AMPL tutorial

This tutorial shows how to solve a model written in AMPL with Penelopt.jl, using AmplNLReader.jl.

Inequality Constraints

Penelopt.jl solves problems of the form minimize f(x) s.t. c(x) = 0. If your AMPL model has inequality constraints, the solver will fail.

1. The AMPL model

In this example, we use the Hock–Schittkowski problem HS6.

var x1 := -1.2;
var x2 := 1;

minimize obj: (1 - x1)^2;

subject to c1: 10 * (x2 - x1^2) = 0;

We save this in a hs6.mod file.

2. Generate the .nl file

AMPL compiles a model into a .nl file that solvers read directly, without needing AMPL itself at solve time:

$ ampl -oghs6 assets/hs6.mod

This produces file hs6.nl.

3. Read the model into Julia

using AmplNLReader

nlp = AmplModel(joinpath(@__DIR__, "assets", "hs6.nl"))

4. Solve with Penelopt

using Penelopt

stats = L2Penalty(nlp; print_level = 1)
┌ Info: 
│ This is Penelopt.jl v0.1.0.
│ Running with linear solver LDLFactorizations.jl v0.10.2.
│ 
│   Problem name: hs6
│    All variables: ████████████████████ 2      All constraints: ████████████████████ 1
│             free: ████████████████████ 2                 free: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│            lower: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0                lower: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│            upper: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0                upper: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│          low/upp: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0              low/upp: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│            fixed: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0                fixed: ████████████████████ 1
│           infeas: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0               infeas: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│             nnzh: ( 66.67% sparsity)   1               linear: ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 0
│                                                     nonlinear: ████████████████████ 1
│                                                          nnzj: (  0.00% sparsity)   2
│                                                      lin_nnzj: (------% sparsity)
│                                                      nln_nnzj: (  0.00% sparsity)   2
│ 
└ 
[ Info: ------------------------------------------------------------------------------------------------------
[ Info: Iter   sIter  Objective       pfeas       dfeas       τ           ptol        dtol        ‖x‖
[ Info: ------------------------------------------------------------------------------------------------------
[ Info: 0      0      +4.8400000e+00  8.15e+00    4.40e+00    1.00e+00    1.00e+00    8.15e-02    1.56e+00
[ Info: 1      7      +5.9658917e-03  2.13e-01    5.60e-02    1.00e+00    2.13e-03    5.60e-04    1.24e+00
[ Info: 2      2      +4.0311094e-07  1.30e-03    4.59e-04    1.00e+00    1.30e-05    4.59e-06    1.41e+00
[ Info: 3      2      +2.3024446e-15  9.81e-10    3.85e-08    1.00e+00    8.05e-08    1.36e-07    1.41e+00
┌ Info: 
│ Number of Iterations: 3
│ 
│ 
│ Objective...........: +2.302444565942107e-15
│ Primal Feasibility..:  9.806964129666085e-10
│ Dual Feasibility....:  3.847362065860079e-08
│ 
│ 
└ EXIT: first_order.
println("status    : ", stats.status)
println("objective : ", stats.objective)
println("solution  : ", stats.solution)
status    : first_order
objective : 2.3024445659421074e-15
solution  : [0.9999999520162052, 0.999999903934343]

See the options reference for the full list of keyword arguments accepted by the solver.