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PracticeLecture15

# PracticeLecture15 - UIUC MATH 482 LECTURE 14 IN CLASS...

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Unformatted text preview: UIUC MATH 482 LECTURE 14 IN CLASS EXERCISES (SPRING 2011) This worksheet is a supplement to Lecture 15. We work through why the primal—dual method is especially simple for the shortest path prob— lem. (This is covered in Section 5.4.) 1. The shortest path problem is min 0’ f subject to .Af==[1,0¢L0,H.,0r (where the 1 is associated to the source node 3) and f 2 0. (a) Why do we not write Af = [1,1,0,0,. . .,0]? (b) Conﬁrm the dual is max 7rs subject to 71'1' — 7Tj S Cij for all arcs (2',j) E E, and 7Tt : Write down the complementary slackness conditions and intepret as best you can. \../ P R D ( e h .Tu d n a m 13 b O r p \./ P R ( e h .Tu W 0 d e mu m 2 3 3. In the general primal—dual setup, you need to solve (RP) and then get a solution to (DRP). However, you can solve (DRP) directly if it is easy enough. (a) Explain why (DRP) is easy to solve. (b) Write down the new solution for 4. (a) Assume you use the solution to (DRP) from the back, and iterate, ﬁnding a new solution for (D) as in 3(b). Why is it that after some iterations, if you ever ﬁnd that the arcs from J admit a path P from s to t, you are optimal (and that path P is the shortest path). (b) Therefore, explain the textbook’s statement that “the primal— dual algorithm reduces the shortest path problem to repeated solution of the simpler problem of ﬁnding a set of nodes reachable from a given node. y z i l l , 4 l i t l i i t . l 5. Solve the following shortest path problems: The four LP’S in the primal—dual method (P) minz 2 (2’3: s.t. Ax = b(2 0) and :10 2 0. (D) maxw = 7r’b 3.1;. 7r’A S c’ (RP) min( = at? Zamej + 1’? = jeJ andmj20f0ra11j€J,xj=0forj¢J,x\$20 (DRP) maxw = 7r’b s.t. 7r’Aj _<_ 0 for j E 1, 2, . . . ,m. J, 7H S 1 forz' Some brief comments. 1(a) Since one of the conservation equations in the shortest path LP is redundant, as discussed in class. 1(b) The interesting conditions are f,_,j(7r,- —— 7r ~— cij) = 0. 2. The (RP) is m—l minC= E i=1 Af+a3a= [1,0,0,...,0]t andijOforallj,fj=0forj¢Jandzr\$ZO The (DRP) is subject to maxw = 7rs subject to 7m — 7rj S 0 for all —> 3') E J and 7l't S 1 3. (a) You can solve by setting 0 1 for all nodes reachable by paths from s using arcs in J o 0 for all nodes from which t is reachable using arcs in J o 1 for all other nodes (b) 91 = min{c,-j — (7r,- — 79)} where the minimum is over all arcs (i—>j)¢Jand7ri—7rj >0. 4. (a) By complementary slackness that each f,_,j 2 0 if —> j) E J. However, it is not hard to see one can ﬁnd an optimal f for (P) by setting fihj = 1 for each -> j) E J that is in our path, and setting all other arcs to 0. Since this is clearly feasible and satisﬁes complementary slackness (Thm 3.4), it is optimal. (b) The point is that in this shortest path algorithm we do not have to do any simplex. Once we know how to solve the simpler problem, it is easy to update J and eventually solve for the shortest path graphically. 5. Compare with the example on page 111—112. ...
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PracticeLecture15 - UIUC MATH 482 LECTURE 14 IN CLASS...

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