Relating transcripted variable with original variable

[DarioSlaifsteinSk]

I'm working with an InfiniteModel and an infeasibility pops up from time to time (random generated data).
KNITRO informs me which variable is the infeasible in the TranscriptionModel here's the print i get from it:

=======================================
           Academic License
       (NOT FOR COMMERCIAL USE)
         Artelys Knitro 13.2.0
=======================================

ERROR: Infeasible variable bound deduced from presolve.
       Variable: x[3997] <==== THIS IS WHAT I NEED
       deduced upper bound =     1.19067145110921e+01 is less than
       deduced lower bound =     1.23712359239063e+01

EXIT: Problem determined to be infeasible with respect to variable bounds.

Final Statistics
----------------
Final objective value               =   1.10033609600000e+07
Final feasibility error (abs / rel) =   2.22e+04 / 1.00e+00
Final optimality error  (abs / rel) =   1.80e+308 / 1.80e+308
# of iterations                     =          0
# of CG iterations                  =          0
# of function evaluations           =          0
# of gradient evaluations           =          0
# of Hessian evaluations            =          0
Total program time (secs)           =       0.105 (     0.016 CPU time)
Time spent in evaluations (secs)    =       0.000

===============================================================================

how do I know the VarRef this x[3997] is associated to? probably something with MathOptInterface right?
I'm gonna post the same thing in the JuMP forum or in the discourse.
Cheers,

[pulsipher]

Yes, the solver works with the transcribed JuMP/MOI model. We provide mappings from the infinite variables to the underlying transcribed JuMP variables, but not vice versa. However, it should be relatively easy to figure out in this case. To get the name of the variable which will let us know which infinite variable and support it corresponds to, we can try something like:

using InfiniteOpt, KNITRO

inf_model = InfiniteModel(KNITRO.Optimizer)
# MODEL GOES HERE

jump_model = optimizer_model(inf_model)
knitro_idx = 3997 # assuming the x[3997] is a naming convention by KNITRO that corresponds to the MOI variable index
jump_vref = VariableRef(jump_model, MOI.VariableIndex(knitro_idx)
println(name(jump_vref))

Based on the printed name, we should find the name of the infinite variable and the corresponding support index.

[DarioSlaifsteinSk]

Okay, so after a bit of back and forth with the formulation (removing what was necessary and so on) this is what I came up with.
The latex_formulation(model) is:

$$ \begin{aligned} das \end{aligned} $$

$$\begin{aligned} \min\quad \mathbb{E}_{t_a \in \mathcal{T}}[\mathcal{C}_{i}] + CAPEX . Q_{bess}\\\ \text{Subject to}\\\ & \underline{P}_g \leq Pg_{i}(t) \leq \overline{P}_g, \quad\forall t \in \mathcal{D}_t \quad \forall i \in [1,N_s]\\\ & 0.01 \leq Q_{bess} \leq 50\\\ & -Q_{bess} .C_{rate} \leq P_{bess,i}(t) \leq Q_{bess}.C_{rate}, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & 0.2 Q_{bess} \leq SoC_{bess,i}(t) \leq 0.95 Q_{bess}, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & \underline{P}_{n_{EV}} \leq P_{n_{EV},i}(t) \leq \overline{P}_{n_{EV}}, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & \underline{P}_{n_{EV}} \leq P_{Tot,n_{EV},i}(t) \leq \overline{P}_{n_{EV}}, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & 11.13 \leq SoC_{n_{EV},i}(t) \leq 52.87, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & SoC_{bess,i}(0.25) = 0.5 Q_{bess} \quad \forall i \in [1,N_s]\\\ & SoC_{n_{EV}=1,i}(t_0) = 33.3899 kWh \quad \forall i \in [1,N_s]\\\ & SoC_{n_{EV}=2,i}(t_0) = 44.5199 kWh \quad \forall i \in [1,N_s]\\\ & \frac{\partial}{\partial t}\left[SoC_{bess,i}(t)\right] + P_{bess,i}(t) = 0, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & \frac{\partial}{\partial t}\left[SoC_{n_{EV},i}(t)\right] + P_{Tot,n_{EV},i}(t) = 0, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\\ & γ_{n_{EV},i}(t) P_{n_{EV},i}(t) + P_{drive,n_{EV},i}(t) - P_{Tot,n_{EV},i}(t) = 0, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s]\\ & \sum_{n_{EV}=1}^{N_{EV}} γ_{f,n_{EV},i}(t).P_{n_{EV},i}(t) + P_{pvMPPT}(t) + P_{bess,i}(t) + P_{g,i}(t) - Ple(t) = 0, \quad\forall t \in \mathcal{D}_t, \forall i \in [1,N_s] \end{aligned}$$

Fixing the battery capacity $Q_{bess}$ to its lower limit and optimizing with KNITRO we get:

=======================================
           Academic License
       (NOT FOR COMMERCIAL USE)        
         Artelys Knitro 14.0.0
=======================================

Knitro presolver has deduced that constraint c[2726] cannot be satisfied.

EXIT: Problem determined to be infeasible with respect to constraint bounds.

Final Statistics
----------------
Final objective value               =   3.04942460000000e+02
Final feasibility error (abs / rel) =   4.72e+00 / 1.00e+00
Final optimality error  (abs / rel) =   1.80e+308 / 1.80e+308
# of iterations                     =          0
# of CG iterations                  =          0
# of function evaluations           =          0
# of gradient evaluations           =          0
# of Hessian evaluations            =          0
Total program time (secs)           =       0.047 (     0.000 CPU time)
Time spent in evaluations (secs)    =       0.000

===============================================================================

So instead of a variable the problem is in a constraint. The debug is quite easy but slightly different:

using JuMP, MathOptInterface, Infi
model = InfiniteModel(KNITRO.Optimizer)
SolvePolicies!(model, data)
jump_model = optimizer_model(model)
julia>  A JuMP Model
Minimization problem with:
Variables: 4609
Objective function type: AffExpr
`AffExpr`-in-`MathOptInterface.EqualTo{Float64}`: 3456 constraints
`AffExpr`-in-`MathOptInterface.GreaterThan{Float64}`: 384 constraints
`AffExpr`-in-`MathOptInterface.LessThan{Float64}`: 1152 constraints
`VariableRef`-in-`MathOptInterface.EqualTo{Float64}`: 1 constraint
`VariableRef`-in-`MathOptInterface.GreaterThan{Float64}`: 2688 constraints
`VariableRef`-in-`MathOptInterface.LessThan{Float64}`: 2688 constraints
Model mode: AUTOMATIC
CachingOptimizer state: ATTACHED_OPTIMIZER
Solver name: Knitro

knitro_idx = 2726; # <== the index we want
# we get the list of constraints
cst_jump = vcat([all_constraints(jump_model, i...) for i in list_of_constraint_types(jump_model)]...);
con_dict = Dict(con => optimizer_index(con).value for con in cst_jump); # and we relate it with the indeces in the optimizer
findall(i -> i==knitro_idx, con_dict) # find the constraints related to our c[2726]
julia> 
3-element Vector{ConstraintRef{Model, C, ScalarShape} where C}:
 SoCev[1,2](support: 37) <= 52.86749999999998
 SoCev[1,1](support: 35) - SoCev[1,1](support: 36) + 0.25 d/dt[SoCev[1,1](t)](support: 36) == 0
 SoCev[1,2](support: 37) >= 11.129999999999995

So, quite a silly modeling mistake of passing a driving power/energy requirement higher than the EV battery capacity.
So just by chaging the $SoC$ limits to $\underline{SoC}, \overline{SoC} = [10 %, 100 %]$ the issue is solved.

Eventhough the solution was easy I haven't seen many places were the full link of InfiniteModel() --> JuMP.model() --> solverModel() is made so this might be useful for newcomers.

[pulsipher]

Glad you figured it out. Noted on the docs. The upcoming release has a fully reworked solution backend and docs to hopefully make things more straightforward.