Very weird output of binary variable?
回答済みAfter optimizing successfully, when I transfer the binary variables as input parameters to a logic function (‘and’, ‘or’ operation etc.), the output is different from when I transfer the values of these binary variables to the exact same function?
The result should be the same! as they are the same inputs to the same function, don't understand why, please help if you know, many thanks.
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Hi,
Can you show us some sample code with outputs of what you are doing? Also what programming language are you using? Are you sure you are using the same precision for these values/comparisons?
Best regards,
Sonja
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Hi Sonja:
Thanks for your reply, I have copied a screen shot here. From the screen shot, we can find that:
print(ttt['A']['A'], x_ij['B']['C'])
the output is: <gurobi.Var T_A_A (value 0.0)> <gurobi.Var x_ij_B_C_ (value 1.0)>
Both are BINARY variables, one var's value is 0, another is 1, while when I calculate the 'or' of these two variable, the output value is still 0?
x = ttt['A']['A'] or x_ij['B']['C']
print(x)output is: <gurobi.Var T_A_A (value 0.0)>
Why is it not 1, as '0 or 1' should be 1?
And if I print(ttt['A']['A'].x or x_ij['B']['C'].x), it is 1 as we expected.
Thanks
Bradley
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My logic experssion and operation is: T[i][j] = T[i][j] or ( T[i][k] and T[k][j] ) with i, j, k in T.keys() loops
T[i][j] are binary variables, it looks like Guroby does not support such logic operation?
If not, how to implement such operation? many thanks.
Bradley
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Hi Bradley,
As you noticed, you need to use the X attributes of Var objects to retrieve their values after optimizing. If you don't use this attribute, you are instead comparing the Python objects themselves. Based on how Python handles these object comparisons, the statement
x = ttt['A']['A'] or x_ij['B']['C']
will assign the Var object ttt['A']['A'] to the Python variable x (as long as ttt['A']['A'] is not None).
Note that this statement does not assign a value of 0 or 1 to x. Rather, it assigns the entire Var object ttt['A']['A'] to x. When you print a Var object after optimizing, the solution value (X attribute) is included in the print statement for convenience.
I hope this helps. Thanks!
Eli
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Hi Eli,
I understand how to get the value of a variable, while what I want is add constraint to T[i][j], T[i][j] are binary variables, and T[i][j] = T[i][j] or ( T[i][k] and T[k][j] ) with i, j, k in T.keys() loops.
Any suggestions on how to implement such logic operation in Gurobi? thanks.
Bradley
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Hi Bradley,
Can you clarify what you are trying to do? It sounds like you want to add a constraint that T[i][j] is equal to one of two values: either T[i][j], or T[i][k]*T[k][j].
If this is your intent, note that it is not meaningful to have a constraint that would set a variable equal to itself. This constraint is always satisfied.
The following linear constraints set a binary variable x equal to the product of two binary variables y and z:
x >= y + z - 1
x <= y
x <= zIf y and/or z are 0, these constraints ensure x equals 0. If y and z both equal 1, then x also equals 1. In your case, this would look like:
model.addConstr(T[i][j] >= T[i][k] + T[k][j])
model.addConstr(T[i][j] <= T[i][k])
model.addConstr(T[i][j] <= T[k][j])However, if you add these constraints for every combination of i, j, and k, you are saying you want T[i][j] to equal the product T[i][k]*T[k][j] for all k; I'm not sure this is what you want.
Thanks!
Eli
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