how to declare a non linear objective function in pyomo? and efficient way of declaring constraints?
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I am trying to declare a non linear objective constraint in Pyomo and everytime I try to solve it using Bonmin Solver, I get the following error:
ERROR: Solver (asl) returned non-zero return code (3221225477)
ERROR: Solver log: Bonmin 1.8.6 using Cbc 2.9.9 and Ipopt 3.12.8 bonmin:
Traceback (most recent call last):
File "", line 2, in
results = opt.solve(model)
File "C:Anacondalibsite-packagespyomooptbasesolvers.py", line 626, in solve
"Solver (%s) did not exit normally" % self.name)
ApplicationError: Solver (asl) did not exit normally
#
My Objective function definition is:
`def obj_func(model):
global summer
summer = 0
global volumer
volumer = 0
for i in range(0,len(data)):
summer += model.x[i]*data.loc[i,'Predicted.Profit']
for j in range(0,len(data)):
volumer += model.x[j]*data.loc[j,'Predicted.Liters.Sold']
return summer/volumer`
I am invoking the solver through SolverFactory,
Lastly I would also like to know the most efficient way of creating constraints as i have constraints of the range 100+ and some of them are non linear. When i try to solve the problem using this model, my memory usage spikes to 100% and my computer hangs.
python constraints nonlinear-optimization pyomo
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
|
show 1 more comment
I am trying to declare a non linear objective constraint in Pyomo and everytime I try to solve it using Bonmin Solver, I get the following error:
ERROR: Solver (asl) returned non-zero return code (3221225477)
ERROR: Solver log: Bonmin 1.8.6 using Cbc 2.9.9 and Ipopt 3.12.8 bonmin:
Traceback (most recent call last):
File "", line 2, in
results = opt.solve(model)
File "C:Anacondalibsite-packagespyomooptbasesolvers.py", line 626, in solve
"Solver (%s) did not exit normally" % self.name)
ApplicationError: Solver (asl) did not exit normally
#
My Objective function definition is:
`def obj_func(model):
global summer
summer = 0
global volumer
volumer = 0
for i in range(0,len(data)):
summer += model.x[i]*data.loc[i,'Predicted.Profit']
for j in range(0,len(data)):
volumer += model.x[j]*data.loc[j,'Predicted.Liters.Sold']
return summer/volumer`
I am invoking the solver through SolverFactory,
Lastly I would also like to know the most efficient way of creating constraints as i have constraints of the range 100+ and some of them are non linear. When i try to solve the problem using this model, my memory usage spikes to 100% and my computer hangs.
python constraints nonlinear-optimization pyomo
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34
|
show 1 more comment
I am trying to declare a non linear objective constraint in Pyomo and everytime I try to solve it using Bonmin Solver, I get the following error:
ERROR: Solver (asl) returned non-zero return code (3221225477)
ERROR: Solver log: Bonmin 1.8.6 using Cbc 2.9.9 and Ipopt 3.12.8 bonmin:
Traceback (most recent call last):
File "", line 2, in
results = opt.solve(model)
File "C:Anacondalibsite-packagespyomooptbasesolvers.py", line 626, in solve
"Solver (%s) did not exit normally" % self.name)
ApplicationError: Solver (asl) did not exit normally
#
My Objective function definition is:
`def obj_func(model):
global summer
summer = 0
global volumer
volumer = 0
for i in range(0,len(data)):
summer += model.x[i]*data.loc[i,'Predicted.Profit']
for j in range(0,len(data)):
volumer += model.x[j]*data.loc[j,'Predicted.Liters.Sold']
return summer/volumer`
I am invoking the solver through SolverFactory,
Lastly I would also like to know the most efficient way of creating constraints as i have constraints of the range 100+ and some of them are non linear. When i try to solve the problem using this model, my memory usage spikes to 100% and my computer hangs.
python constraints nonlinear-optimization pyomo
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
I am trying to declare a non linear objective constraint in Pyomo and everytime I try to solve it using Bonmin Solver, I get the following error:
ERROR: Solver (asl) returned non-zero return code (3221225477)
ERROR: Solver log: Bonmin 1.8.6 using Cbc 2.9.9 and Ipopt 3.12.8 bonmin:
Traceback (most recent call last):
File "", line 2, in
results = opt.solve(model)
File "C:Anacondalibsite-packagespyomooptbasesolvers.py", line 626, in solve
"Solver (%s) did not exit normally" % self.name)
ApplicationError: Solver (asl) did not exit normally
#
My Objective function definition is:
`def obj_func(model):
global summer
summer = 0
global volumer
volumer = 0
for i in range(0,len(data)):
summer += model.x[i]*data.loc[i,'Predicted.Profit']
for j in range(0,len(data)):
volumer += model.x[j]*data.loc[j,'Predicted.Liters.Sold']
return summer/volumer`
I am invoking the solver through SolverFactory,
Lastly I would also like to know the most efficient way of creating constraints as i have constraints of the range 100+ and some of them are non linear. When i try to solve the problem using this model, my memory usage spikes to 100% and my computer hangs.
python constraints nonlinear-optimization pyomo
python constraints nonlinear-optimization pyomo
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
asked Nov 23 '18 at 15:33
linearprogrammerlinearprogrammer
156
156
Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34
|
show 1 more comment
Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34
Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
You should check the solver output to see why it's failing. You can print this output to the screen by adding the tee=True option when calling the solver:
SolverFactory('bonmin').solve(model, tee=True)
Also, when declaring Pyomo Constraints and Objectives, you should avoid using +=. It can often lead to significant performance degradation when building Pyomo expressions. Here is the recommended way to write your objective function:
def obj_func(model):
return sum(model.x[i]*data.loc[i,'Predicted.Profit'] for i in model.I)/sum(model.x[j]*data.loc[j,'Predicted.Liters.Sold'] for j in model.J)
model.Objective(rule=_obj_func)
where model.I and model.J are either Pyomo Set components or Python lists.
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
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oldest
votes
You should check the solver output to see why it's failing. You can print this output to the screen by adding the tee=True option when calling the solver:
SolverFactory('bonmin').solve(model, tee=True)
Also, when declaring Pyomo Constraints and Objectives, you should avoid using +=. It can often lead to significant performance degradation when building Pyomo expressions. Here is the recommended way to write your objective function:
def obj_func(model):
return sum(model.x[i]*data.loc[i,'Predicted.Profit'] for i in model.I)/sum(model.x[j]*data.loc[j,'Predicted.Liters.Sold'] for j in model.J)
model.Objective(rule=_obj_func)
where model.I and model.J are either Pyomo Set components or Python lists.
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
add a comment |
You should check the solver output to see why it's failing. You can print this output to the screen by adding the tee=True option when calling the solver:
SolverFactory('bonmin').solve(model, tee=True)
Also, when declaring Pyomo Constraints and Objectives, you should avoid using +=. It can often lead to significant performance degradation when building Pyomo expressions. Here is the recommended way to write your objective function:
def obj_func(model):
return sum(model.x[i]*data.loc[i,'Predicted.Profit'] for i in model.I)/sum(model.x[j]*data.loc[j,'Predicted.Liters.Sold'] for j in model.J)
model.Objective(rule=_obj_func)
where model.I and model.J are either Pyomo Set components or Python lists.
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
add a comment |
You should check the solver output to see why it's failing. You can print this output to the screen by adding the tee=True option when calling the solver:
SolverFactory('bonmin').solve(model, tee=True)
Also, when declaring Pyomo Constraints and Objectives, you should avoid using +=. It can often lead to significant performance degradation when building Pyomo expressions. Here is the recommended way to write your objective function:
def obj_func(model):
return sum(model.x[i]*data.loc[i,'Predicted.Profit'] for i in model.I)/sum(model.x[j]*data.loc[j,'Predicted.Liters.Sold'] for j in model.J)
model.Objective(rule=_obj_func)
where model.I and model.J are either Pyomo Set components or Python lists.
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
You should check the solver output to see why it's failing. You can print this output to the screen by adding the tee=True option when calling the solver:
SolverFactory('bonmin').solve(model, tee=True)
Also, when declaring Pyomo Constraints and Objectives, you should avoid using +=. It can often lead to significant performance degradation when building Pyomo expressions. Here is the recommended way to write your objective function:
def obj_func(model):
return sum(model.x[i]*data.loc[i,'Predicted.Profit'] for i in model.I)/sum(model.x[j]*data.loc[j,'Predicted.Liters.Sold'] for j in model.J)
model.Objective(rule=_obj_func)
where model.I and model.J are either Pyomo Set components or Python lists.
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
answered Nov 26 '18 at 22:23
Bethany NicholsonBethany Nicholson
1,1061412
1,1061412
add a comment |
add a comment |
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Look for the logs. They are written to some place. But it sounds just like a memory error and then it's hard to recommend someting different than solving a smaller model or buy ram. It looks like a ML task where general nlp solvers often do not scale.
– sascha
Nov 23 '18 at 19:01
What do you mean by "It looks like a ML task where general nlp solvers often do not scale"?
– linearprogrammer
Nov 23 '18 at 19:06
Are you doing machine learning with big data? Then dont expect to be able to use general nonlinear opt
– sascha
Nov 23 '18 at 19:12
No, I have dataset with size (178848,40). I am able to run CBC very easily but it is failing with BONMIN or COUENNE
– linearprogrammer
Nov 23 '18 at 19:19
That's pretty big and you should not compare linear opt with nonlinear opt.
– sascha
Nov 23 '18 at 19:34