Linear and Quadratic terms in CPLEX Objective Function

Question

I would like to minimize an objective function which is rather simple, but I am somehow having problems making the correct calls to from the Python API to CPLEX

I looked at how to use set_quadratic and set_quadratic_coefficients here but that didn't result in a solution to my problem.

My objective function has a set of linear variables and a set of quadratic variables

varCoefs = [1]*(numB + numQ)
varLower = [0]*(numB + numQ)
varNames = [(x,"b%s"%x) for x in range( numB )]
varNames += [(len(varNames) + x,"q%s"%x) for x in range( numQ )]

varCoefs += [10]*len(deltas)
varLower += [1]*len(deltas)
varNames += [(len(varNames) + x,"delta%s"%x) for x in range( len(deltas) )]

varCoefs += [0]*len(target.v)
varLower += [0]*len(target.v)

sContent = [(len(varNames) + x,"s%s"%x) for x in range( len(target.v) )]
varNames += sContent

varCoefs += [-1]
varLower += [0]
varNames += [(len(varNames),'mu')]


problem.variables.add(obj = varCoefs, lb = varLower)
problem.variables.set_names(varNames)

# problem.objective.set_quadratic_coefficients([[['s%s' % x], [1]] for x in range( len(target.v) )])

problem.objective.set_quadratic(
    [cplex.SparsePair(ind=[sContent[x][0]], val=[1]) for x in range( len(target.v) )]
    )

Everything works up to the last call add the quadratic terms. At which point CPLEX throws the following error CPLEX Error 1226: Array entry 13919 not ascending. twice, ignoring the command, and the Python code continues.

I looked up the error, but that didn't seem to help me either.

I did try re-writing the above to add the variables by name and lower bound first... and then call set_linear and set_quadratic afterward, but that doesn't help either.

What am I missing here?

rkersh · Answer 1 · 2016-08-15T18:31:38.880

If you're calling set_quadratic with a quadratic objective function that is separable, it corresponds to CPXXcopyqpsep. If you're calling set_quadratic with a quadratic objective function that is inseparable, it corresponds to CPXXcopyquad. I agree that the error you're getting is not particularly useful, but it makes a bit more sense if you know where it's coming from in the Callable C Library.

With that said, here's a complete example, using your snippet, with some dummy inputs:

import cplex

class MockTarget(object):
    pass

# Dummy data for testing

numB = 3
numQ = 3
deltas = [0.1, 0.1, 0.1]
problem = cplex.Cplex()

target = MockTarget()
target.v = [1, 2, 3]

# Build the problem

varCoefs = [1]*(numB + numQ)
varLower = [0]*(numB + numQ)
varNames = [(x,"b%s"%x) for x in range( numB )]
varNames += [(len(varNames) + x,"q%s"%x) for x in range( numQ )]

varCoefs += [10]*len(deltas)
varLower += [1]*len(deltas)
varNames += [(len(varNames) + x,"delta%s"%x) for x in range( len(deltas) )]

varCoefs += [0]*len(target.v)
varLower += [0]*len(target.v)

sContent = [(len(varNames) + x,"s%s"%x) for x in range( len(target.v) )]
varNames += sContent

varCoefs += [-1]
varLower += [0]
varNames += [(len(varNames),'mu')]


problem.variables.add(obj = varCoefs, lb = varLower)
problem.variables.set_names(varNames)

# Print without quadratic terms so you can see the progression.
problem.write('test1.lp')

# Separable Q

qsepvec = []
for tpl in varNames:
    if tpl in sContent:
        qsepvec.append(1.0)
    else:
        qsepvec.append(0.0)
print qsepvec

problem.objective.set_quadratic(qsepvec)

problem.write('test2.lp')

# Inseparable Q (overwrites previous Q)

qmat = []
for tpl in varNames:
    if tpl in sContent:
        sp = cplex.SparsePair(ind=[tpl[0]], val=[1.0])
        qmat.append(sp)
    else:
        sp = cplex.SparsePair(ind=[], val=[])
        qmat.append(sp)
print qmat

problem.objective.set_quadratic(qmat)

problem.write('test3.lp')

I've written that out in long form rather than using list comprehensions to make it a bit more clear. The contents of the the LP files are below:

test1.lp:

\ENCODING=ISO-8859-1
\Problem name: 

Minimize
 obj: b0 + b1 + b2 + q0 + q1 + q2 + 10 delta0 + 10 delta1 + 10 delta2 + 0 s0
      + 0 s1 + 0 s2 - mu
Bounds
      delta0 >= 1
      delta1 >= 1
      delta2 >= 1
End

test2.lp

\ENCODING=ISO-8859-1
\Problem name: 

Minimize
 obj: b0 + b1 + b2 + q0 + q1 + q2 + 10 delta0 + 10 delta1 + 10 delta2 + 0 s0
      + 0 s1 + 0 s2 - mu + [ s0 ^2 + s1 ^2 + s2 ^2 ] / 2
Bounds
      delta0 >= 1
      delta1 >= 1
      delta2 >= 1
End

test3.lp

\ENCODING=ISO-8859-1
\Problem name: 

Minimize
 obj: b0 + b1 + b2 + q0 + q1 + q2 + 10 delta0 + 10 delta1 + 10 delta2 + 0 s0
      + 0 s1 + 0 s2 - mu + [ s0 ^2 + s1 ^2 + s2 ^2 ] / 2
Bounds
      delta0 >= 1
      delta1 >= 1
      delta2 >= 1
End

You can see that test2.lp and test3.lp are the same (the later overwrites the former, but does the same thing). Hopefully that makes it a bit easier to understand. In general, using this technique of printing out LP's for very simple problems, is one of the more useful debugging techniques.

You should also check out the python examples that are shipped with CPLEX. For example, qpex1.py, miqpex1.py, indefqpex1.py.

Thanks for the explanation. However, there are two issues with your solution: test1 contains no quadratic terms, test3 is actually a replicate of test 2 since you aren't actually passing `qmat` to `set_quadratic` and if you did pass `qmat` you'd obtain the same error as I have. However, I already solved the problem a different way. See solution below. — Constantine, Aug 15 '16 at 18:03
I exported test1.lp with no quadratic constraints on purpose; so you could see the progression. Sorry about my mistake at the end there. I'll correct that. — rkersh, Aug 15 '16 at 18:07
Thanks again, but I really dislike the idea of having to explicitly check conditionals in order to create a large data structure with thousands of empty elements. — Constantine, Aug 15 '16 at 23:44
You can certainly optimize this ... I made no attempt to do so; opting for clarity instead. Do you need help with that? Do you have a separable Q or not? — rkersh, Aug 15 '16 at 23:55
Like I said, I solved the problem by adding the quadratic terms first and then the linear, thus avoid creating any unnecessary data structures, or checking conditionals. Thanks yet again. — Constantine, Aug 15 '16 at 23:58
Sure, that makes sense. I was just checking. Selecting your own answer as the accepted answer seems appropriate IMO. — rkersh, Aug 16 '16 at 00:04

Constantine · Accepted Answer · 2016-08-15T18:16:16.543

I solved the problem by adding the quadratic terms first, setting their coefficients, and then adding the linear terms in a separate call see below.

problem.objective.set_sense(problem.objective.sense.minimize)

varLower = [0]*len(target.v)
varNames = ["s%s"%x for x in range( len(target.v) )]

problem.variables.add(names=varNames, lb=varLower)

problem.objective.set_quadratic(
    [[[x],[1]] for x in range( len(target.v) )]
    )

varCoefs = [-1]
varLower = [0]
varNames = ['mu']


varCoefs += [1]*(numB + numQ)
varLower += [0]*(numB + numQ)
varNames += ["b%s"%x for x in range( numB )]
varNames += ["q%s"%x for x in range( numQ )]

varCoefs += [10]*len(deltas)
varLower += [1]*len(deltas)
varNames += ["delta%s"%x for x in range( len(deltas) )]

problem.variables.add(names=varNames, lb=varLower, obj=varCoefs)

However, I would still like to know why it works this way, and not the other way.

I believe the reason this works is because, as the documentation says for `set_quadratic`, you have to call it "with a list with length equal to the number of variables in the problem." — rkersh, Aug 15 '16 at 18:55

Linear and Quadratic terms in CPLEX Objective Function

2 Answers2