Pulp constraint with absolute value >= minimum ignored

Question

I am trying to constrain my problem by making the absolute difference of two LpVariables more than 0. Is there anything obvious i am missing in my implementation?

This is an implementation i found on http://lpsolve.sourceforge.net/5.1/absolute.htm which ive implemented it in the code below. I tried changing the value for M but it does not make a difference.

When i enter the actual result into the formula (such that q1x-q2x = 0) i find that this does not fulfill the constraints. However pulp says the solution is optimal.

prob = LpProblem("myProblem", LpMinimize)

q1x = LpVariable("q1x",1,8)

q2x = LpVariable("q2x",1,8) 

B12 = LpVariable("B12",0,1)

M=8

prob += (q1x-q2x) + M * B12 >= 1

prob += -(q1x-q2x) + M * (1-B12) >= 1

prob += q1x+q1y

Expected :

 q1x = 1; q2x = 2 or "infeasible"

Actual:

q1x = 1; q2x = 1 "optimal"

Answer 1

The very first "constraint" that is added to the LpProblem class will always be considered the objective function, as described in this test problem.

Also i am assuming you want your decision variables q1x and q2x to be integer which you have to be explicit about when defining your variables

Try this:

prob = LpProblem('myProblem', LpMinimize)

q1x = LpVariable("q1x",lowBound = 1, upBound = 8, cat = "Integer")

q2x = LpVariable("q2x",lowBound = 1, upBound = 8, cat = "Integer")

B12 = LpVariable("B12", lowBound = 0, upBound = 1, cat = "Integer")

# B12 can be also be defined by: 
#B12 = LpVariable("B12", cat = "Binary")

M = 8

prob += q1x + q2x 

prob += (q1x-q2x) + M * B12 >= 1

prob += -(q1x-q2x) + M * (1-B12) >= 1



prob.solve()

for v in prob.variables():
    print(v.name, " = ",  v.varValue)
print("Status:", LpStatus[prob.status])