python - 如何使用 PuLP 的 Gurobi 求解器设置 MIP 启动(初始解决方案)？

Question

我正在使用 Python 中的 PuLP 模块来制定混合整数程序。我正在尝试研究如何通过 PuLP 接口(interface)设置 MIP 启动(即程序启动的可行解决方案)。

有关如何设置 MIP 开始 的详细信息 here

PuLP 包的开发者声称您可以通过PuLP 接口(interface)访问完整的 Gurobi 模型 here

下面粘贴了两个完整的模型。我已将它们做得尽可能小，同时防止 gurobi 求解器使用启发式算法找到最佳值。

我试图在两个模型中设置一个初始解(最优值)，但在 PuLP 模型中它被忽略了，但在 gurobipy 模型中它被忽略了按预期工作。

如何通过 PuLP 界面设置 Gurobi 求解的初始解？

from pulp import *

prob = LpProblem("min example",LpMinimize)

x1=LpVariable("x1",0,None,LpInteger)
x2=LpVariable("x2",0,None,LpInteger)
x3=LpVariable("x3",0,None,LpInteger)
x4=LpVariable("x4",0,None,LpInteger)

# Objective function
prob += 3*x1 + 5*x2 + 6*x3 + 9*x4

# A constraint
prob += -2*x1 + 6*x2 -3*x3 + 4*x4 >= 2, "Con1"
prob += -5*x1 + 3*x2 + x3 + 3*x4 >= -2, "Con2"
prob += 5*x1 - x2 + 4*x3 - 2*x4 >= 3, "Con3"

# Choose solver, and set it to problem, and build the Gurobi model
solver = pulp.GUROBI()
prob.setSolver(solver)
prob.solver.buildSolverModel(prob)

# Attempt to set an initial feasible solution (in this case to an optimal solution)
prob.solverModel.getVars()[0].start = 1
prob.solverModel.getVars()[1].start = 1
prob.solverModel.getVars()[2].start = 0
prob.solverModel.getVars()[3].start = 0

# Solve model
prob.solve()

# Status of the solution is printed to the screen
print "Status:", LpStatus[prob.status]

# Each of the variables is printed with it's resolved optimum value
for v in prob.variables():
    print v.name, "=", v.varValue

# Optimised objective function value is printed to the screen
print "OF = ", value(prob.objective)

哪个返回:

Optimize a model with 3 rows, 4 columns and 12 nonzeros
Coefficient statistics:
  Matrix range    [1e+00, 6e+00]
  Objective range [3e+00, 9e+00]
  Bounds range    [0e+00, 0e+00]
  RHS range       [2e+00, 3e+00]
Found heuristic solution: objective 12
Presolve removed 0 rows and 1 columns
Presolve time: 0.00s
Presolved: 3 rows, 3 columns, 9 nonzeros
Variable types: 0 continuous, 3 integer (0 binary)

Root relaxation: objective 7.400000e+00, 1 iterations, 0.00 seconds

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

     0     0    7.40000    0    1   12.00000    7.40000  38.3%     -    0s
H    0     0                       8.0000000    7.40000  7.50%     -    0s

Explored 0 nodes (1 simplex iterations) in 0.00 seconds
Thread count was 8 (of 8 available processors)

Optimal solution found (tolerance 1.00e-04)
Best objective 8.000000000000e+00, best bound 8.000000000000e+00, gap 0.0%
('Gurobi status=', 2)
Status: Optimal
x1 = 1.0
x2 = 1.0
x3 = -0.0
x4 = -0.0
OF =  8.0

其次，我可以使用 gurobipy 模块实现相同的模型，但在这种情况下，实际使用的是 MIP 启动:

from gurobipy import *

m = Model("min example")
m.modelSense = GRB.MINIMIZE

objFcnCoeffs = [3, 5, 6, 9]
xVars = []
for i in range(4):
    xVars.append(m.addVar(vtype=GRB.INTEGER, obj=objFcnCoeffs[i], name="Open%d" % i))

# Update model to integrate new variables
m.update()

# Constraints
m.addConstr(-2*xVars[0] + 6*xVars[1] -3*xVars[2] + 4*xVars[3] >= 2, "Con1")
m.addConstr(-5*xVars[0] + 3*xVars[1] + xVars[2] + 3*xVars[3] >= -2, "Con2")
m.addConstr(5*xVars[0] - xVars[1] + 4*xVars[2] - 2*xVars[3] >= 3, "Con3")


# Attempt to set an initial feasible solution (in this case to an optimal solution)
startValues = [1, 1, 0, 0]
for i in range(4):
    xVars[i].start = startValues[i]

# Solve model
m.optimize()

# Print solution
print('\nTOTAL COSTS: %g' % m.objVal)
for i in range(4):
    print('\n xVar[%s] = %g' % i, xVars[i])

哪个返回:

Optimize a model with 3 rows, 4 columns and 12 nonzeros
Coefficient statistics:
  Matrix range    [1e+00, 6e+00]
  Objective range [3e+00, 9e+00]
  Bounds range    [0e+00, 0e+00]
  RHS range       [2e+00, 3e+00]
Found heuristic solution: objective 12
Presolve removed 0 rows and 1 columns
Presolve time: 0.00s
Presolved: 3 rows, 3 columns, 9 nonzeros

Loaded MIP start with objective 8

Variable types: 0 continuous, 3 integer (0 binary)

Root relaxation: infeasible, 0 iterations, 0.00 seconds

Explored 0 nodes (0 simplex iterations) in 0.00 seconds
Thread count was 8 (of 8 available processors)

Optimal solution found (tolerance 1.00e-04)
Best objective 8.000000000000e+00, best bound 8.000000000000e+00, gap 0.0%

TOTAL COSTS: 8

 xVar[0] = 1

 xVar[1] = 1

 xVar[2] = 0

 xVar[3] = 0

Answer 1

你正在像这样设置起始值

prob.solverModel.getVars()[0].start = 1

然后您将通过此调用求解模型

prob.solve().

原始的prob没有改变，如果你调用

prob.solver.callSolver(prob)

Gurobi 将使用起始向量。