1. Partitioning Graphs of Supply and Demand Generalization of Knapsack Problem Takao Nishizeki Tohoku University

2. Graph Supply Vertices Demand Vertices Supply Vertices and Demand Vertices

3. Graph 4 6535 4 26107 8 12 1513 25 Each Supply Vertex has a number, called Supply. Each Demand Vertex has a number, called Demand Supply Demand Supply Vertices Demand Vertices

4. Desired Partition partition G into connected components so that 4 6535 4 26107 8 12 1513 5 8 6 4 25

5. Desired Partition partition G into connected components so that 4 6535 4 26107 8 12 1513 25

6. Desired Partition partition G into connected components so that (a) each component has exactly one supply vertex, (b) supply is no less than the sum of demands in the component. (a) each component has exactly one supply vertex, (b) supply is no less than the sum of demands in the component. 4 6535 4 26107 8 12 1513 25 4 6535 4 26107 8 Desired partition Sum ： 23 Sum ： 13 Sum ： 12

7. Maximum Partition Problem (Max PP) 4 6535 4 26107 8 1513 20 No desired partition

8. 4 6535 4 26107 8 1513 20 (a) each component has exactly one supply vertex, (b) supply is no less than the sum of demands in the component. (a) each component has exactly one supply vertex, (b) supply is no less than the sum of demands in the component. A partition of a graph must satisfy 4 6535 4 26107 8 no supply vertex Maximum Partition Problem (Max PP) (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component. (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component.

9. (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component. (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component. 4 6535 4 26107 8 1513 20 4 6535 4 26107 8 finds a partition that maximizes the “fulfillment.” Sum ： 17 Sum ： 7 Sum ： 5 Sum ： 12 Maximum Partition Problem (Max PP) (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component if there is a supply vertex. (a) each component has at most one supply vertex, (b) supply is no less than the sum of demands in the component if there is a supply vertex. A partition of a graph must satisfy sum of demands in all components with supply vertices

10. finds a partition that maximizes the “fulfillment.” The fulfillment of this partition 17 + 5 + 12 + 7 = 41 The fulfillment of this partition 17 + 5 + 12 + 7 = 41 4 6535 4 26107 8 1513 20 4 6535 4 26107 8 Maximum Partition Problem (Max PP) Sum ： 17 Sum ： 7 Sum ： 5 Sum ： 12

11. Sum ： 19 Sum ： 15 Sum ： 8 Sum ： 12 4 6535 4 26107 8 1513 20 finds a partition that maximizes the “fulfillment.” 6 4 6535 4 2107 8 The fulfillment of this partition 19 + 8 + 12 + 15 = 54 The fulfillment of this partition 19 + 8 + 12 + 15 = 54 Maximum fulfillment Maximum Partition Problem (Max PP)

12. Complexity Status 7 436 2394 5 15 25 Trees NP-hard Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b. Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b. 345711 b = 13 max subset sum problem (simple ver. of Knapsack) given set A

13. C Complexity Status Trees NP-hard Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b. Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b. 7 436 2394 5 15 25 345711 b = 13 max subset sum problem (simple ver. of Knapsack) given set A

14. Related Result Max PP for Stars with one supply at center Fully Polynomial-Time Approximation Scheme (FPTAS) [Ibarra and Kim ’75] FPTAS: for any , 0 <  < 1, the algorithm finds an approximation solution such that APPRO > (1–  ) OPT in time polynomial in both n and 1/ . FPTAS: for any , 0 <  < 1, the algorithm finds an approximation solution such that APPRO > (1–  ) OPT in time polynomial in both n and 1/ . max subset sum problem (NP-hard) 345711 13

15. Related Result Max PP for Stars with one supply at center Fully Polynomial-Time Approximation Scheme (FPTAS) [Ibarra and Kim ’75] max subset sum problem (NP-hard) 345711 13 good approximation for larger classes good approximation for larger classes ? ? 7 436 2394 5 15 25 4 6535 4 26107 8 12 1513 25

16. Our Results (approximabiliy) 4 6535 4 26107 8 12 1513 25 General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP PTAS: for any , 0 <  < 1, the algorithm finds an approximation solution such that APPRO > (1–  ) OPT in time polynomial in n. (1/  :regarded as a constant.) PTAS: for any , 0 <  < 1, the algorithm finds an approximation solution such that APPRO > (1–  ) OPT in time polynomial in n. (1/  :regarded as a constant.) No FPTAS unless P=NP

17. Our Results (approximabiliy) 4 6535 4 26107 8 12 1513 25 General graphs Trees (2) FPTAS (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP NP-hard 7 436 2394 5 15 25

18. Our Results (approximabiliy) 4 6535 4 26107 8 12 1513 25 General graphs (2) FPTAS (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard 7 436 2394 5 15 25

19. (1) MAXSNP-hardness Max PP is MAXSNP-hard for bipartite graphs. L-reduction: preserves approximability L-reduction Max PP for bipartite graphs 4 4 7 4 4 7 4 4 7 1 1 1 3-occurrence MAX3SAT each variable appears at most 3 times error ratio:  error ratio:  ’

20. (1) MAXSNP-hardness 3-occurrence MAX3SAT (MAXSNP-hard) instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes # of satisfied clauses variables: v w x y z at most 3 times

21. (1) MAXSNP-hardness instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes # of satisfied clauses variables: v w x y z at most 3 times 3-occurrence MAX3SAT (MAXSNP-hard)

22. (1) MAXSNP-hardness instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes # of satisfied clauses variables: v w x y z 3-occurrence MAX3SAT (MAXSNP-hard)

23. (1) MAXSNP-hardness variable gadget 4 4 7 xx variable x 4 4 7 xx x = false 4 4 7 xx x = true 4+4 = 8 > 7

24. (1) MAXSNP-hardness 1 1 1 4 4 7 x x 4 4 7 y y 4 4 7 z z variable xvariable yvariable z

25. (1) MAXSNP-hardness 1 1 1 4 4 7 x x 4 4 7 y y 4 4 7 z z variable xvariable yvariable z

26. (1) MAXSNP-hardness 1 1 1 4 4 7 x x 4 4 7 y y 4 4 7 z z variable xvariable yvariable z

27. (1) MAXSNP-hardness 4 4 7 4 4 7 4 4 7 1 1 1 x x y y z z variable xvariable yvariable z at most 3 7 – 4 = 3 enough power

28. (1) MAXSNP-hardness 4 4 7 4 4 7 4 4 7 1 1 1 x x y y z z variable xvariable yvariable z

29. (1) MAXSNP-hardness 4 4 7 4 4 7 4 4 7 1 1 1 x x y y z z x = truey = falsez = false

30. (1) MAXSNP-hardness 4 4 7 4 4 7 4 4 7 1 1 1 x x y y z z variable xvariable yvariable z

31. (1) MAXSNP-hardness 4 4 7 4 4 7 4 4 7 1 1 1 x x y y z z x = falsey = falsez = false not satisfied

32. (1) MAXSNP-hardness Max PP is MAXSNP-hard for bipartite graphs. L-reduction: preserves approximability 3-occurrence MAX3SAT Max PP for bipartite graphs L-reduction 4 4 7 4 4 7 4 4 7 1 1 1 each variable appears at most 3 times

33. Our Results (approximabiliy) 4 6535 4 26107 8 12 1513 25 General graphs (2) FPTAS (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard Pseudo-poly.-time algorithm 7 436 2394 5 15 25

34. 7 436 2394 5 15 25 Max PP is NP-hard even for trees. Max PP can be solved for a tree T in time O(F 2 n) if the supplies and demands are integers. Max PP can be solved for a tree T in time O(F 2 n) if the supplies and demands are integers. 2+3+4+ ･･･ +4+5= 43 15+25= 40 F= min{43,40}=40 max fulfillment ≦ F sum of all demands sum of all supplies F = min Pseudo-Polynomial-Time Algorithm

35. 1520 8275 4335 4369 2724 25 root 3 20 477 282436 52394 5 15 20 25 Dynamic Programming Pseudo-Polynomial-Time Algorithm

36. 5 9 TvTv 5 9 3520 10 TvTv 3 9 v 5 5 20 v 35 10 optimal partition of T v optimal partition of T 520 5 10 3 9 T fulfillment = 18fulfillment = 20 Pseudo-Polynomial-Time Algorithm

37. TvTv 5 9 5 9 3520 10 v 520 10 3 fulfillment: 10 5 9 TvTv 5 9 3520 10 v 3520 10 fulfillment: 18 Dynamic Programming 20 - (10+5+3) = 2 marginal power 10 20-10= Pseudo-Polynomial-Time Algorithm

38. 5 9 TvTv 5 9 3520 10 v 520 10 3 fulfillment: 15 TvTv 5 9 5 9 3520 10 v 3520 10 20-(10+3)= 7 fulfillment: 13 20-(10+5)=5 Dynamic Programming Pseudo-Polynomial-Time Algorithm

39. TvTv 3 9 v 5 5 10 20 optimal partition of T 520 5 10 3 9 Pseudo-Polynomial-Time Algorithm

40. 1 2 3 marginal power 0 F 1 2 3 fulfillment staircase,non-increasing Pseudo-Polynomial-Time Algorithm F points

41. TvTv 15 v 5 4 5 53 optimal partition of T TvTv 15 v 5 4 5 53 TvTv v 5 53 5 4 fulfillment: 9 5+4 = 9 deficient power Pseudo-Polynomial-Time Algorithm

42. TvTv 15 v 5 53 5 4 fulfillment: 8 8 TvTv 15 v 5 5 fulfillment: 12 12 3 5 4 TvTv 15 v 5 53 5 4 fulfillment: 9 9 TvTv 15 v 5 53 fulfillment: 5 5 5 4 TvTv 15 v 5 4 5 53 optimal partition of T TvTv 15 v 5 4 5 53 Pseudo-Polynomial-Time Algorithm

43. TvTv 15 v 5 53 5 4 fulfillment: 8 8 TvTv 15 v 5 5 fulfillment: 12 12 3 5 4 TvTv 15 v 5 53 5 4 fulfillment: 9 9 TvTv 15 v 5 53 fulfillment: 5 5 5 4 1 2 3 deficient power 0 F 1 2 3 fulfillment staircase, non-decreasing Pseudo-Polynomial-Time Algorithm F points

44. 3 20 477 282436 52394 5 15 20 25 leaves Pseudo-Polynomial-Time Algorithm

45. 3 20 477 282436 52394 5 15 20 25 3 20 477 282436 52394 5 15 20 25 marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment marginal power 0 F 12 3 fulfillment O(F 2 ) time marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment

46. root 3 20 477 282436 52394 5 15 20 25 Dynamic Programming marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment max fulfillment Pseudo-Polynomial-Time Algorithm

47. root 3 20 477 282436 52394 5 15 20 25 Computation time for each vertex ･･････････････ O(F 2 ) Dynamic Programming There are n vertices. Pseudo-Polynomial-Time Algorithm

48. Computation time O(F 2 n) Computation time O(F 2 n) The algorithm takes polynomial time if F is bounded by a polynomial in n. There are n vertices. Computation time for each vertex ･･････････････ O(F 2 ) Pseudo-Polynomial-Time Algorithm

49. (2) FPTAS For any , 0<  <1, the algorithm finds a partition of a tree T such that OPT – APPRO <  OPT in time polynomial in both n and 1/ . For any , 0<  <1, the algorithm finds a partition of a tree T such that OPT – APPRO <  OPT in time polynomial in both n and 1/ . Let all demands and supply be positive real numbers. n : # of vertices

50. The algorithm is similar to the previous algorithm. 0 fulfillment marginal power t sampled fulfillment (2) FPTAS

51. 0 fulfillment marginal power t The algorithm is similar to the previous algorithm. (2) FPTAS OPT – APPRO < 2nt total error < error/merge × # of merge < 2t × n sampled fulfillment OPT APPRO

52. Error m d : max demand 7 436 2394 5 15 25 9 m d = 9 (2) FPTAS OPT – APPRO < 2nt

53. Error m d : max demand (2) FPTAS OPT – APPRO < 2nt m d ≦ OPT ≦  OPT OPT – APPRO <  OPT error ratio

54. m d : max demand Computation time marginal power 0 t Sampled fulfillment (2) FPTAS Error OPT – APPRO <  OPT

55. Conclusions 4 6535 4 26107 8 12 1513 25 General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees (2) FPTAS NP-hard 7 436 2394 5 15 25

56. Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand vertices that are supplied power. 10 3, 507, 308, 200 demandprofit

57. Variant (Generalization of Knapsack Problem) 10 3, 507, 308, 200 demands = 3+7 = 10 profits = 50+30 = 80 Sum 10 3, 507, 308, 200 demands = 8 profits = 200 Sum max find a partition which maximizes sum of profits of all demand vertices that are supplied power.

58. Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand vertices that are supplied power. FPTAS for KP [Ibarra and Kim ’75] FPTAS for trees 7, 305, 45 10 8, 20 6, 12 2, 10 15 10 3, 507, 308, 200

59. Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand vertices that are supplied power. FPTAS for KP [Ibarra and Kim ’75] FPTAS for trees 7, 305, 45 10 8, 20 6, 12 2, 10 15 extendable to partial k-trees (graphs with bounded treewidth) if there is exactly one supply 20 7, 85, 25 8, 3 6, 12 2, 10 3, 5 8, 13 10 3, 507, 308, 200

60. Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand vertices that are supplied power. FPTAS for KP [Ibarra and Kim ’75] FPTAS for trees 7, 305, 45 10 8, 20 6, 12 2, 10 15 extendable to partial k-trees (graphs with bounded treewidth) if there is exactly one supply 10 3, 507, 308, 200 20 7, 85, 25 8, 3 6, 12 2, 10 3, 5 8, 13

61. Thank You! Thank You!

63. (2b) Parallel connection G1G1 :terminal Series-Parallel Graphs (recursively) (1) A single edge is a SP graph. (2a) Series connection G2G2 G1G1 G2G2 G1G1 G2G2 G1G1 G2G2

64. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 :terminal SP Graph & Decomposition Tree P S S P SS e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

65. e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 SP Graph & Decomposition Tree :terminal

66. e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 SP Graph & Decomposition Tree :terminal Series connection S

67. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 SS :terminal Series connection e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree SP Graph & Decomposition Tree

68. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 P SS :terminal Parallel connection SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

69. e1e1 e2e2 e3e3 e7e7 e6e6 e4e4 e5e5 S P SS :terminal Series connection SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

70. e1e1 e2e2 e3e3 e7e7 e6e6 e4e4 e5e5 S S P SS :terminal Series connection SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

71. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 P S S P SS :terminal Parallel connection SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

72. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 P S S P SS :terminal SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 e6e6 e4e4 e5e5 e7e7 e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7

73. e1e1 e2e2 e3e3 e6e6 e4e4 e5e5 e7e7 P S S P SS DP algorithm :terminal SP Graph & Decomposition Tree e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree

74. Pseudo-Polynomial-Time Algorithm Suppose that a SP graph has exactly one supply. Max PP for such a SP graph can be solved in time O(F 2 n) if the demands and the supply are integers. 3 6 10 7 5 6 2 F : sum of all demands F = 3+7+6+2+6+5 = 29 pseudo-polynomial- time algorithm (2) FPTAS max fulfillment ≦ F n: # of vertices

75. Decomposition treeCorresponding SP subgraphs G 12 G1G1 G2G2 S P S DP table What should we store in the table for Max PP ? 1 3 2 4 7 1 3 2 3 2 4 7 5 5 Pseudo-Polynomial-Time Algorithm

80. G If G had more than one supply, Max PP would find a partition.

81. Pseudo-Polynomial-Time Algorithm G Max PP finds only one component with the supply

82. Pseudo-Polynomial-Time Algorithm G two terminals are supplied power and contained in same component

83. Pseudo-Polynomial-Time Algorithm G Max PP finds only one component with the supply

84. Pseudo-Polynomial-Time Algorithm G two terminals will be supplied power, but are contained in different components

85. Pseudo-Polynomial-Time Algorithm G Max PP finds only one component with the supply

86. Pseudo-Polynomial-Time Algorithm G this terminal will be supplied power this terminal is never supplied power

87. Pseudo-Polynomial-Time Algorithm G Max PP finds only one component with the supply

88. Pseudo-Polynomial-Time Algorithm G two terminals are never supplied power

89. Pseudo-Polynomial-Time Algorithm both are/will be supplied power both are/will be supplied power one is/will be supplied, but the other is never supplied one is/will be supplied, but the other is never supplied both are never supplied same component different components

90. If the component has the supply vertex, then the component may have the “marginal” power. 10 32 4 32 + 32 4 Pseudo-Polynomial-Time Algorithm

91. If the component has the supply vertex, then the component may have the “marginal” power. marginal power 10 – (3+2) = 5 Pseudo-Polynomial-Time Algorithm 10 32

92. If the component has no supply vertex, the component may have the “deficient” power. 32 7 + 32 20 7 32 Pseudo-Polynomial-Time Algorithm

93. If the component has no supply vertex, the component may have the “deficient” power. deficient power 3+7+2 = 12 Pseudo-Polynomial-Time Algorithm 32 7

94. both are/will be supplied power both are/will be supplied power same component one is/will be supplied, but the other is never supplied one is/will be supplied, but the other is never supplied both are never supplied different components

95. Pseudo-Polynomial-Time Algorithm both are/will be supplied power both are/will be supplied power 0 F 1 2 3 fulfillment marginal power fulfillment 1 2 3 marginal power 5 staircase,non-increasing

96. Pseudo-Polynomial-Time Algorithm both are/will be supplied power both are/will be supplied power deficient power fulfillment 1 2 3 deficient power 0 F 1 2 3 fulfillment staircase, non-decreasing

97. Pseudo-Polynomial-Time Algorithm both are/will be supplied power both are/will be supplied power same component one is/will be supplied, but the other is never supplied one is/will be supplied, but the other is never supplied both are never supplied different components marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment marginal power 0 F 12 3 fulfillment marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment marginal power 0 F 1 2 3 fulfillment deficient power 0 F 1 2 3 fulfillment

98. Decomposition treeCorresponding SP subgraphs G 12 G1G1 G2G2 S P S DP table 1 3 2 4 7 1 3 2 3 2 4 7 5 5 Pseudo-Polynomial-Time Algorithm size O(F ) size O(F ) time O(F 2 ) time O(F 2 )

99. Pseudo-Polynomial-Time Algorithm P S S P SS e 1 e 2 e 3 e 4 e 5 e 6 e 7 Decomposition tree O(F 2)O(F 2) O(F 2)O(F 2) # of nodes = O(n) O( F 2 n )O( F 2 n ) O( F 2 n )O( F 2 n ) n : # of vertices in a given SP graph

100. Pseudo-Polynomial-Time Algorithm Suppose that a SP graph has exactly one supply. Max PP for such a SP graph can be solved in time O(F 2 n) if the demands and the supply are integers. 3 6 10 7 5 6 2 F : sum of all demands F = 3+7+6+2+6+5 = 29 pseudo-polynomial- time algorithm (2) FPTAS max fulfillment ≦ F n: # of vertices

101. (2) FPTAS For any , 0<  <1, the algorithm finds a component with the supply vertex such that APPRO > (1–  ) OPT in time polynomial in both n and 1/ . For any , 0<  <1, the algorithm finds a component with the supply vertex such that APPRO > (1–  ) OPT in time polynomial in both n and 1/ . Let all demands and supply be positive real numbers. n : # of vertices

102. (2) FPTAS The algorithm is similar to the previous algorithm. 0 fulfillment marginal power t sampled fulfillment

103. 0 fulfillment marginal power t (2) FPTAS The algorithm is similar to the previous algorithm. APPRO > (1–  ) OPT m d : max demand

105. Open Problem Is there an approximation algorithm for SP graphs having more than one supply? FPTAS or PTAS ? constant-factor ? 3 2 52 8 6 3

106. Open Problem both are/will be supplied power both are/will be supplied power one is/will be supplied, but the other is never supplied one is/will be supplied, but the other is never supplied both are never supplied same component different components

107. Open Problem both are/will be supplied power different components The two components must become ONE component. j k deficient power = j + k 1 2 3 deficient power 0 F 1 2 3 fulfillment j + k (1D)

108. Open Problem j k deficient power = j + k If a given graph has more than one supply. 2 7 2 92 2 7 2 29

110. Our Results 3 6 10 7 5 6 2 Max subset sum problem FPTAS [Ibarra and Kim, 1975] Max PP our FPTAS straightforward ? 3 7 10 5 6 6 2

111. Our Results 3 10 7 5 6 6 2 Max subset sum problem FPTAS [Ibarra and Kim, 1975] Max PP our FPTAS straightforward ? 7 10 5 6 63 2 Sum: 10 margin: 2 Sum: 8 The graph structure plays crucial role. 2 2

113. Every demand vertex must be supplied from exactly one supply vertex. Assumption of the Problem Max PP is applied to Power delivery networks. It is difficult to synchronize two or more sources.

115. Power delivery network Application of Max PP Power delivery problem feeder supply vertex = load demand vertex = cable segment edge =

116. Power delivery network Power delivery problem Determine whether there exists a switching so that all loads can be supplied. If not all loads can be supplied, we want to maximize the sum of loads supplied power. If not all loads can be supplied, we want to maximize the sum of loads supplied power. Application of Max PP Max Partition Problem