What constrained regression function shuld I use?

I have a regression model log (r(i)) = a + b * log(A(i)) where A(i) is a vector and each element is known. Log is the nature log.

I need to find out a, b, and each element of r(i) such that the sum of r(i) equals to a constant k and the sum of error, i.e. sum(square[log (r(i)) – (a + b * log(A(i)))]) is minimized. Both a and b are scalars.

What regression model can I choose?

Answer 1

[Simon Wang]

Also, if this really works, do I need to have specific toolboxes? Do I need to an implement specific solver?

Thanks!

Simon

Answer 2

[Torsten]

Choose a and b such that

exp(a)*(A(1)^b+A(2)^b+...+A(n)^b)=k

Then sum (exp(a)*A(i)^b) = k is satisfied.

Now define r(i) = exp(a) * A(i)^b, and you are done.

Best wishes

Torsten.

Answer 3

[Simon Wang]

Thanks!

exp(a)*(A(1)^b+A(2)^b+...+A(n)^b)=k and sum (exp(a)*A(i)^b) = k are the same equation.

So there are two unknowns a and b and only one equation, I did not see the way to get the unique a and b.

Note:

1) a and b are scalars

2) Condition is not used: sum(square[log (r(i)) – (a + b * log(A(i)))]) is minimized

Thanks!

Simon

[Torsten]

Choose b=1, a=log(k/(A(1)+A(2)+...+A(n))) and define r(i)=exp(a)*A(i).

Then sum(square[log (r(i)) – (a + b * log(A(i)))]) is minimized (because it equals 0) and sum r(i)=k.

Best wishes

Torsten.

Answer 4

[Simon Wang]

THank you for the response. I think it should work and this leads me to change my question a little bit because it will make more sense in real world scenarios. I will post another question and close this one. Thank you very much for the help.