Lecture 12
October 20, 2024
Linear Programs (LPs) are:
LPs must have optimal solutions at intersections of constraints: simplex method.
Text: VSRIKRISH to 22333
Linear models come up frequently because we can linearize nonlinear functions.
When we linearize components of an mathematical program, this is called the linear relaxation of the original problem.
E = 0:0.01:1
plot(E, E.^2, legend=false, grid=false, xlabel="Efficiency", ylabel="Cost", color=:black, yticks=false, xlims=(0, 1), ylims=(0, 1), left_margin=8mm, linewidth=3)
xticks!([0.65, 0.95])
xlims!((0, 1.05))
scatter!([0.65, 0.95], [0.65, 0.95].^2, markersize=10, color=:blue)
plot!(E, 1.6 .* E .- 0.6175, color=:blue, linestyle=:dash, linewidth=3)
plot!(size=(600, 500))A solution will be found at one of the corner points of the feasible polytope.
This means that at this solution, one or more constraints are binding: if we relaxed the constraint by weakening it, we could improve the solution.
x1 = 0:1200
x2 = 0:1400
f1(x) = (600 .- 0.9 .* x) ./ 0.5
f2(x) = 1000 .- x
p = plot(0:667, min.(f1(0:667), f2(0:667)), fillrange=0, color=:lightblue, grid=true, label="Feasible Region", xlabel=L"x_1", ylabel=L"x_2", xlims=(-50, 1200), ylims=(-50, 1400), framestyle=:origin, minorticks=4, right_margin=4mm, left_margin=4mm)
plot!(0:667, f1.(0:667), color=:brown, linewidth=3, label=false)
plot!(0:1000, f2.(0:1000), color=:red, linewidth=3, label=false)
annotate!(400, 1100, text(L"0.9x_1 + 0.5x_2 = 600", color=:purple, pointsize=18))
annotate!(1000, 300, text(L"x_1 + x_2 = 1000", color=:red, pointsize=18))
plot!(size=(600, 500))
Z(x1,x2) = 230 * x1 + 120 * x2
contour!(0:660,0:1000,(x1,x2)->Z(x1,x2), levels=5, c=:devon, linewidth=2, colorbar = false, clabels = true)
scatter!(p, [667], [0], markersize=10, z=2, label="Optimal Solution", markercolor=:orange)The binding constraints are
but not
The marginal cost of a constraint is the amount by which the solution would improve if the constraint capacity was relaxed by one unit.
This is also referred to as the shadow price (these are also called the dual variables of the constraint).
Non-zero shadow prices tell us that the constraint is binding, and their values rank which constraints are most influential.
Lagrange Multipliers are a way to incorporate equality constraints into an “unconstrained” form for an optimization problem.
\[\begin{align*} \max_{x,y}\qquad &f(x,y) \\ \text{subject to:}\qquad &g(x,y) = 0 \end{align*}\]
Then: \[\mathcal{L}(x, y, \lambda) = f(x,y) - \lambda g(x,y)\]
Original Problem
\[ \begin{aligned} & \min &&f(x_1, x_2) \notag \\\\ & \text{subject to:} && x_1 \geq A \notag \\ & && x_2 \leq B \notag \end{aligned} \]
With Dummy Variables
\[ \begin{aligned} & \min &&f(x_1, x_2) \notag \\\\ & \text{subject to:} && x_1 - S_1^2 = A \notag \\ & && x_2 + S_2^2 = B \notag \end{aligned} \]
Then the Lagrangian function becomes:
\[ \mathcal{L}(\mathbf{x}, S_1, S_2, \lambda_1, \lambda_2) = f(\mathbf{x}) - \lambda_1(x_1 - S_1^2 - A) - \lambda_2(x_2 + S_2^2 - B) \]
where \(\lambda_1\), \(\lambda_2\) are penalties for violating the constraints.
The \(\lambda_i\) are the eponymous Lagrange multipliers.
Next step: locate possible optima where the partial derivatives of the Lagrangian are zero.
\[\frac{\partial \mathcal{L}(\cdot)}{\partial \cdot} = 0\]
This is actually many equations, even though our original problem was low-dimensional, and can be slow to solve.
The shadow prices are the Lagrange Multipliers of the optimization problem.
If our inequality constraints \(X \geq A\) and \(X \leq B\) are written as \(X - S_1^2 = A\) and \(X + S_2^2 = B\):
\[\mathcal{L}(X, S_1 S_2, \lambda_1, \lambda_2) = Z(X) - \lambda_1(X - S_1^2 - A) - \lambda_2(X + S_2^2 - B),\]
\[\Rightarrow \qquad \frac{\partial \mathcal{L}}{\partial A} = \lambda_1, \qquad \frac{\partial \mathcal{L}}{\partial B} = \lambda_2.\]
Source: Wikipedia
Adapted from Perez-Arriaga, Ignacio J., Hugh Rudnick, and Michel Rivier (2009)
Capacity expansion involves adding resources to generate or transmit electricity to meet anticipated demand (load) in the future.
Typical objective: Minimize cost
But other constraints may apply, e.g. reducing CO2 emissions or increasing fuel diversity.
In general, we have many fuel options:
| Generator | Fixed Cost ($) | Variable Cost ($/MW) |
|---|---|---|
| Geothermal | 450000 | 0 |
| Coal | 220000 | 21 |
| NG CCGT | 82000 | 25 |
| NG CT | 65000 | 35 |
NY_demand = DataFrame(CSV.File("data/capacity_expansion/2020_hourly_load_NY.csv"))
rename!(NY_demand, :"Time Stamp" => :Date)
demand = NY_demand[:, [:Date, :C]]
rename!(demand, :C => :Demand)
@df demand plot(:Date, :Demand, xlabel="Date", ylabel="Demand (MWh)", label=:false, xrot=45, bottom_margin=15mm)
plot!(size=(1200, 500))The chronological demand curve makes it hard to understand what levels of load occur with lower or greater frequency.
Instead, we can sort demand from high to low, which creates a load duration curve.
We want to find the installed capacity of each technology that meets demand at all hours at minimal total cost.
Although…
What are our variables?
| Variable | Meaning |
|---|---|
| \(x_g\) | installed capacity (MW) from each generator type \(g \in \mathcal{G}\) |
| \(y_{g,t}\) | production (MWh) from each generator type \(g\) in hour \(t \in \mathcal{T}\) |
| \(NSE_t\) | non-served energy (MWh) in hour \(t \in \mathcal{T}\) |
What is our objective?
\[ \begin{align} \min_{x, y, NSE} Z &= {\color{red}\text{investment cost} } + {\color{blue}\text{operating cost} } \\ &= {\color{red} \sum_{g \in \mathcal{G}} \times \text{FixedCost}_g x_g} + \\ & \qquad{\color{blue} \sum_{t \in \mathcal{T}} \sum_{g \in \mathcal{G}} \text{VarCost}_g \times y_{g,t} + \sum_{t \in \mathcal{T}} \text{NSECost} \times NSE_t} \end{align} \]
\[ \begin{align} \min_{x, y, NSE} \quad & \sum_{g \in \mathcal{G}} \text{FixedCost}_g \times x_g + \sum_{t \in \mathcal{T}} \sum_{g \in \mathcal{G}} \text{VarCost}_g \times y_{g,t} & \\ & \quad + \sum_{t \in \mathcal{T}} \text{NSECost} \times NSE_t & \\[0.5em] \text {subject to:} \quad & \sum_{g \in \mathcal{G}} y_{g,t} + NSE_t \geq d_t \qquad \forall t \in \mathcal{T} \\[0.5em] & y_{g,t} \leq x_g \qquad \qquad \qquad\qquad \forall g \in {G}, \forall t \in \mathcal{T} \\[0.5em] & x_g, y_{g,t}, NSE_t \geq 0 \qquad \qquad \forall g \in {G}, \forall t \in \mathcal{T} \end{align} \]
Real problems can get much more complex, particularly if we try to model making decisions under renewable or load uncertainty.
JUMP.jl# define sets
G = 1:nrow(gens[1:end-2, :])
T = 1:nrow(demand)
NSECost = 9000
gencap = Model(HiGHS.Optimizer)
# define variables
@variables(gencap, begin
x[g in G] >= 0
y[g in G, t in T] >= 0
NSE[t in T] >= 0
end)
@objective(gencap, Min,
sum(gens[G, :FixedCost] .* x) + sum(gens[G, :VarCost] .* sum(y[:, t] for t in T)) + NSECost * sum(NSE)
)
@constraint(gencap, load[t in T], sum(y[:, t]) + NSE[t] >= demand.Demand[t])
@constraint(gencap, availability[g in G, t in T], y[g, t] <= x[g])
optimize!(gencap)Wednesday: Economic Dispatch
Next Week: Managing Air Pollution