1. Introduction to Maximum Flows
Class 8
Introduction to Maximum Flows
Obtain a network, and use the same network to illustrate the shortest path problem for communication networks, the max flow problem, the minimum cost flow problem, and the multicommodity flow problem. This will be a very efficient way of introducing the four problems. (Perhaps under 10 minutes of class time.)

2. The Max Flow Problem G = (N,A) xij = flow on arc (i,j)
uij = capacity of flow in arc (i,j)
s = source node
t = sink node
Maximize v
Subject to Sj xij Sk xki = for each i  s,t
Sj xsj = v
0 £ xij £ uij for all (i,j) Î A.
Introduce myself and the Tas.
It is critical that students sign up for the virtual campus web site. Hand out instructions on how to use it. Make sure that everyone signs up on the list.
Also, create a class list that is separate from the web site.
Ask for a show of where students come from. There should be four major contingencies:
Grad EECS, ORC, Transportation, Rest

3. Maximum Flows
We refer to a flow x as maximum if it is feasible and maximizes v. Our objective in the max flow problem is to find a maximum flow. s 1 2 t 10 , 8 8, 7 1,
10, 6 6, 5
A max flow problem. Capacities and a non-optimum flow.

4. The feasibility problem: find a feasible flow
retailers 1 2 3 4 5 6 7 8 9
warehouses 6 5 4 6 7 5
Is there a way of shipping from the warehouses to the retailers to satisfy demand?

5. Transformation to a max flow problem
warehouses
warehouses
retailers 6 5 4 1 6 5 4 s 6 6 7 5 t 6 7 5 2 7 3 8 4 9 5
There is a 1-1 correspondence with flows from s to t with 24 units (why 24?) and feasible flows for the transportation problem.

6. The feasibility problem: find a matching1 2 3 4 5 6 7 8
persons
tasks
Is there a way of assigning persons to tasks so that each person is assigned a task, and each task has a person assigned to it?

7. Transformation to a maximum flow problem
persons
tasks 1 5 t 1 s 1 2 6 3 7 4 8
Does the maximum flow from s to t have 4 units?

8. The Residual Network for a flow x1 2 t 10 , 8 8, 7 1,
10, 6 6, 5 i j x u ij i j
u - x x ij s 1 2 t 4 8 5 6 7
We let rij (resp. rji) denote the residual capacity of arc (i,j) (resp. (j,i)) in G(x). Keep only arcs with positive res. capacity.
The Residual Network G(x)

9. A Useful Idea: Augmenting Paths
An augmenting path is a path from s to t in the residual network.
The residual capacity of the augmenting path P is d(P) = min{rij : (i,j)  P}.
To augment along P is to send d(P) units of flow along each arc of the path. We modify x and the residual capacities appropriately.
rij := rij - d(P) and rji := rji + d(P) for (i,j)  P. s 1 2 t 4 8 5 6 7 s 1 2 t 4 8 6

10. The Ford Fulkerson Maximum Flow Algorithm
Begin
x := 0;
create the residual network G(x);
while there is some directed path from s to t in G(x) do
begin
let P be a path from s to t in G(x);
 := d(P);
send  units of flow along P;
update the r's;
end
end {the flow x is now maximum}.
Ford-Fulkerson Algorithm Animation

11. Proof of Correctness of the Algorithm
Assume that all data are integral.
Lemma: At each iteration all residual capacities are integral.
Proof. It is true at the beginning. Assume it is true after the first k-1 augmentations, and consider augmentation k along path P.
The residual capacity D of P is the smallest residual capacity on P, which is integral.
After updating, we modify residual capacities by 0, or D, and thus residual capacities stay integral.

12. Theorem. The Ford-Fulkerson Algorithm is finite
Proof. The capacity of each augmenting path is at least 1.
The augmentation reduces the residual capacity of some arc (s, j) and does not increase the residual capacity of (s, i) for any i.
So, the sum of the residual capacities of arcs out of s keeps decreasing, and is bounded below by 0.
Number of augmentations is O(nU), where U is the largest capacity in the network.

13. How do we know when a flow is optimal?1 2 t 10 , 9 8, 8 1,
10, 7 6, 6
METHOD 1. There is no augmenting path in the residual network.
reachable from s in G(x) s 1 2 t 8 7 6 9 3
not reachable from s in G(x)

14. Method 2: Cut Duality Theorys 1 2 t 10 , 9 8, 8 1,
10, 7 6, 6
An (s,t)-cut in a network G = (N,A) is a partition of N into two disjoint subsets S and T such that s  S and t  T, e.g., S = { s, 1 } and T = { 2, t }.
The capacity of a cut (S,T) is
CAP(S,T) = iSjTuij

15. Weak Duality Theorem for the Max Flow Problem
Theorem. If x is any feasible flow and if (S,T) is an (s,t)-cut, then the flow v(x) from source to sink in x is at most CAP(S,T).
PROOF. We define the flow across the cut (S,T) to be
Fx(S,T) = iSjT xij iSjT xji s 1 2 t 10 , 9 8, 8 1,
10, 7 6, 6
If S = {s}, then the flow across (S, T) is = 15.

16. Flows Across Cuts s 1 2 t 10 , 9 8, 8 1,
10, 7 6, 6
If S = {s,1}, then the flow across (S, T) is = 15. s 1 2 t 10 , 9 8, 8 1,
10, 7 6, 6
If S = {s,2}, then the flow across (S, T) is = 15.

17. More on Flows Across Cuts
Claim: Let (S,T) be any s-t cut. Then Fx(S,T) = v = flow into t.
Proof. Add the conservation of flow constraints for each node i Î S - {s} to the constraint that the flow leaving s is v. The resulting equality is Fx(S,T) = v.
Sj xij Sk xki = for each i  S\s Sj xsj = v i j
i  S
j  T
xij i j
i  T
j  S
-xij

18. More on Flows Across Cuts
Claim: The flow across (S,T) is at most the capacity of a cut.
Proof. If i  S, and j  T, then xij  uij. If i  T, and k  S, then xik  0.
Fx(S,T) = iSkT xik iSkT xki
Cap(S,T) = iSkT uik iSkT 0

19. Max Flow Min Cut Theorem
Theorem. (Optimality conditions for max flows). The following are equivalent.
1. A flow x is maximum.
2. There is no augmenting path in G(x).
3. There is an s-t cutset (S, T) whose capacity is the flow value of x.
Corollary. (Max-flow Min-Cut). The maximum flow value is the minimum value of a cut.
Proof of Theorem. 1 Þ 2. (not 2 Þ not 1)
Suppose that there is an augmenting path in G(x). Then x is not maximum.

20. Continuation of the proof.
3 Þ 1. Let v = Fx(S, T) be the flow from s to t. By assumption, v = CAP(S, T). By weak duality, the maximum flow is at most CAP(S, T). Thus the flow is maximum.
2 Þ 3. Suppose there is no augmenting path in G(x).
Claim: Let S be the set of nodes reachable from s in G(x). Let T = N\S. Then there is no arc in G(x) from S to T.
Thus i Î S and k Î T Þ xik= uik
i Î T and k Î S Þ xik= 0.

21. Final steps of the proof
Thus i Î S and k Î T Þ xik = uik
i Î T and k Î S Þ xik= 0.
Set S
Set T 1 2 3 4 s t
There is no arc from S to T in G(x)
x12 = u12
x43 = 0
If follows that
Fx(S,T) = iSkT xik iSkT xki
= iSkT uik iSkT = CAP(S,T)

22. Review
Corollary. If the capacities are finite integers, then the Ford-Fulkerson Augmenting Path Algorithm terminates in finite time with a maximum flow from s to t.
Corollary. If the capacities are finite rational numbers, then the Ford-Fulkerson Augmenting Path Algorithm terminates in finite time with a maximum flow from s to t. (why?)
Corollary. To obtain a minimum cut from a maximum flow x, let S denote all nodes reachable from s in G(x).
Remark. This does not establish finiteness if uij =  or if capacities may be irrational.

23. A simple and very bad example1 2 t M

24. After 1 augmentation 1
M-1 M 1 s t 1
M-1 M 1 2

25. After two augmentations1
M-1
M-1 1 1 s t 1
M-1
M-1 1 1 2

26. After 3 augmentations 1
M-2
M-1 2 1 s t 1
M-2
M-1 1 2 2

27. And so on

28. After 2M augmentations 1 M M s t 1 M M 2

29. An even worse example
In Exercise 6.48, there is an example that takes an infinite number of augmentations on irrational data, and does not converge to the correct flow.
But we shall soon see how to solve max flows in a polynomial number of operations, even if data can be irrational.

30. Summary and Extensions
1. Augmenting path theorem
2. Ford-Fulkerson Algorithm
3. Duality Theory.
4. Computational Speedups.

31. Application 6.4 Scheduling on Uniform Parallel Machines
Jobs j in J to be scheduled on M uniform parallel machines
j has Processing time p
Release time r
Due Date d
Suppose there are 2 parallel machines 2 1 4 3
No schedule is possible unless preemption is allowed

32. Application 6.4 Scheduling on Uniform Parallel Machines
Job(j) 1 2 3 4
Processing
1.5 3
4.5 5
Time
Release 2 2 4
Time
Due Date 5 4 7 9
Suppose there are 2 parallel machines 2 1 4 3
No schedule is possible unless preemption is allowed

33. Application 6.4 Scheduling on Uniform Parallel Machines
Job(j) 1 2 3 4
Processing
1.5 3
4.5 5
Time
Release 2 2 4
Time
Due Date 5 4 7 9
Suppose there are 2 parallel machines 2 1 4 3 4 3
A feasible schedule with preemption allowed (jobs 3 and 4 are split)

34. Transformation into a maximum flow problem
red arcs: the amount of processing of job j during t time units is at most t.
green arcs: the amount of processing during t time units is at most Mt.
0-2 2 1 4
1.5
2-4 4 2 3 2 s
4.5
4-5 1 t 3 5 4
5-7 4 4 2
blue arcs: the amount of processing of job j over all time units is pj.
7-9

35. A maximum flow
There is a feasible schedule if there is a maximum flow that saturates each of the arcs out of s.
0-2 1 1 .5
4,2
1.5
2-4 2 1
4,4 2 3 2 s
4.5
4-5
2,2 .5 t 3 5 2
4,4 1
5-7 4 2
4,2 2
Flow decomposition can be used to transform flows into schedules
7-9