Power Systems

Originally Contributed by: Yury Dvorkin and Miles Lubin

This tutorial demonstrates how to formulate basic power systems engineering models in JuMP.

We will consider basic "economic dispatch" and "unit commitment" models without taking into account transmission constraints.

For this tutorial, we use the following packages:

using JuMP
import DataFrames
import GLPK
import Plots
import StatsPlots

Economic dispatch

Economic dispatch (ED) is an optimization problem that minimizes the cost of supplying energy demand subject to operational constraints on power system assets. In its simplest modification, ED is an LP problem solved for an aggregated load and wind forecast and for a single infinitesimal moment.

Mathematically, the ED problem can be written as follows:

\[\min \sum_{i \in I} c^g_{i} \cdot g_{i} + c^w \cdot w,\]

where $c_{i}$ and $g_{i}$ are the incremental cost ($/MWh) and power output (MW) of the $i^{th}$ generator, respectively, and $c^w$ and $w$ are the incremental cost ($/MWh) and wind power injection (MW), respectively.

Subject to the constraints:

  • Minimum ($g^{\min}$) and maximum ($g^{\max}$) limits on power outputs of generators: $g^{\min}_{i} \leq g_{i} \leq g^{\max}_{i}.$
  • Constraint on the wind power injection: $0 \leq w \leq w^f,$ where $w$ and $w^f$ are the wind power injection and wind power forecast, respectively.
  • Power balance constraint: $\sum_{i \in I} g_{i} + w = d^f,$ where $d^f$ is the demand forecast.

Further reading on ED models can be found in A. J. Wood, B. F. Wollenberg, and G. B. Sheblé, "Power Generation, Operation and Control", Wiley, 2013.

Define some input data about the test system.

We define some thermal generators:

function ThermalGenerator(
    min::Float64,
    max::Float64,
    fixed_cost::Float64,
    variable_cost::Float64,
)
    return (
        min = min,
        max = max,
        fixed_cost = fixed_cost,
        variable_cost = variable_cost,
    )
end

generators = [
    ThermalGenerator(0.0, 1000.0, 1000.0, 50.0),
    ThermalGenerator(300.0, 1000.0, 0.0, 100.0),
]
2-element Vector{NamedTuple{(:min, :max, :fixed_cost, :variable_cost), NTuple{4, Float64}}}:
 (min = 0.0, max = 1000.0, fixed_cost = 1000.0, variable_cost = 50.0)
 (min = 300.0, max = 1000.0, fixed_cost = 0.0, variable_cost = 100.0)

A wind generator

WindGenerator(variable_cost::Float64) = (variable_cost = variable_cost,)

wind_generator = WindGenerator(50.0)
(variable_cost = 50.0,)

And a scenario

function Scenario(demand::Float64, wind::Float64)
    return (demand = demand, wind = wind)
end

scenario = Scenario(1500.0, 200.0)
(demand = 1500.0, wind = 200.0)

Create a function solve_ed, which solves the economic dispatch problem for a given set of input parameters.

function solve_ed(generators::Vector, wind, scenario)
    # Define the economic dispatch (ED) model
    ed = Model(GLPK.Optimizer)
    # Define decision variables
    # power output of generators
    N = length(generators)
    @variable(ed, generators[i].min <= g[i = 1:N] <= generators[i].max)
    # wind power injection
    @variable(ed, 0 <= w <= scenario.wind)
    # Define the objective function
    @objective(
        ed,
        Min,
        sum(generators[i].variable_cost * g[i] for i in 1:N) +
        wind.variable_cost * w,
    )
    # Define the power balance constraint
    @constraint(ed, sum(g[i] for i in 1:N) + w == scenario.demand)
    # Solve statement
    optimize!(ed)
    # return the optimal value of the objective function and its minimizers
    return (
        g = value.(g),
        w = value(w),
        wind_spill = scenario.wind - value(w),
        total_cost = objective_value(ed),
    )
end
solve_ed (generic function with 1 method)

Solve the economic dispatch problem

solution = solve_ed(generators, wind_generator, scenario);

println("Dispatch of Generators: ", solution.g, " MW")
println("Dispatch of Wind: ", solution.w, " MW")
println("Wind spillage: ", solution.wind_spill, " MW")
println("Total cost: \$", solution.total_cost)
Dispatch of Generators: [1000.0, 300.0] MW
Dispatch of Wind: 200.0 MW
Wind spillage: 0.0 MW
Total cost: $90000.0

Economic dispatch with adjustable incremental costs

In the following exercise we adjust the incremental cost of generator G1 and observe its impact on the total cost.

function scale_generator_cost(g, scale)
    return ThermalGenerator(g.min, g.max, g.fixed_cost, scale * g.variable_cost)
end

start = time()
c_g_scale_df = DataFrames.DataFrame(
    # Scale factor
    scale = Float64[],
    # Dispatch of Generator 1 [MW]
    dispatch_G1 = Float64[],
    # Dispatch of Generator 2 [MW]
    dispatch_G2 = Float64[],
    # Dispatch of Wind [MW]
    dispatch_wind = Float64[],
    # Spillage of Wind [MW]
    spillage_wind = Float64[],
    # Total cost [$]
    total_cost = Float64[],
)
for c_g1_scale in 0.5:0.1:3.0
    # Update the incremental cost of the first generator at every iteration.
    new_generators = scale_generator_cost.(generators, [c_g1_scale, 1.0])
    # Solve the ed problem with the updated incremental cost
    sol = solve_ed(new_generators, wind_generator, scenario)
    push!(
        c_g_scale_df,
        (c_g1_scale, sol.g[1], sol.g[2], sol.w, sol.wind_spill, sol.total_cost),
    )
end
print(string("elapsed time: ", time() - start, " seconds"))
elapsed time: 0.18093395233154297 seconds
c_g_scale_df

26 rows × 6 columns

scaledispatch_G1dispatch_G2dispatch_windspillage_windtotal_cost
Float64Float64Float64Float64Float64Float64
10.51000.0300.0200.00.065000.0
20.61000.0300.0200.00.070000.0
30.71000.0300.0200.00.075000.0
40.81000.0300.0200.00.080000.0
50.91000.0300.0200.00.085000.0
61.01000.0300.0200.00.090000.0
71.11000.0300.0200.00.095000.0
81.21000.0300.0200.00.0100000.0
91.31000.0300.0200.00.0105000.0
101.41000.0300.0200.00.0110000.0
111.51000.0300.0200.00.0115000.0
121.61000.0300.0200.00.0120000.0
131.71000.0300.0200.00.0125000.0
141.81000.0300.0200.00.0130000.0
151.91000.0300.0200.00.0135000.0
162.01000.0300.0200.00.0140000.0
172.1300.01000.0200.00.0141500.0
182.2300.01000.0200.00.0143000.0
192.3300.01000.0200.00.0144500.0
202.4300.01000.0200.00.0146000.0
212.5300.01000.0200.00.0147500.0
222.6300.01000.0200.00.0149000.0
232.7300.01000.0200.00.0150500.0
242.8300.01000.0200.00.0152000.0
252.9300.01000.0200.00.0153500.0
263.0300.01000.0200.00.0155000.0

Modifying the JuMP model in-place

Note that in the previous exercise we entirely rebuilt the optimization model at every iteration of the internal loop, which incurs an additional computational burden. This burden can be alleviated if instead of re-building the entire model, we modify a specific constraint(s) or the objective function, as it shown in the example below.

Compare the computing time in case of the above and below models.

function solve_ed_inplace(
    generators::Vector,
    wind,
    scenario,
    scale::AbstractVector{Float64},
)
    obj_out = Float64[]
    w_out = Float64[]
    g1_out = Float64[]
    g2_out = Float64[]
    # This function only works for two generators
    @assert length(generators) == 2
    ed = Model(GLPK.Optimizer)
    N = length(generators)
    @variable(ed, generators[i].min <= g[i = 1:N] <= generators[i].max)
    @variable(ed, 0 <= w <= scenario.wind)
    @objective(
        ed,
        Min,
        sum(generators[i].variable_cost * g[i] for i in 1:N) +
        wind.variable_cost * w,
    )
    @constraint(ed, sum(g[i] for i in 1:N) + w == scenario.demand)
    for c_g1_scale in scale
        @objective(
            ed,
            Min,
            c_g1_scale * generators[1].variable_cost * g[1] +
            generators[2].variable_cost * g[2] +
            wind.variable_cost * w,
        )
        optimize!(ed)
        push!(obj_out, objective_value(ed))
        push!(w_out, value(w))
        push!(g1_out, value(g[1]))
        push!(g2_out, value(g[2]))
    end
    df = DataFrames.DataFrame(
        scale = scale,
        dispatch_G1 = g1_out,
        dispatch_G2 = g2_out,
        dispatch_wind = w_out,
        spillage_wind = scenario.wind .- w_out,
        total_cost = obj_out,
    )
    return df
end

start = time()
inplace_df = solve_ed_inplace(generators, wind_generator, scenario, 0.5:0.1:3.0)
print(string("elapsed time: ", time() - start, " seconds"))
elapsed time: 0.2261791229248047 seconds

Adjusting specific constraints or the objective function is faster than re-building the entire model.

inplace_df

26 rows × 6 columns

scaledispatch_G1dispatch_G2dispatch_windspillage_windtotal_cost
Float64Float64Float64Float64Float64Float64
10.51000.0300.0200.00.065000.0
20.61000.0300.0200.00.070000.0
30.71000.0300.0200.00.075000.0
40.81000.0300.0200.00.080000.0
50.91000.0300.0200.00.085000.0
61.01000.0300.0200.00.090000.0
71.11000.0300.0200.00.095000.0
81.21000.0300.0200.00.0100000.0
91.31000.0300.0200.00.0105000.0
101.41000.0300.0200.00.0110000.0
111.51000.0300.0200.00.0115000.0
121.61000.0300.0200.00.0120000.0
131.71000.0300.0200.00.0125000.0
141.81000.0300.0200.00.0130000.0
151.91000.0300.0200.00.0135000.0
162.01000.0300.0200.00.0140000.0
172.1300.01000.0200.00.0141500.0
182.2300.01000.0200.00.0143000.0
192.3300.01000.0200.00.0144500.0
202.4300.01000.0200.00.0146000.0
212.5300.01000.0200.00.0147500.0
222.6300.01000.0200.00.0149000.0
232.7300.01000.0200.00.0150500.0
242.8300.01000.0200.00.0152000.0
252.9300.01000.0200.00.0153500.0
263.0300.01000.0200.00.0155000.0

Inefficient usage of wind generators

The economic dispatch problem does not perform commitment decisions and, thus, assumes that all generators must be dispatched at least at their minimum power output limit. This approach is not cost efficient and may lead to absurd decisions. For example, if $d = \sum_{i \in I} g^{\min}_{i}$, the wind power injection must be zero, i.e. all available wind generation is spilled, to meet the minimum power output constraints on generators.

In the following example, we adjust the total demand and observed how it affects wind spillage.

demand_scale_df = DataFrames.DataFrame(
    demand = Float64[],
    dispatch_G1 = Float64[],
    dispatch_G2 = Float64[],
    dispatch_wind = Float64[],
    spillage_wind = Float64[],
    total_cost = Float64[],
)

function scale_demand(scenario, scale)
    return Scenario(scale * scenario.demand, scenario.wind)
end

for demand_scale in 0.2:0.1:1.5
    new_scenario = scale_demand(scenario, demand_scale)
    sol = solve_ed(generators, wind_generator, new_scenario)
    push!(
        demand_scale_df,
        (
            new_scenario.demand,
            sol.g[1],
            sol.g[2],
            sol.w,
            sol.wind_spill,
            sol.total_cost,
        ),
    )
end

demand_scale_df

14 rows × 6 columns

demanddispatch_G1dispatch_G2dispatch_windspillage_windtotal_cost
Float64Float64Float64Float64Float64Float64
1300.00.0300.00.0200.030000.0
2450.0150.0300.00.0200.037500.0
3600.0300.0300.00.0200.045000.0
4750.0450.0300.00.0200.052500.0
5900.0600.0300.00.0200.060000.0
61050.0750.0300.00.0200.067500.0
71200.0900.0300.00.0200.075000.0
81350.01000.0300.050.0150.082500.0
91500.01000.0300.0200.00.090000.0
101650.01000.0450.0200.00.0105000.0
111800.01000.0600.0200.00.0120000.0
121950.01000.0750.0200.00.0135000.0
132100.01000.0900.0200.00.0150000.0
142250.01000.01050.0200.00.0165000.0
dispatch_plot = StatsPlots.@df(
    demand_scale_df,
    Plots.plot(
        :demand,
        [:dispatch_G1, :dispatch_G2],
        labels = ["G1" "G2"],
        title = "Thermal Dispatch",
        legend = :bottomright,
        linewidth = 3,
        xlabel = "Demand",
        ylabel = "Dispatch [MW]",
    ),
)

wind_plot = StatsPlots.@df(
    demand_scale_df,
    Plots.plot(
        :demand,
        [:dispatch_wind, :spillage_wind],
        labels = ["Dispatch" "Spillage"],
        title = "Wind",
        legend = :bottomright,
        linewidth = 3,
        xlabel = "Demand [MW]",
        ylabel = "Energy [MW]",
    ),
)

Plots.plot(dispatch_plot, wind_plot)

This particular drawback can be overcome by introducing binary decisions on the "on/off" status of generators. This model is called unit commitment and considered later in these notes.

For further reading on the interplay between wind generation and the minimum power output constraints of generators, we refer interested readers to R. Baldick, "Wind and Energy Markets: A Case Study of Texas," IEEE Systems Journal, vol. 6, pp. 27-34, 2012.

Unit commitment

The Unit Commitment (UC) model can be obtained from ED model by introducing binary variable associated with each generator. This binary variable can attain two values: if it is "1", the generator is synchronized and, thus, can be dispatched, otherwise, i.e. if the binary variable is "0", that generator is not synchronized and its power output is set to 0.

To obtain the mathematical formulation of the UC model, we will modify the constraints of the ED model as follows:

\[g^{\min}_{i} \cdot u_{t,i} \leq g_{i} \leq g^{\max}_{i} \cdot u_{t,i},\]

where $u_{i} \in \{0,1\}.$ In this constraint, if $u_{i} = 0$, then $g_{i} = 0$. On the other hand, if $u_{i} = 1$, then $g^{min}_{i} \leq g_{i} \leq g^{max}_{i}$.

For further reading on the UC problem we refer interested readers to G. Morales-Espana, J. M. Latorre, and A. Ramos, "Tight and Compact MILP Formulation for the Thermal Unit Commitment Problem," IEEE Transactions on Power Systems, vol. 28, pp. 4897-4908, 2013.

In the following example we convert the ED model explained above to the UC model.

function solve_uc(generators::Vector, wind, scenario)
    uc = Model(GLPK.Optimizer)
    N = length(generators)
    @variable(uc, generators[i].min <= g[i = 1:N] <= generators[i].max)
    @variable(uc, 0 <= w <= scenario.wind)
    @constraint(uc, sum(g[i] for i in 1:N) + w == scenario.demand)
    # !!! New: add binary on-off variables for each generator
    @variable(uc, u[i = 1:N], Bin)
    @constraint(uc, [i = 1:N], g[i] <= generators[i].max * u[i])
    @constraint(uc, [i = 1:N], g[i] >= generators[i].min * u[i])
    @objective(
        uc,
        Min,
        sum(generators[i].variable_cost * g[i] for i in 1:N) +
        wind.variable_cost * w +
        # !!! new
        sum(generators[i].fixed_cost * u[i] for i in 1:N)
    )
    optimize!(uc)
    status = termination_status(uc)
    if status != OPTIMAL
        return (status = status,)
    end
    return (
        status = status,
        g = value.(g),
        w = value(w),
        wind_spill = scenario.wind - value(w),
        u = value.(u),
        total_cost = objective_value(uc),
    )
end
solve_uc (generic function with 1 method)

Solve the economic dispatch problem

solution = solve_uc(generators, wind_generator, scenario)

println("Dispatch of Generators: ", solution.g, " MW")
println("Commitments of Generators: ", solution.u)
println("Dispatch of Wind: ", solution.w, " MW")
println("Wind spillage: ", solution.wind_spill, " MW")
println("Total cost: \$", solution.total_cost)
Dispatch of Generators: [1000.0, 300.0] MW
Commitments of Generators: [1.0, 1.0]
Dispatch of Wind: 200.0 MW
Wind spillage: 0.0 MW
Total cost: $91000.0

Unit commitment as a function of demand

After implementing the UC model, we can now assess the interplay between the minimum power output constraints on generators and wind generation.

uc_df = DataFrames.DataFrame(
    demand = Float64[],
    commitment_G1 = Float64[],
    commitment_G2 = Float64[],
    dispatch_G1 = Float64[],
    dispatch_G2 = Float64[],
    dispatch_wind = Float64[],
    spillage_wind = Float64[],
    total_cost = Float64[],
)

for demand_scale in 0.2:0.1:1.5
    new_scenario = scale_demand(scenario, demand_scale)
    sol = solve_uc(generators, wind_generator, new_scenario)
    if sol.status == OPTIMAL
        push!(
            uc_df,
            (
                new_scenario.demand,
                sol.u[1],
                sol.u[2],
                sol.g[1],
                sol.g[2],
                sol.w,
                sol.wind_spill,
                sol.total_cost,
            ),
        )
    end
    println("Status: $(sol.status) for demand_scale = $(demand_scale)")
end
Status: OPTIMAL for demand_scale = 0.2
Status: OPTIMAL for demand_scale = 0.3
Status: OPTIMAL for demand_scale = 0.4
Status: OPTIMAL for demand_scale = 0.5
Status: OPTIMAL for demand_scale = 0.6
Status: OPTIMAL for demand_scale = 0.7
Status: OPTIMAL for demand_scale = 0.8
Status: OPTIMAL for demand_scale = 0.9
Status: OPTIMAL for demand_scale = 1.0
Status: OPTIMAL for demand_scale = 1.1
Status: OPTIMAL for demand_scale = 1.2
Status: OPTIMAL for demand_scale = 1.3
Status: OPTIMAL for demand_scale = 1.4
Status: INFEASIBLE for demand_scale = 1.5
uc_df

13 rows × 8 columns

demandcommitment_G1commitment_G2dispatch_G1dispatch_G2dispatch_windspillage_windtotal_cost
Float64Float64Float64Float64Float64Float64Float64Float64
1300.00.01.00.0300.00.0200.030000.0
2450.00.01.00.0300.0150.050.037500.0
3600.01.01.0100.0300.0200.00.046000.0
4750.01.01.0250.0300.0200.00.053500.0
5900.01.01.0400.0300.0200.00.061000.0
61050.01.01.0550.0300.0200.00.068500.0
71200.01.01.0700.0300.0200.00.076000.0
81350.01.01.0850.0300.0200.00.083500.0
91500.01.01.01000.0300.0200.00.091000.0
101650.01.01.01000.0450.0200.00.0106000.0
111800.01.01.01000.0600.0200.00.0121000.0
121950.01.01.01000.0750.0200.00.0136000.0
132100.01.01.01000.0900.0200.00.0151000.0
commitment_plot = StatsPlots.@df(
    uc_df,
    Plots.plot(
        :demand,
        [:commitment_G1, :commitment_G2],
        labels = ["G1" "G2"],
        title = "Committment",
        legend = :bottomright,
        linewidth = 3,
        xlabel = "Demand [MW]",
        ylabel = "Commitment decision {0, 1}",
    ),
)

dispatch_plot = StatsPlots.@df(
    uc_df,
    Plots.plot(
        :demand,
        [:dispatch_G1, :dispatch_G2, :dispatch_wind],
        labels = ["G1" "G2" "Wind"],
        title = "Dispatch [MW]",
        legend = :bottomright,
        linewidth = 3,
        xlabel = "Demand",
        ylabel = "Dispatch [MW]",
    ),
)

Plots.plot(commitment_plot, dispatch_plot)

Nonlinear economic dispatch

As a final example, we modify our economic dispatch problem in two ways:

  • The thermal cost function is user-defined
  • The output of the wind is only the square-root of the dispatch
import Ipopt

"""
    thermal_cost_function(g)

A user-defined thermal cost function in pure-Julia! You can include
nonlinearities, and even things like control flow.

!!! warning
    It's still up to you to make sure that the function has a meaningful
    derivative.
"""
function thermal_cost_function(g)
    if g <= 500
        return g
    else
        return g + 1e-2 * (g - 500)^2
    end
end

function solve_nonlinear_ed(
    generators::Vector,
    wind,
    scenario;
    silent::Bool = false,
)
    model = Model(Ipopt.Optimizer)
    if silent
        set_silent(model)
    end
    register(model, :tcf, 1, thermal_cost_function; autodiff = true)
    N = length(generators)
    @variable(model, generators[i].min <= g[i = 1:N] <= generators[i].max)
    @variable(model, 0 <= w <= scenario.wind)
    @NLobjective(
        model,
        Min,
        sum(generators[i].variable_cost * tcf(g[i]) for i in 1:N) +
        wind.variable_cost * w,
    )
    @NLconstraint(model, sum(g[i] for i in 1:N) + sqrt(w) == scenario.demand)
    optimize!(model)
    return (
        g = value.(g),
        w = value(w),
        wind_spill = scenario.wind - value(w),
        total_cost = objective_value(model),
    )
end

solution = solve_nonlinear_ed(generators, wind_generator, scenario)
(g = [847.3509933774712, 648.6754966887423], w = 15.788781193899027, wind_spill = 184.211218806101, total_cost = 190455.298013245)

Now let's see how the wind is dispatched as a function of the cost:

wind_cost = 0.0:1:100
wind_dispatch = Float64[]
for c in wind_cost
    sol = solve_nonlinear_ed(
        generators,
        WindGenerator(c),
        scenario;
        silent = true,
    )
    push!(wind_dispatch, sol.w)
end

Plots.plot(
    wind_cost,
    wind_dispatch,
    xlabel = "Cost",
    ylabel = "Dispatch [MW]",
    label = false,
)

Tip

This tutorial was generated using Literate.jl. View the source .jl file on GitHub.