1. 3 t 4 1 2 s2s2 s1s1 2 1 4 23 3 1 3 3 3 1 the black number next to an arc is its capacity

2. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 5 1 3 11 the red number next to an arc is its unit flow cost

3. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 5 1 3 11 b s 1 = 2 b s 2 = 2 b t =-4

4. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 b s 1 = 2 b s 2 = 2 b t =-4 2 2 1 1 2 3 5 1 2 1 the black number next to an arc is its capacity the red number next to an arc is its unit flow cost the green number next to an arc is its current flow

5. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} 2 2 1 1 2 3 5 1 2 1

6. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} On C in the order given: {1,2} and {2,4} are forward arcs {1,4} is a backward arc 2 2 1 1 2 3 5 1 2 1

7. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 1 5 33 13 2 11 For Cycle C={1,2,4,1} define +1 for all forward arcs on C h(C)= -1 for all backward arcs on C 0 for all arcs not on C 2 2 1 2 3 5 1 3 1

8. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 1 5 33 13 2 11 For Cycle C={1,2,4,1} define +1 for all forward arcs on C h(C)= -1 for all backward arcs on C 0 for all arcs not on C 2 2 1 2 3 5 1 3 1 Ah(C)=0

9. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 1 5 33 13 2 11 For Cycle C={1,2,4,1} define +1 for all forward arcs on C h(C)= -1 for all backward arcs on C 0 for all arcs not on C 2 2 1 2 3 5 1 3 1 Ah(C)=0  f+ өh(C) is feasible for all ө small enough: i.e., such that 0 ≤ f+ өh(C) ≤ u

10. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 1 5 33 13 2 11 For Cycle C={1,2,4,1} define +1 for all forward arcs on C h(C)= -1 for all backward arcs on C 0 for all arcs not on C 2 2 1 2 3 5 1 3 1 Ah(C)=0  f+ өh(C) is feasible for all ө such that 0 ≤ f+ өh(C) ≤ u h(C) is called a simple circulation

11. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} Push ө unit of flow over C: Δf 12 = өh 12 = +ө, Δf 24 = өh 24 = +ө, Δf 14 = өh 14 = -ө 2 2 1 1 2 3 5 1 2 1

12. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} Push ө unit of flow over C: Δf 12 = өh 12 = +ө, Δf 24 = өh 24 = +ө, Δf 14 = өh 14 = -ө f 12 + ө ≤ 1, hence (f 12 =0) ө ≤ 1 f 24 + ө ≤ 4, hence (f 24 =2) ө ≤ 2 Thus, ө ≤ 1 f 14 - ө ≥ 0, hence (f 14 =1) ө ≤ 1 2 2 1 1 2 3 5 1 2 1

13. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} Push ө unit of flow over C: Δf 12 = өh 12 = +ө, Δf 24 = өh 24 = +ө, Δf 14 = өh 14 = -ө Total change in cost: ө(5 + 2 – 6) = ө 2 2 1 1 2 3 5 1 2 1

14. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,2,4,1} Push ө unit of flow over C: Δf 12 = өh 12 = +ө, Δf 24 = өh 24 = +ө, Δf 14 = өh 14 = -ө Total change in cost: ө(5 + 2 – 6) = ө Not a good change! 2 2 1 1 2 3 5 1 2 1

15. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,3,4,1} Push Ө unit of flow over C: h 13 =+1, h 43 =-1, h 14 = -1 2 2 1 1 2 3 5 1 2 1

16. 3 t 4 1 2 s2s2 s1s1 2 11 42 2563 13 15 33 13 2 11 Cycle C={1,3,4,1} Push Ө unit of flow over C: h 13 =+1, h 43 =-1, h 14 = -1 2 2 1 1 2 3 5 1 2 1 unit cost change: c 13 – c 43 – c 14 = 1 – 1 – 6 = – 6  negative cost cycle Ө ≤ 1

17. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 b s 1 =2 b s 2 =2 b t =-4 2 2 2 4 5 1 2 2 the instance slightly changed and omitting capacities The Network Simplex Method

18. A basic (feasible) solution related to the Tree T 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 4 5 1 2 2

19. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 1 2 2 A basic feasible solution related to the Tree T C 14 = 6 + 1 – 1 = 6

20. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 4 5 1 2 2 A basic feasible solution related to the Tree T C 14 = 6 + 1 – 1 = 6 C 34 = 1 + 1 = 2

21. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 4 5 1 2 2 A basic feasible solution related to the Tree T C 14 = 6 + 1 – 1 = 6 C 34 = 1 + 1 = 2 C 4t = 1 – 5 – 1 = – 5

22. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 4 5 1 2 2 A basic feasible solution related to the Tree T C 14 = 6 + 1 – 1 = 6 C 34 = 1 + 1 = 2 C 4t = 1 – 5 – 1 = – 5 negative reduced cost

23. 3 t 4 1 s2s2 s1s1 2 1 2 56 1 5 3 1 1 2 2 2 4 5 1 2 2 A basic feasible solution related to the Tree T C 14 = 6 + 1 – 1 = 6 C 34 = 1 + 1 = 2 C 4t = 1 – 5 – 1 = – 5 negative reduced cost The cycle is unsaturated, hence will really decrease cost. Θ*=1

24. 3 t 4 1 s2s2 s1s1 2 1 2 4 5 36 2 1 2 5 1 3 1 3 1 2 2 2 4 5 4 1 2 2 2 The residual graph 3 t 4 1 s2s2 s1s1 2 1 2 5 36 2 1 5 1 3 1 1 2 1 2 2 2 4 5 1 2 2 -1 2 -5 4 -3 2 -2 2

25. 3 t 4 1 s2s2 s1s1 2 1 2 4 5 36 2 1 2 5 1 3 1 3 1 2 2 2 4 5 4 1 2 2 2 3 t 4 1 s2s2 s1s1 2 1 2 5 36 2 1 5 1 3 1 1 2 1 2 2 2 4 5 1 2 2 -1 2 -5 4 -3 2 -2 2 directed cycle in the residual graph = undirected unsaturated cycle in the graph

26. 3 t 4 1 s2s2 s1s1 2 1 2 4 5 36 2 1 2 5 1 3 1 3 1 2 2 2 4 5 4 1 2 2 2 directed negative cycle in the residual graph = undirected unsaturated negative cost cycle in the graph 3 t 4 1 s2s2 s1s1 2 1 2 5 36 2 1 5 1 3 1 1 2 1 2 2 2 4 5 1 2 2 -1 2 -5 4 -3 2 -2 2