I feel a bit silly to keep replying to myself, but... On Thu, 31 Mar 2005 13:25:34 -0500, Kohei Yoshida <[EMAIL PROTECTED]> wrote: ... > Consider this scenario: > > A1: = 2*G5*G6 + 3*B3 > G5: = B2 > G6: = 20 > B2 and B3 are decision variables.
This is not a good example either, because to get the coefficient for the first term, setting B2=1 and B3=0 will just do. The same for the 2nd term. So let's scratch this scenario. > Things get really complicated if you consider: > > A1: = G5*(G6 + B3) + B4 > G5: = 20 > G6: = B2 > B2, B3 and B4 are all decision variables. The same is true with this one. By setting 1 to each of the decision variables with all the others set to 0, you can get the coefficient for each variable. So, this is not a good counter-example either. > A1: = G5*(G6 + B3) + B4 > G5: = B5 > G6: = B2 > B2, B3, B4 and B5 are all decision variables. With this one, since this is not linear, the linearity test should fail. I still don't know wether the linearity test can be done by merely changing the cell values. But this seems like the only case where the descent parser may be necessary. But again, I may still be over-complicating the problem... But how about this scenario: A1: = B2*B2*2 + B3 B2 and B3 are decision variables By setting B2=1 and B3=0, you will still get 2 as the "coefficient" for B2, but this is really not linear. I guess I should re-think my approach a bit. Meanwhile, I'm open to any other ideas. Regards, Kohei --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
