Check a model for contradictions that need no solver — check

Compares requirements that are FORCED by the declarations against ceilings that CANNOT be raised, per commodity, region and year. A finding is a contradiction in the data: no solver setting, starting point or tolerance will make such a model feasible, and an infeasible solve will not say which constraint was to blame.

Two checks run:

aux-from-stock: exogenous capacity (@capacity$stock) consumes an auxiliary commodity through @aeff$cap2ainp. The stock is given, so the consumption is unavoidable; if the commodity's supply is bounded and cannot be imported or traded in, a shortfall is a contradiction.
demand-vs-supply: a declared demand exceeds the bounded supply of its commodity in the same region and year, again with no import and no inbound route.

Usage

check_feasibility(object, years = NULL, verbose = TRUE)

Arguments

object: a model, scenario or repository.
years: milestone years to check. NULL (default) checks every year present in the declarations.
verbose: print the report. The findings are returned either way.

Value

A data.frame of findings – check, commodity, region, year, required, available, shortfall, forced_by – with zero rows when nothing is contradicted. Returned invisibly when verbose.

What a clean result does NOT mean

It does not mean the model is feasible. Everything that depends on a DECISION is the solver's question and is not asked here: whether capacity can serve the load, whether a bound binds once the model optimises, whether the dispatch closes in every timeslice, whether a corridor is big enough. This function proves the absence of one kind of contradiction, not the presence of a solution.

Examples

if (FALSE) { # \dontrun{
check_feasibility(mod)
# or, before a solve that would otherwise fail without saying why:
interpolate_model(mod, name = "BASE", check = TRUE)
} # }