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<p class="MsoNormal">I am using the command-line version of CLP to solve a series of LP’s, and so far I am very happy with the performance. Some take 40 minutes to solve, but most take 20 seconds to solve.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">I may have to solve 12 times more LP’s than I currently am solving now, so I am considering using some type of warm start to speed up the solution.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Specifically, I can ensure that a large set of the LP’s I need to solve have the same structure… the same number of rows, columns and the same non-zero pattern in A matrix. Each of the LP’s in a set will have some differences in the coefficients
– both the RHS, the A matrix and the objective function.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">I am considering using the “BasisOut/BasisIn” parameters of the command line CLP to do the warm start: Solve the first LP with the BasisOut argument, and then the subsequent LP’s with the BasisIn argument. I’ve done some semi-manual experiments
and it seems to work, but before I go to the effort of fully implementing the idea I am hoping that others could give some advice.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">A few questions:<o:p></o:p></p>
<p class="MsoListParagraph" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><![if !supportLists]><span style="mso-list:Ignore">(1)<span style="font:7.0pt "Times New Roman"">
</span></span><![endif]>Does anyone have experience performing a warm start using the command-line version of CLP? If so, do you have any advice that you could share?<o:p></o:p></p>
<p class="MsoListParagraph" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><![if !supportLists]><span style="mso-list:Ignore">(2)<span style="font:7.0pt "Times New Roman"">
</span></span><![endif]>Does anyone have specific advice on the “BasisOut/BasisIn” strategy? For example, while the non-zero patterns will be the same, I cannot mathematically guarantee that the basis from one LP in the set will form a valid basis in another
(the input basis could imply a negative variable, although that is unlikely). Does CLP reject the basis in that case and start cold, or is it more clever? Any other advice on this strategy will also be most welcomed.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Thanks in advance,<o:p></o:p></p>
<p class="MsoNormal">Marc <o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
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