Hungarian Algorithm with Unbalanced Matrix

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Febrina Febrina

• December 03, 2021 16:22

I have an unbalanced matrix and I have added a dummy column,

I have run the code like below, but with these constraints, I didn't get the expected result

Where I need each station (ST) will definitely get 1 person (P) with the lowest value, it's okay if the person didn't get any station.

    ST1  ST2  ST3  D
P1   5    3    2   3
P2   6    2    1   9
P3   3    1    6   2
P4   8    4    5   2

#Defining the Variable
X = {}
for i in I:
for j in J:
X[i,j] = m.addVar(vtype= GRB.BINARY)

#Objective Function
m.setObjective(quicksum(durmin[i][j]*X[i,j] for i in I for j in J), GRB.MINIMIZE)

#Constraint-1
for i in I:
m.addConstr(quicksum(X[i,j] for j in J) <= 1)
#Constraint-2
for j in J:
m.addConstr(quicksum(X[i,j] for i in I) >= 1)

0

• Official comment

  

  Simranjit Kaur

  Gurobi Staff

  • December 06, 2024 00:22

  This post is more than three years old. Some information may not be up to date. For current information, please check the Gurobi Documentation or Knowledge Base. If you need more help, please create a new post in the community forum. Or why not try our AI Gurobot?.
• 

  Jaromił Najman

  Gurobi Staff

  • December 06, 2021 09:45

  Could you please elaborate more on the task you are trying to achieve?

  What are the sets \(I\) and \(J\)? What is \(\texttt{durmin}\)? What is the expected result? Why would you need to add a dummy column? Is it to guarantee uniqueness of the solution?

  Best regards,
  Jaromił

  0

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