Linear Program (LP) ―― Gurobi Optimizer

使用Gurobi optimizer求解…

$max f(X) = 2x_1 + 3x_2$

$x_1 + 2*x_2 \leq 8$
$4*x_1 \leq16$
$4*x_2 \leq 12$
$x_1, x_2 \geq 0$

# Import lib from gurobipy import * # Create model model_1 = Model() # Add decision variables to a model x1 = model_1.addVar(lb=0, vtype=GRB.CONTINUOUS, name="x1") x2 = model_1.addVar(lb=0, vtype=GRB.CONTINUOUS, name="x2") # Set objective function f1 = 2*x1 + 3*x2 model_1.setObjective(f1, GRB.MAXIMIZE) # Add constraints to a model ct1 = model_1.addConstr(lhs=x1+2*x2, sense=GRB.LESS_EQUAL, rhs=8, name='ct1') ct2 = model_1.addConstr(lhs=4*x1, sense=GRB.LESS_EQUAL, rhs=16, name="ct2") ct2 = model_1.addConstr(lhs=4*x2, sense=GRB.LESS_EQUAL, rhs=12, name="ct3") # Optimize model_1.optimize() # Print data if model_1.status == GRB.Status.OPTIMAL:     print('='*25)     # Print objVal     print('==> objVal:' ,model_1.objVal)     # Print var value     print('==> x1:', x1, 'x2:', x2)

Academic license - for non-commercial use only Optimize a model with 3 rows, 2 columns and 4 nonzeros Coefficient statistics:   Matrix range     [1e+00, 4e+00]   Objective range  [2e+00, 3e+00]   Bounds range     [0e+00, 0e+00]   RHS range        [8e+00, 2e+01] Presolve removed 2 rows and 0 columns Presolve time: 0.01s Presolved: 1 rows, 2 columns, 2 nonzeros  Iteration    Objective       Primal Inf.    Dual Inf.      Time        0    1.6000000e+01   2.000000e+00   0.000000e+00      0s        1    1.4000000e+01   0.000000e+00   0.000000e+00      0s  Solved in 1 iterations and 0.02 seconds Optimal objective  1.400000000e+01 ========================= ==> objVal: 14.0 ==> x1: <gurobi.Var x1 (value 4.0)> x2: <gurobi.Var x2 (value 2.0)>

$minf(X) = -3x_1 + x_2 + x_3$

$x_1 - 2x_2 + x_3 \leq 11$
$-4x_1 + x_2 + 2x_3 \geq 3$
$-2x_1 + x_3 = 1$
$x_1, x_2, x_3 \geq 0$

Add decision variables to a model in a loop.
Build a linear expression by using quicksum() function.

# Create model model_2 = Model() # Add decision variables to a model in a loop x_names = ['x1', 'x2', 'x3'] x = {} for x_name in x_names:     x[x_name] = model_2.addVar(name=x_name) # Set objective function f2 = -3*x['x1'] + x['x2'] + x['x3'] model_2.setObjective(f2, GRB.MINIMIZE) # Add constraints to a model model_2.addConstr(lhs=x['x1']-2*x['x2']+x['x3'], sense=GRB.LESS_EQUAL, rhs=11) model_2.addConstr(lhs=-4*x['x1']+x['x2']+2*x['x3'], sense=GRB.GREATER_EQUAL, rhs=3) model_2.addConstr(lhs=-2*x['x1']+x['x3'], sense=GRB.EQUAL, rhs=1) # Optimize model_2.optimize() # Print data if model_2.status == GRB.Status.OPTIMAL:     print('='*25)     #     print('==> Objective function:', f2)     # Print ObjVal     print('==> ObjVal:', model_2.ObjVal)     # Pirnt variable value     print('==> x1:',x['x1'], 'x2:',x['x2'], 'x3:',x['x3'])

Optimize a model with 3 rows, 3 columns and 8 nonzeros Coefficient statistics:   Matrix range     [1e+00, 4e+00]   Objective range  [1e+00, 3e+00]   Bounds range     [0e+00, 0e+00]   RHS range        [1e+00, 1e+01] Presolve removed 3 rows and 3 columns Presolve time: 0.01s Presolve: All rows and columns removed Iteration    Objective       Primal Inf.    Dual Inf.      Time        0   -2.0000000e+00   0.000000e+00   0.000000e+00      0s  Solved in 0 iterations and 0.01 seconds Optimal objective -2.000000000e+00 ========================= ==> Objective function: <gurobi.LinExpr: -3.0 x1 + x2 + x3> ==> ObjVal: -2.0 ==> x1: <gurobi.Var x1 (value 4.0)> x2: <gurobi.Var x2 (value 1.0)> x3: <gurobi.Var x3 (value 9.0)>

$minf(X) = 2x_1 + 3x_2 + 5x_3 + 2x_4 + 3x_5$

$x_1 + x_2 + 2x_3 + x_4 + 3x_5 \geq 4$
$2x_1 - x_2 + 3x_3 + x_4 + x_5 \geq 3$
$x_1, x_2, x_3, x_4, x_5 \geq 0$

Add decision variables to a model by using Model.addVars() function.
Build a linear expression by using quicksum() function.

# Creatw model model_3 = Model(name='model_3') # Add decision variables to a model by using Model.addVars() x_vector = [('x1'), ('x2'), ('x3'), ('x4'), ('x5')] x = model_3.addVars(x_vector, name=x_vector) # Set objective function f3 = quicksum([ 2*x['x1'], 3*x['x2'], 5*x['x3'], 2*x['x4'], 3*x['x5'] ]) model_3.setObjective(f3) # Add constraints to a model model_3.addConstr(lhs=quicksum([ 1*x['x1'], 1*x['x2'], 2*x['x3'], 1*x['x4'], 3*x['x5'] ]) , sense=GRB.GREATER_EQUAL, rhs=4) model_3.addConstr(lhs=quicksum([ 2*x['x1'], -1*x['x2'], 3*x['x3'], 1*x['x4'], 1*x['x5'] ]) , sense=GRB.GREATER_EQUAL, rhs=3) # OPtimize model_3.optimize() # Print data if model_3.status == GRB.Status.OPTIMAL:     print('='*25)     #      print('==> Objective function:', model_3.getObjective())     # ObjVal     print('==> ObjVal:', model_3.ObjVal)      # Print optimal variable value     for v in model_3.getVars():         print('==> ' , (v.varName, v.x) )

Optimize a model with 2 rows, 5 columns and 10 nonzeros Coefficient statistics:   Matrix range     [1e+00, 3e+00]   Objective range  [2e+00, 5e+00]   Bounds range     [0e+00, 0e+00]   RHS range        [3e+00, 4e+00] Presolve removed 0 rows and 1 columns Presolve time: 0.01s Presolved: 2 rows, 4 columns, 8 nonzeros  Iteration    Objective       Primal Inf.    Dual Inf.      Time        0    0.0000000e+00   1.750000e+00   0.000000e+00      0s        2    5.0000000e+00   0.000000e+00   0.000000e+00      0s  Solved in 2 iterations and 0.01 seconds Optimal objective  5.000000000e+00 ========================= ==> Objective function: <gurobi.LinExpr: 2.0 x1 + 3.0 x2 + 5.0 x3 + 2.0 x4 + 3.0 x5> ==> ObjVal: 5.0 ==>  ('x1', 1.0) ==>  ('x2', 0.0) ==>  ('x3', 0.0) ==>  ('x4', 0.0) ==>  ('x5', 1.0)

文章来源: https://blog.csdn.net/weixin_39124421/article/details/89033380

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linear