Proving that a Valid Inequality is Facet-defining

1. Proving that a Valid Inequality is Facet-defining • Ref: W, p144-147. • X  Z+n. For simplicity, assume conv(X) bounded and full-dimensional. Consider example, X = {(x, y)R+m B1: i=1m xi  my}. conv(X) is full-dimensional: Consider the m+2 affinely independent points (0, 0), (0, 1), (ei, 1), i = 1, … , m. • Problem 1: Given X  Z+n and a valid inequality x  0 for X, show that the inequality defines a facet of conv(X). Ex: Show that xi  y is facet-defining. • Approach 1: (Use definition) Find n points x1, … , xn X satisfying x = 0, and then prove that these n points are affinely independent. Ex: Consider m+1 points: (0, 0), (ei, 1), and (ei+ej, 1) for j  i.

2. Approach 2: (indirect but useful way, see Thm 3.5, 3.6) • Select t  n points x1, … , xt  X satisfying x = 0. Suppose that all these points lie on a generic hyperplanex = 0. • Solve the linear equation system j=1n jxjk = 0 for k = 1, … , t in the n+1 unknowns (, 0). • If the only solution is (, 0) = (, 0) for  0, then the inequality x  0 is facet-defining. Ex: Show xi  y is facet-defining. Select points (0, 0), (ei, 1), (ei+ej, 1) for j  i that are feasible and satisfy xi= y. As (0, 0) lies on i=1m ixi + m+1y = 0, 0 = 0. As (ei, 1) lies on the hyperplanei=1m ixi + m+1y = 0, i = -m+1. As (ei+ej, 1) lies on the hyperplanei=1m ixi- iy= 0, j = 0 for j  i. So the hyperplane is ixi - iy= 0, and xi  y is facet-defining.

3. If conv(X) is not full-dimensional, we use Thm 3.6 in the previous slides. Refer Proposition 3.5 on p.274 for a possible application of Thm 3.6.

4. 4. Describing Polyhedra by Extreme Points and Extreme Rays • Prop 4.1: If P = {xRn: Ax  b}   and rank(A) = n-k, P has a face of dimension k and no proper face of lower dimension. Pf) For any face F  P, rank(AF=, bF=)  n-k  dim(F)  k. (Prop. 2.4) Show  F with dim(F) = k. Let F be a face of minimum dimension ( > 0). (If k=0, nothing to prove) Let x* be an inner point of F, dim(F) > 0   y  x*  F. Consider z() = x* + (y – x*),   R1. Suppose z() intersect aix = bi for some i  MF. Choose * = min {|i|: i  MF, z(i) lines in aix = bi}, and * = |i*|. Then *  0 (x* is an inner point)  Fi* = {xP: AF=x = bF=, ai*x = bi*}   is a face of P of smaller dimension than F, which is a contradiction. Hence z() not intersectaix = bi for any i  MF.  Ax* + A(y-x*)  b    R1  A(y-x*) = 0  y  F  F = {y: Ay = Ax*}  dim(F) = k since rank(A) = n-k. 

5. Frequently we assume P  R+n  rank(A) = n  P has zero-dimensional faces if P  . Assume rank(A) = n hereafter. • Def 4.1: x  P is an extreme point of P if there do not exist x1, x2  P, x1  x2 such that x = (1/2)x1 + (1/2)x2. • Prop 4.2: x is an extreme point of P  x is a zero-dimensional face of P. Pf) () Suppose x is zero-dimensional face  rank(Ax=) = n. (Prop 2.4) Let (A, b) be submatrix of (Ax=, bx=) with A: nn and rank n  x = A-1b. If x = (1/2)x1 + (1/2)x2, x1, x2  P, then since Axi  b, i = 1, 2, we have Ax1 = Ax2 = b ( Ax = (1/2)Ax1 + (1/2)Ax2 = b, Ax1  b, Ax2  b ) • x1 = x2 = x, so x is an extreme point. () If xP is not a zero-dimensional face of P, then rank(Ax=) < n. (Prop 2.4)   y  0 such that Ax=y = 0. For small  > 0, let x1 = x + y, x2 = x - y, x1, x2  P. Then x = (1/2)x1 + (1/2)x2, hence x is not an extreme point. 

6. Def 4.2: Let P0 = {r  Rn: Ar  0}. (recession cone, characteristic cone of P) If P = {x  Rn: Ax  b}  , then r  P0 \ {0} is called a ray of P. • r  Rn, r  0 is a ray of P   x  P, {y  Rn: y = x + r,   R+1}  P. • Note: Cone K is called pointed if K  (-K) = {0}. K  (-K) is called lineality space of cone K. For P = {x  Rn: Ax  b}, if rank(A) = n, P0  (-P0) = {r  Rn: Ar  0, -Ar 0} = {0}. Hence P0 is guaranteed to be pointed. • Def 4.3: A ray r of P is an extreme ray if there do not exist r1, r2 P0, r1 r2, R+1such that r = (1/2)r1+ (1/2)r2.

7. Prop 4.3: If P  , r extreme ray of P if and only if {r:   R+1} is one-dimensional face of P0. Pf) Let Ar= = {ai: iM, air = 0}. If {r: R+1} is a one-dimensional face of P0, rank(Ar=) = n-1  solutions of Ar=y = 0 are y = r, R1. If r = (1/2)r1 + (1/2)r2, get contradiction as in Prop 4.2. If r  P0 and rank(Ar=) < n-1, then nullity of Ar=  2. •  r*  r,   R1 such that Ar=r* = 0. Then r = (1/2)r1 + (1/2)r2, where r1 = r + r*, r2 = r - r*. Hence r is not an extreme ray, contradiction.  • Cor 4.4: A polyhedron has a finite number of extreme points and extreme rays. • Question: Given P = {xRn: Ax  b, x  0}  , how can we identify the extreme rays of P0? • Thm 4.5: If P , rank(A) = n, and max{cx: xP} is finite, then there is an optimal solution that is an extreme point. Pf) Set of optimal solution is face F = {xP: cx = c0}. By Prop. 4.1, F contains (n – rank(A))-dimensional face. By Prop. 4.2, F contains an extreme point. 

8. Thm 4.6: extreme points xk,  c  Zn such that xk is the unique optimal solution of max{cx: x  P}. Pf) Let Mxk= be equality set of xk. Let c* = iM=ai, c = c* for some >0 to get integer vector c. Then  xP\{xk}, cx = iM=aix <  bi =  aixk = cxk  (Compare with earlier Proposition regarding face.) • Thm 4.7: P  , rank(A) = n, max{cx: x  P} unbounded, then P has an extreme ray r* with cr* > 0. Pf) {uR+m: uA = c} =  from duality of LP  By Farkas,  r  Rn such that Ar  0, cr > 0. Consider max{cr: Ar  0, cr  1} = 1. By Thm 4.5,  optimal extreme point solution r*. Equality set of r* is Ar*=r = 0 and cr = 1  rank(Ar*=) = n – 1.  r* extreme ray of P (Prop 4.3)

9. Thm 4.8: (Affine) Minkowski’sThm: finitely constrained  finitely generated. P = {xRn: Ax  b}  , rank(A) = n (existence of extreme point guaranteed) • P = {xRn: x = kK kxk + jJ jrj,  k=1, k0, kK, j0, jJ}, Where xk: extreme points of P, rj: extreme rays of P. Pf) Let Q = {xRn: x = kK kxk + jJ jrj,  k=1, k0, kK, j0, jJ}. Q  P is clear Suppose  y  P \ Q (i.e. yP, but yQ). Show contradiction. Then not exist ,  satisfying kK kxk + jJ jrj = y - kK k = -1 k  0 for kK, j  0 for jJ By Farkas’ lemma,  (, 0)  Rn+1 such that xk - 0  0 for kK, rj  0 for jJ and y - 0> 0. Consider LP max{x: xP} If LP has a finite optimal solution, then  an extreme point optimal solution. Have xk - 0  0, but y - 0 > 0 (y > xk  k), contradiction. If unbounded,  extreme ray rj with rj > 0 (Thm 4.7), contradiction. Hence there does not exist such y, i.e. Q = P 

10. Consider Primal-Dual pair of LP z = max{cx: xP}, P = {xR+n: Ax  b} w = min{ub: uQ}, Q = {uR+m: uA  c} {xk, kK} extreme points of P, {rj, jJ} extreme rays of P0 {ui, iI} extreme points of Q, {vt, tT} extreme rays of Q0 • Thm of the alternatives: •  x such that x  0, Ax  b •  u such that u  0, uA  0, ub < 0 Pf) Consider primal-dual pair (P) max 0x, Ax  b, x  0 (D) min ub, uA  0, u  0

11. Thm 4.9: • The following are equivalent: • The primal problem is feasible, that is, P  ; • vtb  0 for all t  T. • The following are equivalent when the primal problem is feasible: • z is unbounded from above; •  rj of P with crj> 0; • the dual problem is infeasible, that is, Q = . • If the primal problem is feasible and z is unbounded, then z = maxkKcxk = w = miniIuib. Pf) I) P   if and only if vb  0  v  R+m with vA  0 (from previous) By Minkowski, Q0 = {vR+m: vA  0} = {vR+m: v = tT tvt, t  0, tT} • vb  0  vQ0 if and only if vtb  0  tT. II) P = {xRn: x = kK kxk + jJ jrj,  k=1, k0, kK, j0, jJ}  . z bounded if and only if crj  0  jJ. b  c: apply (I) to dual (in negation form) III) From strong duality and Minkowski’s theorem to P and Q. 

12. Note: More general form of Minkowski’sthm Decomposition Thm: Suppose P = {xRn: Ax  b}   Then P = S + K + Q, where S + K is the cone {xRn: Ax  0} S = {xRn: Ax = 0} is the lineality space of S + K K is a pointed cone. K + Q is a pointed polyhedron. Q is a polytope given by the convex hull of extreme points of K + Q.

13. Projection of a polyhedron: Projection of (x, y) Rn  Rp on H = {(x, y): y = 0} is (x, 0). Consider projection of P  Rn Rponto y = 0 as a projection from the (x, y)-space to the x-space, denoted by projx(P). (x such that (x, y)P for some yRp) • Thm 4.10: Let P = {(x, y) Rn  Rp: Ax + Gy  b}, then projx(P) = {xRn: vt(b-Ax)  0  tT}, where {vt}tT are extreme rays of Q = {vR+m: vG = 0} Pf) H = {(x, y)Rn Rp: y = 0} • ProjH(P) = {(x, 0) Rn  Rp: (x, y)P for some yRp}  {yRp:Gy (b-Ax)}    v  0, vG= 0, v(b-Ax) < 0 infeasible   v  0, vG= 0, we have v(b-Ax)  0  v(b-Ax)  0 p’Ax p’b for all v  Q  vt(b-Ax)  0 for all vt T (vtAx vtb )

14. For Thm 4.10, use the thmof the alternatives: •  x such that Ax  b •  u such that u  0, uA= 0, ub < 0 Pf) Consider primal-dual pair (P) max 0x, Ax  b (D) min ub, uA= 0, u  0 • Cor 4.11: Projection of polyhedron is polyhedron.

15. Cor 4.12: If P = {(x, y) Rn  Rp: Ax + Gy  b} and {xRn : Dx d}, where D is qn, then Q = projx(P) if and only if: • For i = 1, … , q, dix  d0i is a valid inequality for P. • For each x*  Q,  y* such that (x*, y*)  P. Pf) I. is equivalent to Q  projx(P). II is equivalent to Q  projx(P).  • Thm 4.13: (Affine Weyl’s theorem) (finitely generated  finitely constrained) If A: m1n, B: m2n, rational matrices and Q = {xRn: x = yA + zB, k=1m1yk = 1, yR+m1, zR+m2}, Then Q is a rational polyhedron. Pf) Q = projx(P), where P = {(x, y, z)Rn R+m1  R+m2 : x – yA – zB = 0, k=1m1yk = 1}  (Recall that we used Fourier-Motzkinelimination in IE531.)