# 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 should hopefully make the implementation as simple or simpler than MathProgBase.

## 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.

Tip

If you haven't used JuMP before, start with the tutorial Getting started with JuMP

### linprog

Here is one way of transitioning from linprog:

using JuMP

function 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 JuMP

function 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

Here is one way of transitioning from quadprog:

using JuMP

function 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