Transitioning from MathProgBase

MathOptInterface is a replacement for MathProgBase.jl. However, it is not a direct replacement.

Transitioning a solver interface

MathOptInterface is more extensive than MathProgBase which may make its implementation seem daunting at first. There are however numerous utilities in MathOptInterface that the simplify implementation process.

For more information, read Implementing a solver interface.

Transitioning the high-level functions

MathOptInterface doesn't provide replacements for the high-level interfaces in MathProgBase. We recommend you use JuMP as a modeling interface instead.

linprog

Here is one way of transitioning from linprog:

using JuMPfunction linprog(c, A, sense, b, l, u, solver)    N = length(c)    model = Model(solver)    @variable(model, l[i] <= x[i=1:N] <= u[i])    @objective(model, Min, c' * x)    eq_rows, ge_rows, le_rows = sense .== '=', sense .== '>', sense .== '<'    @constraint(model, A[eq_rows, :] * x .== b[eq_rows])    @constraint(model, A[ge_rows, :] * x .>= b[ge_rows])    @constraint(model, A[le_rows, :] * x .<= b[le_rows])    optimize!(model)    return (        status = termination_status(model),        objval = objective_value(model),        sol = value.(x)    )end

mixintprog

Here is one way of transitioning from mixintprog:

using JuMPfunction mixintprog(c, A, rowlb, rowub, vartypes, lb, ub, solver)    N = length(c)    model = Model(solver)    @variable(model, lb[i] <= x[i=1:N] <= ub[i])    for i in 1:N        if vartypes[i] == :Bin            set_binary(x[i])        elseif vartypes[i] == :Int            set_integer(x[i])        end    end    @objective(model, Min, c' * x)    @constraint(model, rowlb .<= A * x .<= rowub)    optimize!(model)    return (        status = termination_status(model),        objval = objective_value(model),        sol = value.(x)    )end

quadprog

Here is one way of transitioning from quadprog:

using JuMPfunction quadprog(c, Q, A, rowlb, rowub, lb, ub, solver)    N = length(c)    model = Model(solver)    @variable(model, lb[i] <= x[i=1:N] <= ub[i])    @objective(model, Min, c' * x + 0.5 * x' * Q * x)    @constraint(model, rowlb .<= A * x .<= rowub)    optimize!(model)    return (        status = termination_status(model),        objval = objective_value(model),        sol = value.(x)    )end