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Unformatted text preview: 110 Chapter5 THE PRIMALDUAL ALGORITHM min c’f
+1 <— Row 5
0 Af= ; (P) f.>_0 where A is the (m — l) X n nodearc 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 PrimalDual Method App/fed to the ShortestPath 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
mz;>0 update 7: and J, and resolve 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 primaldual 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 shortestpath 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 primaldual 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 shortestpath 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.
 Spring '08
 Staff

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