The Effects of Transformations on the Optimal Set in Interval Linear - PowerPoint PPT Presentation

The Effects of Transformations on the Optimal Set in Interval Linear Programming 1 Department of Applied Mathematics, Faculty of Mathematics and Physics, Charles University 2 Faculty of Informatics and Statistics, University of Economics, Prague The 14th International Symposium on Operations Research in Slovenia (SOR’17), Bled Elif Garajová 1 (with Milan Hladík 1 & Miroslav Rada 2 )

approximation and rounding estimating the future inexact measurements Interval Linear Programming €25 6 c €27 1 a 5 0 05 g b 3 14159 1 Consider a linear programming problem . . . minimize c T x subject to Ax ≤ b

Interval Linear Programming 1 approximation and rounding b ≈ 3 . 14159 Consider a linear programming problem . . . minimize c T x subject to Ax ≤ b estimating the future €25 . 6 ≤ c ≤ €27 . 1 inexact measurements a = 5 ± 0 . 05 g

Interval Linear Programming 1 approximation and rounding b = [ 3 . 141592 , 3 . 141593 ] Consider an interval linear programming problem . . . minimize c T x subject to A x ≤ b estimating the future c = [ 25 . 6 , 27 . 1 ] inexact measurements a = [ 4 . 95 , 5 . 05 ]

Interval Linear Programming: Definitions • An interval linear program is a family of linear programs interval program, if x is a feasible/optimal solution for 2 minimize c T x subject to x ∈ M ( A , b ) , where A ∈ A , b ∈ b , c ∈ c and M ( A , b ) is the feasible set. • A linear program in the family is called a scenario . • A vector x is a (weakly) feasible/optimal solution to the some scenario with A ∈ A , b ∈ b , c ∈ c .

Interval Linear Programming: Example maximize x 2 subject to • What are the possible feasible solutions? • Which solutions are optimal for some scenario? • What is the set/range of all optimal values? 3 [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 1 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 0 . 5 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 0 . 33 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 0 . 25 − 1

Interval Linear Programming: Example maximize 0 1 x 1 x 2 0 1 -1 4 3 2 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 0 . 25 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 0 . 33 − 1

Interval Linear Programming: Example 2 1 x 1 x 2 0 1 -1 4 maximize 3 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 0 . 5 − 1

Interval Linear Programming: Example maximize 1 1 x 1 x 2 0 1 -1 4 3 2 1 -1 -2 -3 -4 subject to x 2 3 Let's traverse through this! [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 − 1

Interval Linear Programming: Example 2 x 1 x 2 0 1 -1 4 3 1 maximize -1 -2 -3 -4 subject to x 2 3 [ − 1 , 1 ] x 1 + x 2 ≤ 0 x 2 ≤ 1 Optimal values: { 0 , 1 }

Overview of Basic Results Feasible solutions: Optimal values: • Best optimal value (inequalities): • Worst optimal value (inequalities): Optimal solutions: • Special cases, approximations, … 4 • Oettli–Prager: x solves A x = b ⇔ | A c x − b c | ≤ A ∆ | x | + b ∆ • Gerlach: x solves A x ≤ b ⇔ A c x − A ∆ | x | ≤ b For each s ∈ {± 1 } n solve min ( c c − D s c ∆ ) T x s. t. ( A c − A ∆ D s ) x ≤ b , D s x ≥ 0 T y ≤ c , A T y ≥ c , y ≤ 0 max b T y s. t. A

x 1 x 2 The solution 0 0 is now optimal, too! ...but the feasible set is the same! (Li, 2015) Dependency Problem (I) max 0 1 x 2 0 x 2 x 1 0 x 2 x 1 s. t. x 1 max s. t. x 1 5 [ 0 , 1 ] x 1 − x 2 = 0 , x 2 ≤ 1 , x 1 , x 2 ≥ 0 . Optimal set: { ( x 1 , x 2 ) ∈ R 2 : x 1 ∈ [ 1 , ∞ ) and x 2 = 1 }

The solution 0 0 is now optimal, too! ...but the feasible set is the same! (Li, 2015) Dependency Problem (I) max max x 1 s. t. s. t. x 1 5 [ 0 , 1 ] x 1 − x 2 = 0 , [ 0 , 1 ] x 1 − x 2 ≤ 0 , x 2 ≤ 1 , [ 0 , 1 ] x 1 − x 2 ≥ 0 , x 1 , x 2 ≥ 0 . x 2 ≤ 1 , x 1 , x 2 ≥ 0 . Optimal set: { ( x 1 , x 2 ) ∈ R 2 : x 1 ∈ [ 1 , ∞ ) and x 2 = 1 }

...but the feasible set is the same! (Li, 2015) Dependency Problem (I) max max x 1 s. t. x 1 s. t. 5 [ 0 , 1 ] x 1 − x 2 = 0 , 1 x 1 − x 2 ≤ 0 , x 2 ≤ 1 , 0 x 1 − x 2 ≥ 0 , x 1 , x 2 ≥ 0 . x 2 ≤ 1 , x 1 , x 2 ≥ 0 . Optimal set: { ( x 1 , x 2 ) ∈ R 2 : x 1 ∈ [ 1 , ∞ ) and x 2 = 1 } The solution ( 0 , 0 ) is now optimal, too!

Dependency Problem (I) max Wei Li. A note on dependency between interval linear systems (2015). max x 1 s. t. s. t. x 1 5 [ 0 , 1 ] x 1 − x 2 = 0 , [ 0 , 1 ] x 1 − x 2 ≤ 0 , x 2 ≤ 1 , [ 0 , 1 ] x 1 − x 2 ≥ 0 , x 1 , x 2 ≥ 0 . x 2 ≤ 1 , x 1 , x 2 ≥ 0 . Optimal set: { ( x 1 , x 2 ) ∈ R 2 : x 1 ∈ [ 1 , ∞ ) and x 2 = 1 } The solution ( 0 , 0 ) is now optimal, too! ...but the feasible set is the same! (Li, 2015)

Dependency Problem (II) Example (Hladík, 2012) All real numbers are now feasible solutions, because we can 6 [ 1 , 2 ] x + − [ 1 , 2 ] x − ≤ 2 , x + , x − ≥ 0 [ 1 , 2 ] x ≤ 2 → Original feasible set: ( −∞ , 2 ] Consider the new scenario 1 x + − 2 x − ≤ 2 , x + , x − ≥ 0… express any real x as x = x + − x − , where x + = max ( 2 x , 0 ) and x − = | x | .

What if A is fixed? Transformations: The General Case 7 min c T x : A x = b , x ≥ 0 transitivity min c T x : min c T x : A x ≤ b , x ≥ 0 A x ≤ b

Transformations: The General Case 7 min c T x : A x = b , x ≥ 0 transitivity min c T x : min c T x : A x ≤ b , x ≥ 0 A x ≤ b What if A is fixed?

b 1 x b 3 x Splitting Equations into Inequalities 0 x optimal for min c T x Ax b b 3 Ax 0 Ax Theorem 1 x optimal for scenario min c T x b 2 Proof idea: is equal to the optimal solution set of the program The optimal solution set of the interval linear program 8 min c T x : Ax = b , x ≥ 0 min c T x : Ax ≤ b 1 , − Ax ≤ − b 2 , x ≥ 0 with b 1 = b 2 = b .

Splitting Equations into Inequalities Theorem 1 The optimal solution set of the interval linear program is equal to the optimal solution set of the program Proof idea: 8 min c T x : Ax = b , x ≥ 0 min c T x : Ax ≤ b 1 , − Ax ≤ − b 2 , x ≥ 0 with b 1 = b 2 = b . x ∗ optimal for scenario min c T x : b 2 ≤ Ax ≤ b 1 , x ≥ 0 ⇒ Ax ∗ = b 3 ∈ b ⇒ x ∗ optimal for min c T x : Ax = b 3 , x ≥ 0

c 2 c 1 Imposing Non-negativity Proof idea: b . 3 x Ax c . Then, x is optimal for min c T c 3 A T y For an optimal x , we have by dual feasibility some y with 9 Theorem 2 min c T Let S denote the optimal solution set of min c T x : Ax ≤ b and let S ′ be the optimal solution set of the program 1 x + − c T 2 x − : Ax + − Ax − ≤ b , x + , x − ≥ 0 with c 1 = c 2 = c . Then, the following properties hold: • If x ∈ S , then there is ( x + , x − ) ∈ S ′ with x = x + − x − . • If ( x + , x − ) ∈ S ′ , then x + − x − ∈ S .

Imposing Non-negativity Theorem 2 min c T Proof idea: 9 Let S denote the optimal solution set of min c T x : Ax ≤ b and let S ′ be the optimal solution set of the program 1 x + − c T 2 x − : Ax + − Ax − ≤ b , x + , x − ≥ 0 with c 1 = c 2 = c . Then, the following properties hold: • If x ∈ S , then there is ( x + , x − ) ∈ S ′ with x = x + − x − . • If ( x + , x − ) ∈ S ′ , then x + − x − ∈ S . For an optimal x ∗ , we have by dual feasibility some y ∗ with A T y ∗ = c 3 ∈ [ c 2 , c 1 ] ⊆ c . Then, x ∗ is optimal for min c T 3 x : Ax ≤ b .

10 Transformations: The Special Case min c T x : Ax = b , x ≥ 0 transitivity min c T x : min c T x : Ax ≤ b Ax ≤ b , x ≥ 0

Transformations: The Special Case 10 min c T x : Ax = b , x ≥ 0 transitivity T h e o r e m 1 min c T x : min c T x : Ax ≤ b Ax ≤ b , x ≥ 0

Transformations: The Special Case Theorem 2 10 min c T x : Ax = b , x ≥ 0 transitivity T h e transitivity o r e m 1 min c T x : min c T x : Ax ≤ b Ax ≤ b , x ≥ 0

Conclusion • In interval linear programming, the basic transformations may change the feasible/optimal set and other properties of a program. • We have shown that the transformations do not affect the optimal set for problems with a fixed coefficient matrix (they may still change other properties!). • Thus, we can directly generalize results concerning the optimal set of a particular type of programs to other types. 11