For a linear optimization of an integral (with integral constraints), I perform a linear programming for the equivalent series. Maximum and minimum of the LP problem tend to be equal as I increase the number of variables.

Does it necessarily mean that there is a unique feasible solution? If yes, it means that the value of the integral is obtained from the constraints analytically, isn't it?
Example: https://math.stackexchange.com/q/457500/88652

  • $\begingroup$ it's not 100% clear what you are solving for. Your $p(x)$ actually depends on $f(x)$ and $C$, isn't it? $\endgroup$ – Dima Pasechnik Aug 14 '13 at 3:10
  • $\begingroup$ Of course it depends on them, so does the integral extrema. But, regardless of the $f(x)$ and $C$, the extrema estimated by LP tend to each other. I have tested it for different $f(x)$ and $C$, with only the conditions mentioned in the example. A very interesting point which may help you: the ratio of maximum to minimum for an arbitrary $f(x)$, is very near to one, except for the values near $C=0$ wherein there is a very high pick. $\endgroup$ – Amir Kazemi Aug 14 '13 at 3:47
  • $\begingroup$ Are you saying that when $C$ is small then the values of maxima and minima are rather different? $\endgroup$ – Dima Pasechnik Aug 15 '13 at 3:38
  • $\begingroup$ Yes Dima, I do not claim that the maximum and minimum are exactly the same. But the ratio of them is very close to one, an asymptote a bit larger than one. This ratio get further from the asymptote as we approach $C=0$, where this ratio is exactly one because $p(x)=0$. There is a discontinuity of the ratio at $C=0$. $\endgroup$ – Amir Kazemi Aug 15 '13 at 3:45
  • $\begingroup$ Do you suggest any other method except LP for solving the problem? $\endgroup$ – Amir Kazemi Aug 15 '13 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.