Add exterior of ellipsoid example (#329)

* Add exterior of ellipsoid example

* Fix

* Fix

* Fixes

Author: blegat

docs/src/tutorials/Polynomial Optimization/bilinear.jl

@@ -58,5 +58,6 @@ end
 # The first level of the hierarchy gives a lower bound of `2100`
 
 model2 = solve(2)
+nothing # hide
 @test objective_value(model2) ≈ 2100 rtol=1e-4 #src
 @test termination_status(model2) == MOI.OPTIMAL #src

docs/src/tutorials/Polynomial Optimization/ellipsoid.jl

@@ -0,0 +1,81 @@
+# # Exterior of ellipsoid
+
+#md # [![](https://mybinder.org/badge_logo.svg)](@__BINDER_ROOT_URL__/generated/Polynomial Optimization/ellipsoid.ipynb)
+#md # [![](https://img.shields.io/badge/show-nbviewer-579ACA.svg)](@__NBVIEWER_ROOT_URL__/generated/Polynomial Optimization/ellipsoid.ipynb)
+# **Adapted from**: Section 3.5 of [F99] and Table 5.1 of [Las09]
+#
+# [F99] Floudas, Christodoulos A. et al.
+# *Handbook of Test Problems in Local and Global Optimization.*
+# Nonconvex Optimization and Its Applications (NOIA, volume 33).
+#
+# [Las09] Lasserre, J. B.
+# *Moments, positive polynomials and their applications*
+# World Scientific, **2009**.
+
+# ## Introduction
+
+# Consider the polynomial optimization problem [F99, Section 3.5]
+
+A = [
+     0  0  1
+     0 -1  0
+    -2  1 -1
+]
+bz = [3, 0, -4] - [0, -1, -6]
+y = [1.5, -0.5, -5]
+
+using Test #src
+using DynamicPolynomials
+@polyvar x[1:3]
+p = -2x[1] + x[2] - x[3]
+using SumOfSquares
+e = A * x - y
+f = e'e - bz'bz / 4
+K = @set sum(x) <= 4 && 3x[2] + x[3] <= 6 && f >= 0 && 0 <= x[1] && x[1] <= 2 && 0 <= x[2] && 0 <= x[3] && x[3] <= 3
+
+# We will now see how to find the optimal solution using Sum of Squares Programming.
+# We first need to pick an SDP solver, see [here](https://jump.dev/JuMP.jl/v1.12/installation/#Supported-solvers) for a list of the available choices.
+
+import Clarabel
+solver = Clarabel.Optimizer
+
+# A Sum-of-Squares certificate that $p \ge \alpha$ over the domain `S`, ensures that $\alpha$ is a lower bound to the polynomial optimization problem.
+# The following function searches for the largest lower bound and finds zero using the `d`th level of the hierarchy`.
+
+function solve(d)
+    model = SOSModel(solver)
+    @variable(model, α)
+    @objective(model, Max, α)
+    @constraint(model, c, p >= α, domain = K, maxdegree = d)
+    optimize!(model)
+    println(solution_summary(model))
+    return model
+end
+
+# The first level of the hierarchy gives a lower bound of `-7``
+
+model2 = solve(2)
+nothing # hide
+@test objective_value(model2) ≈ -6 rtol=1e-4 #src
+@test termination_status(model2) == MOI.OPTIMAL #src
+
+# The second level improves the lower bound
+
+model4 = solve(4)
+nothing # hide
+@test objective_value(model4) ≈ -74/13 rtol=1e-4 #src
+@test termination_status(model4) == MOI.OPTIMAL #src
+
+# The third level improves it even further
+
+model6 = solve(6)
+nothing # hide
+@test objective_value(model6) ≈ -4.06848 rtol=1e-4 #src
+@test termination_status(model6) == MOI.OPTIMAL #src
+
+# The fourth level finds the optimal objective value as lower bound.
+
+model8 = solve(8)
+nothing # hide
+@test objective_value(model8) ≈ -4 rtol=1e-4 #src
+@test termination_status(model8) == MOI.OPTIMAL #src

docs/src/tutorials/Polynomial Optimization/min_univariate.jl

@@ -66,18 +66,21 @@ end
 # The first level of the hierarchy gives a lower bound of `-7``
 
 model4 = solve(4)
+nothing # hide
 @test objective_value(model4) ≈ -7 rtol=1e-4 #src
 @test termination_status(model4) == MOI.OPTIMAL #src
 
 # The second level improves the lower bound
 
 model5 = solve(5)
+nothing # hide
 @test objective_value(model5) ≈ -20/3 rtol=1e-4 #src
 @test termination_status(model5) == MOI.OPTIMAL #src
 
 # The third level finds the optimal objective value as lower bound...
 
 model7 = solve(7)
+nothing # hide
 @test objective_value(model7) ≈ -5.5080 rtol=1e-4 #src
 @test termination_status(model7) == MOI.OPTIMAL #src