1. Global Price Updates Help A.V Goldberg and R. Kennedy Advanced Algorithms Seminar Instructor: Prof. Haim Kaplan Presented by: Orit Nissan-Messing

2. Motivation Solving two “new” problems:  Bipartite matching problem  Assigment problem Using Push & Relabel methods Improve complexity using “Global Updates”

3. Some definitions… Bipartite Graph G = (X U Y, E) Undirected graph The sets of vertices V can be divided into two disjoint sets (X, Y) where there is no edge e Є E such that both endpoint of e are in the same set.

4. Bipartite graph example: X Y

5. Bipartite Matching problem Matching in G is a subset of edges M that have no node in common. Cardinality of matching is |M|. Bipartite matching problem is to find the matching with the maximum cardinality. Usage: matching tasks to processors.

6. Bipartite matching to Max Flow Given Ĝ = (X U Y,Ê) Build G for the maximum flow problem by:  Adding nodes s and t.  V = X U Y U {s,t}  For each v  X, placing (s,v) in E with capacity 1.  For each v  Y, placing (v,t) in E with capacity 1.  For each edge {v,w}  Ê, placing (v,w) in E with capacity 1.  Reverse arcs will have zero capacity. G is called a matching network

7. Building a matching network S t 1 1 1

8. Any integral flow in G can be “converted” into a matching in Ĝ.  The matching are the arcs (x,y) with f = 1. Maximum flow in G corresponds to a maximum matching in Ĝ.

9. Pseudoflow – one more time pseudoflow satisfies the following :  f(u,v) = - f(v,u)  f(u,v) ≤ u(u,v) Excess flow e f (v) e f (v) = ∑ (u,v) Є E f(u,v) A flow is a pseudoflow such that there is no excess at any node.

10. Some more definitions Distance labeling d is valid w.r.t f if:  d(t) = 0  d(s) = n  For every arc (v,w)  E f : d(v) ≤ d(w) + 1 Admissible arc (v,w) : d(v) = d(w) + 1. G A is acyclic for any valid distance labeling

11. Basic algorithm Init :  d(s) = n, d(t) = 0  for every v  V d(v) = 0;  saturate all arcs out of s. Apply sequence of Push and Relabels until f is a flow.

12. Push and Relabel operations Push :  Send a unit of flow from v to w. Relabel:  d(v) ← min (v,w)  Ef { d(w) +1 } Limitation on invocation as we know already…

13. Analysis of the basic algorithm All arcs have unit capacity, therefore each push is a saturating push 2n ≥ d(v) ≥ 0 for all nodes v. Relabel increases d(v).  Number of Relabel operations per node is O(n).  The work done in Relabel operations is O(nm).

14. Analysis - Cont If v is relabeled k times then the number of Pushes from v is at most (k+1)*degree (v). The number of Push operation are O(nm). Selecting an arc to push flow on will be done using the “current arc method”. Current arc

15. d w (v) definition For any node v and w:  d w (v) = bfs distance from v to w in G f of the current preflow f. If w is unreachable from v then d w (v) = ∞.

16. Global Update definition Global Update: for every node v setting d(v) = min { d t (v), n+d s (v)}. The global update also sets the “current arc” of every node to be the first arc. Can be done in O(m). d(v) can not be decreased...

17. Global update – When? At Init. After each Push and Relabel operation if :  Since the last update, at least one unit of excess has reached s or t.  And  Since the last update, the algorithm has done at least m work in Push and Relabel operations.

18. Minimum Distance Discharge – The idea Selecting a unit of excess at active node with minimum d(v). Process this unit of excess until:  Relabel occurs.  The excess reached t or s. If after the Relabel the node still has the minimum d(v) continue with this node.

19. Minimum Distance Discharge algorithm with Global Update Init :  saturate all arcs out of s.  Global update. Apply sequence of Push, Relabels and Global updates by the MDD order until f is a flow.

20. Minimum distance discharge – how is it implemented? Building buckets B 0,….., B 2n. Bucket B i holds the active nodes with d(v) = i. μ - index of the bucket from which we selected the recent active node (unit of excess).

21. Minimum distance discharge – how is it implemented? - Cont After each Global Update:  Current arc is set to the nodes first arc.  The buckets are built again - O(n).  μ is reset to 0.

22. Some definitions for the analysis Γ(f,d) – the minimum distance label of an active node w.r.t pseudoflow f and distance label d. Γ max – The maximum Γ value reached by the algorithm so far.

23. Minimum Distance Discharge - Analysis Lemma 4.1: Between two consecutive Global Updates, the algorithm does O(n) work in examining empty buckets.

24. Minimum Distance Discharge - Analysis Proof:  μ decreases only when it is set to zero after an update.

25. Minimum Distance Discharge - Analysis Proof:  μ decreases only when it is set to zero after an update. d=3 d=2 d=1 T μ = 3 Active node

26. Minimum Distance Discharge - Analysis Proof:  μ decreases only when it is set to zero after an update.  There are 2n+1 buckets.

27. Minimum Distance Discharge - Analysis We will divide the analysis into 4 stages:  Γ max Є [0,k]  Γ max Є [k,n]  Γ max Є [n,n+k]  Γ max Є [n+k,2n] K will be chosen later.

28. Stage 1 and 3 Lemma 4.2: The Minimum Distance Discharge Algorithm expends O(km) work during the periods when Γ max Є [0,k] and Γ max Є [n,n+k]

29. Stage 1 and 3 – Proof If Γ max Є [0,k] then Γ Є [0,k]. If Γ max Є [n,n+k] then Γ Є [n,n+k], since no excess can reached t anymore. Each node can be relabeled at most k+1 times Relabels and Pushes require O(km). During O(km) work, at most k Global Updates can be done. Global Updates require O(km) work during each stage. => Stage 1 and 3 require O(km) work.

30. Flow Decomposition Lemma 4.3: (Without a proof) Any Integral pseudoflow f in the augmented residual graph of an integral flow g in a matching network can be decomposed into cycles and simple paths that are pairwise node disjoint except at the endpoints of the paths, such that each element in the decomposition carries one unit of flow. Each Path is from a node v with e f (v) 0

31. Example for 4.3 2 0 Augmented residual graph for flow f (f=2). All Arcs have unit capacity on both sides -2 s t

32. Stage 2 Definition: residual flow value total excess that can reach t in G f. Lemma 4.4 If Γ max ≥ K (>2), the residual flow value is at most n/(k-1) if G is a matching network.

33. Stage 2 Proof: Let’s look at the first time Γ(f,d) ≥ K. Assume f* is the maximum flow in G. f’ = f* - f. -f’ is a pseudoflow in the augmented residual graph of f* => can be decomposed into cycles and paths as in Lemma 4.3

34. Stage 2 - cont From –f’ decomposition we can find f’ decomposition. Γ≥ k therefore any path in G f from an active node to t must contain at least k+1 nodes. Since each such path is node disjoint => there are at most (n-2)/(k-1) such paths. The amount of excess that can reach t is at most n/(k-1).

35. Stage 4 Same idea as in Stage 2. For Γ max ≥ n+ k (>n+2) the residual flow value is at most n/(k-1) if G is a matching network.

36. Stage 2 and Stage 4 Lemma 4.6 Between any two consecutive global update operations, the algorithm does Θ(m) work.

37. Stage 2 and Stage 4 - Cont Proof: It’s sufficient to show that the work done in moving a unit of excess to s or t after global update is O(m). For each node v, at least one of d t (v), d s (v) is finite. After global update there is at least one admissible arc leaving each node. So the first unit of excess being processed after global update can reach t or s without any Relabel.

38. Stage 2 and Stage 4 - Cont Choosing a unit of excess to process is done in O(n). The path to s or t can be at most n. The work of pushing the excess to its destination is O(n) + the work in finding the admissible arc leaving the node. O(n + m ) = O(m).

39. Complexity of Minimum Distance Discharge Algorithm Lemma 4.7: The Minimum Distance Discharge algorithm with Global Updates computed maximum flow in a matching network (and hence a maximum cardinality bipartite matching) in O(m√n).

40. Complexity of Minimum Distance Discharge Algorithm Proof:  The amount of work during stage 1 and 3 is O(km).  The amount of excess processed during stage 2 and 4 is at most 2n/(k-1)  Moving this excess takes O(mn/k).  O(km + mn/k) = O(m√n) when k = √n.

41. Bad family 1 1 1 -3 d(v) = 7 0 0 0 0 0 0 relabel

42. Bad family 1 1 1 -3 7 1 0 0 0 0 0 relabel

43. Bad family 1 1 1 -3 7 1 1 0 0 0 0 relabel

44. Bad family 0 1 1 -3 7 1 1 1 0 0 0 1 relabel

45. Bad family 0 1 1 -3 7 1 1 1 0 0 1 0 1 0

46. Bad family 0 0 1 -3 7 1 1 1 0 0 1 0 relabel 1 1

47. Bad family 0 0 1 -3 7 1 1 1 1 0 1 0 2 0

48. Bad family 0 0 1 -3 7 1 1 2 1 0 1 0 2 0 relabel

49. Bad family 0 0 0 -3 7 1 1 2 1 0 1 1 2 0 relabel

50. Bad family 1 0 0 -3 7 1 1 2 1 0 2 0 2 0 relabel

51. Bad family 0 0 0 -3 7 2 1 2 1 0 2 0 2 1 relabel

52. Bad family 0 1 0 -3 7 2 1 2 2 0 2 0 2 0 relabel

53. Bad family 0 0 0 -3 7 2 3 2 2 0 2 1 2 0 relabel

54. So what Global Update would have changed? 0 0 0 -3 7 2 3 2 2 0 3 1 2 0 After some ping pong Relabels it would set d(v) to the right values such that there is a path to s or t

55. 0 0 0 -3 7 8 8 8 9 0 9 1 2 0

56. Second problem : Assignment Problem Weight of matching M is the sum of the weights of the edges in M. The Assignment problem is to find a maximum cardinality matching with minimum weight. Assumption cost are integers in the range [0,..C]

57. Assignment problem to Minimum Cost Circulation problem Given Ĝ weighted bipartite graph Build G for the Minimum Cost Circulation problem by:  Adding nodes s and t  For each v  X, placing (s,v) in E with capacity 1 and cost -nC  For each v  Y, placing (v,t) in E with capacity 1 and cost 0  For each edge {v,w}  Ê, placing (v,w) in E with capacity 1 and the cost as in Ĝ.  Reverse arcs will have zero capacity and obey the cost symmetry constraint.  Adding n/2 arcs (t,s) in E with 0 cost and capacity 1.

58. Building G for the Minimum Cost Circulation Problem S t 2 -45 0 2 5 3 4 1 0 -2 u=0

59. As in the matching problem, Any integral circulation in G can be “converted” into a matching in Ĝ.  The matching are the arcs (x,y) with f = 1. Minimum circulation in G corresponds to a maximum matching of minimum weight in Ĝ.

60. Back to definitions Price function P: V →R. Reduced Cost of arc (v,w) C p (v,w) = p(v) + c(v,w) – p(w) U = X U {t} E u : the set of arcs whose tail node is in U.

61. Definition example 2 0 2 -12 0 0 0 12

62. ε-optimality A pseudoflow f is ε-optimal w.r.t price function p, if for every residual arc a  a Є E u  C p (a) ≥ 0  a Є E u  C p (a) ≥ - 2ε A pseudoflow f is ε-optimal if it’s ε-optimal with respect to some price function p. When the arc cost are int and ε < 1/n, any ε-optimal circulation is optimal.

63. Admissibility a Є E f is admissible:  a Є E u  C p (a) < ε  a Є E u  C p (a) < -ε

64. Succssive Approximation Algorithm – The idea  Starting with ε = C  Running in iterations until ε < 1/n.  Each iteration refines ε (divide by a constant α) and saturate all arcs with C p < 0 thus making f pseudoflow ε–optimal.  Making f an ε–optimal circulation by Pushs and Relabels.  Number of iterations = 1+ log α (nC)

65. Cost scaling Algorithm Min-Cost(V,E,u,c)  Init: ε ← C, p(v) = 0, f(a) = 0  While ε ≥ 1/n (ε,f,p) ← Refine(ε,f,p)

66. Refine Procedure Init:  ε ← ε/α  All a Є E with C p (a) < 0, f(a) =1 While f is not a circulation (ε optimal )  Apply Relabel and Push operations. Return (ε,f,p)

67. Relabel and Push Push(v,w)  Send a unit of flow from v to w Relabel(v)  If v Є U P(v) = max (v,w) Є E f {p(w) – c(v,w) }  Else P(v) = max (v,w) Є E f {p(w) – c(v,w) - 2 ε }

68. Relabel and Push Relabel and Push preserve ε optimality. Every Push is a saturating push. Each Relabel decreases p(v) by at least ε. After Relabel, no admissible arc enters v. Ga is acyclic.

69. Relabel Analysis Based on the analysis of minimum cost circulation:  Every Relabel decreases a node price by at least ε.  Each iteration, p(v) decreases by O(nε).  Or more accurately by at most 2αnε.

70. 2 0 2 -12 0 0 0 ε = 1 admissible: a Є Eu Cp (a) < ε a Є Eu Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

71. -2 1 1 2 0 2 12 0 0 0 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

72. -2 0 11 2 0 2 12 0 0 0 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} Relabel : -2 ReducedCost (cost)

73. -2 0 11 2 -2 (0) 4(2) 12 -2 (0) 0 0 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} ReducedCost (cost) P = -2

74. 0 10 2 -2 (0) 4(2) 12 -2 (0) 0 0 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} P = -2 Relabel : -4

75. 0 10 2 -2 (0) 0(2) 12 -2 (0) 0 0 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} P = -2 P = -4

76. 0 01 2 -2 (0) 12 2(0) 0 0 P= -4 0 (-2) P= -2 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

77. 1 00 0(2) 0 (0) 12 2(0) 0 0 P= -4 0 (-2) P= -2 Relabel : -2 admissible: a Є Eu Cp (a) < ε Else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

78. 0 01 0(2) 0 (0) 12 2(0) 0 0 P= -4 0 (-2) P= -2 Relabel : -4 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

79. 0 01 0(2) -2 (0) 12 4(0) 0 0 P= -4 -2 (-2) P= -4 P = -2 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

80. 0 10 0(2) -2 (0) 12 4(0) 0 0 P= -4 2 (2) P= -4 P = -2 Relabel : -6 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

81. 0 10 0(2) -2 (0) 12 4(0) 0 0 P= -6 0 (2) P= -4 P = -2 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

82. 0 01 0(2) -2 (0) 12 4(0) 0 0 P= -6 0 (-2) P= -4 P = -2 admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε}

83. 1 00 0(2) 2 (0) 12 4(0) 0 0 P= -6 0 (-2) P= -4 P = -2 Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε

84. 0 0 1 0 0(-2) 2 (0) 12 (4(0 0 0 P= -6 0 (-2) P= -4 P = -2 Relabel(v) v Є U P(v) = max {p(w) – c(v,w) } Else P(v) = max {p(w) – c(v,w) - 2ε} admissible: a Є Eu Cp (a) < ε else Cp (a) < -ε Relabel : -2

85. This is our matching with minimum cost. 0 0 0 0 0 0 (-2) 2 (0) 12 4(0) 2(0) 0 P= -6 0 (-2) P= -4 P = -2

86. Global Updates The goal : Changing p(v) such as there is a path from every node with excess e f (v) > 0 to some node e f (w) < 0 in G a.

87. Length function Define length function ℓ(a) = ceil ( C p (a)/ ε) if a Є E u otherwise ℓ(a) = 1 + ceil ( C p (a)/ ε) d ℓ (v) – the distance of node v from nodes with deficit. From CS-TR-95-1556

88. Global update intuition Global update intuition: decreasing p(v) by d ℓ (v) * ε for each node would preserve ε optimality and G a acyclicity and ensuring that there is an admissible path for every node with excess to a node with deficit.

89. Example of Global Update 0 01 0(2) -2 (0) 12 4(0) 0 0 P= -6 0 (2) P= -4 P = -2 ε = 1 dℓ = 1, p = -3 d ℓ = 0, p = - 4 dℓ = 0, p = 0 dℓ = 1, p = -1 dℓ = 12, p = -18 This is our set of deficit node

90. But.. But it costs too much.. There is no bound on price change for every node. We would like Global Update to take o(m) They applied a variation of Dial where the scanned nodes are those who can be reached from an active node, active nodes has a bound on the price change.

91. Global update – When? At Init. After each Push and Relabel operation if :  Since the last update, at least one unit of excess has canceled some deficit.  And  Since the last update, the algorithm has done at least m work in Push and Relabel operations.

92. Price Change δ(v) = ceil ( -p(v) / ε ) Setting at the beginning of refine iteration p(v) = 0 for all nodes will make δ(v) to be the price change of node v during the iteration.

93. Minimum Change Discharge – The idea Selecting a unit of excess at active node with minimum δ(v). Process this unit of excess until:  Relabel occurs.  A deficit is canceled. If after the Relabel and the node still has the minimum δ(v) continue with this node.

94. Minimum change discharge – how is it implemented? Building buckets B 0,….., B 2αn. Bucket B i holds the active nodes with δ(v) = i. μ - index of the bucket from which we selected the recent active node (unit of excess).

95. Definitions for Refine Analysis Γ(f,p) – the minimum change price of an active node w.r.t pseudoflow f and p. Γ max – The maximum value reached by Γ so far in the iteration.

96. Minimum Change Discharge - Analysis Lemma 9.1: (same as Lemma 4.1) Between two consecutive Global Updates, the algorithm does O(n) work in examining empty buckets.

97. Refine Analysis: small δ Lemma 9.2: The algorithm does O(km) work in the course of Relabel operations on nodes v obeying δ(v) ≤ k and Push operations from those nodes.

98. Minimum Change Discharge - Analysis Proof:  V can be relabeled at most k+1 times while δ(v) ≤ k.  The work for Relabels on those nodes O(km).  The work for Pushes is also O(km).

99. Refine Analysis Define E to be total excess  E = ∑ e f (v) > 0 e f (v) Lemma 9.4: (will be proved later) At any point during the execution of refine other than the first E* Γ max ≤ 2(( 5 + α)n -1)

100. Refine Analysis : Γ max ≥ K Corollary 9.5  When Γ max ≥ K then E = O(n/k) Can be derived from Lemma 9.4 E* Γ max ≤ 2(( 5 + α)n -1)

101. Refine analysis: Γ max ≤ k big δ Lemma 9.6: While Γ max ≤ k, the amount of work done in relabeling at nodes v with δ(v) > k and pushes from those nodes is O(n^2/k)

102. Refine analysis: Γ max ≤ k Proof:  Define a node with δ(v) > k when Γ max ≤ k a “bad node”.  Find a bound on Relabel operations on bad nodes.  A node v is being processed for one of 2 reasons: It was selected from a bucket δ(v) ≤ Γ max - can not be a bad node. A unit of excess was pushed to it.

103. Refine analysis: Γ max ≤ k Proof continue: A unit of excess just pushed to v :  After v is relabeled, a different node w with δ(w) = Γ will be chosen (by MCD).  The excess will remain at v until Γ max ≥ δ(v). When Γ max ≥ K then the total excess remaining is O(n/k).

104. Refine analysis: Γ max ≤ k Proof continue: Since the excess will have to remain at v until Γ max ≥ K,The numbers of relabeling of bad nodes is O(n/k) The degree of each node is at most n. The work in pushes and relabeling bad nodes is O(n^2/k)

105. Minimum Change Discharge - Analysis Lemma 9.7: Between any two consecutive global update operations, the algorithm does O(m) work.

106. Proof: It’s sufficient to show that the work done in canceling a deficit after global update is O(m). From Global Update “definition” – after global update there is a path from a node with excess to a node with deficit which does not require any Relabel operation.

107. The work of pushing the excess to its destination is O(n). The amount of work in choosing the unit of excess to process in O(n)

108. Minimum Change Discharge - Analysis Theorem 9.8: The Minimum Change Discharge algorithm with global updates computes a minimum cost circulation in a matching network in O(m√n* log (nC)) time

109. Minimum Change Discharge - Analysis Proof: During a refine :  Γ max ≤ k: The algorithm takes O(km+n^2/k)  Γ max ≥ k: Total excess remaining is O(n/k). The work done in pushing an excess to a deficit is O(m) The algorithm takes O(km/n)

110. Minimum Change Discharge - Analysis Proof Continue: The work done in Refine : O(km/n + n^2/k + km). Choosing k to be √n and α =2  Refine : O(m√n) Minimum change discharge with Global Update takes: O(m√n* log (nC))

111. Lemma 9.4 Lemma 9.4: At any point during the execution of refine other than the first E* Γ max ≤ 2(( 5 + α)n -1)

112. Flow Decomposition Lemma 9.3: Given a matching network G and an integral circulation g, any integral pseudoflow f in G g can be decomposed into :  Cycles and  Paths, each from a node u with a deficit to a node v with excess. Where all the elements are pairwiase node disjoint except at s,t and the endpoints of the paths, and each element carries one unit of flow.

113. Refine Analysis Lemma 9.4 Proof: c : the (reduced) arc cost at the beginning of this refine. G :the augmented residual graph at the same time. E * Γ max ≤ ∑ e f (v) > 0 δ(v)e f (v)

114. Lemma 9.4 Prove: -∑ e f (v) > 0 p(v)e f (v) = c p (f) – c(f)  C p (f) – C(f) = ∑ f(v,w)>0 (p(v)-p(w))f(v,w)  Let D be the decomposition of f into paths and cycles.  Cycles do not effect the above equation.  ∑ f(v,w)>0 (p(v)-p(w))f(v,w) = ∑p(u)-p(v) when (u  v) Є D

115. Lemma 9.4  Nodes with deficit never relabled, thus p(u) = 0.  C p (f) – C(f) = -∑ p(v) when (u  v) Є D  C p (f) – C(f) = -∑ e f (v) > 0 p(v)e f (v)

116. Lemma 9.4 C p (f) – C(f): Finding a bound C p (f) – C(f): C(f): For a circulation g, the residual graph has exactly n arcs a Є Eu There are at most n negative cost arcs, each has cost at least - 2αε C(f) ≥ - 2αεn

117. Lemma 9.4, Cp(f): Cp(f): f(a) >0  a reverse Є Ef From f ε-optimality:  a reverse Є Ef  C p (a reverse) ≥ -2ε. f(a) > 0  C p (a) ≤ 2ε. So how many arcs with positive flow are there?

118. Lemma 9.4, Cp(f):  C p (f) = ∑ d Є D Cp(d )  Define V(d) : interior of a path W(d) the set containing the initial and final arcs of d. If d is a cycle then V(d) = d W(d) = Ø.  C p(f) = ∑ d Є D C p(V(d)) + ∑ d Є D C p(W(d))

119. Lemma 9.4 Counting the number of arcs in that equation:  Total number of arc not incident to s or t in path and cycles interiors is at most n by node disjoint.  Total number of arc incident to s or t are at most 2n-1  Total excess is at most n  initial and final arcs of paths are no more than 2n. Arc with positive flow has reduced cost at most 2ε. Cp(f) ≤ (n+ 2n-1+2n) 2ε = (5n-1) 2ε Cp(f) – c(f) ≤ 2(5 + α)n -1) ε