1 / 13

Flow x feasible for capacity constraints

4. -4. Flow x feasible for capacity constraints x is feasible for b(1)=4 & b(4)=-4 . Is x min-cost-flow?. No, because: residual G(x) has nega-cycle W ( with c(W)=-1<0 ). nega-cycle W ={(3,4) (4,2) (2,3)} with c(W)=-1 and b = min { 4,3,2 }  [i,j]  W: x ij :=x ij +2.

tyrone-larson
Télécharger la présentation

Flow x feasible for capacity constraints

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 4 -4 Flow x feasible for capacity constraints x is feasible for b(1)=4 & b(4)=-4 . Is x min-cost-flow? No, because: residual G(x) has nega-cycle W ( with c(W)=-1<0 )

  2. nega-cycle W ={(3,4) (4,2) (2,3)} with c(W)=-1 andb = min {4,3,2} [i,j] W: xij:=xij +2

  3. nega-cycle W ={(3,4) (4,2) (2,1) (1,3)} with c(W)=-1 andb = min {2, 1, 3, 2} [i,j] W: xij:=xij +1

  4. Algorithm CYCLE (*for MCF*); Determine a feasible solution x ; (with maxflow) While nega-cycle in G(x) do identify such a nega-cycle W b := min { rij : {i,j}  W} augment x along W withb update G(x) ; O(nm) } O(n) Initial flow x has cost cx <= mUC O(nm2 UC) time 10.5 withminimum mean cycles O(nm2 log n) iterations  O(n2m3log n)

  5. Let pi be a value on node i, for every node i pi pj 5 reduced cost: cp(i,j):= c(i,j)-pi +pj 2 3 -1 P=1-2-4-3 with c(P): 12+7+9=18 'reduced cost' cp(P): -3+18+2 For any path P: reduced cost cp(P) = -pa+c(P)+pb For any cycle W: reduced cost cp(W)=c(W) x=x* for c no c - nega cycle in G(x)  no cp - nega cycle in G(x)  x=x* for cp

  6. no c - nega cycle in G(x)  x=x* for c i.e. x not optimal  there is a c - nega cycle in G(x) Proof Suppose cx*<cx. By equality x*=x+(x*-x) The difference flow x':=x*-x is a circular flow in G(x) -with all b(i)=0 We have c(x*)=cx+c(x*-x)=cx+cx', so cx'<0 Further x' is sum of O(m) cycles Wq ,so cx' = Sq c(Wq) Thus at least one cycles must have c-negative cost

  7. Theorem 1 x=x* for c no c - nega cycle in G(x)  no cp - nega cycle in G(x)  x=x* for cp Theorem 2 x=x*  p such that cp(i,j)>=0 for every [i,j]  G(x) proof Assume x=x*. (No nega-cycle in G(x), so possible to) Compute c-shortest distances di w.r.t. arbitrary node s in G(x). For pi:=-di we have cp(i,j)=c(i,j)-pi+pj = c(i,j)+di-dj Now shortest path relations dj <=di +c(i,j) for [i,j] translate to cp(i,j)>=0 in G(x)  Suppose for some p all reduced costs cp(i,j)>=0 in G(x) All cycles W in G(x) have cp(W)>=0, so (theorem 1) x is optimal

  8. cij , rij and new values pi -2 -2 2 -3 -0 -2 0 -2 2 1 0 1 new reduced costs cP new d(i), Pst 2 -4 4 3 0 2 Pstwithb:= min {4,2,5,4,|-4|}}; Algorithm shortest-path-MCFLet x be MCF flow for pseudo-values bx(i);Define pi s.t. cp>=0 in G(x) ; (*e.g. x=0 andp=0*)  i : e(i):=b(i)-bx(i); E:={i: e(i)>0}, D:={j: e(j)<0}; while E non-empty do select s in E and t in D in G(x): cp -shortest distances d from s and Pst; b:= min {rij : [i,j] in Pst}  {e(s)}  {-e(t)}; augment x along Pst with b; update G(x), E, D; p:=p-d ; (*N.B. again cp >=0 in G(x) *)

  9. 0 -2 2 1 0 1 p=(0,-2,-2,-3) 0 0 p*=(0,-2,-2,-3)–(0,0,1,1)= (0,0,-2,-3,-4) Pstwithb:= min {4,2,3, 2,|-2|}=2; Algorithm shortest-path-MCFLet x be MCF flow for pseudo-values bx(i);Define pi s.t. cp>=0 in G(x) ; (*e.g. x=0 andp=0*)  i : e(i):=b(i)-bx(i); E:={i: e(i)>0}, D:={j: e(j)<0}; repeat select s in E and t in D in G(x): cp -shortest distances d from s and Pst; b:= min {rij : [i,j] in Pst}  {e(s)}  {-e(t)}; augment x along Pst with b; update G(x), E, D; p:=p-d ; (*N.B. again cp >=0 in G(x) *) until E empty cp*

  10. Algorithm shortest-path-MCFLet x be MCF flow for pseudo-values bx(i);Define pi s.t. cp>=0 in G(x) ; (*e.g. x=0 andp=0*)  i : e(i):=b(i)-bx(i); E:={i: e(i)>0}, D:={j: e(j)<0}; repeat select s in E and t in D in G(x): cp -shortest distances d from s and Pst; b:= min {rij : [i,j] in Pst}  {e(s)}  {-e(t)}; x' := x with b augmentation along Pst ; p' := p-d ; (*N.B. again c p'>=0 in G(x) *) x:=x' ; p' :=p, update G(x), E, D; until E empty ; case 2: [i,j] in G(x') but not in G(x) This means: x(j,i)=0 and x'(j,i)=b (or xij=uij and x'ij=uij-b), so so [j,i] on Pst and di = dj + cp(j,i) Now cp'(i,j)=c(i,j)-(pi-di)+(pj-dj)= =cp(i,j)+di-dj = -cp(j,i)+di-dj=0 Assume cp>=0 in G(x).Why: cp'>=0 in G(x') Proof: We check it for all arcs [i,j] of G(x') case 1: [i,j] is in G(x) and in G(x') cp'(i,j)=c(i,j)-(pi-di)+(pj-dj)=cp(i,j)+di-dj >=0, because with d as distance in network G(x), cp , we have dj <= di +cp(i,j) for [i,j] in G(x)

  11. Algorithm shortest-path-MCFLet x be MCF flow for pseudo-values bx(i);Define pi s.t. cp>=0 in G(x) ; (*e.g. x=0 andp=0*)  i : e(i):=b(i)-bx(i); E:={i: e(i)>0}, D:={j: e(j)<0}; repeat select s in E and t in D in G(x): cp -shortest distances d from s and Pst; b:= min {rij : [i,j] in Pst}  {e(s)}  {-e(t)}; x' := x with b augmentation along Pst ; p' := p-d ; (*N.B. again c p'>=0 in G(x) *) x:=x' ; p' :=p, update G(x), E, D; until E empty ; Complexiteit per iteratie O(m+n log n) (thepath-distance computation dominates). per iteratie Si in E ei decreases by at least one unit initieelSi in E ei <= Sbi <= nB with B largest supply  O(nB (m+n log n) ) is O(n3) for B=1 in assignment problem

More Related