1. CONSTRAINT GENERATION (BENDERS DECOMPOSITION)

2. Problem formulation Min f (x, y) Min cTx + f (y) Subject to Subject to
x ε X, y ε Y
x : nice variables
y : annoying variables
Min cTx + f (y)
Subject to
A x – F (y) ≥ b
x ≥ 0 , y ε Y

3. Numerical example

4. Projection on the annoying variables y space
Min cTx + f (y)
Subject to
A x – F (y) ≥ b
x ≥ 0 , y ε Y
Miny εY {f (y) +
Minx ≥ 0{cTx :A x ≥ b + F (y)}}

5. Numerical example

6. X(y) = {x : A x ≥ b + F (y), x ≥ 0 }
Equivalent problem
Miny εY {f (y) +
Minx ≥ 0{cTx :A x ≥ b + F (y)}}
Assume that X(y) is not empty for all y εY
Feasible domain
X(y) = {x : A x ≥ b + F (y), x ≥ 0 }

7. Problem transformations
Miny єY {f (y) + Minx ≥ 0{cTx :A x ≥ b + F (y}}
Additional transformation using linear programming duality
DUAL
(Dy) Max (b + F (y))Tu
Subject to
AT u ≤ c
u ≥ 0
PRIMAL
(Py) Min cTx
Subject to
A x ≥ b + F (y)
x ≥ 0

8. Numerical example

9. Miny єY {f (y) + Minx ≥ 0{cTx :A x ≥ b + F (y}}
More transformations
Miny єY {f (y) + Minx ≥ 0{cTx :A x ≥ b + F (y}}
DUAL
(Dy) Max (b + F (y))Tu
Subject to
AT u ≤ c
u ≥ 0
MaxekεE {(b + F (y))Tek}

11. Miny єY {f (y) + Max ekεE {(b + F (y))Tek}} Subject to
More transformations z
Miny єY {f (y) + Max ekεE {(b + F (y))Tek}}
Subject to
Max ekεE {(b + F (y))Tek} = z
MaxekεE {(b + F (y))Tek} = z
(b + F (y))Tek ≤ z, 1 ≤ k ≤ r

12. Equivalent transformed problem
Min {f (y) + z}
Subject to
(b + F (y))Tek ≤ z ≤ k ≤ r
y ε Y

14. Relaxation Min {f (y) + z} Subject to (b + F (y))Tek ≤ z 1 ≤ k ≤ r
y ε Y
Denote
R subset of {1, 2, …, r}
k ε R

16. for different subsets R
Solution approach
Solve a sequence of relaxed problems
Min {f (y) + z}
Subject to
(b + F (y))Tek ≤ z k ε R
y ε Y
for different subsets R

18. Constraint generation
Mechanism
1. to verify if the optimal solution y’ and z’ of a
relaxed problem is feasible (hence optimal)
for the problem
or if not,
2. to generate additional constraints to define a
new relaxed problem

19. Solving the dual problem
Mechanism
solve the dual problem
(Dy’) Max (b + F (y’))Tu
Subject to
AT u ≤ c
u ≥ 0
to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ and z’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Indeed
(EP) Min {f (y’) + z’}
Subject to
(b + F (y’))Tek ≤ (b + F (y’))Tek’≤ z
1 ≤ k ≤ r
y ε Y

20. Solving the dual problem
Mechanism solve the dual problem (Dy’) Max (b + F (y’))Tu Subject to AT u ≤ c u ≥ 0 to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Otherwize if (b + F (y’))T ek’ z’
then add the new constraint
(b + F (y’))T ek’ ≤ z’
to obtain a new the relaxion

21. Solving the dual problem
New relaxation
Min {f (y) + z}
Subject to
(b + F (y))Tek ≤ z k ε R
(b + F (y))Tek’ ≤ z
y ε Y
If (b + F (y’))T ek’ ≤ z’
solution y’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Otherwize if (b + F (y’))T ek’ z’
then add the new constraint
(b + F (y’))Tu’ ≤ z’
to obtain a new the relaxion

25. Solving the dual problem
Mechanism
solve the dual problem
Max (b + F (y’))Tu
Subject to
AT u ≤ c
u ≥ 0
to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ and z’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Indeed
(EP) Min {f (y’) + z’}
Subject to
(b + F (y’))Tek ≤ (b + F (y’))Tek’≤ z
1 ≤ k ≤ r
y ε Y

27. Additional note: If the primal problem has an interesting structure then we can modify the mechanish
Mechanism
solve the dual problem
(Dy’) Max (b + F (y’))Tu
Subject to
AT u ≤ c
u ≥ 0
to generate an optimal solution ek’
Alternate mechanism

28. Convergence of the relaxation approach Convergence of the method

29. Graphic illustration

30. Advantages Solve a sequence of relaxed (smaller) problems
Mechanism to verify if the optimal solution of relaxed problem is feasible (hence optimal) for the original problem: solving the dual problems D (y’)
Same mechanism generates additional constraints to modify the relaxed problem and to get closer to the original one

31. Drawback
DUAL METHOD:
the solution y, z is feasible only once the
procedure is completed

32. Extensions
Case where the set X(y) is empty for some y εY
Generalized version to deal with the problem
Min f (x, y)
Subject to
F (x, y) ≤ 0
x ε X, y ε Y

33. Variant of Benders decomposition using upper bound UB and lower bound LB of the optimal value of (P)

34. for different subsets R
Solution approach
Solve a sequence of relaxed problems
Min {f (y) + z}
Subject to
(b + F (y))Tek ≤ z k ε R
y ε Y
for different subsets R

35. Constraint generation
Mechanism
1. to verify if the optimal solution y’ and z’ of a
relaxed problem is feasible (hence optimal)
for the problem
or if not,
2. to generate additional constraints to define a
new relaxed problem

36. Solving the dual problem
Mechanism
solve the dual problem
(Dy’) Max (b + F (y’))Tu
Subject to
AT u ≤ c
u ≥ 0
to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ and z’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Indeed
Min {f (y’) + z’}
Subject to
(b + F (y’))Tek ≤ (b + F (y’))Tek’≤ z
1 ≤ k ≤ r
y ε Y

37. Solving the dual problem
Mechanism
solve the dual problem
Max (b + F (y’))Tu
Subject to
AT u ≤ c
u ≥ 0
to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ and z’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Indeed
Min {f (y’) + z’}
Subject to
(b + F (y’))Tek ≤ (b + F (y’))Tek’≤ z
1 ≤ k ≤ r
y ε Y

38. Solving the dual problem
Mechanism solve the dual problem (Dy’) Max (b + F (y’))Tu Subject to AT u ≤ c u ≥ 0 to generate an optimal solution ek’
If (b + F (y’))T ek’ ≤ z’
solution y’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Otherwize if (b + F (y’))T ek’ z’
then add the new constraint
(b + F (y’))T ek’ ≤ z’
to obtain a new the relaxion

39. Solving the dual problem
New relaxation
Min {f (y) + z}
Subject to
(b + F (y))Tek ≤ z k ε R
(b + F (y))Tek’ ≤ z
y ε Y
If (b + F (y’))T ek’ ≤ z’
solution y’ is feasible
(hence optimal) for the equivalent problem and we have an optimal solution for the problem
Otherwize if (b + F (y’))T ek’ z’
then add the new constraint
(b + F (y’))Tu’ ≤ z’
to obtain a new the relaxion