Jump using `max(x...)` where x is a matrix?

Specific Domains Optimization (Mathematical)
,
1

I need to revive a previously asked question. I have a problem that involves evaluating the exp function over an array of variables and I want to avoid huge values by centering those variables before appplying exp. I get a bounds error:

using JuMP
using Ipopt

m = Model(Ipopt.Optimizer)
fmax(y...) = max(y...)
JuMP.register(m, :fmax, N, fmax, autodiff=true)
@variable(m,x[i=1:2, j=1:3])
@NLexpression(m, xbar[i=1:2, j=1:3], x[i,j] - fmax(x...))  # normalize by largest value in x
@NLexpression(m, p[i=1:2, j=1:3], exp(xbar[i,j]) / sum(xbar[k,l] for k=1:2, l=1:3))
@NLobjective(m, Max, sum(p[i,j] for i=1:2, j=1:3))
optimize!(m)

ERROR: BoundsError: attempt to access 2-element Vector{Float64} at index [1:6]

it is strange because it seems the subexpression is built correctly:

julia> m[:xbar]
2×3 Matrix{NonlinearExpression}:
 subexpression[1]: x[1,1] - fmax(x[1,1], x[2,1], x[1,2], x[2,2], x[1,3], x[2,3])  …  subexpression[5]: x[1,3] - fmax(x[1,1], x[2,1], x[1,2], x[2,2], x[1,3], x[2,3])
 subexpression[2]: x[2,1] - fmax(x[1,1], x[2,1], x[1,2], x[2,2], x[1,3], x[2,3])     subexpression[6]: x[2,3] - fmax(x[1,1], x[2,1], x[1,2], x[2,2], x[1,3], x[2,3])

or, at least to me that looks like exactly what I wanted to achieve. What 2-element vector are we talking abou here?
thanks!

2

When I tried to run this, it became clear that the value of N is not defined.
I think you have N = 2 set somewhere.
Setting N = 3 gives 3 in place of 2 here:
ERROR: BoundsError: attempt to access 3-element Vector{Float64} at index [1:6]
while setting N = 6 misses this error but ultimately returns from IPOPT:
EXIT: Invalid number in NLP function or derivative detected.

1 Like
3

Not the answer to your question but

I don’t have my laptop right now, but it seems to me maximum(x) is more appropriate than max(x...)

2 Likes
4

Oh! How dumb, N = 6 in my case!
And no, it’s precisely impossible to use maximum that’s why we have to splat the vector into max

5

Are you sure that fmax(y…) = maximum(y) won’t work?

6

sure.

using JuMP
using Ipopt

m = Model(Ipopt.Optimizer)
N = 6
fmax(y...) = maximum(y...)
JuMP.register(m, :fmax, N, fmax, autodiff=true)
@variable(m,x[i=1:2, j=1:3])
@NLexpression(m, xbar[i=1:2, j=1:3], x[i,j] - fmax(x...))  # normalize by largest value in x
@NLexpression(m, p[i=1:2, j=1:3], exp(xbar[i,j]) / sum(xbar[k,l] for k=1:2, l=1:3))
@NLobjective(m, Max, sum(p[i,j] for i=1:2, j=1:3))
optimize!(m)

ERROR: Unable to register the function :fmax because it does not support differentiation via ForwardDiff.

Common reasons for this include:

 * the function assumes `Float64` will be passed as input, it must work for any
   generic `Real` type.
 * the function allocates temporary storage using `zeros(3)` or similar. This
   defaults to `Float64`, so use `zeros(T, 3)` instead.
7

I am wondering is it possible to use the max relaxation in this context?

Something like this,

@variable(m, xmax)
@variable(m, x[i=1:2, j=1:3])
for x_var in x
    @constraint(m, x_var <= xmax)
end