Uncertainty in the objective function

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

In the previous tutorial, An introduction to SDDP.jl, we created a stochastic hydro-thermal scheduling model. In this tutorial, we extend the problem by adding uncertainty to the fuel costs.

Previously, we assumed that the fuel cost was deterministic: $50/MWh in the first stage, $100/MWh in the second stage, and $150/MWh in the third stage. For this tutorial, we assume that in addition to these base costs, the actual fuel cost is correlated with the inflows.

Our new model for the uncertainty is given by the following table:

In stage t, the objective is now to minimize:

fuel_multiplier * fuel_cost[t] * thermal_generation

Creating a model

To add an uncertain objective, we can simply call @stageobjective from inside the SDDP.parameterize function.

using SDDP, HiGHS

model = SDDP.LinearPolicyGraph(
    stages = 3,
    sense = :Min,
    lower_bound = 0.0,
    optimizer = HiGHS.Optimizer,
) do subproblem, t
    # Define the state variable.
    @variable(subproblem, 0 <= volume <= 200, SDDP.State, initial_value = 200)
    # Define the control variables.
    @variables(subproblem, begin
        thermal_generation >= 0
        hydro_generation >= 0
        hydro_spill >= 0
        inflow
    end)
    # Define the constraints
    @constraints(
        subproblem,
        begin
            volume.out == volume.in + inflow - hydro_generation - hydro_spill
            thermal_generation + hydro_generation == 150.0
        end
    )
    fuel_cost = [50.0, 100.0, 150.0]
    # Parameterize the subproblem.
    Ω = [
        (inflow = 0.0, fuel_multiplier = 1.5),
        (inflow = 50.0, fuel_multiplier = 1.0),
        (inflow = 100.0, fuel_multiplier = 0.75),
    ]
    SDDP.parameterize(subproblem, Ω, [1 / 3, 1 / 3, 1 / 3]) do ω
        JuMP.fix(inflow, ω.inflow)
        @stageobjective(
            subproblem,
            ω.fuel_multiplier * fuel_cost[t] * thermal_generation
        )
    end
end

A policy graph with 3 nodes.
 Node indices: 1, 2, 3

Training and simulating the policy

As in the previous two tutorials, we train and simulate the policy:

SDDP.train(model)

simulations = SDDP.simulate(model, 500)

objective_values =
    [sum(stage[:stage_objective] for stage in sim) for sim in simulations]

using Statistics

μ = round(mean(objective_values), digits = 2)
ci = round(1.96 * std(objective_values) / sqrt(500), digits = 2)

println("Confidence interval: ", μ, " ± ", ci)
println("Lower bound: ", round(SDDP.calculate_bound(model), digits = 2))

-------------------------------------------------------------------
         SDDP.jl (c) Oscar Dowson and contributors, 2017-23
-------------------------------------------------------------------
problem
  nodes           : 3
  state variables : 1
  scenarios       : 2.70000e+01
  existing cuts   : false
options
  solver          : serial mode
  risk measure    : SDDP.Expectation()
  sampling scheme : SDDP.InSampleMonteCarlo
subproblem structure
  VariableRef                             : [7, 7]
  AffExpr in MOI.EqualTo{Float64}         : [2, 2]
  VariableRef in MOI.GreaterThan{Float64} : [5, 5]
  VariableRef in MOI.LessThan{Float64}    : [1, 2]
numerical stability report
  matrix range     [1e+00, 1e+00]
  objective range  [1e+00, 2e+02]
  bounds range     [2e+02, 2e+02]
  rhs range        [2e+02, 2e+02]
-------------------------------------------------------------------
 iteration    simulation      bound        time (s)     solves  pid
-------------------------------------------------------------------
         1   9.375000e+03  3.958333e+03  4.221201e-03        12   1
        40   1.125000e+04  1.062500e+04  1.103120e-01       642   1
-------------------------------------------------------------------
status         : simulation_stopping
total time (s) : 1.103120e-01
total solves   : 642
best bound     :  1.062500e+04
simulation ci  :  1.169368e+04 ± 2.551313e+03
numeric issues : 0
-------------------------------------------------------------------

Confidence interval: 10483.75 ± 723.49
Lower bound: 10625.0