1. EMIS 8374 Solving Max Flow Problems with Non-Zero Lower Bounds Updated 18 March 2008

2. Not Always Feasible 2
(3,5) 1 s
(0,2) 3
(, u) t

3. Example Max Flow Problem
(, u)
(0,2) 2 4
(0,5)
(0,4) 1 6
(3,6)
(5,6) t s
(0,5)
(0,6) 3 5
(0,7)
(7,8)

4. Step 1: Convert to Circulation Problem
(0,2) 2 4
(0,5)
(0,4) 1 6
(3,6)
(5,6)
(0,5)
(0,6) 3 5
(0,7)
(7,8)
(0,)

5. Preflow Satisfies capacity constraints
Does not have to satisfy node-balance constraints

6. Step 2: Put a Preflow of pij = ij on (i, j)2 4 1 6 3 5 3 5 7

7. Step 3: Calculate excess e(i) for each i
(3+0) – (0+0) = 3
(0+0) – 5 = -5 2 4 1 6 3 5 3 5 7
(5+0) – (0+7) = -2
7 – (0+3) = 4

8. Network with preflow and excess3 -5 2 4 1 6 3 5 3 5 7 4 -2

9. Step 4: Let u'ij = uij - pij for each arc and remove preflow3 -5
(0,2) 2 4
(0,5)
(0,4) 1 6
(0,3)
(0,1)
(0,5)
(0,6) 3 5
(0,7)
(0,1) 4 -2
(0,)

10. Step 5: Add s' and t' t' s' 2 4 1 6 3 5 (0,3) (0,5) (0,2) (0,2) (0,4)
(0,1)
(0,5)
(0,6) 3 5
(0,7)
(0,1)
(0,)

11. Max s'-t' Flow Problem If e(j) < 0 then let cs'j = -e(j)
If e(i) > 0 then let cit' = e(i)
Let x be a maximum s'-t' flow. If this flow saturates the source- and sink-adjacent arcs, then it can be added to the preflow to create a feasible flow in the original network.
Otherwise, the original problem is infeasible.

12. Max s'-t' Flow Problem
Suppose that the max flow x saturates arc (s',4):
The flow x is balanced at all nodes.
xs'j = 5 = -e(4).
If we then remove s' and (s',4), node 4 will become unbalanced. It will send out 5 more units than it takes in.
Adding back the preflow p54 = 5 will rebalance node 4 .

13. Max s'-t' Flow Problem
Suppose that the max flow x saturates arc (2, t'):
The flow x is balanced at all nodes.
xt'j = 3 = e(2).
If we then remove t' and (2, t'), node 2 will become unbalanced. It will receive 3 more units than it sends out.
Adding back the preflow p23 = 3 will rebalance the node.

14. Step 6: Find Max s'-t' Flow x
(3,3)
(x, u)
(5,5) s'
(2,2) 2 4
(2,2)
(4,4)
(3,4)
(1,5) 1 6
(0,3)
(0,1)
(0,5)
(4,6) 3 5
(2,7)
(0,1)
(5,)

15. Step 7: Remove s' and t' 2 4 1 6 3 5 (x, u) (2,2) (3,4) (1,5) (0,3)
(0,1)
(0,5)
(4,6) 3 5
(2,7)
(0,1)
(5,)

16. Original Network with Flow x
(0+0) – (1+2) = -3
(2+3) – 0 = 5 2 2 4 3 1 1 6 5 4 3 5 2
(2+0) – (0+0) = 2
0 – (4+0) = -4

17. Step 8: Balance the Flow by Adding Preflow p2 2 4 3 1 1 6
0+3=3
0+5=5 5 4 3 5 2
0+7=7

18. Step 9: Remove Arc from t to s
We now have a feasible flow (v = 5) for the original problem.
(, x + p, u)
(0,2,2) 2 4
(0,1,5)
(0,3,4) 1 6 s
(3,3,6) t
(5,5,6)
(0,0,5)
(0,4,6) 3 5
(0,2,7)
(7,7,8)

19. Step 10: Construct the Residual Network2 2 4 1 4 3 t 1 6 s 3 1 5 2 2 3 5 5 4 1

20. Step 11: Use Augmenting Path Algorithm to Find the Max Flow2 2 4 1 4 3 t 1 6 s 3 1 5 2 2 3 5 5 4 1

21. Augmenting Path in the Residual Network5 2 2 4 1 3 1 t 1 6 s 3 1 4 2 1 6 3 5 4 1

22. Residual Network 5 2 2 4 4 1 t 1 6 s 3 1 4 1 1 6 3 5 6 1

23. Maximum Flow 2 4 1 6 3 5 (, x + p, u) v = 10 (0,2,2) (0,5,5) (0,4,4)s
(0,4,5)
(3,3,6) t
(5,6,6)
(0,5,6) 3 5
(0,6,7)
(7,8,8)

24. Example 2 2
(3,5)
(, u) 1 s
(0,2) 3 t

25. Example 2: Circulation & Preflow-3 2 2
(3,5) 3 3 1
(0,2) 1 3
(0,) 3

26. Example 2: Max s'-t' Flow Prob.
(0,3) t' 2
(0,2)
(0,3) 1
(, u)
(0,2) 3
(0,)

27. Example 2: Max s'-t' Flow s' t' 2 1 3 (2,3) (x, u) (0,2)
Source and sink arcs not saturated.
Problem is infeasible.
(2,3) 1
(2,2) 3
(2,)