Linear Programming Formulation

First, you will learn how to solve the following Linear Program using Gurobi Python interface.

$$ \begin{aligned} \text{Min} \,\, 10x + 26y & \\ 11x + 3y & \ge 21 \\ 6x + 20y & \ge 39 \\ x \ge 0 ,& y \ge 0 \\ x, y \in \mathbb{Z} \end{aligned} $$

The following should be the first line of any of your python + gurobi programs. This is to ensure that the program can access the gurobi functions and classes.

In [1]:
from gurobipy import *

Next we create a new gurobi model and name it lp. Note that you can change the name of the model ls as well as name of the gurobi object which in this case is m.

In [2]:
# Create a new Gurobi Model
m = Model("mip")

Next we create two positive variables $x$ and $y$ and add it to the model.

In [3]:
# Create two new variables
x = m.addVar(lb=0, vtype=GRB.INTEGER, name ="x")
y = m.addVar(lb=0, vtype=GRB.INTEGER, name ="y")

We then add the objective function to the model. The objective function in this case is minimize. If you wanted to maximize, you would set the objective function parameter to GRB.MAXIMIZE.

In [4]:
# Set the objective function
m.setObjective(10*x + 26*y, GRB.MINIMIZE)

We then add two constraints named c0 and c1. The earlier versions of gurobi have a m.update() command which is not needed in version 7.

In [5]:
#Add Constraints
m.addConstr(11*x + 3*y >= 21, "c0")
m.addConstr(6*x + 20*y >= 39, "c1")
Out[5]:
<gurobi.Constr *Awaiting Model Update*>

Next we solve the model.

In [6]:
# Solve the model
m.optimize()

Optimize a model with 2 rows, 2 columns and 4 nonzeros
Variable types: 0 continuous, 2 integer (0 binary)
Coefficient statistics:
  Matrix range     [3e+00, 2e+01]
  Objective range  [1e+01, 3e+01]
  Bounds range     [0e+00, 0e+00]
  RHS range        [2e+01, 4e+01]
Found heuristic solution: objective 182
Presolve time: 0.01s
Presolved: 2 rows, 2 columns, 4 nonzeros
Variable types: 0 continuous, 2 integer (0 binary)

Root relaxation: objective 5.526733e+01, 2 iterations, 0.00 seconds

    Nodes    |    Current Node    |     Objective Bounds      |     Work
 Expl Unexpl |  Obj  Depth IntInf | Incumbent    BestBd   Gap | It/Node Time

     0     0   55.26733    0    2  182.00000   55.26733  69.6%     -    0s
H    0     0                      66.0000000   55.26733  16.3%     -    0s
     0     0     cutoff    0        66.00000   64.00007  3.03%     -    0s

Cutting planes:
  MIR: 1

Explored 0 nodes (3 simplex iterations) in 0.21 seconds
Thread count was 4 (of 4 available processors)

Solution count 2: 66 182 
Pool objective bound 66

Optimal solution found (tolerance 1.00e-04)
Best objective 6.600000000000e+01, best bound 6.600000000000e+01, gap 0.0000%

If the model was solved to optimality, we print the objective function and optimal solution. We show two ways to access the optimal solution below.

In [7]:
if m.status == GRB.Status.OPTIMAL:
    print('Obj Function:', m.objVal)
    for v in m.getVars():
        print(v.varName, v.x)
# Another way to print the variable
    print("Optimal Solution:")
    print(x.varName, x.x)
    print(y.varName, y.x)        
else:
    print(m.status)

Obj Function: 66.0
x 4.0
y 1.0
Optimal Solution:
x 4.0
y 1.0
In [ ]: