How to caculate a bass model.

y=m*(1-exp(-(p+q)*x))/(1+(q/q)*exp(-(p+q)*x))

x=1：8

y=[9824,9390,8402,6973,6719,5820,5377,6163]

How to caculate the value of the m p and q.

[jose antonio valles]

m is the people o the market = 5000, p is the innovation = 0.003 and q is the imitation = 0.5

Answer 1

[John D'Errico]

Totally impossible.

y = m*(1-exp(-(p+q)*x))/(1+(q/q)*exp(-(p+q)*x))

First, I notice the q/q term in there. sorry, but unless q==0, that ratio is 1. And if q is zero, then q/q is undefined.

So your model now reduces to

y = m*(1-exp(-(p+q)*x))/(1+exp(-(p+q)*x))

What are p and q? pick any value for the sum, and there are infinitely many ways to choose p and q, all equally good, all equally valid. So at best, we can estimate the sum, p+q. But you can never infer the value of p versus q.

So now your model reduces to

y = m*(1-exp(-r*x))/(1+exp(-r*x))

Are we any closer to something you can use?

x=1:8;
y=[9824,9390,8402,6973,6719,5820,5377,6163];
plot(x,y,'o')

Seriously? You want to fit that model to what is not much more than a straight line, with one point at the end that blips up? Is this a practical joke?

The basic functional form for the function you propose fitting is what could be broadly called a sigmoidal shape. The m coefficient out front does nothing except scale the curve in y. And the value of r is no more than a stretch on the x-axis.

fun = @(x,r) (1-exp(-r*x))./(1+exp(-r*x))
fun =
  function_handle with value:
    @(x,r)(1-exp(-r*x))./(1+exp(-r*x))
ezplot(@(x) fun(x,1))

So that is one of many models that can be used to fit a curve that transitions smoothly between two constant levels, and is asymptotic to the two levels as x goes from -inf to +inf.

Now, you want to fit that data to this model? I'll just tell you that no, you will never get any meaningful fit to that data from this model.

I'm sorry. I have no idea where you got that data from, but it is essentially useless to fit anything more than a straight line. And I have no idea why you think the model you have chosen would be appropriate. It is simply impossible to estimate parameters for that model, nor do I have any idea why you would even think that model might be appropriate for this data.

[Yuehao Sui]

Thx your answer, i got a mistake the q/q should be q/p.

Thanks a lot

[John D'Errico]

Even at that, q/p does not change the fundamental shape of the model. (Unless p and q have opposite signs, in which case the fundamental shape of the curve is more like a hyperbola. ANd that would fit your data even more poorly.)

So as long as p and q have the same sign, you still have a curve with the same fundamental shape as what I plotted. And nothing you do to vary q, p, and m will make that curve fit your data.

[jose antonio valles]

% This algorithm calculates the Bass Model

% input data

n = input('Enter Market sise: ');

p = input('Enter the innovation factor: ');

q = input('Enter yhe imitation factor: ');

v=0

for i=1:10

s=(p*n)+(q-p)*v-(q/n)*v^2

v=v+s

% output data

fprintf('the Bass model is: %.2f\n',s)

end