Solving many scenarios¶

How to solve one model for several sets of data in one call, and read the answers as one table. The reference is Sweeps and rolling horizons.

Put the scenarios in one table¶

Give the table that differs a scenario column that the spec does not declare. EachCoordinate solves once for each label:

import polars as pl

import specsolve as sps

generators = pl.read_csv('examples/tutorial/data/generators.csv')
load = pl.read_csv('examples/tutorial/data/load.csv')
sources = {
    'generator': generators['generator'],
    'snapshot': load['snapshot'],
    'p_max': generators.select('generator', value='p_max'),
    'cost': generators.select('generator', value='cost'),
}

loads = pl.concat(
    [
        load.with_columns(scenario=pl.lit('mild')),
        load.with_columns(scenario=pl.lit('cold'), value=1.3 * pl.col('value')),
    ]
)
scenarios = sps.solve_over('examples/dispatch.yaml', sources | {'load': loads}, sps.EachCoordinate('scenario'))
print(scenarios.record.select('scenario', 'objective'))

shape: (2, 2)
┌──────────┬───────────┐
│ scenario ┆ objective │
│ ---      ┆ ---       │
│ str      ┆ f64       │
╞══════════╪═══════════╡
│ cold     ┆ 16530.0   │
│ mild     ┆ 10500.0   │
└──────────┴───────────┘

A table without the column is the same in every scenario.

Read the answers¶

Every reader adds the scenario column:

gas = scenarios.primal('p').filter(pl.col('generator') == 'gas')
print(gas.pivot(on='scenario', index='snapshot', values='value'))

shape: (4, 3)
┌──────────┬───────┬───────┐
│ snapshot ┆ cold  ┆ mild  │
│ ---      ┆ ---   ┆ ---   │
│ i64      ┆ f64   ┆ f64   │
╞══════════╪═══════╪═══════╡
│ 0        ┆ 0.0   ┆ 0.0   │
│ 1        ┆ 76.0  ┆ 40.0  │
│ 2        ┆ 154.0 ┆ 100.0 │
│ 3        ┆ 37.0  ┆ 10.0  │
└──────────┴───────┴───────┘

record has one row for each scenario. There is no total objective, and a scenario with an integer variable has no duals.

Vary more than one table¶

Give each table that differs the same column. Here gas also costs 80 in the cold scenario:

costs = pl.concat(
    [
        sources['cost'].with_columns(scenario=pl.lit('mild')),
        sources['cost'].with_columns(
            scenario=pl.lit('cold'),
            value=pl.when(pl.col('generator') == 'gas').then(80.0).otherwise(pl.col('value')),
        ),
    ]
)
both = sps.solve_over(
    'examples/dispatch.yaml', sources | {'load': loads, 'cost': costs}, sps.EachCoordinate('scenario')
)
print(both.record.select('scenario', 'objective'))

shape: (2, 2)
┌──────────┬───────────┐
│ scenario ┆ objective │
│ ---      ┆ ---       │
│ str      ┆ f64       │
╞══════════╪═══════════╡
│ cold     ┆ 24540.0   │
│ mild     ┆ 10500.0   │
└──────────┴───────────┘

Solve them in parallel¶

The scenarios do not depend on each other, so they can run on a pool of workers. Each worker needs the memory of one scenario. Running a sweep in parallel is the recipe.