Mixing Inequalities and Maximal Lattice-Free Triangles

Transcription

1 Mixing Inequalities and Maximal Lattice-Free Triangles Santanu S. Dey Laurence A. Wolsey Georgia Institute of Technology; Center for Operations Research and Econometrics, UCL, Belgium. 2

2 Outline Mixing Inequalities 2 Connection With Lattice-free Triangles 3 Result: Strengthening Mixing Inequalities For Use In Simplex Tableau

3 Mixing Inequalities.

4 The mixing set [Günlük and Pochet (2)] y + x b y + x 2 b 2 y + x 3 b y + x n b n y R + x i Z i n

5 The mixing set [Günlük and Pochet (2)] y + x b y + x 2 b 2 y + x 3 b y + x n b n y R + x i Z i n Mixing Inequality is facet-defining for the Mixing Set: y n ( b i b i )( b i x i ) () i= where b i = b i b i +, b i b i and b =.

6 5 The mixing set appears as a substructure in many problems The mixing inequality can be used to derive facets for: Production Planning (Constant capacity lot-sizing) 2 Capacitated Facility Location 3 Capacitated Network Design

7 5 The mixing set appears as a substructure in many problems The mixing inequality can be used to derive facets for: Production Planning (Constant capacity lot-sizing) 2 Capacitated Facility Location 3 Capacitated Network Design Can we use mixing inequalities for general problems?

8 6 Rearranging the mixing set for simplicity ( ) ( x + ) ( x 2 + ) y ( b b 2 ) x, x 2 Z, y R + Let r i = b i (mod ). We assume < r < r 2 <. Mixing Inequality: y (r 2 r )( b 2 x 2 ) + r ( b x )

9 6 Rearranging the mixing set for simplicity ( ) ( x + ) ( x 2 + ) y ( b b 2 ) x, x 2 Z, y R + Let r i = b i (mod ). We assume < r < r 2 <. Mixing Inequality: y (r 2 r )( b 2 x 2 ) + r ( b x ) Introduce non-negative slack variables: ( ) ( ) ( x + x 2 + ) y + ( ) ( y + ) y 2 = ( b b 2 ) x, x 2 Z, y, y, y 2 R + Mixing Inequality: r 2 D y + r D y + r 2 r D y 2 where D = (r 2 r )( r 2 ) + r ( r ).

10 Using mixing inequalities for general simplex tableau x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + 8 y + 8 y y 2 Idea: Rewrite/Relax rows of simplex tableau to look like the Mixing Set u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 +

11 Using mixing inequalities for general simplex tableau x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + 8 y + 8 y y 2 Idea: Rewrite/Relax rows of simplex tableau to look like the Mixing Set ( ) ( ) ( ).5 u + u 2 + u u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + (.9.9 ) ( ) ( ) u 4 + u u.4 4 = ( ).4.6

12 Using mixing inequalities for general simplex tableau x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + 8 y + 8 y y 2 Idea: Rewrite/Relax rows of simplex tableau to look like the Mixing Set ( ) ( ) ( ).5 u + u 2 + u ( ) ( u }{{} + x ) u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ( u }{{} 2 + x 2 (.9.9 ) (.5u 3 +.9u 4 ) + }{{} y ) ( ) ( ) u 4 + u u.4 4 = ( ) (.4u 3 +.4u 4 ) = }{{} y 2 ( ).4.6 ( ).4.6

13 Using mixing inequalities for general simplex tableau x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + 8 y + 8 y y 2 Idea: Rewrite/Relax rows of simplex tableau to look like the Mixing Set ( ) ( ) ( ).5 u + u 2 + u ( ) ( u }{{} + x ) u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ( u }{{} 2 + x 2 (.9.9 ) (.5u 3 +.9u 4 ) + }{{} y ) ( ) ( ) u 4 + u u.4 4 = ( (.5u u 4 ) + () + 5 (.4u u 4 ) ) (.4u 3 +.4u 4 ) = }{{} y 2 ( ).4.6 ( ).4.6

14 8 They are many possible ways to relax simplex tableau u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 +

15 8 They are many possible ways to relax simplex tableau ( u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ) ( ) ( ) ( ) ( ) ( ) ( ) u + u 2 + u. 3 + u u 3 + u.4 4 =.6

16 8 They are many possible ways to relax simplex tableau ( ( u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ) ( ) ( ) ( ) ( ) ( ) ( ) u + u 2 + u. 3 + u u 3 + u.4 4 =.6 ) ( ) ( ) ( ) ( ) ( ).4 (u + u 3 ) + u }{{} 2 + (.u }{{} 3 +.9u 4 ) +.6u }{{} 3 +.4u }{{} 4 = }{{}.6 x x 2 y y y 2

17 8 They are many possible ways to relax simplex tableau ( ( u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ) ( ) ( ) ( ) ( ) ( ) ( ) u + u 2 + u. 3 + u u 3 + u.4 4 =.6 ) ( ) ( ) ( ) ( ) ( ).4 (u + u 3 ) + u }{{} 2 + (.u }{{} 3 +.9u 4 ) +.6u }{{} 3 +.4u }{{} 4 = }{{}.6 x x 2 y y y 2 8 (.u 3 +.9u 4 ) + 8 (.6u 3) (.3u 4)

18 9 They are many possible ways to relax simplex tableau ( ( u + u 2 +.5u 3 +.9u 4 =.4 u + u 2 +.u 3 +.5u 4 =.6 u Z 4 + ) ( ) ( ) ( ) ( ) ( ) ( ) u + u 2 + u. 3 + u u 3 + u.4 4 =.6 ) ( ) ( ) ( ) ( ) (.4 (u + u 3 ) + u }{{} 2 + (.u }{{} 3 +.9u 4 ) +.6u }{{} 3 +.4u }{{} 4 = }{{}.6 x x 2 y y y 2 8 (.u 3 +.9u 4 ) + 8 (.6u 3) (.3u 4)

19 The best cut For the simplex tableau: x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + n i= ( a i a i 2 ) u i + ( m c j j= c j 2 8 y + 8 y y 2 ) (.4 v j =.6 ). u i Z +, v j R + (2)

20 The best cut For the simplex tableau: x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + n i= ( a i a i 2 ) u i + ( m c j j= c j 2 The best cut by relaxation of the simplex tableau is where, i= 8 y + 8 y y 2 ) (.4 v j =.6 ). u i Z +, v j R + (2) n m φ (a, i a2)u i i + π (c j, cj 2 )v j (3) j=

21 The best cut For the simplex tableau: x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + n i= ( a i a i 2 ) u i + ( m c j j= c j 2 The best cut by relaxation of the simplex tableau is where, i= 8 y + 8 y y 2 ) (.4 v j =.6 ). u i Z +, v j R + (2) n m φ (a, i a2)u i i + π (c j, cj 2 )v j (3) j= φ (a, a 2 ) = min y 8 + y y 8 2 s.t. x + y y = a x 2 + y y 2 = a 2 x, x 2 Z, y, y, y 2 R +

22 The best cut For the simplex tableau: x + y y =.4 x 2 + y y 2 =.6 x, x 2 Z, y, y, y 2 R + n i= ( a i a i 2 ) u i + ( m c j j= c j 2 The best cut by relaxation of the simplex tableau is where, i= 8 y + 8 y y 2 ) (.4 v j =.6 ). u i Z +, v j R + (2) n m φ (a, i a2)u i i + π (c j, cj 2 )v j (3) j= φ (a, a 2 ) = min y 8 + y y 8 2 s.t. x + y y = a x 2 + y y 2 = a 2 x, x 2 Z, y, y, y 2 R + π (c, c 2 ) = min y 8 + y y 8 2 s.t. y y = c y y 2 = c 2 y, y, y 2 R +

23 Closed form for φ and π Let F(a, a 2 ) = (a (mod ), a 2 (mod )).

24 Closed form for φ and π Let F(a, a 2 ) = (a (mod ), a 2 (mod )). Proposition Consider two rows of a simplex tableau: n i= a ix i + m j= c iy i = b, x i Z +, y i R +. Then the inequality n i= φ (F(a i ))x i + m i= π (c i )y i is valid inequality where, φ (w, w 2 ) = σ ( w ) + σ 2 ( w 2 ) (w, w 2 ) R σ 3 (w ) + σ 2 ( w 2 ) (w, w 2 ) R 2 σ ( w ) + σ 2 ( w 2 ) (w, w 2 ) R 3 σ ( w ) + σ 4 (w 2 ) (w, w 2 ) R 4 σ 3 (w ) + σ 2 ( w 2 ) (w, w 2 ) R 5 σ ( w ) + σ 4 (w 2 ) (w, w 2 ) R 6 where (r, r 2 ) = F(c) and σ =, σ r 2 (r ) 2 r 2 (r 2 r ) 2 = r σ 3 =, σ r r 2 (r ) 2 r 2 (r 2 r ) 4 = 2 +r and r 2 (r ) 2 r 2 (r 2 r ) r r 2 r r 2 (r ) 2 r 2 (r 2 r ), (4) π (c) = lim h φ(ch) h (5)

25 Closed form for φ and π Let F(a, a 2 ) = (a (mod ), a 2 (mod )). Proposition Consider two rows of a simplex tableau: n i= a ix i + m j= c iy i = b, x i Z +, y i R +. Then the inequality n i= φ (F(a i ))x i + m i= π (c i )y i is valid inequality where, φ (w, w 2 ) = σ ( w ) + σ 2 ( w 2 ) (w, w 2 ) R σ 3 (w ) + σ 2 ( w 2 ) (w, w 2 ) R 2 σ ( w ) + σ 2 ( w 2 ) (w, w 2 ) R 3 σ ( w ) + σ 4 (w 2 ) (w, w 2 ) R 4 σ 3 (w ) + σ 2 ( w 2 ) (w, w 2 ) R 5 σ ( w ) + σ 4 (w 2 ) (w, w 2 ) R 6 where (r, r 2 ) = F(c) and σ =, σ r 2 (r ) 2 r 2 (r 2 r ) 2 = r σ 3 =, σ r r 2 (r ) 2 r 2 (r 2 r ) 4 = 2 +r and r 2 (r ) 2 r 2 (r 2 r ) r r 2 r r 2 (r ) 2 r 2 (r 2 r ), (4) π (c) = lim h φ(ch) h (5) Observation: The closed form of φ depends only on the fractional part of columns of integer variables.

26 Closed form for φ contd. 2

27 Is it possible to get a stronger general-purpose mixing-type cut? 3

28 3 Is it possible to get a stronger general-purpose mixing-type cut? More precisely: Yes!

29 3 Is it possible to get a stronger general-purpose mixing-type cut? More precisely: Yes! (φ, π ) represents a valid inequality for the infinite group relaxation of mixed integer programs.

30 3 Is it possible to get a stronger general-purpose mixing-type cut? More precisely: Yes! (φ, π ) represents a valid inequality for the infinite group relaxation of mixed integer programs. We show that there exist functions (φ M, π ) that strictly dominate (φ, π ) and are extreme for the infinite group relaxation of mixed integer programs.

31 3 Is it possible to get a stronger general-purpose mixing-type cut? More precisely: Yes! (φ, π ) represents a valid inequality for the infinite group relaxation of mixed integer programs. We show that there exist functions (φ M, π ) that strictly dominate (φ, π ) and are extreme for the infinite group relaxation of mixed integer programs. Upshot: Better cut coefficients can be obtained using the function (φ M, π ) that improve upon the coefficients obtained by (φ, π ). We proceed step-by-step in the following slides.

32 The Framework: Infinite Group Relaxation.

33 5 Infinite relaxation of two-rows of simplex tableau The mixed integer relaxation: MG(I 2, R 2, f ): ax(a) + wy(w) + f = a I 2 w R 2 ( ) ( ) x B + x B2 x B, x B2 Z, x(a) Z +, y(w) R + x, y have finite support, (6) where I 2 = {(a, a 2 ) R 2 a, a 2 }, i.e. set of all columns of two fractions.

34 5 Infinite relaxation of two-rows of simplex tableau The mixed integer relaxation: MG(I 2, R 2, f ): ax(a) + wy(w) + f = a I 2 w R 2 ( ) ( ) x B + x B2 x B, x B2 Z, x(a) Z +, y(w) R + x, y have finite support, (6) where I 2 = {(a, a 2 ) R 2 a, a 2 }, i.e. set of all columns of two fractions. Assuming all the nonbasic variables are continuous: MG(, R 2, f ): ( ) ( ) wy(w) + f = x B + x B2 w R 2 x B, x B2 Z, y(w) R + y has a finite support. (7)

35 Problem statement precisely Now that we have defined the infinite relaxation, we proceed to ask the following precise questions... 6

36 Problem statement precisely Now that we have defined the infinite relaxation, we proceed to ask the following precise questions... The function (φ, π ) represents a valid inequality for MG(I 2, R 2, f ) as: If ( x, ȳ) satisfies: ( ) ( ) ax(a) + wy(w) + f = x B + x B2 a I 2 w R 2 then ( x, ȳ) satisfies: a I 2 φ (a)x(a) + x B, x B2 Z, x(a) Z +, y(w) R +, w R 2 π (w)y(w) (8) Question: Do there exist functions (φ, π ), φ : I 2 R + and π : R 2 R + such that: φ φ and π π and φ (a)x(a) + π (w) (9) a I 2 w R 2 is a valid inequality? 6

37 Analysis of π : The Strength of Continuous Coefficients.

38 8 Maximal lattice-free convex sets [Lovász(989)] Definition A set S is called a maximal lattice-free convex set in R 2 if it is convex and interior(s) Z 2 =, 2 There exists no convex set S satisfying (), such that S S. [Borozan and Cornuéjols (27), Andersen, Louveaux, Weismantel, and Wolsey (27)] Theorem For the system MG(, R 2, f ), an inequality of the form w R 2 π(w)y(w) is un-dominated, if the set P(π) = {w R 2 π(w f ) } () is a maximal lattice-free convex set.

39 9 Strength of continuous coefficients, i.e., π π (w, w 2 ) = min 8 y + 8 y y 2 s.t. y y = w y y 2 = w 2 y, y, y 2 R +

40 Strength of continuous coefficients, i.e., π π (w, w 2 ) = min 8 y + 8 y y 2 s.t. y y = w We construct P(π ) : {w R 2 π(w f ) }: y y 2 = w 2 y, y, y 2 R +.5 (f, f 2 )

41 Strength of continuous coefficients, i.e., π π (w, w 2 ) = min 8 y + 8 y y 2 s.t. y y = w We construct P(π ) : {w R 2 π(w f ) }: y y 2 = w 2 y, y, y 2 R (f, f 2 ) For all r, P(π ) is a maximal lattice-free triangle. π is undominated by any inequality: i.e. π : R 2 R + such that π π, π (w) < π (w) for some w R 2 where π is valid inequality for MG(, R 2, r).

42 2 π is an extreme inequality for MG(, R 2, r) [Cornuéjols and Margot(28)] Theorem If P(π) is a maximal lattice-free triangle, then π represents an extreme inequality for MG(, R 2, f ), i.e, π, π 2 : R 2 R + such that π and π 2 represent valid inequalities, π π 2, and π = 2 π + 2 π 2. Corollary The function representing coefficient of continuous variable obtained using mixing inequalities, i.e., π, is extreme for MG(, R 2, r). We cannot improve the coefficient of continuous variables in the mixing cut.

43 Analysis of φ : The Strength of Integer Coefficients.

44 It is possible to strengthen φ φ (a, a 2 ) = min y 8 + y y 8 2 s.t. x + y y = a x 2 + y y 2 = a 2 x, x 2 Z, y, y, y 2 R + π (c, c 2 ) = min y 8 + y y 8 2 s.t. y y = c y y 2 = c 2 y, y, y 2 R +

45 It is possible to strengthen φ φ (a, a 2 ) = min y 8 + y y 8 2 s.t. x + y y = a x 2 + y y 2 = a 2 x, x 2 Z, y, y, y 2 R + π (c, c 2 ) = min y 8 + y y 8 2 s.t. y y = c y y 2 = c 2 y, y, y 2 R + We can rewrite: φ (a, a 2 ) = min x,x 2 Z(π (a x, a 2 x 2 )). ()

46 22 It is possible to strengthen φ φ (a, a 2 ) = min y 8 + y y 8 2 s.t. x + y y = a x 2 + y y 2 = a 2 x, x 2 Z, y, y, y 2 R + π (c, c 2 ) = min y 8 + y y 8 2 s.t. y y = c y y 2 = c 2 y, y, y 2 R + We can rewrite: φ (a, a 2 ) = min x,x 2 Z(π (a x, a 2 x 2 )). () () is the fill-in function [Gomory and Johnson (972)]. [D. and Wolsey (28)] Theorem If P(π) is a lattice-free triangle with non-integral vertices and exactly one integer point in the interior on each side, then φ : I 2 R + defined as in () is not an undominated inequality. a function φ : I 2 R + such that φ represents a valid inequality and φ < φ.

47 A stronger inequality φ M 23

48 Comparing φ M with φ 24

49 25 Main result: (φ M, π ) is extreme inequality for MG(I 2, R 2, r) Theorem Given < r < r 2 < such that r + r 2, the functions (φ M, π ) represent an extreme inequality for MG(I 2, R 2, r) Note: The condition r + r 2 is not very serious, since given any two rows of tableau such that r + r 2 >, multiple both the rows with. Then for the resulting rows, r + r 2. Steps in Proof: Prove function results in a valid inequality:

50 25 Main result: (φ M, π ) is extreme inequality for MG(I 2, R 2, r) Theorem Given < r < r 2 < such that r + r 2, the functions (φ M, π ) represent an extreme inequality for MG(I 2, R 2, r) Note: The condition r + r 2 is not very serious, since given any two rows of tableau such that r + r 2 >, multiple both the rows with. Then for the resulting rows, r + r 2. Steps in Proof: Prove function results in a valid inequality: Show that φ M (u) + φ M (v) φ M (u + v) u, v I 2, i.e., prove φ M : I 2 R + is subadditive.

51 25 Main result: (φ M, π ) is extreme inequality for MG(I 2, R 2, r) Theorem Given < r < r 2 < such that r + r 2, the functions (φ M, π ) represent an extreme inequality for MG(I 2, R 2, r) Note: The condition r + r 2 is not very serious, since given any two rows of tableau such that r + r 2 >, multiple both the rows with. Then for the resulting rows, r + r 2. Steps in Proof: Prove function results in a valid inequality: Show that φ M (u) + φ M (v) φ M (u + v) u, v I 2, i.e., prove φ M : I 2 R + is subadditive. 2 Prove function is un-dominated:

52 25 Main result: (φ M, π ) is extreme inequality for MG(I 2, R 2, r) Theorem Given < r < r 2 < such that r + r 2, the functions (φ M, π ) represent an extreme inequality for MG(I 2, R 2, r) Note: The condition r + r 2 is not very serious, since given any two rows of tableau such that r + r 2 >, multiple both the rows with. Then for the resulting rows, r + r 2. Steps in Proof: Prove function results in a valid inequality: Show that φ M (u) + φ M (v) φ M (u + v) u, v I 2, i.e., prove φ M : I 2 R + is subadditive. 2 Prove function is un-dominated: Easily verified that, φ M (u) + φ M (r u) = u I 2. Therefore, by [Johnson s (974)] φ M is an un-dominated inequality.

53 25 Main result: (φ M, π ) is extreme inequality for MG(I 2, R 2, r) Theorem Given < r < r 2 < such that r + r 2, the functions (φ M, π ) represent an extreme inequality for MG(I 2, R 2, r) Note: The condition r + r 2 is not very serious, since given any two rows of tableau such that r + r 2 >, multiple both the rows with. Then for the resulting rows, r + r 2. Steps in Proof: Prove function results in a valid inequality: Show that φ M (u) + φ M (v) φ M (u + v) u, v I 2, i.e., prove φ M : I 2 R + is subadditive. 2 Prove function is un-dominated: Easily verified that, φ M (u) + φ M (r u) = u I 2. Therefore, by [Johnson s (974)] φ M is an un-dominated inequality. 3 Prove function is extreme.

54 26 Proving φ M is a subadditive function [D. and Richard (28)] Proposition Let φ be a continuous, piecewise linear and nonnegative function on I 2 such that φ(u) + φ(r u) =. Let V and E be the set of vertices and edges of φ. Then φ is subadditive iff φ(v ) + φ(v 2 ) φ(v + v 2 ) φ(e ) + φ(e 2 ) φ(v 3 ) where e q, e 2 q 2, e + e 2 = v 3, v 3 V(φ) V (φ) v, v 2 V(φ), q, q 2 E(φ). To check subadditivity of a piecewise linear function we need to check only the function at the vertices and edges.

55 Proving φ M is a subadditive function: Too many cases!.5.5 ~ 28 Cases ~ 44 Cases.5 v + v 2 v v 2.5 v v 2 v +v After checking these case we prove that the function φ M is a valid inequality.

56 28 A value-function interpretation of φ M In fact, by the subadditivity of φ M we obtain the following result: Proposition If r + r 2, then for the problem: ( ) ( ) ( ) x + x + + r r 2 2 r +r 2 2 The following inequality is valid: ( ) ( ) ( ) ( ) r x 3 + y + y + y 2 = r2 x, x 2 Z, x 3 Z +, y, y, y 2 R + (r 2 r )( r 2 ) x 3 + r 2 2D D y + r D y + r 2 r D y 2

57 A value-function interpretation of φ M In fact, by the subadditivity of φ M we obtain the following result: Proposition If r + r 2, then for the problem: ( ) ( ) ( ) x + x + + r r 2 2 r +r 2 2 The following inequality is valid: ( ) ( ) ( ) ( ) r x 3 + y + y + y 2 = r2 x, x 2 Z, x 3 Z +, y, y, y 2 R + (r 2 r )( r 2 ) x 3 + r 2 2D D y + r D y + r 2 r D y 2 Then φ M can be obtained as follows: φ M (a, a 2 ) = min (r 2 r )( r 2 ) x 3 + r 2 2D D y + r D y + r 2 r D y 2 s.t. x + (+ r r 2 )x 3 + y y = a 2 x 2 + r + r 2 x 3 + y y 2 = a 2 2 x, x 2 Z, x 3 Z +, y, y, y 2 R + Without the terms corresponding to x 3 the above reduces to the problem corresponding to φ. 28

58 29 Proving (φ M, π ) is extreme inequality for MG(I 2, R 2, r) [D. and Wolsey (28)] Theorem Let π : R 2 R + be a extreme inequality for MG(, R 2, r). 2 Let u I 2 and define V = max n Z+,n { π(w) F(u n n + w) = r}.... Lifting. 3 Define φ : I 2 R + as φ(v) = min n Z+ {nv + π(w) F(u n + w) = v}.... Fill-in. If (φ, π) is an un-dominated inequality for MG(I 2, R 2, r), then (φ, π) is an extreme inequality for MG(I 2, R 2, r).

59 29 Proving (φ M, π ) is extreme inequality for MG(I 2, R 2, r) [D. and Wolsey (28)] Theorem Let π : R 2 R + be a extreme inequality for MG(, R 2, r). 2 Let u I 2 and define V = max n Z+,n { π(w) F(u n n + w) = r}.... Lifting. 3 Define φ : I 2 R + as φ(v) = min n Z+ {nv + π(w) F(u n + w) = v}.... Fill-in. If (φ, π) is an un-dominated inequality for MG(I 2, R 2, r), then (φ, π) is an extreme inequality for MG(I 2, R 2, r). In our case, u = (+ r r 2 2, r +r 2 2 ) and (r 2 r )( r 2 ) 2D = max π (w) n Z+,n { F(u n + w) = r} n

60 29 Proving (φ M, π ) is extreme inequality for MG(I 2, R 2, r) [D. and Wolsey (28)] Theorem Let π : R 2 R + be a extreme inequality for MG(, R 2, r). 2 Let u I 2 and define V = max n Z+,n { π(w) F(u n n + w) = r}.... Lifting. 3 Define φ : I 2 R + as φ(v) = min n Z+ {nv + π(w) F(u n + w) = v}.... Fill-in. If (φ, π) is an un-dominated inequality for MG(I 2, R 2, r), then (φ, π) is an extreme inequality for MG(I 2, R 2, r). In our case, u = (+ r r 2 2, r +r 2 2 ) and and (r 2 r )( r 2 ) 2D φ M (a, a 2 ) = min = max π (w) n Z+,n { F(u n + w) = r} n (r 2 r )( r 2 ) x 3 + r 2 2D D y + r D y + r 2 r D y 2 s.t. x + ( + r r 2 )x 3 + y y = a 2 x 2 + r + r 2 x 3 + y y 2 = a 2 2 x, x 2 Z, x 3 Z +, y, y, y 2 R +

61 3 Discussion We illustrated techniques to use mixing inequalities for general two-rows of a simplex tableau.

62 3 Discussion We illustrated techniques to use mixing inequalities for general two-rows of a simplex tableau. 2 We showed that when applying the mixing inequalities for general two-rows of a simplex tableau, the inequality can be strengthened.

63 3 Discussion We illustrated techniques to use mixing inequalities for general two-rows of a simplex tableau. 2 We showed that when applying the mixing inequalities for general two-rows of a simplex tableau, the inequality can be strengthened. 3 A new class of extreme inequality for two-row mixed integer infinite group problem.

64 3 Discussion We illustrated techniques to use mixing inequalities for general two-rows of a simplex tableau. 2 We showed that when applying the mixing inequalities for general two-rows of a simplex tableau, the inequality can be strengthened. 3 A new class of extreme inequality for two-row mixed integer infinite group problem. Challenges: The proof of validity of the stronger inequality is not elegant: More importantly the proof does not extend to more rows of a mixing inequalities.

65 Thank You. 3