Julia - JuMP 变量数组上的 Gurobi 回调

如何解决Julia - JuMP 变量数组上的 Gurobi 回调

在 Gurobi 和 JuMP 0.21 中，有关如何通过回调访问变量的详细记录here：

using JuMP,Gurobi,Test

model = direct_model(Gurobi.Optimizer())
@variable(model,0 <= x <= 2.5,Int)
@variable(model,0 <= y <= 2.5,Int)
@objective(model,Max,y)
cb_calls = Cint[]
function my_callback_function(cb_data,cb_where::Cint)
    # You can reference variables outside the function as normal
    push!(cb_calls,cb_where)
    # You can select where the callback is run
    if cb_where != GRB_CB_MIPSOL && cb_where != GRB_CB_MIPNODE
        return
    end
    # You can query a callback attribute using GRBcbget
    if cb_where == GRB_CB_MIPNODE
        resultP = Ref{Cint}()
        GRBcbget(cb_data,cb_where,GRB_CB_MIPNODE_STATUS,resultP)
        if resultP[] != GRB_OPTIMAL
            return  # Solution is something other than optimal.
        end
    end
    # Before querying `callback_value`,you must call:
    Gurobi.load_callback_variable_primal(cb_data,cb_where)
    x_val = callback_value(cb_data,x)
    y_val = callback_value(cb_data,y)
    # You can submit solver-independent MathOptInterface attributes such as
    # lazy constraints,user-cuts,and heuristic solutions.
    if y_val - x_val > 1 + 1e-6
        con = @build_constraint(y - x <= 1)
        MOI.submit(model,MOI.LazyConstraint(cb_data),con)
    elseif y_val + x_val > 3 + 1e-6
        con = @build_constraint(y + x <= 3)
        MOI.submit(model,con)
    end
    if rand() < 0.1
        # You can terminate the callback as follows:
        GRBterminate(backend(model))
    end
    return
end
# You _must_ set this parameter if using lazy constraints.
MOI.set(model,MOI.RawParameter("LazyConstraints"),1)
MOI.set(model,Gurobi.CallbackFunction(),my_callback_function)
optimize!(model)
@test termination_status(model) == MOI.OPTIMAL
@test primal_status(model) == MOI.FEASIBLE_POINT
@test value(x) == 1
@test value(y) == 2

即，您将使用 x_val = callback_value(cb_data,x)。但是，当您有一个特定索引不是从 1 开始的变量数组时，您应该怎么做，即我的变量不在向量中，但由于以下原因声明：

@variable(m,x[i=1:n,j=i+1:n],Bin)

我是否应该在 x 的两个维度上使用双循环访问 callback_value 并多次调用 j？如果是这样，row_number() 的索引将不相同，不是吗？

解决方法

使用广播：

x_val = callback_value.(Ref(cb_data),x)

或者在需要该值时调用 callback_value(cb_data,x[i,j])。

例如：

using JuMP,Gurobi
model = Model(Gurobi.Optimizer)
@variable(model,0 <= x[i=1:3,j=i+1:3] <= 2.5,Int)
function my_callback_function(cb_data)
    x_val = callback_value.(Ref(cb_data),x)
    display(x_val)
    for i=1:3,j=i+1:3
        con = @build_constraint(x[i,j] <= floor(Int,x_val[i,j]))
        MOI.submit(model,MOI.LazyConstraint(cb_data),con)
    end
end
MOI.set(model,MOI.LazyConstraintCallback(),my_callback_function)
optimize!(model)

收益

julia> optimize!(model)
Gurobi Optimizer version 9.1.0 build v9.1.0rc0 (mac64)
Thread count: 4 physical cores,8 logical processors,using up to 8 threads
Optimize a model with 0 rows,3 columns and 0 nonzeros
Model fingerprint: 0x5d543c3a
Variable types: 0 continuous,3 integer (0 binary)
Coefficient statistics:
  Matrix range     [0e+00,0e+00]
  Objective range  [0e+00,0e+00]
  Bounds range     [2e+00,2e+00]
  RHS range        [0e+00,0e+00]
JuMP.Containers.SparseAxisArray{Float64,2,Tuple{Int64,Int64}} with 3 entries:
  [1,2]  =  -0.0
  [2,3]  =  -0.0
  [1,3]  =  -0.0
JuMP.Containers.SparseAxisArray{Float64,2]  =  2.0
  [2,3]  =  2.0
  [1,3]  =  2.0
JuMP.Containers.SparseAxisArray{Float64,3]  =  -0.0
Presolve time: 0.00s
Presolved: 0 rows,3 columns,0 nonzeros
Variable types: 0 continuous,3 integer (0 binary)
JuMP.Containers.SparseAxisArray{Float64,3]  =  -0.0
Found heuristic solution: objective 0.0000000

Explored 0 nodes (0 simplex iterations) in 0.14 seconds
Thread count was 8 (of 8 available processors)

Solution count 1: 0 

Optimal solution found (tolerance 1.00e-04)
Best objective 0.000000000000e+00,best bound 0.000000000000e+00,gap 0.0000%

User-callback calls 31,time in user-callback 0.14 sec