Pinned topic C++: Get LPrelaxation value at root node without stopping branchandcut
I need to get the objective value of the LPrelaxation at the root node (I only need the objective and not the values of the variables itself). I've seen some posts on this topic (i.e. https://www.ibm.com/developerworks/forums/thread.jspa?messageID=14475701�), however, in my case stopping the branchandcut at the root node is not an option.
Thank you for your help,
Markus

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20120907T22:09:03ZThis is the accepted answer. This is the accepted answer.You could attach a SolveCallback with an internal flag that is initially set true. It would call solve() and then, if the flag were true, call getObjValue(), record the result and set the flag to false.
Paul
Mathematicians are like Frenchmen: whenever you say something to them, they translate it into their own language, and at once it is something entirely different. (Goethe) 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20120917T20:39:04ZThis is the accepted answer. This is the accepted answer.If you remove this line
IloCplex::BranchCallbackI::abort();
from the code example in the quoted thread then the solve does not stop at the root node. Like Paul suggested you should also add a flag that is initially true. If the callback's main() function is executed for the first time you set the flag to false and store the objective function value somewhere. In subsequent invocations of the main() function you just don't do anything. Notice however that adding a control callback (branch callback or solve callback) will disable dynamic search and may therefore hurt performance.
What is the problem with interrupting the solve at the root node? Consider the following code sequence:
cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node. if ( solve() ) { double rootObjective = cplex.getBestObjValue(); cplex.setParam(IloCplex::IntParam::NodeLim, 210000000); solve(); // Continue solve. }
This will interrupt the solve after the root node. You can query the dual bound and invoke solve() again. Since you did not modify the problem the second solve() will continue just where the first one stopped.Updated on 20140324T22:52:26Z at 20140324T22:52:26Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20120927T13:19:03ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20120917T20:39:04Z
If you remove this line
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">IloCplex::BranchCallbackI::abort(); </pre>
from the code example in the quoted thread then the solve does not stop at the root node. Like Paul suggested you should also add a flag that is initially true. If the callback's main() function is executed for the first time you set the flag to false and store the objective function value somewhere. In subsequent invocations of the main() function you just don't do anything. Notice however that adding a control callback (branch callback or solve callback) will disable dynamic search and may therefore hurt performance.
What is the problem with interrupting the solve at the root node? Consider the following code sequence:
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node. if ( solve() ) { double rootObjective = cplex.getBestObjValue(); cplex.setParam(IloCplex::IntParam::NodeLim, 210000000); solve(); // Continue solve. } </pre>
This will interrupt the solve after the root node. You can query the dual bound and invoke solve() again. Since you did not modify the problem the second solve() will continue just where the first one stopped.
This is precisely what I was looking for. My other thread can be cleaned up, if needed. However, I have a query. Your code reads:
cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node.
if ( solve() ) {
double rootObjective = cplex.getBestObjValue();
cplex.setParam(IloCplex::IntParam::NodeLim, 210000000);
solve(); // Continue solve.
}
Now, what is the difference between the above and having the first line as:
cplex.setParam(IloCplex::IntParam::NodeLim, 1);//Note the 1 instead of 0?
Also, in your original code, is the getBestObjValue() function going to return the LP solution or any feasible integer solution that CPLEX found while preprocessing/applying heuristics, etc.?
Thanks. 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20121001T09:43:41ZThis is the accepted answer. This is the accepted answer. UserCplex
 20120927T13:19:03Z
Hi Daniel,
This is precisely what I was looking for. My other thread can be cleaned up, if needed. However, I have a query. Your code reads:
cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node.
if ( solve() ) {
double rootObjective = cplex.getBestObjValue();
cplex.setParam(IloCplex::IntParam::NodeLim, 210000000);
solve(); // Continue solve.
}
Now, what is the difference between the above and having the first line as:
cplex.setParam(IloCplex::IntParam::NodeLim, 1);//Note the 1 instead of 0?
Also, in your original code, is the getBestObjValue() function going to return the LP solution or any feasible integer solution that CPLEX found while preprocessing/applying heuristics, etc.?
Thanks.
>
> cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node.
> if ( solve() ) {
> double rootObjective = cplex.getBestObjValue();
> cplex.setParam(IloCplex::IntParam::NodeLim, 210000000);
> solve(); // Continue solve.
> }
>
> Now, what is the difference between the above and having the first line as:
>
> cplex.setParam(IloCplex::IntParam::NodeLim, 1);//Note the 1 instead of 0?
>
The difference is subtle and not important in this case. NodeLim=1 stops after the root node is completely done, a first branching decision has been made, the next node is selected (but not yet processed) etc. NodeLim=0 just stops a little earlier.
> Also, in your original code, is the getBestObjValue() function going to return the LP solution or any feasible integer solution that CPLEX found while preprocessing/applying heuristics, etc.?
>
getBestObjValue() returns the dual bound, which in this case (the root node) is the LP relaxation at the root node.
Integer feasible solutions are accessed via getIncumbentObjValue() etc. 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130122T18:49:11ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20121001T09:43:41Z
> This is precisely what I was looking for. My other thread can be cleaned up, if needed. However, I have a query. Your code reads:
>
> cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node.
> if ( solve() ) {
> double rootObjective = cplex.getBestObjValue();
> cplex.setParam(IloCplex::IntParam::NodeLim, 210000000);
> solve(); // Continue solve.
> }
>
> Now, what is the difference between the above and having the first line as:
>
> cplex.setParam(IloCplex::IntParam::NodeLim, 1);//Note the 1 instead of 0?
>
The difference is subtle and not important in this case. NodeLim=1 stops after the root node is completely done, a first branching decision has been made, the next node is selected (but not yet processed) etc. NodeLim=0 just stops a little earlier.
> Also, in your original code, is the getBestObjValue() function going to return the LP solution or any feasible integer solution that CPLEX found while preprocessing/applying heuristics, etc.?
>
getBestObjValue() returns the dual bound, which in this case (the root node) is the LP relaxation at the root node.
Integer feasible solutions are accessed via getIncumbentObjValue() etc.
In the Cplex manual it says that subsequent calls to solve() have their own TiLim. Therefore, we need to check the solution status; otherwise if the first solve ends with a solution status 11, 107, or 108 (btw, are there any other easier way to check this? ) then we don't have a bound (probably it will give infinity as a bound or worse, an error; this is okay though) but the problem is that Cplex is going to try to solve the problem again, although the time we allow already passed, etc.
This is a plausible scenario for me and my solution is the following, please correct me if anything is wrong:
cplex.setParam(IloCplex::TiLim, My.TiLim); // Original time limit cplex.setParam(IloCplex::NodeLim, 0); // Stop after root node to get root relaxation time and objective if ( cplex.solve() ) { Root.Time = cplex.getTime(); Root.Status = cplex.getCplexStatus(); Root.Bound = cplex.getBestObjValue(); // *Does this line produce an error if the time limit is reached?* if ( Root.Status != IloCplex::AbortTimeLim  ( Root.Status == IloCplex::NodeLimInfeas  Root.Status == IloCplex::NodeLimFeas ) ) { // *Are there any other possible statuses to consider? Or what is the best way to check this?* Root.Bound = cplex.getBestObjValue(); // Get Objective cplex.setParam(IloCplex::NodeLim, 210000000); // Set Node limit to default value cplex.setParam(IloCplex::TiLim, My.TiLim  Root.Time ); // Modify the time limit cplex.solve(); // Continue solve. } }

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130125T09:36:10ZThis is the accepted answer. This is the accepted answer. hllsen
 20130122T18:49:11Z
Maybe out of the scope of this question however I am going to use the same method to obtain the root relaxation bound+time and I wondered what happens if the first solve() ends due to time limit.
In the Cplex manual it says that subsequent calls to solve() have their own TiLim. Therefore, we need to check the solution status; otherwise if the first solve ends with a solution status 11, 107, or 108 (btw, are there any other easier way to check this? ) then we don't have a bound (probably it will give infinity as a bound or worse, an error; this is okay though) but the problem is that Cplex is going to try to solve the problem again, although the time we allow already passed, etc.
This is a plausible scenario for me and my solution is the following, please correct me if anything is wrong:
<pre class="jivepre">cplex.setParam(IloCplex::TiLim, My.TiLim); // Original time limit cplex.setParam(IloCplex::NodeLim, 0); // Stop after root node to get root relaxation time and objective if ( cplex.solve() ) { Root.Time = cplex.getTime(); Root.Status = cplex.getCplexStatus(); Root.Bound = cplex.getBestObjValue(); // *Does this line produce an error if the time limit is reached?* if ( Root.Status != IloCplex::AbortTimeLim  ( Root.Status == IloCplex::NodeLimInfeas  Root.Status == IloCplex::NodeLimFeas ) ) { // *Are there any other possible statuses to consider? Or what is the best way to check this?* Root.Bound = cplex.getBestObjValue(); // Get Objective cplex.setParam(IloCplex::NodeLim, 210000000); // Set Node limit to default value cplex.setParam(IloCplex::TiLim, My.TiLim  Root.Time ); // Modify the time limit cplex.solve(); // Continue solve. } }
</pre>
 cplex.solve() will only return true if an integer feasible solution is available. So if CPLEX does not find an integer feasible solution before the time limit expires or the root node is complete (whichever comes first) your code will probably not work as expected.
 Since cplex.solve() returns true only if a feasible solution is available, status IloCplex::NodeLimInfeas cannot happen (at least not in the current state of your code).
 cplex.getTime() does only return a timestamp, not the time CPLEX spent so far. So you need to do record the timestamp before you call solve() and then calculate the different of this time stamp to cplex.getTime() to get the elapsed time. In the current code My.TiLim  Root.Time will not produce the remaining time budget.

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130202T20:27:17ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20130125T09:36:10Z
I think there are some flaws in your code:
 cplex.solve() will only return true if an integer feasible solution is available. So if CPLEX does not find an integer feasible solution before the time limit expires or the root node is complete (whichever comes first) your code will probably not work as expected.
 Since cplex.solve() returns true only if a feasible solution is available, status IloCplex::NodeLimInfeas cannot happen (at least not in the current state of your code).
 cplex.getTime() does only return a timestamp, not the time CPLEX spent so far. So you need to do record the timestamp before you call solve() and then calculate the different of this time stamp to cplex.getTime() to get the elapsed time. In the current code My.TiLim  Root.Time will not produce the remaining time budget.
Presolve has eliminated 0 rows and 1 columns... Presolve has eliminated 0 rows and 28767 columns... Presolve has eliminated 222 rows and 28767 columns... Tried aggregator 1 time. Presolve has eliminated 222 rows and 28767 columns... MIP Presolve eliminated 222 rows and 28767 columns. Reduced MIP has 37903 rows, 3773734 columns, and 224162878 nonzeros. Reduced MIP has 3773734 binaries, 0 generals, 0 SOSs, and 0 indicators. Presolve time = 222.52 sec. Dual steepestedge pricing selected. Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap 0 0 1.00000e+037 0 0 Root node processing (before b&c): Real time = 1495.41 Parallel b&c, 7 threads: Real time = 0.00 Sync time (average) = 0.00 Wait time (average) = 0.00  Total (root+branch&cut) = 1495.41 sec.
Updated on 20140324T22:40:57Z at 20140324T22:40:57Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130202T20:46:19ZThis is the accepted answer. This is the accepted answer. hllsen
 20130202T20:27:17Z
Thank you Daniel for the comments. It was a rough week so I couldn't find the time to write back and I already applied your suggestions. But, now I'm having a different problem with the cplex time limit: Although TiLim parameter set to 600 seconds, it takes 1495 seconds for cplex.solve() to stop and extra ~900 seconds to release the memory. I know the model is huge but is this kind of timing results expected?
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">Presolve has eliminated 0 rows and 1 columns... Presolve has eliminated 0 rows and 28767 columns... Presolve has eliminated 222 rows and 28767 columns... Tried aggregator 1 time. Presolve has eliminated 222 rows and 28767 columns... MIP Presolve eliminated 222 rows and 28767 columns. Reduced MIP has 37903 rows, 3773734 columns, and 224162878 nonzeros. Reduced MIP has 3773734 binaries, 0 generals, 0 SOSs, and 0 indicators. Presolve time = 222.52 sec. Dual steepestedge pricing selected. Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap 0 0 1.00000e+037 0 0 Root node processing (before b&c): Real time = 1495.41 Parallel b&c, 7 threads: Real time = 0.00 Sync time (average) = 0.00 Wait time (average) = 0.00  Total (root+branch&cut) = 1495.41 sec. </pre>
Also, Cplex uses 15gigs during the presolve operation but even if the working memory limit is set to 15gigs and NodeFileInd is at its default, cplex claims more than 25gigs while starting the B&B. 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T06:34:16ZThis is the accepted answer. This is the accepted answer. hllsen
 20130202T20:27:17Z
Thank you Daniel for the comments. It was a rough week so I couldn't find the time to write back and I already applied your suggestions. But, now I'm having a different problem with the cplex time limit: Although TiLim parameter set to 600 seconds, it takes 1495 seconds for cplex.solve() to stop and extra ~900 seconds to release the memory. I know the model is huge but is this kind of timing results expected?
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">Presolve has eliminated 0 rows and 1 columns... Presolve has eliminated 0 rows and 28767 columns... Presolve has eliminated 222 rows and 28767 columns... Tried aggregator 1 time. Presolve has eliminated 222 rows and 28767 columns... MIP Presolve eliminated 222 rows and 28767 columns. Reduced MIP has 37903 rows, 3773734 columns, and 224162878 nonzeros. Reduced MIP has 3773734 binaries, 0 generals, 0 SOSs, and 0 indicators. Presolve time = 222.52 sec. Dual steepestedge pricing selected. Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap 0 0 1.00000e+037 0 0 Root node processing (before b&c): Real time = 1495.41 Parallel b&c, 7 threads: Real time = 0.00 Sync time (average) = 0.00 Wait time (average) = 0.00  Total (root+branch&cut) = 1495.41 sec. </pre>
How many threads do you use? If you leave the threads parameter at its default value, how many CPUs do you have? 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T06:37:45ZThis is the accepted answer. This is the accepted answer. hllsen
 20130202T20:46:19Z
I forgot to mention that while releasing the memory MyProgram.exe does not use that much CPU (may be this is usual?) and after some point CPU usage drops to zero but the memory usage fluctuates as if Cplex is solving/doing something and releases memory "slowly" packet by packet.
Also, Cplex uses 15gigs during the presolve operation but even if the working memory limit is set to 15gigs and NodeFileInd is at its default, cplex claims more than 25gigs while starting the B&B.
> I forgot to mention that while releasing the memory MyProgram.exe does not use that much CPU (may be this is usual?) and after some point CPU usage drops to zero but the memory usage fluctuates as if Cplex is solving/doing something and releases memory "slowly" packet by packet.
>
How does the code look that releases the memory? Do you release both the IloModel and the IloCplex instance? Or only one of them? If you release the IloModel instance, does it help to add
cplex.extract(IloModel(env));
before calling model.end()? There is quite some notification going on between IloCplex and an extracted IloModel and this may slow down deletion of a model (you can search this forum or the FAQs for this issue). If you release the IloCplex and the IloModel instance then try to first release the IloCplex instance and then the IloModel instance.
> Also, Cplex uses 15gigs during the presolve operation but even if the working memory limit is set to 15gigs and NodeFileInd is at its default, cplex claims more than 25gigs while starting the B&B.
>
A potential reason for this may be that a copy of the model is created for each thread. How many threads do you use? Does the memory consumption go down if you use fewer threads?Updated on 20140324T22:40:49Z at 20140324T22:40:49Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T06:43:57ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20130205T06:37:45Z
> hllsen wrote:
> I forgot to mention that while releasing the memory MyProgram.exe does not use that much CPU (may be this is usual?) and after some point CPU usage drops to zero but the memory usage fluctuates as if Cplex is solving/doing something and releases memory "slowly" packet by packet.
>
How does the code look that releases the memory? Do you release both the IloModel and the IloCplex instance? Or only one of them? If you release the IloModel instance, does it help to add
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.extract(IloModel(env)); </pre>
before calling model.end()? There is quite some notification going on between IloCplex and an extracted IloModel and this may slow down deletion of a model (you can search this forum or the FAQs for this issue). If you release the IloCplex and the IloModel instance then try to first release the IloCplex instance and then the IloModel instance.
> Also, Cplex uses 15gigs during the presolve operation but even if the working memory limit is set to 15gigs and NodeFileInd is at its default, cplex claims more than 25gigs while starting the B&B.
>
A potential reason for this may be that a copy of the model is created for each thread. How many threads do you use? Does the memory consumption go down if you use fewer threads?
cplex.extract(IloModel(env));
just try
cplex.clearModel();
Both statements will "disconnect" the big model from the IloCplex instance so that the IloCplex instance is no longer notified of all the changes that are made to the model.Updated on 20140324T22:40:39Z at 20140324T22:40:39Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T09:24:27ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20130205T06:34:16Z
What version of CPLEX do you use? Overshooting the time limit by so much is not expected and sounds like a bug. Is it possible to attach a .SAV file for the model here or send to me daniel(dot)junglas(at)de(dot)ibm(dot)com?
How many threads do you use? If you leave the threads parameter at its default value, how many CPUs do you have? 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T09:33:18ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20130205T06:43:57Z
Sorry, what I suggested results in a memory leak. Instead of
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.extract(IloModel(env)); </pre>
just try
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.clearModel(); </pre>
Both statements will "disconnect" the big model from the IloCplex instance so that the IloCplex instance is no longer notified of all the changes that are made to the model.env.end();
I'll let you know the results when I'll be able to try the clearModel() trick and less number of threads.Updated on 20140324T22:40:34Z at 20140324T22:40:34Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20130205T09:34:22ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20130205T06:43:57Z
Sorry, what I suggested results in a memory leak. Instead of
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.extract(IloModel(env)); </pre>
just try
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.clearModel(); </pre>
Both statements will "disconnect" the big model from the IloCplex instance so that the IloCplex instance is no longer notified of all the changes that are made to the model.env.end();
I'll let you know the results when I'll be able to try the clearModel() trick and less number of threads.Updated on 20140324T22:40:29Z at 20140324T22:40:29Z by ironman 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20131205T05:24:55ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20120917T20:39:04Z
If you remove this line
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">IloCplex::BranchCallbackI::abort(); </pre>
from the code example in the quoted thread then the solve does not stop at the root node. Like Paul suggested you should also add a flag that is initially true. If the callback's main() function is executed for the first time you set the flag to false and store the objective function value somewhere. In subsequent invocations of the main() function you just don't do anything. Notice however that adding a control callback (branch callback or solve callback) will disable dynamic search and may therefore hurt performance.
What is the problem with interrupting the solve at the root node? Consider the following code sequence:
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node. if ( solve() ) { double rootObjective = cplex.getBestObjValue(); cplex.setParam(IloCplex::IntParam::NodeLim, 210000000); solve(); // Continue solve. } </pre>
This will interrupt the solve after the root node. You can query the dual bound and invoke solve() again. Since you did not modify the problem the second solve() will continue just where the first one stopped.Hi,
I have a question regarding your code below. The code works perfectly fine however when I fill the user cut pool of cplex with some valid inequalities, the second solve command gives an exception (bad access). Unless I give enough node limit (or time limit) so that the optimal solution is achieved, I cannot avoid the exception.
Is there a particular reason for that? (e.g., am I allowed to do this while using the cplex's user cut pool??)

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20131205T06:39:28ZThis is the accepted answer. This is the accepted answer. semihatakan
 20131205T05:24:55Z
Hi,
I have a question regarding your code below. The code works perfectly fine however when I fill the user cut pool of cplex with some valid inequalities, the second solve command gives an exception (bad access). Unless I give enough node limit (or time limit) so that the optimal solution is achieved, I cannot avoid the exception.
Is there a particular reason for that? (e.g., am I allowed to do this while using the cplex's user cut pool??)
<pre class="jivepre" dir="ltr" style="marginright: 2em; marginbottom: 0.5em; marginleft: 2em; padding: 10px 10px 10px 1em; borderwidth: 1px 1px 1px 2px; borderstyle: solid; bordercolor: rgb(238, 238, 238) rgb(238, 238, 238) rgb(238, 238, 238) rgb(204, 204, 204); outline: 0px; verticalalign: baseline; fontfamily: 'Courier New', Courier, monospace; lineheight: 18px; backgroundcolor: rgb(238, 238, 238) !important; overflow: auto !important;"> </pre>Adding cuts to the user cut pool should not be a problem here. Do you have a stack trace for that bad access exception? How exactly do you add the user cuts? Is it possible that you end() one of the cuts before you start the second solve()?

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20131205T07:08:30ZThis is the accepted answer. This is the accepted answer. DanielJunglas
 20131205T06:39:28Z
Adding cuts to the user cut pool should not be a problem here. Do you have a stack trace for that bad access exception? How exactly do you add the user cuts? Is it possible that you end() one of the cuts before you start the second solve()?
Hi Daniel,
Thanks a lot for the fast response.
The code for adding the cuts are below, valid_inequalities is an IloConstraintArray:
cplex.addUserCuts( valid_inequalities );
valid_inequalities.endElements();
valid_inequalities.end();
I do end them just after adding them to cplex, but this is correct right? The cuts are added after the model is extracted to cplex. I do not use any usercut callbacks. Then, I call the following sequence of solves:
cplex.setParam(IloCplex::NodeLim, 0);
solved = cplex.solve();
cplex.solve();
The second solve gives an exception unless I clear all the usercuts (or solve to optimality). There is nothing else between these solves (nothing is end()'ed). Also, the same number of user cuts is displayed in the CPLEX log for the second solve, but then the exception occurs. The last message I can see is the "Parallel Mode: none, ....".
About the stack trace, I don't know what it means :/. However, I followed the location of the bug (maybe this is stack trace, again don't know..)
IloBool solve() { return IloAlgorithm::solve(); } // The message is: EXC_BAD_ACCESS (code=1, address=0x10572e888)
Going deeper, the exception seemed to be in IloAlgorithm::callSolve() const:
0x10002828c: movq %rax, %r14 // exception is at this line
Do these help? There is one more step, but it doesn't seem human readable.
Just to let you know, I managed to get the info that I want through MIPINFOCALLBACK, so this is not urgent. But still curious of what might have caused the exception...
Thanks a lot
Updated on 20131205T07:10:39Z at 20131205T07:10:39Z by semihatakan 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20131210T06:48:52ZThis is the accepted answer. This is the accepted answer. semihatakan
 20131205T07:08:30Z
Hi Daniel,
Thanks a lot for the fast response.
The code for adding the cuts are below, valid_inequalities is an IloConstraintArray:
cplex.addUserCuts( valid_inequalities );
valid_inequalities.endElements();
valid_inequalities.end();
I do end them just after adding them to cplex, but this is correct right? The cuts are added after the model is extracted to cplex. I do not use any usercut callbacks. Then, I call the following sequence of solves:
cplex.setParam(IloCplex::NodeLim, 0);
solved = cplex.solve();
cplex.solve();
The second solve gives an exception unless I clear all the usercuts (or solve to optimality). There is nothing else between these solves (nothing is end()'ed). Also, the same number of user cuts is displayed in the CPLEX log for the second solve, but then the exception occurs. The last message I can see is the "Parallel Mode: none, ....".
About the stack trace, I don't know what it means :/. However, I followed the location of the bug (maybe this is stack trace, again don't know..)
IloBool solve() { return IloAlgorithm::solve(); } // The message is: EXC_BAD_ACCESS (code=1, address=0x10572e888)
Going deeper, the exception seemed to be in IloAlgorithm::callSolve() const:
0x10002828c: movq %rax, %r14 // exception is at this line
Do these help? There is one more step, but it doesn't seem human readable.
Just to let you know, I managed to get the info that I want through MIPINFOCALLBACK, so this is not urgent. But still curious of what might have caused the exception...
Thanks a lot
So far I have failed to reproduce this problem here. Doing things like you described works without a flaw for me. If you still need to get to the bottom of that issue, could you create a minimal example that shows the crash and post that here or send it directly to daniel(dot)junglas(at)de(dot)ibm(dot)com?

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170201T20:04:04ZThis is the accepted answer. This is the accepted answer. SystemAdmin
 20120917T20:39:04Z
If you remove this line
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">IloCplex::BranchCallbackI::abort(); </pre>
from the code example in the quoted thread then the solve does not stop at the root node. Like Paul suggested you should also add a flag that is initially true. If the callback's main() function is executed for the first time you set the flag to false and store the objective function value somewhere. In subsequent invocations of the main() function you just don't do anything. Notice however that adding a control callback (branch callback or solve callback) will disable dynamic search and may therefore hurt performance.
What is the problem with interrupting the solve at the root node? Consider the following code sequence:
<pre class="java dw" dataeditorlang="java" datapbcklang="java" dir="ltr">cplex.setParam(IloCplex::IntParam::NodeLim, 0); // Stop after root node. if ( solve() ) { double rootObjective = cplex.getBestObjValue(); cplex.setParam(IloCplex::IntParam::NodeLim, 210000000); solve(); // Continue solve. } </pre>
This will interrupt the solve after the root node. You can query the dual bound and invoke solve() again. Since you did not modify the problem the second solve() will continue just where the first one stopped.Hi, I tried to follow this suggestion but whenever I try
PCSVAL.setParam(IloCplex::NodeLim, 0); PCSVAL.solve(); PrintLPSol(file_name); PCSVAL.setParam(IloCplex::NodeLim, 2100000000); PCSVAL.solve();
after the root node is solved, cplex restarts from beginning contradicting your statement: "Since you did not modify the problem the second solve() will continue just where the first one stopped. ".
Have any idea of what I am missing ?
Thanks for the help.
Best Regards. 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170202T06:27:39ZThis is the accepted answer. This is the accepted answer. pegonzalez
 20170201T20:04:04Z
Hi, I tried to follow this suggestion but whenever I try
after the root node is solved, cplex restarts from beginning contradicting your statement: "Since you did not modify the problem the second solve() will continue just where the first one stopped. ".
Have any idea of what I am missing ?
Thanks for the help.
Best Regards.What do you mean by "it starts from the beginning"? Does it do presolve again? Can you show the log output of the first and second solve? What is the return value of PCSVAL.getStatus() after the first solve? Are you sure you are not changing anything on the PCSVAL object in PrintLPSol? What exactly are you doing in this function?

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170202T12:07:29ZThis is the accepted answer. This is the accepted answer. DanielJunglas
 20170202T06:27:39Z
What do you mean by "it starts from the beginning"? Does it do presolve again? Can you show the log output of the first and second solve? What is the return value of PCSVAL.getStatus() after the first solve? Are you sure you are not changing anything on the PCSVAL object in PrintLPSol? What exactly are you doing in this function?
Hi Daniel,
when I said it starts from beginning, I mean that cplex restarts the branchandcut from the root node, as it can be seen in the partial log below:
Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.35 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 User cuts applied: 5 Root node processing (before b&c): Real time = 5.68 sec. (6092.01 ticks) Sequential b&c: Real time = 0.00 sec. (0.00 ticks)  Total (root+branch&cut) = 5.68 sec. (6092.01 ticks) Status: Unknown Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.42 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 0 2 1044.8895 350 1044.9789 1695 0 0
After modifying the code as you asked to get the status, I verified that it returns "Unknown"
PCSVAL.setParam(IloCplex::NodeLim, 0); PCSVAL.solve(); cout<<"Status: "<<PCSVAL.getStatus()<<endl; PrintLPSol(file_name); PCSVAL.setParam(IloCplex::NodeLim, 2100000000); PCSVAL.solve();
The function PrintLPSol just write the values of the variables and the relaxation cost, gathered through an user callback, in a file:
void Exato::PrintLPSol(char* file_name){ ofstream myfile; myfile.open (file_name, ios::out); myfile<<"LPOBJ: "<<LP_OBJ<<endl; myfile<<endl; for (ArcList::iterator it = p>begin_arcs(); it!=p>end_arcs(); ++it) { Arc * arc = *it; if(LP_Sol_X[arc>get_i()][arc>get_j()]>1E6){ myfile <<"LP_X["<<arc>get_i()<<","<<arc>get_j()<<"]: "<<LP_Sol_X[arc>get_i()][arc>get_j()]<<" \t"<<arc>get_dist()<<endl; } } myfile<<endl; for (ArcList::iterator it = p>begin_arcs(); it!=p>end_arcs(); ++it) { Arc * arc = *it; if(LP_Sol_Y[arc>get_i()][arc>get_j()]>1E6){ myfile<<"LP_Y["<<arc>get_i()<<","<<arc>get_j()<<"]: "<<LP_Sol_Y[arc>get_i()][arc>get_j()]<<endl; } } myfile<<endl; for (int i=0; i<p>get_nvertices()+1; ++i) { if(LP_Sol_Y[i][i]>1E6){ myfile<<"LP_Y["<<i<<","<<i<<"]: "<<LP_Sol_Y[i][i]<<endl; } } myfile.close(); }
Sorry for the imprecisions and thanks for the help.
Hope it is clearer now.
Best regards 
Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170202T12:32:31ZThis is the accepted answer. This is the accepted answer. pegonzalez
 20170202T12:07:29Z
Hi Daniel,
when I said it starts from beginning, I mean that cplex restarts the branchandcut from the root node, as it can be seen in the partial log below:
<pre class="javascript dw" dataeditorlang="js" datapbcklang="javascript" dir="ltr">Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.35 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 User cuts applied: 5 Root node processing (before b&c): Real time = 5.68 sec. (6092.01 ticks) Sequential b&c: Real time = 0.00 sec. (0.00 ticks)  Total (root+branch&cut) = 5.68 sec. (6092.01 ticks) Status: Unknown Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.42 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 0 2 1044.8895 350 1044.9789 1695 0 0 </pre>After modifying the code as you asked to get the status, I verified that it returns "Unknown"
<pre class="javascript dw" dataeditorlang="js" datapbcklang="javascript" dir="ltr">PCSVAL.setParam(IloCplex::NodeLim, 0); PCSVAL.solve(); cout<<"Status: "<<PCSVAL.getStatus()<<endl; PrintLPSol(file_name); PCSVAL.setParam(IloCplex::NodeLim, 2100000000); PCSVAL.solve(); </pre>The function PrintLPSol just write the values of the variables and the relaxation cost, gathered through an user callback, in a file:
<pre class="javascript dw" dataeditorlang="js" datapbcklang="javascript" dir="ltr">void Exato::PrintLPSol(char* file_name){ ofstream myfile; myfile.open (file_name, ios::out); myfile<<"LPOBJ: "<<LP_OBJ<<endl; myfile<<endl; for (ArcList::iterator it = p>begin_arcs(); it!=p>end_arcs(); ++it) { Arc * arc = *it; if(LP_Sol_X[arc>get_i()][arc>get_j()]>1E6){ myfile <<"LP_X["<<arc>get_i()<<","<<arc>get_j()<<"]: "<<LP_Sol_X[arc>get_i()][arc>get_j()]<<" \t"<<arc>get_dist()<<endl; } } myfile<<endl; for (ArcList::iterator it = p>begin_arcs(); it!=p>end_arcs(); ++it) { Arc * arc = *it; if(LP_Sol_Y[arc>get_i()][arc>get_j()]>1E6){ myfile<<"LP_Y["<<arc>get_i()<<","<<arc>get_j()<<"]: "<<LP_Sol_Y[arc>get_i()][arc>get_j()]<<endl; } } myfile<<endl; for (int i=0; i<p>get_nvertices()+1; ++i) { if(LP_Sol_Y[i][i]>1E6){ myfile<<"LP_Y["<<i<<","<<i<<"]: "<<LP_Sol_Y[i][i]<<endl; } } myfile.close(); } </pre>Sorry for the imprecisions and thanks for the help.
Hope it is clearer now.
Best regardsThis is expected if you set the node limit to 0 in the first call. Try setting it to 1 instead, then CPLEX should not redo the root node.

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170202T12:40:48ZThis is the accepted answer. This is the accepted answer. DanielJunglas
 20170202T12:32:31Z
This is expected if you set the node limit to 0 in the first call. Try setting it to 1 instead, then CPLEX should not redo the root node.
Unfortunately even after changing the node limit to 1 it keeps redoing it.
Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.34 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 0 2 1044.8895 350 1044.9789 1695 0 0 Elapsed time = 7.82 sec. (7866.82 ticks, tree = 0.01 MB, solutions = 0) User cuts applied: 5 Root node processing (before b&c): Real time = 7.72 sec. (7763.17 ticks) Sequential b&c: Real time = 0.10 sec. (103.95 ticks)  Total (root+branch&cut) = 7.82 sec. (7867.12 ticks) Status: Unknown Lazy constraint(s) or lazy constraint callback is present. Disabling dual reductions (CPX_PARAM_REDUCE) in presolve. Clique table members: 72149. MIP emphasis: balance optimality and feasibility. MIP search method: traditional branchandcut. Parallel mode: none, using 1 thread. Root relaxation solution time = 3.41 sec. (1851.88 ticks) Nodes Cuts/ Node Left Objective IInf Best Integer Best Bound ItCnt Gap Variable B NodeID Parent Depth 0 0 1043.6695 339 1043.6695 1626 0 0 1044.8400 368 User: 3 1670 0 0 1044.8685 365 User: 1 1685 0 0 1044.8895 350 User: 1 1695 0 2 1044.8895 350 1044.9789 1695 0 0 Elapsed time = 7.89 sec. (7866.80 ticks, tree = 0.01 MB, solutions = 0) 1 3 1057.9440 336 1058.1151 1922 y[48][48] N 1 0 1

Re: C++: Get LPrelaxation value at root node without stopping branchandcut
20170211T18:54:27ZThis is the accepted answer. This is the accepted answer. pegonzalez
 20170202T12:40:48Z
Unfortunately even after changing the node limit to 1 it keeps redoing it.
This is unexpected. I tried hard but could not reproduce the problem here. Whatever I tried, the second solve correctly continued after the root node for me.
Could you create a minimal example to reproduce the problem?