Discrete (and Continuous) Optimization Solutions of Exercises 7 WI4 ...

Discrete (and Continuous) OptimizationSolutions of Exercises 7WI4 131Kees RoosTechnische Universiteit DelftFaculteit Electrotechniek, Wiskunde en InformaticaAfdeling Informatie, Systemen en Algoritmieke-mail: C.Roos@ewi.tudelft.nlURL: http://www.isa.ewi.tudelft.nl/˜roosNovember – December, A.D. 2003

Course Schedule1. Formulations (18 pages)2. Optimality, Relaxation, and Bounds (10 pages)3. Well-solved Problems (13 pages)4. Matching and Assigments (10 pages)5. Dynamic Programming (11 pages)6. Complexity and Problem Reduction (8 pages)7. Branch and Bound (17 pages)8. Cutting Plane Algorithms (21 pages)9. Strong Valid Inequalities (22 pages)10. Lagrangian Duality (14 pages)11. Column Generation Algorithms (16 pages)12. Heuristic Algorithms (15 pages)13. From Theory to Solutions (20 pages)Optimization Group 1

Exercise 7.8.1Consider the enumeration tree for a minimization problem, as shown left below.1. Give tightest possible lower and upper bounds on the optimal value z.2. Which nodes can be pruned and which must be explored further?Solution: Starting at the lower nodes we tighten the bounds as indicated: in each node thenew bound is the minimum of the corresponding bound for its immediate successors. Theresult is as indicated in the tree at the right below. The nodes that can be pruned are 8 (byinfeasibility), 7 (by optimality), 4 (by optimality), 6 (by bound). Nodes 3 and 5 need furtherinvestigation.∞31328127317314300258infeasible52732226635∞3283112831731314310278infeasibleOptimization Group 2∞52732227635

Exercise 7.8.3 (part (ii))(ii). Solve the instancez = max17x 1 + 10x 2 + 25x 3 + 17x 45x 1 + 3x 2 + 8x 3 + 7x 4 ≤ 12x ∈ {0,1} 4 .The variables are already ordered so that c ja jis nonincreasing. Hence, by part (i), the solution of the linearrelaxation is x = (1,1, 1 2 ,0) with objective value 391 2 . We branch on the variable x 3. The correspondingsubsets are:S 0 = { x ∈ {0,1} 4 : 5x 1 + 3x 2 + 7x 4 ≤ 12, x 3 = 0 }S 1 = { x ∈ {0,1} 4 : 5x 1 + 3x 2 + 7x 4 ≤ 4, x 3 = 1 } .We first explore S 1 . It is clear that the only feasible solution is x = (0,1,1,0) with objective value 35.Thus we have already the incumbent x = (0,1,1,0), and the lower bound z = 35. Node S 1 is pruned byoptimality.The solution of the linear relaxation for node S 0 is x = (1,1,0, 4 7 ) with objective value 365 . Branching on7x 4 we get the subsetsS 10 = { x ∈ {0,1} 4 : 5x 1 + 3x 2 ≤ 12, x 3 = 0, x 4 = 0 }S 11 = { x ∈ {0,1} 4 : 5x 1 + 3x 2 ≤ −3, x 3 = 0, x 4 = 1 } .Node S 11 is pruned by infeasibility. The optimal solution for the (integer) subproblem for node S 10 is x =(1,1,0,0) with objective value 27. This node can be pruned by bound, since 27 < z = 35.Since no active nodes are left, the current incumbent x = (0,1,1,0), with objective value z = 35, is theoptimal solution.Optimization Group 4

Solve the 5-city STSP instance with distance matrix⎛(c ij ) =Exercise 7.8.5 ((i))⎜⎝- 10 2 4 610 - 9 3 12 9 - 5 64 3 5 - 26 1 6 2 -To obtain a lower bound for z we first find a minimal weight 1-tree. It has weight z = 12, which is a lowerbound for z.110 24 692 31 53 64 521-tree for S16⎞⎟⎠ .S¬ {2,5}12{2,5}S 0 S 1We can also easily obtain an upper bound: it happens that the 1-tree contains a Hamilton path from node 3 tonode 2; adding edge {2,3} we get a tour of length 18; it also contains a Hamilton path from node 3 to node5; adding edge {3,5} we get a tour of length 16. So we put ¯z = 16.We choose to branch on the cheapest edge in the minimal weight 1-tree, namely edge {2,5}. Then the set Sof all tours T is decomposed intoS 0 = {T ∈ S : {2,5} /∈ T } , S 1 = {T ∈ S : {2,5} ∈ T } .Optimization Group 5

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 0 . One may easily verify that a 1-tree of minimal weightnot containing arc {2,5} has weight z 0 = 16. Hence this node is pruned by bound.110 24 692 353 64 5216S¬ {2,5}12{2,5}S 0S 10 S 11S 016S 1¬ {4,5} {4,5}12Thus we proceed with node S 1 . We already found that a minimal weight 1-tree containingarc {2,5} has weight z = 12. At this stage we decide to branch on edge {4,5}:S 10 = {T ∈ S 0 : {4,5} /∈ T } , S 11 = {T ∈ S 0 : {4,5} ∈ T } .Optimization Group 6

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 10 . Now a minimal weight 1-tree has weight 15.110 24 692 31 53 64 516S¬ {2,5}12{2,5}S 10¬ {1,3} {1,3}15S 10 S100 S 101S 016S 1¬ {4,5} {4,5}12S 11At this stage we branch on edge {1,3}:S 100 = {T ∈ S 10 : {1,3} /∈ T } , S 101 = {T ∈ S 10 : {1,3} ∈ T } .Optimization Group 7

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 100 . Now a minimal weight 1-tree has weight 19. Sonode S 100 is pruned by bound.1104 692 31 53 64 5S 01616S¬ {2,5}12{2,5}S 10S 1¬ {4,5} {4,5}12¬ {1,3} {1,3}15S 11S 100S10019S 101Optimization Group 8

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 101 . Now a minimal weight 1-tree has weight 15.110 24 692 31 53 64 5S 016S 101S 1001916S¬ {2,5}12{2,5}S 10S 1¬ {4,5} {4,5}12¬ {1,3} {1,3}15S 101S 11¬ {2,4} {2,4}15S 1010 S 1011At this stage we branch on edge {2,4}:S 1010 = {T ∈ S 10 : {2,4} /∈ T } , S 1011 = {T ∈ S 10 : {2,4} ∈ T } .Optimization Group 9

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 1010 . Now a minimal weight 1-tree has weight 18. Thisnode is pruned by bound.110 24 692 314 556S 016S 1010S 1001916S¬ {2,5}12{2,5}S 10S 1¬ {4,5} {4,5}12¬ {1,3} {1,3}15S 101S 11¬ {2,4} {2,4}15S 1010 S 1011Optimization Group 10

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 1011 . The minimal weight 1-tree has weight 15.110 24 692 31 53 6S 016S 1004 519S 101116S¬ {2,5}12{2,5}S 10S 1¬ {4,5} {4,5}12¬ {1,3} {1,3}15S 101S 11¬ {2,4} {2,4}15S 1010 S 1011¬ {1,4}15{1,4}S 10110 S 10111We branch on edge {1,4}:S 10110 = {T ∈ S 1011 : {1,4} /∈ T } , S 10111 = {T ∈ S 1011 : {1,4} ∈ T } .Optimization Group 11

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 10110 . The minimal weight 1-tree is a tour of length 17.The node is pruned by bound.110 2692 31 53 6S 1004 519S 1011016S¬ {2,5}12{2,5}S 016S 10S 1¬ {4,5} {4,5}12¬ {1,3} {1,3}15S 101S 11¬ {2,4} {2,4}15S 1010 S 1011¬ {1,4} {1,4}15S 1011017S 10111Optimization Group 12

Exercise 7.8.5 ((i)) (cont.)We proceed with exploring node S 10111 . The minimal weight 1-tree is a tour of length 15.16¬ {2,5}S12{2,5}1¬ {4,5} {4,5}10 216 124 6S910 S2 3¬ {1,3} 11{1,3}161 5S3 6100 S¬ {2,4} 101{2,4}19 164 5S 1010 S¬ {1,4} 1011{1,4}S1610111S 0S 1S 10110 S 1011117 16The prescribed edges form a path 3 → 1 → 4 → 2 → 5 of length 10. There is only oneway to complete this path to a tour, namely by adding edge {3,5}. The tour has length 16.Thus for the current node 16 is lower bound. Hence this node can be pruned by bound.Optimization Group 13

Exercise 7.8.5 ((i)) (cont.)Node that node S 10 is now also pruned by bound. We proceed with exploring node S 11 , theonly remaining active node. The minimal weight 1-tree is a tour of length 12.110 24 692 31 53 64 52We branch on edge {1,3}:S¬ {2,5}12{2,5}S 016S 10S 11 S110 S 1111616S 1¬ {4,5} {4,5}12S 11¬ {1,3} {1,3}12S 110 = {T ∈ S 11 : {1,3} /∈ T } , S 111 = {T ∈ S 11 : {1,3} ∈ T } .Optimization Group 14

Exercise 7.8.5 ((i)) (cont.)We explore node S 110 . The minimal weight 1-tree is a tour of length 15.1104 692 31 53 64 52S 01616S¬ {2,5}12{2,5}S 1016S 110S 1010S 1¬ {4,5} {4,5}12S 110S 11¬ {1,3} {1,3}12¬ {1,4} {1,4}15S 11117S 1011infeasibleWe branch on edge {1,4}:S 1100 = {T ∈ S 110 : {1,4} /∈ T } , S 1101 = {T ∈ S 110 : {1,4} ∈ T } .One easily sees that a minimal weight 1-tree in node S 1100 has weight 17, so this node ispruned by bound.In node S 1101 the edges {2,5}, {4,5} {1,4} are prescribed. These edges form the path1 → 4 → 5 → 2. To complete a tour one has to add the edges {1,3} and {3,5}, butedge {1,3} is note present in this node. So this node is pruned by infeasibility.Optimization Group 15

Exercise 7.8.5 ((i)) (cont.)We explore node S 111 . The minimal weight 1-tree is a tour of length 12.S 11116S1¬ {2,5}12{2,5}10 24 6S 0 S 19¬ {4,5} {4,5}2 316 121 5S3 610 S 11¬ {1,3} {1,3}16 124 52S 110 S 11117¬ {1,4}12{1,4}S 1110 S 1111We branch on edge {1,4}:S 1110 = {T ∈ S 111 : {1,4} /∈ T } , S 1111 = {T ∈ S 111 : {1,4} ∈ T } .In node S 1111 the edges {2,5}, {4,5} {1,3} and {1,4} are prescribed. These edgesform the path 3 → 1 → 4 → 5 → 2. To complete a tour one has to add edge {2,3},which gives a tour of length 18. So this node can be pruned by bound.18Optimization Group 16

Solve the 4-city STSP instance with distance matrix⎛Exercise 7.8.5 ((ii))(c ij ) =⎝- 7 6 33 - 6 92 3 - 17 9 4 2ny B&B using an assignment relaxation to obtain bounds. To obtain a lower bound for z we first find a minimalweight matching in the corresponding bipartite graph. This matching is depicted below. The proof of itsoptimality is provided by the indicated values for the dual variables u and v. It has weight z = 13, which is alower bound for z.⎞⎠211’ 210232’ 33’ 413S13044’ 1Minimal weight matching for SWe find that z = 13 is lower bound for the length of an optimal tour. It happens, however, that the assignmentis a tour: 1 → 4 → 3 → 2 → 1. This yields the upper bound ¯z = 13 for the length of an optimal tour.Since z = ¯z, the tour is optimal, and the search tree contains only one node!Optimization Group 18

Exercise 7.8.5 (extra)After the success in part (ii) we might try to solve the 5-city TSP instance with distance matrix⎛⎞- 10 2 4 610 - 9 3 1(c ij ) = ⎜⎝2 9 - 5 6 ⎟4 3 5 - 2⎠ .6 1 6 2 -also by using assignment relaxations to obtain bounds in the B&B method.The minimal weight matching in the corresponding bipartite graph is depicted below. The proof of its optimalityis provided by the indicated values for the dual variables u and v. It has weight z = 10, which is a lower boundfor z.011’ 2122’ 1033’ 2S¬(1,3)10(1,3)244’ 2S 0 S 10 5 5’ 0Minimal weight matching for SWe branch on arc (1,3).Optimization Group 19

Exercise 7.8.5 (extra)We proceed with exploring node S 0 . A minimal weight matching not containing arc (1,3)has weight z 0 = 13.21121’ 22’ 1S¬(1,3)10(1,3)0 3 3’ 32 4 4’ 20 5 5’ 0S 2 S 3¬(4,3) (4,3)Minimal weight matching for S 0S 4 S 5S 0¬(2,5)13(2,5)S 1We branch in node S 0 on arc (2,5), and in node S 3 on arc (4,3).Optimization Group 20

Exercise 7.8.5 (extra)We proceed with exploring node S 4 . A minimal weight matching not containing arcs (1,3)and (4,3), but containing arc (2,5) has weight z 4 = 16.2101231’ 22’ 13’ 6S¬(1,3)10(1,3)S 0¬(2,5)13(2,5)S 1S 2 S 32 4 4’ 2¬(4,3) (4,3)160 55’ 0S 416Minimal weight matching for S 4S 5¬(5,4) (5,4)S 6 S 7The minimal weight matching in node S 4 is a tour: 3 → 1 → 4 → 2 → 5 → 3 of length16. Hence, z 4 = ¯z 4 = 16.We proceed by branching in node S 5 on arc (5,4).Optimization Group 21

Exercise 7.8.5 (extra)We proceed with exploring node S 6 . A minimal weight matching not containing arcs (1,3)and (5,4), but containing arcs (2,5) and (4,3) has weight z 6 =?.2101231’ 22’ 13’ 6S¬(1,3)10(1,3)S 0¬(2,5)13(2,5)S 120454’ 25’ 0S 416Minimal weight matching for S 4S 2 S 3¬(4,3) (4,3)16S 5¬(5,4) (5,4)S 6 S 7And so on! It looks like if the search tree becomes rather large again! I give up. If somebodycan show me a solution with a relatively small search tree I will be glad!Optimization Group 22