2. 1 Network Flow & Linear Programming Jeff Edmonds York University COSC 3101 Lecture 5 Optimization Problems Network Flow Def nNetwork Flow Def n Simple Example Application: Matching Flow Strategy Hill Climbing Augmenting Flow Primal-Dual Hill Climbing Min Cut Max Flow = Min Cut Running Time Linear Programming

3. 2 Ingredients: Instances: The possible inputs to the problem. Solutions for Instance: Each instance has an exponentially large set of solutions. Cost of Solution: Each solution has an easy to compute cost or value. Specification : The input is one instance. : A valid solution with optimal cost. (minimum or maximum) Optimization Problems

4. 3 Instance: A Network is a directed graph G Edges represent pipes that carry flow Each edge has a maximum capacity c A source node s out of which flow leaves A sink node t into which flow arrives Goal: Max Flow Network Flow

5. 4 Instance: A Network is a directed graph G Edges represent pipes that carry flow Each edge has a maximum capacity c A source node s out of which flow leaves A sink node t into which flow arrives Network Flow

6. 5 For some edges/pipes, it is not clear which direction the flow should go in order to maximize the flow from s to t. Hence we allow flow in both directions.

7. 6 Network Flow u v 1719 244 This edge/pipe allows download at a rate of 1719kbps OR upload at a rate of 244kbps Not both simultaneously

8. 7 Network Flow u v 75 10 This edge/pipe allows flow to the right at a rate of 75 L/sec OR to the left right at a rate of 10 L/sec

9. 8 Network Flow u v 15/75 0/10 This edge/pipe allows flow to the right at a rate of 75 L/sec OR to the left right at a rate of 10 L/sec Currently the flow is to the right at 15 L/sec

10. 9 Network Flow u v 0/75 5/10 This edge/pipe allows flow to the right at a rate of 75 L/sec OR to the left right at a rate of 10 L/sec Currently the flow is to the left at 5 L/sec

11. 10 Solution: The amount of flow F through each edge. Flow can’t exceed capacity i.e. F  c. Unidirectional flow F  0 and F = 0 or F = 0 and F  0 Some texts: F = -F Network Flow

12. 11 Solution: The amount of flow F through each edge. Flow F can’t exceed capacity c. Unidirectional flow No leaks, no extra flow. Network Flow

13. 12 Solution: The amount of flow F through each edge. Flow F can’t exceed capacity c. Unidirectional flow No leaks, no extra flow. For each node v: flow in = flow out  u F =  w F Except for s and t. Network Flow

14. 13 Value of Solution: Flow from s into the network minus flow from the network back into s. rate(F) =  u F = flow from network into t minus flow back in. =  u F -  v F -  v F What about flow back into s? Goal: Max Flow Network Flow

15. 14 A network with its edge capacities Network Flow What is the maximum that can flow from s to t?

16. 15 A network with its edge capacities Network Flow The max total rate of the flow is 1+2-0 = 3. flow/capacity = 2/5 Prove that total can not be higher.

17. 16 No more flow can be pushed along the top path because the edge is at capacity. Similarly, the edge. No flow is pushed along the bottom path because this would decrease the total from s to t. Network Flow

18. 17 is a minimum cut Its capacity is the sum of the capacities crossing the cut = 1+2 = 3. is not included in because it is going in the wrong direction. Network Flow

19. 18 The edges crossing forward across the cut are at capacity those crossing backwards have zero flow. This is always true. Network Flow

20. 19 The maximum flow is 1+2=3 The minimum cut is 1+2=3. These are always equal. Network Flow

21. 20 An Application: Matching SamMary BobBeth JohnSue FredAnn Who loves whom. Who should be matched with whom so as many as possible matched and nobody matched twice? 3 matches Can we do better? 4 matches

22. 21 An Application: Matching st c = 1 Total flow out of u  flow into u  1 Boy u matched to at most one girl. 1 c = 1 Total flow into v = flow out of v  1 Girl v matched to at most one boy. 1 u v

23. 22 Network Flow Strategy: Push flow into s. Must make decisions. Get stuck and must backtrack. Difficult and time consuming. s c=100 f=100 c=100 c=1 f=100

24. 23 Network Flow Strategy: Find a path for a single drop. Push as much flow through as fits. w = augment = Min  Path c

25. 24 Network Flow Strategy: Find a path for a single drop. Push as much flow through as fits. w = augment = Min  Path c flow/capacity = 20/21

26. 25 Network Flow Given Flow F Construct Augmenting Graph G F Find path P using BFS, DFS, or generic search algorithm Let w be the max amount flow can increase along path P. Increase flow along path P by w. i.e newF = oldF + w × P +w

27. 26 Network Flow Given Flow F Construct Augmenting Graph G F Find path P using BFS, DFS, or generic search algorithm No path Stop.

28. 27 Hill Climbing We have a valid solution. (not necessarily optimal) Take a step that goes up. measure progress Value of our solution. Problems: Exit Can't take a step that goes up. Running time? Initially have the “zero Local Max Global Max Can our Network Flow Algorithm get stuck in a local maximum? Make small local changes to your solution to construct a slightly better solution. If you take small step, could be exponential time.

29. 28 Network Flow Previous Input Previous Output Same Input Can our Network Flow Algorithm get stuck in local max? Worse Output Need only one example. Yes!

30. 29 Network Flow

31. 30 Network Flow Previous Input Previous Output Same Input Yes! Our Network Flow Algorithm can get stuck. Worse Output Need only one example.

32. 31 Hill Climbing Avoiding getting stuck in a local maximum Good ExecutionBad Execution Made better choices of direction Hard Back up an retry Exponential time Define a bigger step

33. 32 Hill Climbing Different Solutions Current Solution Alg defines to where alg can step i.e. what small local changes can be made to current solution This defines the topography

34. 33 Hill Climbing Different Solutions Current Solution Define a slightly bigger step This defines the topography Perhaps removes some local maxima

35. 34 Network Flow Previous Input Previous Output Same Input Worse Output Mistake? Putting 2 through this edge

36. 35 Network Flow Mistake? Putting 2 through this edge We need to decrease the flow in this edge. But if we decrease the total flow then the algorithm might run exponentially or forever. We need to decrease the flow in this edge AND increase the total flow.

37. 36 Network Flow Let us focus on how much we can CHANGE the rate of flow through a given edge.

38. 37 Network Flow u v 15/75 0/10 This edge/pipe allows flow to the right at a rate of 75 L/sec OR to the left right at a rate of 10 L/sec Currently the flow is to the right at 15 L/sec

39. 38 Network Flow u v 15/75 0/10 Currently the flow is to the right at 15 L/sec u v F  [-10,75] Eqv Flow Graph F = 15 15

40. 39 u v 0/75 10/10 Currently the flow is to the left at 10 L/sec u v F  [-10,75] Eqv Flow Graph F = -10 15 Network Flow

41. 40 Network Flow Add 5 flow to right u v F  [-10,75] Eqv Flow Graph F = 15 15 Current Flow 20 u v F  [-10,75] Eqv Flow Graph F = 20 Resulting Flow

42. 41 Network Flow Add -5 flow to right u v F  [-10,75] Eqv Flow Graph F = 15 15 Current Flow 10 u v F  [-10,75] Eqv Flow Graph F = 10 Resulting Flow Equivalent Add 5 flow to left

43. 42 Network Flow Add -20 flow to right u v F  [-10,75] Eqv Flow Graph F = 15 15 Current Flow -5 u v F  [-10,75] Eqv Flow Graph F = -5 Resulting Flow Add 20 flow to left Equivalent

44. 43 Network Flow -5 u v F  [-10,75] Eqv Flow Graph F = -5 Resulting Flow u v -5/75 0/10 Flow Graph u v 0/75 5/10 Flow Graph Equivalent

45. 44 Network Flow 15+Δ ≤ 75 F +Δ ≤ c Δ ≤ c -F Walking Δ u v 15+ Δ /75 0/10 New Flow Graph u v F  [-10,75] Eqv Flow Graph F = 15 15 75-15 = 60 Allowed change to the right

46. 45 Network Flow 15-Δ ≥ -10 F -Δ ≥ -c Δ ≤ F + c u v 15- Δ /75 0/10 New Flow Graph Walking Δ u v F  [-10,75] Eqv Flow Graph F = 15 Allowed Flow ≠ Allowed Change in Flow 15 Allowed change to the left 15+10=25

47. 46 Network Flow Δ ≤ F + c u v F  [-10,75] Eqv Flow Graph F = 15 Δ ≤ c -F u v Augmentation Graph 15 75-15 = 60 Allowed change to the right Allowed change to the left 15+10=25

48. 47 Network Flow Given Flow F Construct Augmenting Graph G F Find path P Old New Where we got stuck before

49. 48 Network Flow Given Flow F Construct Augmenting Graph G F Find path P Let w be the max amount flow can increase along path P. Increase flow along path P by w. i.e newF = oldF + w × P +w -w ?

50. 49 Network Flow Given Flow F Construct Augmenting Graph G F Find path P using BFS, DFS, or generic search algorithm No path Stop.

51. 50 Network Flow Previous Input Previous Output Same Input Worse Output Same Output

52. 51

53. 52 An Application: Matching SamMary BobBeth JohnSue FredAnn Who loves whom. Who should be matched with whom so as many as possible matched and nobody matched twice? 3 matches Can we do better? 4 matches

54. 53 An Application: Matching st Flow st Augmentation Graph Augmentation Path Alternates adding edge removing edge adding edge removing edge adding edge Extra edge added st New Flow

55. 54 An Application: Matching SamMary BobBeth JohnSue FredAnn Who loves whom. Who should be matched with whom so as many as possible matched and nobody matched twice? 3 matches Can we do better? 4 matches

56. 55 Network Flow Can our Network Flow Algorithm get stuck in local max? Need to prove for every input network for every choice of augmenting paths Maximum Flow is found! No! How?

57. 56 Primal-Dual Hill Climbing Mars settlement has hilly landscape and many layers of roofs.

58. 57 Primal-Dual Hill Climbing Primal Problem: Exponential # of locations to stand. Find a highest one. Dual problem: Exponential # of roofs. Find a lowest one.

59. 58 Primal-Dual Hill Climbing Prove: Every roof is above every location to stand.  R  L height(R)  height(L)  height(R min )  height(L max ) Is there a gap?

60. 59 Primal-Dual Hill Climbing Prove: For every location to stand either: the alg takes a step up or the alg gives a reason that explains why not by giving a ceiling of equal height. i.e.  L [  L’ height(L’)  height(L) or  R height(R) = height(L)] or But  R  L height(R)  height(L) No Gap

61. 60 Primal-Dual Hill Climbing Prove: For every location to stand either: the alg takes a step up or the alg gives a reason that explains why not by giving a ceiling of equal height. i.e.  L [  L’ height(L’)  height(L) or  R height(R) = height(L)] or Can't go up from this location and no matching ceiling. Can't happen! ?

62. 61 Primal-Dual Hill Climbing Prove: For every location to stand either: the alg takes a step up or the alg gives a reason that explains why not by giving a ceiling of equal height. i.e.  L [  L’ height(L’)  height(L) or  R height(R) = height(L)] or No local maximum!

63. 62 Primal-Dual Hill Climbing Claim: Primal and dual have the same optimal value. height(R min ) = height(L max ) Proved:  R  L, height(R)  height(L) Proved: Alg runs until it provides L alg and R alg height(R alg ) = height(L alg ) No Gap height(R min )  height(R alg )  height(L alg )  height(L max ) height(R min )  height(L max ) L alg witness that height(L max ) is no smaller. R alg witness that height(L max ) is no bigger. Exit

64. 63 A network with its edge capacities Network Flow What is the maximum that can flow from s to t?

65. 64 A network with its edge capacities Network Flow The max total rate of the flow is 1+2-0 = 3. flow/capacity = 2/5 Prove that total can not be higher.

66. 65 No more flow can be pushed along the top path because the edge is at capacity. Similarly, the edge. No flow is pushed along the bottom path because this would decrease the total from s to t. Network Flow

67. 66 is a minimum cut Its capacity is the sum of the capacities crossing the cut = 1+2 = 3. is not included in because it is going in the wrong direction. Network Flow

68. 67 The edges crossing forward across the cut are at capacity those crossing backwards have zero flow. This is always true. Network Flow

69. 68 The maximum flow is 1+2=3 The minimum cut is 1+2=3. These are always equal. Network Flow

70. 69 Primal-Dual Network Flow Primal Problem: Max Flow Dual Problem: Min Cut What are the roofs to the flows?

71. 70 Instance: A Network is a directed graph G Special nodes s and t. Edges represent pipes that carry flow Each edge has a maximum capacity c Min Cut s t

72. 71 Instance: A Network is a directed graph G Special nodes s and t. Edges represent pipes that carry flow Each edge has a maximum capacity c Min Cut

73. 72 Solution: C = partition of nodes with s  U, t  V. Min Cut s t U V York UC Berkeley = Canada = USA

74. 73 Min Cut s t York UC Berkeley Ontario Solution: C = partition of nodes with s  U, t  V.

75. 74 Min Cut s t York UC Berkeley Toronto Solution: C = partition of nodes with s  U, t  V.

76. 75 Min Cut s t York UC Berkeley York Solution: C = partition of nodes with s  U, t  V.

77. 76 Min Cut s t York UC Berkeley UCB Solution: C = partition of nodes with s  U, t  V.

78. 77 Min Cut s t York UC Berkeley Berkeley Solution: C = partition of nodes with s  U, t  V.

79. 78 Min Cut s t York UC Berkeley U V Solution: C = partition of nodes with s  U, t  V.

80. 79 Value Solution C= : cap(C) = how much can flow from U to V =  u  U,v  V c Min Cut s t U V u v Goal: Min Cut

81. 80 Theorem: For all Networks Max F rate(F) = Min C cap(C) Prove:  F,C rate(F)  cap(C) Max Flow = Min Cut U V u v Prove:  flow F, alg either finds a better flow F or finds cut C such that rate(F) = cap(C) Alg stops with an F and C for which rate(F) = cap(C) F witnesses that the optimal flow can't be less C witnesses that it can't be more. Exit

82. 81 cap(C) = how much can flow from U to V =  u  U,v  V c rate(F,C) = Flow from U to V minus flow V to U =  u  U,v  V F - F Lemma: rate(F) = rate(F,C)  cap(C) Max Flow = Min Cut Prove:  flow F, cut C rate(F)  cap(C) s t U V York UC Berkeley = Canada = USA + - F  c & F  0 No leaks, no extra flow.

83. 82 Lemma:  F,C rate(F,C) = rate(F) Proof: By induction on the size of U. Base case: C = rate(F) = Flow from s into the network minus flow from the network back into s  =  u F -  v F  = rate(F,C) where C = Max Flow = Min Cut s t York UC Berkeley + -

84. 83 Lemma:  F,C rate(F,C) = rate(F) Proof: By induction on the size of C. Base case: C = Inductive step: Move nodes across C one at a time. Max Flow = Min Cut Flow into x = Flow out. Hence, flow across cut does not change. rate(F) = rate(F,C)  cap(C)

85. 84 Theorem: For all Networks Max F rate(F) = Min C cap(C) Prove:  F,C rate(F)  cap(C) Max Flow = Min Cut U V u v Prove:  flow F, alg either finds a better flow F or finds cut C such that rate(F) = cap(C)

86. 85 Given Flow F Construct Augmenting Graph G F Find path P Let w be the max amount flow can increase along path P. Increase flow along path P by w. i.e newF = oldF + w × P +w -w Max Flow = Min Cut

87. 86 Given Flow F Construct Augmenting Graph G F Find path P using BFS, DFS, or generic search algorithm No path Stop Max Flow = Min Cut

88. 87 Let F alg be this final flow. Let cut C alg =, where U are the nodes reachable from s in the augmented graph and V not. Claim: rate(F alg ) = cap(C alg ) Max Flow = Min Cut Bad example because U might contain more than s.

89. 88 Max Flow = Min Cut Prove: rate(F alg ) = cap(C alg ) Prove: rate(F alg,C alg ) = cap(C alg ) =

90. 89 cap(C) = how much can flow from U to V =  u  U,v  V c rate(F,C) = Flow from U to V minus flow V to U =  u  U,v  V F - F Max Flow = Min Cut Prove: rate(F alg,C alg ) = cap(C alg ) + F /c u v F<u,v>/c<u,v>F<u,v>/c<u,v> Flow Graph U V v’u’ -

91. 90 Max Flow = Min Cut F /c u v F<u,v>/c<u,v>F<u,v>/c<u,v> Flow Graph U V v’u’ need equal need zero + - cap(C) = how much can flow from U to V =  u  U,v  V c rate(F,C) = Flow from U to V minus flow V to U =  u  U,v  V F - F Prove: rate(F alg,C alg ) = cap(C alg )

92. 91 Max Flow = Min Cut u v F<u,v>/c<u,v>F<u,v>/c<u,v> Flow Graph c<u,v>-F<u,v>c<u,v>-F<u,v> F +c U V v’u’ F /c u Augmentation Graph U V v’ v u’ reachable from s = = not not edges =0 need equal + need zero - cap(C) = how much can flow from U to V =  u  U,v  V c rate(F,C) = Flow from U to V minus flow V to U =  u  U,v  V F - F Prove: rate(F alg,C alg ) = cap(C alg )

93. 92 Theorem: For all Networks Max F rate(F) = Min C cap(C) Prove:  F,C rate(F)  cap(C) Max Flow = Min Cut U V u v Prove:  flow F, alg either finds a better flow F or finds cut C such that rate(F) = cap(C) Alg stops with an F and C for which rate(F) = cap(C) F witnesses that the optimal flow can't be less C witnesses that it can't be more.

94. 93 Hill Climbing Problems: Can our Network Flow Algorithm get stuck in a local maximum? Local Max Global Max No!

95. 94 Hill Climbing Problems: Running time? If you take small step, could be exponential time.

96. 95 Network Flow

97. 96 Network Flow Add flow 1

98. 97 Network Flow Add flow 1

99. 98 Hill Climbing Problems: Running time? If each iteration you take the biggest step possible, Alg is poly time in number of nodes and number of bits in capacities. If each iteration you take path with the fewest edges Alg is poly time in number of nodes

100. 99 Taking the biggest step possible

101. 100 Linear Programming

102. 101 A Hotdog A combination of pork, grain, and sawdust, …

103. 102 Constraints: Amount of moistureAmount of moisture Amount of protein,Amount of protein, …

104. 103 The Hotdog Problem Given today’s prices, what is a fast algorithm to find the cheapest hotdog?

105. 104 Abstract Out Essential Details Cost: 29, 8, 1, 2 Amount to add: x 1, x 2, x 3, x 4 pork grainwater sawdust 3x 1 + 4x 2 – 7x 3 + 8x 4 ≤ 12 2x 1 - 8x 2 + 4x 3 - 3x 4 ≤ 24 -8x 1 + 2x 2 – 3x 3 - 9x 4 ≤ 8 x 1 + 2x 2 + 9x 3 - 3x 4 ≤ 31 Constraints: moisturemoisture protein,protein, … 29x 1 + 8x 2 + 1x 3 + 2x 4 Cost of Hotdog:

106. 105 29x 1 + 8x 2 + 1x 3 + 2x 4 Subject to: Minimize: Abstract Out Essential Details 3x 1 + 4x 2 – 7x 3 + 8x 4 ≤ 12 2x 1 - 8x 2 + 4x 3 - 3x 4 ≤ 24 -8x 1 + 2x 2 – 3x 3 - 9x 4 ≤ 8 x 1 + 2x 2 + 9x 3 - 3x 4 ≤ 31

107. 106 A Fast Algorithm For decades people thought that there was no fast algorithm. Then one was found! Theoretical Computer Science finds new algorithms every day. 3x 1 + 4x 2 – 7x 3 + 8x 4 ³ 12 2x 1 - 8x 2 + 4x 3 - 3x 4 ³ 24 -8x 1 + 2x 2 – 3x 3 - 9x 4 ³ 8 x 1 + 2x 2 + 9x 3 - 3x 4 ³ 31 29x 1 + 8x 2 + 1x 3 + 2x 4 Subject to: Minimize: 

108. 107 Network Flow as a Linear Program Given an instance of Network Flow: > express it as a Linear Program: The variables: Maximize: Subject to: Flows f for each edge.  : F  c. (Flow can't exceed capacity)  v:  u F =  w F (flow in = flow out) rate(F) =  u F -  v F

109. 108

110. 109 Primal Dual

111. 110 End

112. 111 Network Flow Locally in each edge, we see how much we can change the flow. Allowed Flow ≠ Allowed Change in Flow People find this hard. This is my first attempt to explain it.

113. 112 Network Flow u v 0/75 0/10 Old Flow Graph Walking 10 u v 10/75 0/10 New Flow Graph

114. 113 Network Flow u v 0/75 0/10 Old Flow Graph Walking 10 u v 0/75 10/10 New Flow Graph

115. 114 Network Flow u v 0/75 0/10 Old Flow Graph Walking -10 u v -10/75 0/10 New Flow Graph Same Walking 10 u v 0/75 10/10 New Flow Graph Not allowed

116. 115 Network Flow Walking -10 u v -10/75 0/10 New Flow Graph Same Walking 10 u v 0/75 10/10 New Flow Graph Not allowed u v F  [-10,75] Eqv Flow Graph F = -10

117. 116 Network Flow 10+Δ ≤ 75 F +Δ ≤ c Δ ≤ c -F Walking Δ u v 10+ Δ /75 0/10 New Flow Graph u v F  [-10,75] Eqv Flow Graph F = 10

118. 117 Network Flow 10-Δ ≥ -10 F -Δ ≥ -c Δ ≤ F + c u v 10- Δ /75 0/10 New Flow Graph Walking Δ u v F  [-10,75] Eqv Flow Graph F = 10

119. 118 Network Flow Δ ≤ F + c u v F  [-10,75] Eqv Flow Graph F = 10 Δ ≤ c -F u v Augmentation Graph

120. 119 Network Flow Locally in each edge, we see how much we can change the flow. Allowed Flow ≠ Allowed Change in Flow People find this hard. This is my second attempt to explain it.

121. 120 Network Flow u v 0/75 0/10 Flow Graph u v 75 10 Augmentation Graph

122. 121 Network Flow 75-21=54 u v 21/75 0/10 Flow Graph u v Augmentation Graph 2 21+2=23 Walking

123. 122 Network Flow 75-21=54 u v 21/75 0/10 Flow Graph u v Augmentation Graph 10 21-10=11 10 Edge shouldn’t have flow in both directions. 0 10 ? Walking 10 Walking -10

124. 123 Network Flow 75-21=54 u v 11/75 0/10 Flow Graph u v Augmentation Graph 10 ? Capacity of 10 not met

125. 124 Network Flow 75-21=54 u v 21/75 0/10 Flow Graph u v Augmentation Graph 21 10 ? Capacity of 10 exceeded 21

126. 125 Network Flow 75-21=54 u v 21/75 0/10 Flow Graph u v Augmentation Graph 21 21-21=0 10 ? Capacity of 10 not met Walking 21 Walking -21

127. 126 Network Flow 75-21=54 u v 21/75 0/10 Flow Graph u v Augmentation Graph 21+10 21-21=0 10 ? 21+10=31 Capacity of 10 met Walking 21 and another 10

128. 127 Network Flow 75-21=54 21+10=31 u v 21/75 0/10 Flow Graph u v Augmentation Graph 25 21-25=-4 4 0 Walking 25

129. 128 u v F<u,v>/c<u,v>F<u,v>/c<u,v> 0/c Flow Graph u v Augmentation Graph F<u,v>+wF<u,v>+w w Walking c<u,v>-F<u,v>c<u,v>-F<u,v> F<u,v>+c<v,u>F<u,v>+c<v,u> c<u,v>c<u,v> F<u,v>F<u,v> c<v,u>c<v,u> Network Flow

130. 129 u v F<u,v>/c<u,v>F<u,v>/c<u,v> 0/c Flow Graph u v Augmentation Graph F<u,v>-wF<u,v>-w c<u,v>-F<u,v>c<u,v>-F<u,v> F<u,v>+c<v,u>F<u,v>+c<v,u> c<u,v>c<u,v> F<u,v>F<u,v> c<u,v>c<u,v> w Walking Network Flow

131. 130

132. 131 Network Flow How much flow can I have in each direction? u v 0/75 0/10 Where can I stand in each direction?

133. 132 Network Flow How much flow can I have in each direction? 75 to the right u v 75/75 0/10 Where can I stand in each direction?

134. 133 Network Flow How much flow can I have in each direction? 75 to the right u v 0/75 10/10 10 to the left Where can I stand in each direction?

135. 134 Network Flow u v 15/75 0/10 Add 5 flow to right Walking 5 u v 20/75 0/10 15 20

136. 135 Network Flow u v 15/75 0/10 Add 75-15=60 flow to right Walking 75-15=60 u v 75/75 0/10 15

137. 136 Network Flow u v 15/75 0/10 How much flow can I add to right How much can I change my position to the right? 75-15=60 15 u v 75-15=60 Augmentation Graph

138. 137 Network Flow u v f/c 0/10 How much flow can I add to right How much can I change my position to the right? 75-15=60 15 f-c u v Augmentation Graph

139. 138 Network Flow u v 15/75 0/10 Add 5 flow to left Walking 5 u v 15/75 5/10 Edge shouldn’t have flow in both directions. 15 10

140. 139 Network Flow u v 15/75 0/10 Add 5 flow to left Walking 5 u v 15-5=10/75 0/10 15 10

141. 140 Network Flow u v 15/75 0/10 Add 15 flow to left Walking 15 u v 15-15=0/75 0/10 15

142. 141 Network Flow u v 15/75 0/10 Add 15+5=20 flow to left Walking 15+5=20 u v -5/75 0/10 Edge shouldn’t have negative flow. 15 -5

143. 142 Network Flow u v 15/75 0/10 Add 15+5=20 flow to left Walking 15+5=20 u v 0/75 5/10 15 -5

144. 143 Network Flow u v 15/75 0/10 Add 15+10=25 flow to left Walking 15+10=25 u v 0/75 10/10 15

145. 144 Network Flow u v 15/75 0/10 How much flow can I add to left How much can I change my position to the left? 15+10=25 15 u v 15+10=25 Augmentation Graph

146. 145 Network Flow u v f/75 0/c How much flow can I add to left How much can I change my position to the left? 15+10=25 15 u v f+c Augmentation Graph

147. 146 u v F<u,v>/c<u,v>F<u,v>/c<u,v> 0/c Flow Graph u v Augmentation Graph F<u,v>+wF<u,v>+w w Walking c<u,v>-F<u,v>c<u,v>-F<u,v> F<u,v>+c<v,u>F<u,v>+c<v,u> c<u,v>c<u,v> F<u,v>F<u,v> c<v,u>c<v,u> Network Flow

148. 147 u v F<u,v>/c<u,v>F<u,v>/c<u,v> 0/c Flow Graph u v Augmentation Graph F<u,v>-wF<u,v>-w c<u,v>-F<u,v>c<u,v>-F<u,v> F<u,v>+c<v,u>F<u,v>+c<v,u> c<u,v>c<u,v> F<u,v>F<u,v> c<u,v>c<u,v> w Walking Network Flow