Lemma 1 for each s v such that x s and y s the net

Info icon This preview shows pages 2–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Lemma 1. For each S V such that x / S and y / S , the net flow out of S and the net flow into S both equal 0. That is, f + ( S ) = f - ( S ). Proof. By the conservation condition, 0 = u S { uv A f ( uv ) - wu A f ( wu ) } = uv A : u S f ( uv ) - wu A : u S f ( wu ) = (total flow on arcs with tails in S ) - (total flow on arcs with heads in S ) = uv A : u S,v V - S f ( uv ) - wu A : w V - S,u S f ( wu ) = f + ( S ) - f - ( S ) .
Image of page 2

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Let N be a network with source x and sink y . Let f be a flow in N . Lemma 2. Net flow out of x equals net flow into y . f + ( x ) - f - ( x ) = f - ( y ) - f + ( y ) . Proof. Consider S = V - { x, y } . Assume first that there is no arc between x and y . Then f + ( S ) = f - ( x ) + f - ( y ) and f - ( S ) = f + ( x ) + f + ( y ) . By the previous lemma, f + ( S ) = f - ( S ), which implies the claim. Proof (continued). If xy is an arc but yx is not, then f + ( S ) = f - ( x ) + ( f - ( y ) - f ( xy )) f - ( S ) = ( f + ( x ) - f ( xy )) + f + ( y ) . Since f + ( S ) = f - ( S ), we get the same equality. Similarly, if yx is an arc but xy is not, or if both xy and yx are arcs, we still get the same equality. Definition 6. The value of the flow f is defined as val ( f ) := f + ( x ) - f - ( x ) = f - ( y ) - f + ( y ) . Maximum flow problem Definition 7. A flow f in N is a maximum flow if val ( f ) val ( g ) for every flow g in N . The value of a maximum flow is the maximum amount of the commodity that can be transported from x to y under the capacity constraints. Problem 1 (Maximum Flow Problem) . Find a maximum flow in a given network. Cuts Intuition: A set of arcs “separating” x and y limits flows from x to y . Definition 8. Let N = ( V, A ) be a network with source x and sink y . Let S be a set of vertices of N containing x but not containing y . The cut associated with S is the set of all arcs uv A , where u S and v V - S . This cut is denoted by ( S, V - S ). Each subset S V such that x S but y / S gives rise to a cut. Definition 9. The capacity of a cut ( S, V - S ) is the sum of the capacities of its arcs cap ( S, V - S ) = X uv A : u S,v V - S c ( uv ) . Lemma 3. For every flow f and cut ( S, V - S ) in N , val ( f ) = f + ( S ) - f - ( S ) . Proof. By the definition of val ( f ) and the conservation condition, val ( f ) = xv A f ( xv ) - wx A f ( wx ) = u S { uv A f ( uv ) - wu A f ( wu ) } = uv A : u S f ( uv ) - wu A : u S f ( wu ) = (total flow on arcs with tails in S ) - (total flow on arcs with heads in S ) = uv A : u S,v V - S f ( uv ) - wu A : w V - S,u S f ( wu ) = f + ( S ) - f - ( S ) .
Image of page 3
Definition 10. An arc a is saturated by f if f ( a ) = c ( a ) and a is f -zero if f ( a ) = 0. Lemma 4. For every flow f and cut ( S, V - S ) of N , val ( f ) cap ( S, V - S ) . Furthermore, val ( f ) = cap ( S, V - S ) iff each arc in ( S, V - S ) is saturated by f and each arc from V - S to S is f -zero. Proof. By the previous lemma, val ( f ) = f + ( S ) - f - ( S ). Since f + ( S ) cap ( S, V - S ) and f - ( S ) 0, val ( f ) cap ( S, V - S ) . To prove the “furthermore” claim, note that val ( f ) = cap ( S, V - S ) iff f + ( S ) = cap ( S, V - S ) and f - ( S ) = 0. We have f + ( S ) = cap ( S, V - S ) iff each arc in ( S, V - S ) is saturated by f , and f - ( S ) = 0 iff each arc from V - S to S is f -zero.
Image of page 4

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern