Using Linear Interpolation to Find Interest

[ainster31]

Homework Statement

Homework Equations

The Attempt at a Solution

I understand how they calculated NPW but how did they use the linear interpolation method?

 

[Ray Vickson]

Homework Statement

Homework Equations

The Attempt at a Solution

I understand how they calculated NPW but how did they use the linear interpolation method?

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 $(x_2,f_2)$:
$$f(x) = f_1 + \frac{f_2-f_1}{x_2-x_1} (x-x_1)$$
This gives us the value of $x_0$, the root of $f(x) = 0,$ as follows:
$$x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}$$
It follows (doing a lot of algebraic simplification) that
$$\frac{x_0 - x_1}{x_2-x_0} = -\frac{f_1}{f_2}$$
In other words, we can find $x_0$ by solving the equation
$$\frac{x - x_1}{x_2-x} = -\frac{f_1}{f_2}$$
The solution is $x = x_0$. That is not how I would do it, but it is correct.

 

[ainster31]

That is not how I would do it, but it is correct.

How would you do it? Just set f(x)=0 and solve for x?

 

[Ray Vickson]

How would you do it? Just set f(x)=0 and solve for x?

I already gave you the formula I would use:
$$x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}$$
and I already explained how I got it.

 

[ainster31]

I already gave you the formula I would use:
$$x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}$$
and I already explained how I got it.

You never explicitly said that you would use that formula so I didn't realize.

Thanks. I get it now.

 

[Chestermiller]

I would use the following:

$$\frac{x-10}{20-10}=\frac{0-764}{-438-764}$$

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

Chet

 

[Ray Vickson]

Homework Statement

Homework Equations

The Attempt at a Solution

I understand how they calculated NPW but how did they use the linear interpolation method?

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.