Math 5490 Network Flows

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1 Math 590 Network Flows Lecture 7: Preflow Push Algorithm, cont. Stephen Billups University of Colorado at Denver Math 590Network Flows p./6

2 Preliminaries Optimization Seminar Next Thursday: Speaker: Ariela Sofer, George Mason University Title: Optimization in Medical Diagnosis and Treatment Location: NC 3, :30-:30 Tonight: Pre-Flow Push Algorithm Overview of Linear Programming Assignments: HWK 7: 7., 7.9 (a,b,c,d,f,g), 7., 7.9 (Due Tuesday, March 8) Read: 9., 9.3 Math 590Network Flows p./6

3 Animation of the algorithm (fig 7.3) e(3)=0 d(3)= 3 5 e()=0 d()= Source 3 Sink e()=0 d()=0 e()=0 d()= Starting labels Math 590Network Flows p.3/6

4 Animation of the algorithm (fig 7.3) e(3)= d(3)= 3 5 e()=0 d()= Source 3 Sink e()=0 d()=0 e()= d()= After preprocess operation: arcs out of source node ((,3) and (,)) have been saturated. Selecting node for push/relabel operation, arc (,) is only admissible arc. Push δ = min{e(), r } = unit along this arc (saturating the arc). Math 590Network Flows p.3/6

5 Animation of the algorithm (fig 7.3) e(3)= d(3)= 3 5 e()=0 d()= Source 3 Sink e()= d()=0 e()= d()= After saturating arc (,) Now, node has no admissible arcs, so set d() = min{d(3) +, d() + } = min{, 5} =. Selecting node 3, push units along arc (3,) (nonsaturating) Math 590Network Flows p.3/6

6 Animation of the algorithm (fig 7.3) e(3)=0 d(3)= 3 e()=0 d()= Source 3 Sink e()=5 d()=0 e()= d()= After pushing all the excess from node 3 along arc (3,) Selecting node, push unit along (,3). Math 590Network Flows p.3/6

7 Animation of the algorithm (fig 7.3) e(3)= d(3)= 3 e()=0 d()= Source Sink e()=5 d()=0 e()=0 d()= After pushing all the excess from node along arc (,3) Selecting node 3, push unit along (3,). Math 590Network Flows p.3/6

8 Animation of the algorithm (fig 7.3) e(3)=0 d(3)= 3 5 e()=0 d()= Source Sink e()=6 d()=0 e()=0 d()= After pushing all the excess from node 3 along arc (3,). Done! Math 590Network Flows p.3/6

9 Correctness of the algorithm When the algorithm terminates, there are no excesses, so the preflow is actually a flow. Furthermore, the label of node s is n, so the residual network contains no path from node s to node t. So the flow is a maximum flow. Math 590Network Flows p./6

10 Complexity The generic pre-flow push algorithm runs in O(n m) time. Before we consider special implementations that improve on this complexity bound, let us first consider how to speed up the generic algorithm. The pre-flow push algorithm often gets the maximal amount of flow to the sink, and then has to spend quite a bit more effort pushing remaining excesses back to the source in order to get a feasible flow. To improve performance, we can do the following:. Detect when the maximum amount of flow has reached the sink.. Convert the resulting preflow to a flow. (This will be homework). To detect when a maximum amount of flow has reached the sink, we maintain a set N of nodes for which the residual network does not contain a path to the sink. Math 590Network Flows p.5/6

11 Discussion Question: How can we decide when to add a node to N? Possibile answers:. When a node has label n.. Occasionally do a reverse breadth first search to get exact distance labels. (How often can you do this?) 3. As was done with the shortest path augmenting flow algorithm, we can keep track of the number of nodes for each distance label (Let numb(k) represent the number of nodes with label k.) If numb(k ) = 0 then there is no path to the sink for any node with label > k. Math 590Network Flows p.6/6

12 Specific Implementations of the Preflow-Push Algorithm We can improve on the complexity of the preflow-push algorithm by specifying particular rules for choosing which active nodes to look at. There are three main variations:. FIFO: Examines the active nodes in FIFO order. This approach runs in O(n 3 ) time.. Highest-label: Always choose the active node with the highest label. Complexity: O(n m / ). Intuition: (See Figure 7.5). 3. Excess scaling: Choose active nodes with sufficiently large excess (typically, at least half of the maximum excess). But also make sure that no excess becomes too large by any push operation. O(nm + n log U). The excess scaling algorithm is similar to the capacity scaling algorithm. This approach ensures that each push carries a sufficiently large amount of flow, thereby keep the number of pushes small. Math 590Network Flows p.7/6

13 Essentials of Linear Programming Stanard form LP Geometry of LPs Dual Problem Duality Theorems Optimality Conditions Math 590Network Flows p.8/6

14 Standard form LP min x subject to c T x Ax = b x 0 where A is an m n matrix, b IR m, c IR n, and x IR n. It is futher assumed that A has full row rank. (If it doesn t, we could find an equivalent set of constraints by eliminating rows). Example min x x + x + 3x 3 subject to x x + 7x 3 = x + x + x 3 = 5 x, x, x 3 0 Math 590Network Flows p.9/6

15 Constraint Region The constraint region for a linear program is a polyhedral region. The classic picture to illustrate this is for the following LP in min c x + c x + c 3 x 3 subject to x + x + x 3 = x, x, x 3 0 The constraint region is shown in the following figure: If an LP has a solution, then it must have a solution on a vertex of the feasible region. (Note: it could have addition solutions that are not vertices). Math 590Network Flows p.0/6

16 The Dual Associated with the (Primal) linear program described above is another linear program called the dual: min x subject to c T x Ax = b x 0 {z } Primal max b T y subject to A T y c {z } Dual Observe that the dual involves the same data as the primal (A, c, and b). But note the following differences: The dual is a maximization problem instead of a minimization problem. The coefficients in the objective function are the right hand sides of the from the primal problem. The right hand sides are the objective coefficients from the primal. Math 590Network Flows p./6

17 Alternative Form of Dual Some of you may be used to seeing the dual problem written as max subject to b T y A T y + s = c s 0 which is derived by adding the slack variable s. Math 590Network Flows p./6

18 Weak Duality Theorem: Given a feasible solution x to the primal problem and a feasible solution y to the dual problem, then b T y c T x. Proof x is feasible for the primal, so Ax = b. y is feasible for the dual, so A T y c. Thus, b T y = x T A T y x T c = c T x, where the inequality above comes from the fact that x 0 and A T y c. Math 590Network Flows p.3/6

19 Strong Duality Theorem: If the primal problem has an optimal solution, so does its dual, and the respective optimal costs are equal. Proof Take Math Math 590Network Flows p./6

20 Complementary Slackness Let x and y be feasible solutions to the primal and dual problems, respectively. These vectors are optimal solutions for the two respective problems if and only if: Proof y i (Ax b) i = 0 for all i (c j (y T A) j )x j = 0 for all j The first set of inequalities is trivial since Ax b = 0. For the second set, by strong duality, we have c T x = y T b. Thus, (c Ay) T x = c T x y T Ax = c T x y T b = 0. But, x 0 and A T y c, (by respective feasibilities). Thus, (c Ay) T x = X j (c j (y T A) j )x j ) is the sum of nonnegative quantities. Since this sum equals zero, then every term of the sum must be zero. Math 590Network Flows p.5/6

21 Four Case Theorem One of the following cases must hold:. Both the primal and dual problems have solutions x and y, respectively, and c T x = b T y.. The primal problem is unbounded and the dual is infeasible. 3. The dual problem is unbounded and the primal is infeasible.. Both problems are infeasible. Proof If either problem has a solution, then the strong duality theorem implies that the first case holds. If either problem is unbounded, the other must be infeasible by the weak duality theorem. The only remaining possibility would be if neither problem has a solution, and neither is unbounded. This means that both problems are infeasible. Math 590Network Flows p.6/6

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