Division Objective function in Pyscipopt
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I am trying to maximize the division of two quantities in PySCIPOpt.
Since it is a division of two linear metrics, it becomes a nonlinear optimization.
My code is somewhat like this:
model = Model()
x={}
for i in range(0,len(data)):
x[i] = model.addVar(vtype = 'B',name = 'x(%s)'%i)
data['Index'] = range(0,len(data))
profit = 0
volume = 0
for index in data['Index']:
profit += x[index] * data['Predicted.Profit'][index]
volume += x[index] * data['Predicted.Liters.Sold'][index]
model.setObjective(profit/volume,"maximize")
As soon as I hit the setObjective command, my kernel gets busy for more than 20 minutes and just keeps working. I want to know is there something wrong with my definition of Objective function? Could I be doing something else? Something more efficient?
By the way, my dataset's shape is 178848x36
edit
The function ran after twenty minutes! But gave an error saying : AssertionError: given coefficients are neither Expr or number but ProdExpr
How do i Bypass this error? Since I wish to optimize the division of decisionVariables! As well as make this objective function declaration more efficient?
Edit2
Is there a way to cast ProdExpr as Expr??
python division nonlinear-optimization scip
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
add a comment |
I am trying to maximize the division of two quantities in PySCIPOpt.
Since it is a division of two linear metrics, it becomes a nonlinear optimization.
My code is somewhat like this:
model = Model()
x={}
for i in range(0,len(data)):
x[i] = model.addVar(vtype = 'B',name = 'x(%s)'%i)
data['Index'] = range(0,len(data))
profit = 0
volume = 0
for index in data['Index']:
profit += x[index] * data['Predicted.Profit'][index]
volume += x[index] * data['Predicted.Liters.Sold'][index]
model.setObjective(profit/volume,"maximize")
As soon as I hit the setObjective command, my kernel gets busy for more than 20 minutes and just keeps working. I want to know is there something wrong with my definition of Objective function? Could I be doing something else? Something more efficient?
By the way, my dataset's shape is 178848x36
edit
The function ran after twenty minutes! But gave an error saying : AssertionError: given coefficients are neither Expr or number but ProdExpr
How do i Bypass this error? Since I wish to optimize the division of decisionVariables! As well as make this objective function declaration more efficient?
Edit2
Is there a way to cast ProdExpr as Expr??
python division nonlinear-optimization scip
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06
add a comment |
I am trying to maximize the division of two quantities in PySCIPOpt.
Since it is a division of two linear metrics, it becomes a nonlinear optimization.
My code is somewhat like this:
model = Model()
x={}
for i in range(0,len(data)):
x[i] = model.addVar(vtype = 'B',name = 'x(%s)'%i)
data['Index'] = range(0,len(data))
profit = 0
volume = 0
for index in data['Index']:
profit += x[index] * data['Predicted.Profit'][index]
volume += x[index] * data['Predicted.Liters.Sold'][index]
model.setObjective(profit/volume,"maximize")
As soon as I hit the setObjective command, my kernel gets busy for more than 20 minutes and just keeps working. I want to know is there something wrong with my definition of Objective function? Could I be doing something else? Something more efficient?
By the way, my dataset's shape is 178848x36
edit
The function ran after twenty minutes! But gave an error saying : AssertionError: given coefficients are neither Expr or number but ProdExpr
How do i Bypass this error? Since I wish to optimize the division of decisionVariables! As well as make this objective function declaration more efficient?
Edit2
Is there a way to cast ProdExpr as Expr??
python division nonlinear-optimization scip
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
I am trying to maximize the division of two quantities in PySCIPOpt.
Since it is a division of two linear metrics, it becomes a nonlinear optimization.
My code is somewhat like this:
model = Model()
x={}
for i in range(0,len(data)):
x[i] = model.addVar(vtype = 'B',name = 'x(%s)'%i)
data['Index'] = range(0,len(data))
profit = 0
volume = 0
for index in data['Index']:
profit += x[index] * data['Predicted.Profit'][index]
volume += x[index] * data['Predicted.Liters.Sold'][index]
model.setObjective(profit/volume,"maximize")
As soon as I hit the setObjective command, my kernel gets busy for more than 20 minutes and just keeps working. I want to know is there something wrong with my definition of Objective function? Could I be doing something else? Something more efficient?
By the way, my dataset's shape is 178848x36
edit
The function ran after twenty minutes! But gave an error saying : AssertionError: given coefficients are neither Expr or number but ProdExpr
How do i Bypass this error? Since I wish to optimize the division of decisionVariables! As well as make this objective function declaration more efficient?
Edit2
Is there a way to cast ProdExpr as Expr??
python division nonlinear-optimization scip
python division nonlinear-optimization scip
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
edited Nov 24 '18 at 12:54
linearprogrammer
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
asked Nov 24 '18 at 12:12
linearprogrammerlinearprogrammer
156
156
@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06
add a comment |
@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06
@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06
add a comment |
1 Answer
1
active
oldest
votes
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. For PySCIPOpt, you need to do that step manually so avoid the creation of new variables and constraints behind the user's back. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set withsetParam('limits/gap', <value>).
– mattmilten
Nov 26 '18 at 11:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. For PySCIPOpt, you need to do that step manually so avoid the creation of new variables and constraints behind the user's back. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set withsetParam('limits/gap', <value>).
– mattmilten
Nov 26 '18 at 11:31
add a comment |
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. For PySCIPOpt, you need to do that step manually so avoid the creation of new variables and constraints behind the user's back. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set withsetParam('limits/gap', <value>).
– mattmilten
Nov 26 '18 at 11:31
add a comment |
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. For PySCIPOpt, you need to do that step manually so avoid the creation of new variables and constraints behind the user's back. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. For PySCIPOpt, you need to do that step manually so avoid the creation of new variables and constraints behind the user's back. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
answered Nov 25 '18 at 13:44
mattmiltenmattmilten
2,33212043
2,33212043
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set withsetParam('limits/gap', <value>).
– mattmilten
Nov 26 '18 at 11:31
add a comment |
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set withsetParam('limits/gap', <value>).
– mattmilten
Nov 26 '18 at 11:31
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
While introduce these steps to my code, I have one more question for you. In your experience, Can SCIP converge into a solution for a nonlinear problem with 200k integer decision variables?
– linearprogrammer
Nov 26 '18 at 10:49
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
Impossible to say. You have to give it a try. If it doesn't terminate but produces good enough solutions, you might want to set a gap limit.
– mattmilten
Nov 26 '18 at 11:05
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
By gap limit, do you mean the number of iterations?
– linearprogrammer
Nov 26 '18 at 11:25
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set with
setParam('limits/gap', <value>).– mattmilten
Nov 26 '18 at 11:31
No, I mean the gap limit as defined here: scip.zib.de/doc/html/FAQ.php#gapdefinition . Can be set with
setParam('limits/gap', <value>).– mattmilten
Nov 26 '18 at 11:31
add a comment |
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@mattmitten any insights??
– linearprogrammer
Nov 25 '18 at 8:20
In general, SCIP does not work with nonlinear objective functions. They are converted into a constraint and a new auxiliary variable works as objective. See here: github.com/SCIP-Interfaces/PySCIPOpt/issues/167
– mattmilten
Nov 25 '18 at 13:41
Please don't forget to accept my answer if it is sufficient.
– mattmilten
Nov 26 '18 at 11:06