User-defined operators with vector outputs

This tutorial was generated using Literate.jl. Download the source as a .jl file.

The purpose of this tutorial is to demonstrate how to write a user-defined operator with a vector-valued output.

Required packages

This tutorial uses the following packages:

using JuMP
import Ipopt
import Test

Motivation

A common situation is to have a user-defined operator like the following that returns multiple outputs (we define function_calls to keep track of how many times we call this method):

function_calls = 0
function foo(x, y)
    global function_calls += 1
    common_term = x^2 + y^2
    term_1 = sqrt(1 + common_term)
    term_2 = common_term
    return term_1, term_2
end

foo (generic function with 1 method)

For example, the first term might be used in the objective, and the second term might be used in a constraint, and often they share work that is expensive to evaluate.

This is a problem for JuMP, because it requires user-defined operators to return a single number. One option is to define two separate functions, the first returning the first argument, and the second returning the second argument.

foo_1(x, y) = foo(x, y)[1]
foo_2(x, y) = foo(x, y)[2]

foo_2 (generic function with 1 method)

However, if the common term is expensive to compute, this approach is wasteful because it will evaluate the expensive term twice. Let's have a look at how many times we evaluate x^2 + y^2 during a solve:

model = Model(Ipopt.Optimizer)
set_silent(model)
@variable(model, x[1:2] >= 0, start = 0.1)
@operator(model, op_foo_1, 2, foo_1)
@operator(model, op_foo_2, 2, foo_2)
@objective(model, Max, op_foo_1(x[1], x[2]))
@constraint(model, op_foo_2(x[1], x[2]) <= 2)
function_calls = 0
optimize!(model)
@assert is_solved_and_feasible(model)
Test.@test objective_value(model) ≈ √3 atol = 1e-4
Test.@test value.(x) ≈ [1.0, 1.0] atol = 1e-4
println("Naive approach: function calls = $(function_calls)")

Naive approach: function calls = 44

Memoization

An alternative approach is to use memoization, which uses a cache to store the result of function evaluations. We can write a memoization function as follows:

"""
    memoize(foo::Function, n_outputs::Int)

Take a function `foo` and return a vector of length `n_outputs`, where element
`i` is a function that returns the equivalent of `foo(x...)[i]`.

To avoid duplication of work, cache the most-recent evaluations of `foo`.
Because `foo_i` is auto-differentiated with ForwardDiff, our cache needs to
work when `x` is a `Float64` and a `ForwardDiff.Dual`.
"""
function memoize(foo::Function, n_outputs::Int)
    last_x, last_f = nothing, nothing
    last_dx, last_dfdx = nothing, nothing
    function foo_i(i, x::T...) where {T<:Real}
        if T == Float64
            if x !== last_x
                last_x, last_f = x, foo(x...)
            end
            return last_f[i]::T
        else
            if x !== last_dx
                last_dx, last_dfdx = x, foo(x...)
            end
            return last_dfdx[i]::T
        end
    end
    return [(x...) -> foo_i(i, x...) for i in 1:n_outputs]
end

Main.memoize

Let's see how it works. First, construct the memoized versions of foo:

memoized_foo = memoize(foo, 2)

2-element Vector{Main.var"#4#7"{Int64, Main.var"#foo_i#5"{typeof(Main.foo)}}}:
 #4 (generic function with 1 method)
 #4 (generic function with 1 method)

Now try evaluating the first element of memoized_foo.

function_calls = 0
memoized_foo[1](1.0, 1.0)
println("function_calls = ", function_calls)

function_calls = 1

As expected, this evaluated the function once. However, if we call the function again, we hit the cache instead of needing to re-compute foo and function_calls is still 1!

memoized_foo[1](1.0, 1.0)
println("function_calls = ", function_calls)

function_calls = 1

Now let's see how this works during a real solve:

model = Model(Ipopt.Optimizer)
set_silent(model)
@variable(model, x[1:2] >= 0, start = 0.1)
@operator(model, op_foo_1, 2, memoized_foo[1])
@operator(model, op_foo_2, 2, memoized_foo[2])
@objective(model, Max, op_foo_1(x[1], x[2]))
@constraint(model, op_foo_2(x[1], x[2]) <= 2)
function_calls = 0
optimize!(model)
@assert is_solved_and_feasible(model)
Test.@test objective_value(model) ≈ √3 atol = 1e-4
Test.@test value.(x) ≈ [1.0, 1.0] atol = 1e-4
println("Memoized approach: function_calls = $(function_calls)")

Memoized approach: function_calls = 22

Compared to the naive approach, the memoized approach requires half as many function evaluations.