All of the examples can be found in Jupyter notebook form here.

Basic Usage

using Convex
using LinearAlgebra
if VERSION < v"1.2.0-DEV.0"
    (I::UniformScaling)(n::Integer) = Diagonal(fill(I.λ, n))
end

using SCS
# passing in verbose=0 to hide output from SCS
solver = SCSSolver(verbose=0)

SCS.SCSSolver(Base.Iterators.Pairs(:verbose=>0))

Linear program

\[\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & A x \leq b\\ & x \geq 1 \\ & x \leq 10 \\ & x_2 \leq 5 \\ & x_1 + x_4 - x_2 \leq 10 \\ \end{array}\]

x = Variable(4)
c = [1; 2; 3; 4]
A = I(4)
b = [10; 10; 10; 10]
p = minimize(dot(c, x)) # or c' * x
p.constraints += A * x <= b
p.constraints += [x >= 1; x <= 10; x[2] <= 5; x[1] + x[4] - x[2] <= 10]
solve!(p, solver)

println(round(p.optval, digits=2))
println(round.(x.value, digits=2))
println(evaluate(x[1] + x[4] - x[2]))

10.0
[1.0; 1.0; 1.0; 1.0]
[1.0]

Matrix Variables and promotions

\[\begin{array}{ll} \mbox{minimize} & \| X \|_F + y \\ \mbox{subject to} & 2 X \leq 1\\ & X' + y \geq 1 \\ & X \geq 0 \\ & y \geq 0 \\ \end{array}\]

X = Variable(2, 2)
y = Variable()
# X is a 2 x 2 variable, and y is scalar. X' + y promotes y to a 2 x 2 variable before adding them
p = minimize(norm(X) + y, 2 * X <= 1, X' + y >= 1, X >= 0, y >= 0)
solve!(p, solver)
println(round.(X.value, digits=2))
println(y.value)
p.optval

1.0000000000226728

Norm, exponential and geometric mean

\[\begin{array}{ll} \mbox{satisfy} & \| x \|_2 \leq 100 \\ & e^{x_1} \leq 5 \\ & x_2 \geq 7 \\ & \sqrt{x_3 x_4} \geq x_2 \end{array}\]

x = Variable(4)
p = satisfy(norm(x) <= 100, exp(x[1]) <= 5, x[2] >= 7, geomean(x[3], x[4]) >= x[2])
solve!(p, SCSSolver(verbose=0))
println(p.status)
x.value

4×1 Array{Float64,2}:
  0.0              
  9.792376428098947
 14.180784526931312
 14.180784526931312

SDP cone and Eigenvalues

y = Semidefinite(2)
p = maximize(lambdamin(y), tr(y)<=6)
solve!(p, SCSSolver(verbose=0))
p.optval

2.999999993193094

x = Variable()
y = Variable((2, 2))
# SDP constraints
p = minimize(x + y[1, 1], isposdef(y), x >= 1, y[2, 1] == 1)
solve!(p, solver)
y.value

2×2 Array{Float64,2}:
 1.06177e-5  1.0      
 1.0         2.93819e6

Mixed integer program

\[\begin{array}{ll} \mbox{minimize} & \sum_{i=1}^n x_i \\ \mbox{subject to} & x \in \mathbb{Z}^n \\ & x \geq 0.5 \\ \end{array}\]

using GLPKMathProgInterface
x = Variable(4, :Int)
p = minimize(sum(x), x >= 0.5)
solve!(p, GLPKSolverMIP())
x.value

4×1 Array{Float64,2}:
 1.0
 1.0
 1.0
 1.0

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