pg111.112oftext - 110 Chapter5 THE PRIMAL-DUAL ALGORITHM...

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Unformatted text preview: 110 Chapter5 THE PRIMAL-DUAL ALGORITHM min c’f +1 <— Row 5 0 Af= ; (P) f.>_0 where A is the (m — l) X n node-arc incidence matrix for a directed graph G = (V, E) with m nodes and n arcs, with the row corresponding to the terminal I omitted (it is redundant); f e R" is the vector of flow (in this problem with components 0 or I), and c e R" is the cost vector. The dual is max 7:, 7:, — 7:, g c” for all arcs (i,j) e E (D) 7:, 2 O for all 1' 7:, = 0 where we must fix 7:, = 0, since its row is omitted from A. The set of admissible arcs is then defined by J: {arcs (i,j) :7t, — 7:, = CU} and the restricted primal becomes m—l min f = Z x? (=1 +l <—— Row 5 0 Af—l— x” = (RP) 0 f, 2 0 for all j f; = 0 1' at J x? 2 0 Finally, the dual of the restricted primal is max w = 7:, 7:, 7:, g 0 for all arcs (i,j) E J (DRP) 7:, g l for all z' V 5.4 The Primal-Dual Method App/fed to the Shortest-Path Problem 111 Now DRP is very easy to solve: since 7:, g l and we wish to maximize 7:“ we try 7:, = 1. If there is no path from 7:, to m, using only arcs in J, then we can propagate the I from s to all nodes reachable by a path from 5 without violating the 7:, — 7:, g 0 constraints, and an optimal solution to DRP is then for all nodes reachable by paths from s using arcs in J l 7? = 0 for all nodes from which 1 is reachable using arcs in J (5.17) l for all other nodes (Notice that this 7? is not unique.) We can then calculate 0, = min {cu — (7r, — 7r,)} arc: (I, J) ¢J such that m-z;>0 update 7: and J, and re-solve DRP. If we get to a point where there is a path from s to 1‘ using arcs in J, 7:, = 0, and we are optimal, since 6 path from s to t using only arcs in J is optimal. Thus, the primal-dual algorithm reduces the shortest- repeated solution of the simpler problem of finding the set 0 from a given node. min = wnnx = Any path problem to f nodes reachable Example 5.1 Figure 5—3 shows a shortest- path problem somewhat more complicated than our previous example. Figure 5—3 An example of the shortest-path problem. The cir- cled numbers are are weights. Since the costs are nonnegative, we can begin with 7: = 0 in D. Figure 5—4 shows the succession of five iterations of the primal-dual method that finally reaches the optimal 7: = (6, 5, 5, 2, 4) and the corresponding are set J, which contains the optimal path. Any path from s to t in the final admissible are set will satisfy complementary slackness and hence be optimal; in this case the optimal path is e2, e5, e5, e7, with cost 6 = 7r,. D I 1 6I = 2 for c K arc e7 . o f:> J=¢ CD 1' x ‘ O . t 0 V 0 I I C D'vr=(0,0,0:00) e Iteration l 3. 2 2 2 :> J = {7} :> W‘ sh 2 2 alg [I l D: 3:012, 2: 2: 2) Iteration 2 [O 4 2 wh reqi = 7 4 :9 J { ,6} :> grow 4 4 § D: 1r=(4, 4, 4, 2, 4) Iteration 3 5 2 o o Letu 5 :(>J={7,6,5,4} :> 1«\ “CO! I \ 61=lfor 5 4 O 0 arce2 we ch cost x D: 7r: (5, 5, 5, 2, 4)‘ DRP: 1? =0, 0, o, 0, 0) tile pr Iteration 4 tion 0 5 2 0 0 “comt 8 VS from l J: 7, 6, 5,4, 4:52 { WQ‘ 5 4 0 0 D: _7r=(6, 5, 5, 2, 4) DRP: 1r- (0, 0,0, 0,0) Iteration 5 . In Figure 5—4 Solvmg a shortest-path problem using the prima‘l- and all dual algorithm. RP and but only We now note some properties of this algorithm that lead to a very simple 1cm, whi ' interpretation of what is happening First, if we define at any point in the The algorithm the set lems tha W = {i : t is reachable from i by admissible arcs} Gen“?! ti . _ algorithm = I“ 7T! = 0} We 1 112 ...
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This note was uploaded on 02/19/2012 for the course MATH 482 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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pg111.112oftext - 110 Chapter5 THE PRIMAL-DUAL ALGORITHM...

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