6. Polyhedral Ties between Linear and Integer Programs

1. 6. Polyhedral Ties between Linear and Integer Programs • P = {xR+n: Ax  b} S = P  Zn, want conv(S) is a rational polyhedron P bounded  S finite or   conv(S) is polyhedron (Weyl) If P unbounded, may exist infinite number of points in S. Can we have Weyl type result? • Thm 6.1: P = {xR+n: Ax  b}  , S = P  Zn, where (A, b) is an integer m(n+1) matrix, then the following statements are true. •  finite {ql}lL of S and rays {rj}jJ of P such thatS = {xR+n: x = lL lql+ jJjrj,  l=1, Z+|L|, Z+|J|} • If P is a cone,  rays {vh}hH of P such that S = {xR+n: x = hH hvh, Z+|H|}.

2. Pf) I.P = {xR+n: x = kKkxk + jJjrj,  k=1, k, j0, for kK and jJ} Assume {rj} for jJ are integer vectors. Let Q = {xZ+n: x = kK kxk + jJ jrj,  k=1, k 0 for kK, 0j<1 for jJ} Q is a finite set, say Q = {qlZ+n: l  L} and Q  S. Observe that xi  S  xi  Z+n and xi ={kKkixk+ jJ(ji - ji)rj}+{jJjirj}, k=1, k, j0, for kK and jJ. • xi = ql(i) + jJjirj, ji= j for all jJ. II. If P is a cone, ql  S implies ql  S    Z+1. Take {vh : hH} = {ql : l  L}  {rj : j  J} 

3. Thm 6.2: P = {xR+n: Ax  b}, S = P  Zn  conv(S) is a rational polyhedron. Pf) Convex combination of points {xiS, iI} can be written as x = iI ixi = iI i(ql(i) + jJ jirj) (iI i = 1) = lL({iI: l(i)=l}i)ql + jJ(iI iji)rj = lL lql + jJ jrj, Where l = {iI: l(i)=l}i  0 for lL, lLl = iIi = 1, and j =iI iji  0 for jJ.  conv(S) = {xR+n: x = lL lql + jJ jrj, lLl= 1, l, j 0 for lL and jJ}, with ql, rj Z+n for lL and jJ. By Thm 4.13 (Weyl), conv(S) is a rational polyhedron.  • Proof extends straightforwardly to mixed-integer sets with rational data. Hence, all of the following results apply to mixed-integer sets. • Extreme rays of P = {xR+n: Ax  b} and conv(P  Zn) coincide.

4. Thm 6.2 suggests that we can solve (IP) max{cx: xS, S = P  Zn by solving the LP (CIP) max{cx: xconv(S)}. • Thm 6.3: Given S = PZn  , P={xR+n: Ax  b} and cRn, it follows that: • IP unbounded  CIP unbounded. • If CIP has a bounded optimal value  It has an optimal solution (namely, an extreme point of conv(S)) that is an optimal solution to IP. • IP has an optimal solution x0  x0 is an optimal solution to CIP. Pf) Let z0 and z* be the optimal values of IP and CIP, respectively. z*  z0 since conv(S)  S. a) z0 =   z* =  z* =    extreme point x0conv(S) and ray rZ+n such that cr > 0. x0 + r  conv(S)    0  x0 + r  S  Z+1  z0 = . b) CIP has an optimal solution   extreme point optimal solution, say x0  x0  S  z0  cx0 = z*  z0 = z* (since z*  z0) c) x0 optimal to IP  CIP not unbounded (from a.) and x0  conv(S)  . Hence, x0 is an optimal solution to CIP (from b.) 

5. Cor 6.4: IP is either infeasible or unbounded or has an optimal solution. • We say that (, 0) is a valid inequality for a set S if x  0 for all xS. • Prop 6.5: If x  0 is valid for S, it is also valid for conv(S). Pf) Consider an xconv(S). Then x = jJ jxj, where xj  S for jJ and jJj = 1 and j  0 for jJ. Hence x = jJj(xj)  jJj0 = 0. 

6. Prop 6.6: x  0 face of dimension k-1 of conv(S)   k affinely independent points x1, … , xk  S such that xi = 0, i = 1, … , k. Pf)  affinely independent x1, … , xk  conv(S) such that xi = 0, i =1, … ,k. Suppose x1S, x1 = jJ jx*j, x*jS, j>0 for all jJ, jJ j=1. x1= 0, x*j  0 for jJ  x*j= 0 for all jJ x1, … , xkaffinely independent   x*j* such that x*j*, x2, … , xkaffinely independent. Replace x1 with x*j* and repeat the procedure.  • Relate IP to LP: z(d) = max{cx: Ax  d, xZ+n} zLP(d) = max{cx: Ax  d, xR+n}

7. Prop 6.7: • zLP(0) = z(0) • z(0) = 0  Q = {uR+n: uA  c}   • z(0) =   Q =  • If Q  , then S = P  Zn =  or z(b) finite • If Q = , then S =  or z(b) =  Pf) Clearly 0  z(0)  zLP(0) (primal feasible) If Q    zLP(0) = 0  z(0) = zLP(0) = 0 If Q =    ray rj of P with crj > 0. can take rj to be integer.  z(0) = zLP(0) =  (a, b, c) If Q    P =  or zLP(b) is finite, hence z(b) finite (d) If Q =   P =  or zLP(b) = zLP(0) =  If S    z(0) = z(b) =  (from c) (e)  • Cor 6.8: • P =   S =  • zLP(b) finite  S =  or z(b) finite • zLP(b) =   S =  or z(b) = 