example9

docs/src/examples/FOBP_example2.2.jl

@@ -64,8 +64,8 @@ model = BilevelModel(Ipopt.Optimizer, mode = BilevelJuMP.ProductMode(1e-9))
 optimize!(model)
 
 @test objective_value(model) ≈ 3 * (3.5 * 8 / 15) + (8 / 15) atol=1e-3
-@test value(x) ≈ 3.5 * 8 / 15 atol=1e-6
-@test value(y) ≈ 8 / 15 atol=1e-6
+@test value(x) ≈ 3.5 * 8 / 15 atol=1e-3
+@test value(y) ≈ 8 / 15 atol=1e-3
 
 # TODO: why are these commented out?    #src
 # @test dual(l2) #≈ [0] atol=atol       #src

docs/src/examples/FOBP_example2.jl

@@ -1,4 +1,4 @@
-# # Foundations of Bilevel Programming: Example 2
+# # Example 2
 # This example is from the book _Foundations of Bilevel Programming_ by Stephan
 # Dempe, Chapter 3.2, Page 25. [url](https://www.springer.com/gp/book/9781402006319)
 # Moving the bound on x to lower level

docs/src/examples/FOBP_example3.jl

@@ -1,4 +1,4 @@
-# # Foundations of Bilevel Programming: Example 3
+# # Example 3
 # This example is from the book _Foundations of Bilevel Programming_ by Stephan
 # Dempe, Chapter 3.7, Page 59. [url](https://www.springer.com/gp/book/9781402006319)
 

docs/src/examples/FOBP_example4.jl

@@ -1,4 +1,4 @@
-# # Foundations of Bilevel Programming: Example 4
+# # Example 4
 # This example is from the book _Foundations of Bilevel Programming_ by Stephan
 # Dempe, Chapter 5.1, Page 127. [url](https://www.springer.com/gp/book/9781402006319)
 

docs/src/examples/FOBP_example5.jl

@@ -1,4 +1,4 @@
-# # Foundations of Bilevel Programming: Example 5
+# # Example 5
 # This example is from the book _Foundations of Bilevel Programming_ by Stephan
 # Dempe, Chapter 8.1, Page 255. [url](https://www.springer.com/gp/book/9781402006319)
 

docs/src/examples/IEGT_example1.jl

@@ -0,0 +1,69 @@
+# # Example 9
+# This example is from the book Investment in Electricity Generation and Transmission. [url](https://www.springer.com/gp/book/9783319294995)
+# Bold point(s): using Dualof() in case we have information on the dual variable(s) of the lower level problem. 
+
+
+# Model of the problem 
+# First level
+# ```math
+# \min 40000x + 8760*(10y_1-\lambda * y_1),\\
+# \notag s.t.\\
+# 0 \leq x \leq 250,\\
+# 0 \leq \lambda \leq 20,\\
+# ```
+# Second level
+# ```math
+# \min 10y_1 + 12y_2 + 15y_3,\\
+# \notag s.t.\\
+# y_1 + y_2 + y_3 = 200,\quad [\lambda]\\
+# y_1 \leq x,\\
+# y_2 \leq 150,\\
+# y_3 \leq 100,\\
+# y[i] \geq 0, \forall i \in I\\
+# ```
+# In this problem, ``\lambda`` in upper level objective function is dual variable of the first constraint is the lower level problem.
+
+using BilevelJuMP
+using Ipopt
+using JuMP
+using Test
+
+model = BilevelModel(Ipopt.Optimizer, mode = BilevelJuMP.ProductMode(1e-9))
+
+# First we need to create all of the variables in the upper and lower problems:
+
+# Upper level variables
+@variable(Upper(model), x, start = 50)
+
+# Lower level variables
+@variable(Lower(model), -1 <= y[i=1:3] <= 300, start = [50, 150, 0][i])
+
+# Then we can add the constraints of the upper problem:
+@constraint(Upper(model), lb0, x >= 0)
+@constraint(Upper(model), ub0, x <= 250)
+
+# Followed by the constraints of the lower problem:
+
+# Right after the definition of a constraint, we are able to define its dual variable. In order to this, we need a reference to the constraint. Here **b** is a reference to the constraint with the dual variable  ``\lambda``.
+@constraint(Lower(model), b, y[1] + y[2] + y[3] == 200)
+@variable(Upper(model), 0 <= lambda <= 20, DualOf(b), start = 15)
+
+@constraint(Lower(model), ub1, y[1] <= x)
+@constraint(Lower(model), ub2, y[2] <= 150)
+@constraint(Lower(model), ub3, y[3] <= 100)
+@constraint(Lower(model), lb[i=1:3], y[i] >= 0)
+
+# At last we have objective function of the upper and lower problem:
+@objective(Upper(model), Min, 40_000x + 8760*(10y[1]-lambda*y[1]))
+@objective(Lower(model), Min, 10y[1] + 12y[2] + 15y[3])
+
+# Now we can solve the problem and verify the solution again that reported by book:
+optimize!(model)
+primal_status(model)
+termination_status(model)
+
+# Auto testing:
+@test objective_value(model) ≈ -190_000 atol=1e-1 rtol=1e-2
+@test value(x) ≈ 50 atol=1e-3 rtol=1e-2
+@test value.(y) ≈ [50, 150, 0] atol=1e-3 rtol=1e-2
+@test value(lambda) ≈ 15 atol=1e-3 rtol=1e-2