Download
1 / 4
40 likes | 114 Vues
If Pij & Pjk then Pik Not(Pij & Pjk) or Pik (see implication in PL) NotPij or NotPjk or Pik 1-xij + 1-xjk + yik >= 1 xij + xjk + (1-yik) <=2 This works with objective function Min sum of yik’s So object function forces yik to be 0
E N D
If Pij & Pjk then Pik Not(Pij & Pjk) or Pik (see implication in PL) NotPij or NotPjk or Pik 1-xij + 1-xjk + yik >= 1 xij + xjk + (1-yik) <=2 This works with objective function Min sum of yik’s So object function forces yik to be 0 Unless both xij,xjk are 1 then constraint forces yik to be 1 So a quadratic term is represented by yik and problem remains linear, i.e. ILP
Graph coloring • Xv,c and Yc • Xv,c + Xu,c <= 1 for all (u,v) edges, c in other words vertices u,v must be assigned different colors if they share an edge. • We need Yc to represent the objective function which is to minimize total number of colors. If Xu,c=1 then Yc =1 • Using PL, Puc=>Pc, 1-Xuc + Yc >=1 • Yc >= Xuc for all u nodes, c
GUB constraint … • Yc <= sum u Xuc for all colors • This provides an upper bound on Yc • Although objective function of sum over c Yc will force Yc’s to zero anyways if the color is not being used.
Set packing problem • Let Xi represent a vertex i of the graph • Select max number of vertices • No two vertices can share an edge • Ax <= 1 • Represents Xi + Xj <=1 for all (i,j) edges
