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Using Linear Interpolation to Find Interest

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    F96Lk18.png

    2. Relevant equations



    3. The attempt at a solution

    tpXoSpS.png

    I understand how they calculated NPW but how did they use the linear interpolation method?
     
  2. jcsd
  3. Feb 24, 2014 #2

    Ray Vickson

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    I have never before seen their version of the linear interpolation method, but it does work. In linear interpolation we fit a linear function ##f = a + bx## through two points ##(x_1,f_1)## and ##(x_2,f_2)##:
    [tex] f(x) = f_1 + \frac{f_2-f_1}{x_2-x_1} (x-x_1)[/tex]
    This gives us the value of ##x_0##, the root of ##f(x) = 0,## as follows:
    [tex] x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}[/tex]
    It follows (doing a lot of algebraic simplification) that
    [tex] \frac{x_0 - x_1}{x_2-x_0} = -\frac{f_1}{f_2} [/tex]
    In other words, we can find ##x_0## by solving the equation
    [tex] \frac{x - x_1}{x_2-x} = -\frac{f_1}{f_2} [/tex]
    The solution is ##x = x_0##. That is not how I would do it, but it is correct.
     
  4. Mar 3, 2014 #3
    How would you do it? Just set f(x)=0 and solve for x?
     
  5. Mar 3, 2014 #4

    Ray Vickson

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    I already gave you the formula I would use:
    [tex] x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}[/tex]
    and I already explained how I got it.
     
  6. Mar 3, 2014 #5
    You never explicitly said that you would use that formula so I didn't realize.

    Thanks. I get it now.
     
  7. Mar 3, 2014 #6
    I would use the following:

    [tex]\frac{x-10}{20-10}=\frac{0-764}{-438-764}[/tex]

    p.s. Check to see how closely 16.36% comes to making the Net present value zero.

    Chet
     
  8. Mar 4, 2014 #7

    Ray Vickson

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    Just so you know: there is a difference between using linear interpolation on ##r## directly, and using linear interpolation on ##x = 1/(1+r)##. The present value is a polynomial in ##x##, but not in ##r##. Linear interpolation on ##x## gives ##x_0 \approx .8609359558##, giving ##r_0 \approx .1615265843##, or about 16.15%. Which answer is closer? Well, the exact solution is ##r = .1594684085##, or about 15.95%. So, in this case at least, interpolation on ##x## produces slightly better results.
     
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