Computing the duals of a mixed-integer program
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This tutorial explains how to compute the duals of a mixed-integer linear program by fixing the discrete variables to their optimal solution and resolving as a linear program.
This tutorial uses the following packages:
using JuMP
import HiGHS
The model
Our example model is the unit commitment example from Unit commitment. The details are unimportant, other than to note that there are two types of continuous variables, g
and w
, representing the quantity of generation from thermal and wind plants, and a discrete variable dispatch
, which is 1
if plant i
is operating, and 0
if not.
We are interested in the "dual" of the power_balance
constraint, because it represents the marginal price of electricity that consumers should pay for their consumption.
generators = [
(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),
]
N = length(generators)
model = Model(HiGHS.Optimizer)
set_silent(model)
@variables(model, begin
generators[i].min <= g[i = 1:N] <= generators[i].max
0 <= w <= 200
dispatch[i = 1:N], Bin
end)
@constraints(model, begin
power_balance, sum(g[i] for i in 1:N) + w == 1500
[i = 1:N], g[i] <= generators[i].max * dispatch[i]
[i = 1:N], g[i] >= generators[i].min * dispatch[i]
end)
@objective(
model,
Min,
sum(
generators[i].fixed_cost * dispatch[i] +
generators[i].variable_cost * g[i] for i in 1:N
)
)
print(model)
Min 1000 dispatch[1] + 50 g[1] + 100 g[2]
Subject to
power_balance : g[1] + g[2] + w = 1500
g[1] ≥ 0
g[2] - 300 dispatch[2] ≥ 0
g[1] - 1000 dispatch[1] ≤ 0
g[2] - 1000 dispatch[2] ≤ 0
g[1] ≥ 0
g[2] ≥ 300
w ≥ 0
g[1] ≤ 1000
g[2] ≤ 1000
w ≤ 200
dispatch[1] binary
dispatch[2] binary
Manually fix the variables
If we optimize this model, we obtain a dual_status
of NO_SOLUTION
:
optimize!(model)
@assert is_solved_and_feasible(model)
dual_status(model)
NO_SOLUTION::ResultStatusCode = 0
This is because HiGHS cannot compute the duals of a mixed-integer program. We can work around this problem by fixing the integer variables to their optimal solution, relaxing integrality, and re-solving as a linear program.
discrete_values = value.(dispatch)
fix.(dispatch, discrete_values; force = true)
unset_binary.(dispatch)
print(model)
Min 1000 dispatch[1] + 50 g[1] + 100 g[2]
Subject to
power_balance : g[1] + g[2] + w = 1500
g[1] ≥ 0
g[2] - 300 dispatch[2] ≥ 0
g[1] - 1000 dispatch[1] ≤ 0
g[2] - 1000 dispatch[2] ≤ 0
dispatch[1] = 1
dispatch[2] = 1
g[1] ≥ 0
g[2] ≥ 300
w ≥ 0
g[1] ≤ 1000
g[2] ≤ 1000
w ≤ 200
Now if we re-solve the problem, we obtain a FEASIBLE_POINT
for the dual:
optimize!(model)
@assert is_solved_and_feasible(model)
dual_status(model)
FEASIBLE_POINT::ResultStatusCode = 1
and a marginal price of electricity of $100/MWh:
dual(power_balance)
100.0
To reset the problem back to a mixed-integer linear program, we need to unfix
and call set_binary
:
unfix.(dispatch)
set_binary.(dispatch)
print(model)
Min 1000 dispatch[1] + 50 g[1] + 100 g[2]
Subject to
power_balance : g[1] + g[2] + w = 1500
g[1] ≥ 0
g[2] - 300 dispatch[2] ≥ 0
g[1] - 1000 dispatch[1] ≤ 0
g[2] - 1000 dispatch[2] ≤ 0
g[1] ≥ 0
g[2] ≥ 300
w ≥ 0
g[1] ≤ 1000
g[2] ≤ 1000
w ≤ 200
dispatch[1] binary
dispatch[2] binary
Use fix_discrete_variables
Manually choosing the variables to relax and fix works for our small example, but it becomes more difficult in problems with a larger number of binary and integer variables. To automate the process we just did manually, JuMP provides the fix_discrete_variables
function:
optimize!(model)
@assert is_solved_and_feasible(model)
dual_status(model)
NO_SOLUTION::ResultStatusCode = 0
undo = fix_discrete_variables(model);
Here undo
is a function that, when called with no arguments, returns the model to the original mixed-integer formulation.
After calling |
print(model)
Min 1000 dispatch[1] + 50 g[1] + 100 g[2]
Subject to
power_balance : g[1] + g[2] + w = 1500
g[1] ≥ 0
g[2] - 300 dispatch[2] ≥ 0
g[1] - 1000 dispatch[1] ≤ 0
g[2] - 1000 dispatch[2] ≤ 0
dispatch[1] = 1
dispatch[2] = 1
g[1] ≥ 0
g[2] ≥ 300
w ≥ 0
g[1] ≤ 1000
g[2] ≤ 1000
w ≤ 200
optimize!(model)
@assert is_solved_and_feasible(model)
dual_status(model)
FEASIBLE_POINT::ResultStatusCode = 1
dual(power_balance)
100.0
Finally, call undo
to revert the reformulation
undo()
print(model)
Min 1000 dispatch[1] + 50 g[1] + 100 g[2]
Subject to
power_balance : g[1] + g[2] + w = 1500
g[1] ≥ 0
g[2] - 300 dispatch[2] ≥ 0
g[1] - 1000 dispatch[1] ≤ 0
g[2] - 1000 dispatch[2] ≤ 0
g[1] ≥ 0
g[2] ≥ 300
w ≥ 0
g[1] ≤ 1000
g[2] ≤ 1000
w ≤ 200
dispatch[1] binary
dispatch[2] binary