Простые примеры

function example_rosenbrock()
    model = Model(Ipopt.Optimizer)
    set_silent(model)
    @variable(model, x)
    @variable(model, y)
    @objective(model, Min, (1 - x)^2 + 100 * (y - x^2)^2)
    optimize!(model)
    Test.@test is_solved_and_feasible(model)
    Test.@test objective_value(model) ≈ 0.0 atol = 1e-10
    Test.@test value(x) ≈ 1.0
    Test.@test value(y) ≈ 1.0
    return
end

example_rosenbrock()

function example_clnlbeam()
    N = 1000
    h = 1 / N
    alpha = 350
    model = Model(Ipopt.Optimizer)
    @variables(model, begin
        -1 <= t[1:(N+1)] <= 1
        -0.05 <= x[1:(N+1)] <= 0.05
        u[1:(N+1)]
    end)
    @objective(
        model,
        Min,
        sum(
            0.5 * h * (u[i+1]^2 + u[i]^2) +
            0.5 * alpha * h * (cos(t[i+1]) + cos(t[i])) for i in 1:N
        ),
    )
    @constraint(
        model,
        [i = 1:N],
        x[i+1] - x[i] - 0.5 * h * (sin(t[i+1]) + sin(t[i])) == 0,
    )
    @constraint(
        model,
        [i = 1:N],
        t[i+1] - t[i] - 0.5 * h * u[i+1] - 0.5 * h * u[i] == 0,
    )
    optimize!(model)
    println("""
    termination_status = $(termination_status(model))
    primal_status      = $(primal_status(model))
    objective_value    = $(objective_value(model))
    """)
    Test.@test is_solved_and_feasible(model)
    return
end

example_clnlbeam()

This is Ipopt version 3.14.16, running with linear solver MUMPS 5.7.3.

Number of nonzeros in equality constraint Jacobian...:     8000
Number of nonzeros in inequality constraint Jacobian.:        0
Number of nonzeros in Lagrangian Hessian.............:     4002

Total number of variables............................:     3003
                     variables with only lower bounds:        0
                variables with lower and upper bounds:     2002
                     variables with only upper bounds:        0
Total number of equality constraints.................:     2000
Total number of inequality constraints...............:        0
        inequality constraints with only lower bounds:        0
   inequality constraints with lower and upper bounds:        0
        inequality constraints with only upper bounds:        0

iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls
   0  3.5000000e+02 0.00e+00 0.00e+00  -1.0 0.00e+00    -  0.00e+00 0.00e+00   0
   1  3.5000000e+02 0.00e+00 0.00e+00  -1.7 0.00e+00    -  1.00e+00 1.00e+00   0
   2  3.5000000e+02 0.00e+00 0.00e+00  -3.8 0.00e+00  -2.0 1.00e+00 1.00e+00T  0
   3  3.5000000e+02 0.00e+00 0.00e+00  -5.7 0.00e+00   0.2 1.00e+00 1.00e+00T  0
   4  3.5000000e+02 0.00e+00 0.00e+00  -8.6 0.00e+00  -0.2 1.00e+00 1.00e+00T  0

Number of Iterations....: 4

                                   (scaled)                 (unscaled)
Objective...............:   3.5000000000000318e+02    3.5000000000000318e+02
Dual infeasibility......:   0.0000000000000000e+00    0.0000000000000000e+00
Constraint violation....:   0.0000000000000000e+00    0.0000000000000000e+00
Variable bound violation:   0.0000000000000000e+00    0.0000000000000000e+00
Complementarity.........:   2.5059035596802450e-09    2.5059035596802450e-09
Overall NLP error.......:   2.5059035596802450e-09    2.5059035596802450e-09


Number of objective function evaluations             = 5
Number of objective gradient evaluations             = 5
Number of equality constraint evaluations            = 5
Number of inequality constraint evaluations          = 0
Number of equality constraint Jacobian evaluations   = 5
Number of inequality constraint Jacobian evaluations = 0
Number of Lagrangian Hessian evaluations             = 4
Total seconds in IPOPT                               = 0.042

EXIT: Optimal Solution Found.
termination_status = LOCALLY_SOLVED
primal_status      = FEASIBLE_POINT
objective_value    = 350.0000000000032

function example_mle()
    n = 1_000
    Random.seed!(1234)
    data = randn(n)
    model = Model(Ipopt.Optimizer)
    set_silent(model)
    @variable(model, μ, start = 0.0)
    @variable(model, σ >= 0.0, start = 1.0)
    @objective(
        model,
        Max,
        n / 2 * log(1 / (2 * π * σ^2)) -
        sum((data[i] - μ)^2 for i in 1:n) / (2 * σ^2)
    )
    optimize!(model)
    @assert is_solved_and_feasible(model)
    println("μ             = ", value(μ))
    println("mean(data)    = ", Statistics.mean(data))
    println("σ^2           = ", value(σ)^2)
    println("var(data)     = ", Statistics.var(data))
    println("MLE objective = ", objective_value(model))
    Test.@test value(μ) ≈ Statistics.mean(data) atol = 1e-3
    Test.@test value(σ)^2 ≈ Statistics.var(data) atol = 1e-2
    # Можно также находить ограниченную ОМП.
    @constraint(model, μ == σ^2)
    optimize!(model)
    @assert is_solved_and_feasible(model)
    Test.@test value(μ) ≈ value(σ)^2
    println()
    println("With constraint μ == σ^2:")
    println("μ                         = ", value(μ))
    println("σ^2                       = ", value(σ)^2)
    println("Constrained MLE objective = ", objective_value(model))
    return
end

example_mle()

μ             = -0.0215521734290741
mean(data)    = -0.021552173429074114
σ^2           = 1.100101397871862
var(data)     = 1.1012026004695599
MLE objective = -1466.6397109231782

With constraint μ == σ^2:
μ                         = 0.6621385003734601
σ^2                       = 0.66213850037346
Constrained MLE objective = -1896.4889420749978

function example_qcp()
    model = Model(Ipopt.Optimizer)
    set_silent(model)
    @variable(model, x)
    @variable(model, y >= 0)
    @variable(model, z >= 0)
    @objective(model, Max, x)
    @constraint(model, x + y + z == 1)
    @constraint(model, x * x + y * y - z * z <= 0)
    @constraint(model, x * x - y * z <= 0)
    optimize!(model)
    Test.@test is_solved_and_feasible(model)
    print(model)
    println("Objective value: ", objective_value(model))
    println("x = ", value(x))
    println("y = ", value(y))
    Test.@test objective_value(model) ≈ 0.32699 atol = 1e-5
    Test.@test value(x) ≈ 0.32699 atol = 1e-5
    Test.@test value(y) ≈ 0.25707 atol = 1e-5
    return
end

example_qcp()

Max x
Subject to
 x + y + z = 1
 x² + y² - z² ≤ 0
 x² - y*z ≤ 0
 y ≥ 0
 z ≥ 0
Objective value: 0.32699283491387243
x = 0.32699283491387243
y = 0.2570658388068964