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DanHouston
Obsidian | Level 7

I was having an issue where optmilp ran for over 24 hours and had not finished solving and optlp would complete in a few minutes. Due to the Big-M and integer drift problem optlp does not really honor one of the constraints (so this may not really be an apples to apples comparison).

 

This is the scaled up version of the sample I posted in thread:

Logistics Optimization Problem (constraining ARCS flowing to NODE) 

 

Attached are the logs and code with one running optlp and the other optmilp. If anyone has any suggestions on improving solve time that would be greatly appreciated.

 

Thanks!

1 ACCEPTED SOLUTION

Accepted Solutions
RobPratt
SAS Super FREQ

Yes, LP and MILP times are not apples to apples because they are solving different problems.  Solving a MILP with 8.6 million variables and 4.3 million constraints in 22 minutes is actually quite impressive.  The log shows that the branch-and-cut tree consists of only one node (the root).  You might try the root node solver options ROOTNODE ALGORITHM=INTERIORPOINT (which can be good for large problems) or ROOTNODE ALGORITHM=NETWORK (which can be good for problems with network structure).  Another option to try is DECOMP METHOD=NETWORK.

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3 REPLIES 3
RobPratt
SAS Super FREQ

Yes, LP and MILP times are not apples to apples because they are solving different problems.  Solving a MILP with 8.6 million variables and 4.3 million constraints in 22 minutes is actually quite impressive.  The log shows that the branch-and-cut tree consists of only one node (the root).  You might try the root node solver options ROOTNODE ALGORITHM=INTERIORPOINT (which can be good for large problems) or ROOTNODE ALGORITHM=NETWORK (which can be good for problems with network structure).  Another option to try is DECOMP METHOD=NETWORK.

DanHouston
Obsidian | Level 7

Thanks. I will definitely give these a shot.

 

Is there a way to address the Big-M Drift problem with the binary variables when using OPTLP?

RobPratt
SAS Super FREQ
The logic of big-M constraints requires integrality. If the binary variable takes a fractional value, the desired logical implication is not enforced.

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