+1)*i/(1+i) = i*(1+i)^N, which is exactly what you got.

    RGV
     
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