Download Assumptions

Maximum Flow Notation
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Maximum Flow Assumptions
1.
Network is directed
2.
All capacities are non-negative integers
3.
Network does not have a path from s to t with all arcs having
infinite capacity
4.
If (i, j)  A then (j, i)  A
 Nonrestrictive because we allow arcs with capacity = 0
5.
Network has no parallel arcs
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Applications
Feasible flow problem (application 6.1)

 j: i , j A
xij 
x
0  xij  uij
As before, we assume that
ji
 j: j ,i A
 bi 
for i  N ,
for all (i, j )  A
 i  N bi   0
Introduce two new nodes
A source node s and a sink node t.
For each node i with b(i)>0, we add an arc (s, i) with capacity b(i).
For each node i with b(i)<0, we add an arc (i, t) with capacity -b(i).
We refer to the new network as the transformed network.
Then we solve a maximum flow problem from node s to node t in the
transformed network. If the maximum flow saturates all the source and sink
arcs, problem (6.2) has a feasible solution; otherwise, it is infeasible.
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Additional Applications
1. Pipeline
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s
t
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Definitions and Notation
Residual Network
Given a flow x, the residual capacity, rij, of an arc (i, j)A is the
maximum additional flow that can be put on (i, j).
Note that there will also be an rji [recall assumption]
rij has two components
(1) uij - xij unused capacity of (i, j)
(2) xji
flow on (j, i) which we can cancel to increase the flow
from i to j
rij = uij - xij + xji
xij, uij
Example
i
j
2
1
4
3
rij
Residual network
i
j
2
1
4
3
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s - t cut
Cut partition N into two sets, S and s = N - S
Notation: [S, s ]
or arcs with one endpoint in S and one in
s-t cut if s  S & t 
s
s
Forward arc of cut if (i, j) i  S
j
Backward arc of cut if (i, j) i 
jS
s
s
set denoted (S,
s)
set denoted ( s , S)
Example
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4
s
t
3
(S,
s ) = (2, 4), (3, 4), (3, t)
( s , S) = 
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Capacity of an s - t cut
 
u S, S 

(i, j )(S , S )
uij
Minimum cut
Residual capacity of s - t cut
r s, s  
 rij
i, j s, s 
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Flow across s - t cut
v
 xij   xij
i, j s,s  i, j s ,s 
Property 6.1 The value of any flow is less than or equal to the capacity of
any s - t cut
Property 6.1 implies that if we find a flow x that equals the capacity of some
cut [S, s ] then x is a maximum flow and [S, s ] is a minimum cut.
We will reference this in the max-flow min cut theorem.
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We can also state property 6.1 in terms of the residual capacity.
Suppose x has a flow value of v
x has a flow value of v + v
v  v 
v  0
u

 
i , j  s ,s
(6.5)
ij
Then 6.5 minus 6.3
Recall that 6.3 says: (v 
 xij   xij )
i, j s, s  i, j s , s 
Gives
v 
 uij  xij    xij
i, j s, s 
i, j s , s 
or
v 
 u
i , j  s , s 
 v 
ij
 xij  
 u
 i , j  s , s 
ij
x
i , j  s , s 
ji
 xij  x ji    rij
 s ,s 
Thus we get
Property 6.2 For any flow x of value v in a network, the additional flow that
can be sent from s to t is less than or equal to the residual capacity of any
s - t cut.
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Generic Path Augmenting Algorithm
Definition: Augmenting Path: directed path from s to t in the residual
network
Residual capacity of an augmenting path: minimum residual
capacity of any arc in the augmenting path.
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2
s
t
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2
s
t
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If the minimum residual capacity along an augmenting path is > 0 then we
can push more flow along that path.
We can build an algorithm that works by finding paths and pushing the
maximum flow on them.
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algorithm augmenting path;
begin
x : = 0;
while G(x) contains a directed path from node s to node t do
begin
identify an augmenting path P from node s to node t;
 : min rij : i, j  P;
augment
 units of flow along P and update G(x);
end;
end;
Figure 6.12 Generic augmenting path algorithm.
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s
t
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4
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2
s
t
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4
1
2
s
t
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4
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2
s
t
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4
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Relationship between the original and residual
networks
We can keep track of xij or rij, because they have an equivalence.
Augmenting path in original:
Path P (need not be directed) from s to t such that xij < uij for forward
arcs and xji > 0 for backward arcs.
Can show equivalence to directed path in residual network.
When we updated the flows we increased the xij values on the forward arcs
and decreased the xji values on the backward arcs.
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If we use rij's how do we get xij's?
Recall rij = uij - xij + xji or uij - rij = xij - xji
if
uij > rij let xij = uij - rij and xji = 0
if
uij < rij let xij = 0 and xji = rij - uij
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Explanation of path augmenting when there are
backward arcs.
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Labeling algorithm & maximum flow minimum cut
theorem
Discuss augmenting path algorithm in more detail.
Specifically
1. How do we know if there are any more augmenting paths?
2. Will the algorithm terminate in an infinite number of steps?
3. Will the algorithm terminate at opt.?
In general the algorithm fans out from s labeling nodes - we can label a node
if we can reach that node in the residual network (that is there is a path with
rij > 0 for all arcs in the path). When we label the sink, stop and augment
flow. If we can't label the sink stop.
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algorithm labeling;
begin
label node t;
while t is labeled do
begin
unlabel all nodes;
set pred (j) : = 0 for each jN;
label node s and set LIST : = {s};
while LIST   and t is unlabeled do
begin
remove a node i from LIST;
for each arc (i, j) in the residual network
emanating from
node i do
if rij > 0 and node j is unlabeled then set
pred (j) : = i , label node j, and add j to LIST;
end;
if t is labeled then augment
end;
end;
procedure augment;
begin
use the predecessor labels to trace back from the sink to the source to
obtain an augmenting path P from node s to node t;
 : min rij : i, j   P;
augment  units of flow along P and update the residual capacities;
end;
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Example
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t
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LIST
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Pred
s
1
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3
4
t
3
4
t
2
s
t
3
LIST
2
4
Pred
s
1
2
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1
2
s
t
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LIST
s
1, 3
2, 3 (Note 2 tries to label only 3)
3
4
4
Pred
s
1
0
s
2
0
1
3
s
4
0
t
0
3
4
The algorithm labels t, augment flow
 = min {2, 3, 5} = 2
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2
s
t
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LIST
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Pred
s
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2
3
4
t
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1
2
s
t
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Can't label anything but s.
Original arcs
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t
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Correctness of the algorithm
When it stops how do we know we have an optimal solution?
When we stop we can't label the sink.
Let S = set of labeled nodes
S = set of unlabeled nodes
sS
t
S
Since no node in
S
can be labeled by a node in S this implies
rij = 0 for all (i, j)  (S,
Since
S)
rij  uij  xij   x ji
xij  uij
& x ji  0
what do we know about xij and xji?
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Plug into equation 6.3
v
x

 
i , j  s , s
ij

x

 
i , j s , s
ij

u

 
i , j s , s
ij
 us, s 
 Current flow = capacity of cut [s, s ]
By Property 6.1 this implies that x is a maximum flow and [s, s ] is a
minimum cut.
This establishes the algorithm correctness and the max-flow min-cut
theorem.
Theorem 6.3 (Max-Flow Min-Cut Theorem). The maximum value of the
flow from a source node s to a sink node t in a capacitated network equals
the minimum capacity among all s-t cuts.
Similarly, we can show
Theorem 6.4 (Augmenting Path Theorem). A flow x* is a maximum flow
if and only if the residual network G(x*) contains no augmenting path.
We can also show
Theorem 6.5 (Integrality Theorem). If all arc capacities are integer, the
maximum flow problem has an integer maximum flow.
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Complexity
0(m) per augmentation
max nU
(increase flow by 1 each time)
0(nmU)
If there are infinite capacity arcs just set uij very large but finite.
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Drawbacks of the Ford-Fulkerson method
1.
2.
3.
Empirically Ford-Fulkerson does well but can construct pathological
problems (see book).
Irrational capacities cause complications
"Forgetfulness" - start labeling over each time - we will discuss
improvements.
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Flows with Lower Bounds
lij  0
max v
s.t.
v i  s

x

x

 ij
 ji 0 i  N  s, t
 j:i, j A  j: j,i A  v i  t

lij  xij  uij
i, j   
Two key steps
1. Show that a feasible flow exists
(lij, uij)
(6, 9)
Infeasible Example
1
(4, 5)
2
3
2. Find the maximum flow
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Consider finding maximum flow first.
Assume we have a feasible flow.
Now with one modification we can use prior algorithms.
Let rij=(uij - xij) + (xji - lji)
Since the flow is feasible => lij  xij  uij = > rij  0
Our algorithms work with rij values (not arc flows, capacities, or lower
bounds) so we can still use them.
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One way to construct maximum flows, xij values, from rij values
For (i, j) let
uij = uij - lij
xij = xij - lij
rij = rij
recall
rij = (uij - xij) + (xji - lji)
equivalently
rij = (uij - xij + xji)
similarly
rji = (uji - xji + xij)
compute x as before
xij = max (uij - rij, 0)
xji = max (uji - rji, 0)
In original variables
xij = lij + max (uij - rij - lij, 0)
xji = lji + max (uji - rji - lji, 0)
Using the modified rij values we will find an optimal solution and can prove
theorem 6.10
Theorem 6.10 (Generalized Max-Flow Min-Cut Theorem). If the
capacity of an s-t cut [s, s ] in a network with both lower and upper bounds
on arc flows is defined by (6.7), the maximum value of flow from node s to
node t equals the minimum capacity among all s-t cuts.
(6.7)
us, s  
u

 
i , j  s ,s
ij

l

 
ij
i , j  s ,s
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How find a feasible flow?
First transform to a circulation problem.
Add
(t, s) with uts = 
Find a flow satisfying
x
ij
 j: i , j 

x
ji
 j: j ,i 
 0 i  N
lij  xij  uij i, j   
(6.11)
let xij = xij - lij
Now
 xij   x ji  bi 
 j :i, j   j : j, i 
i, j   A
  u ij  lij
0  xij
bi  
Note
l
ji
 j: j ,i 

l
i  N
(6.2)
ij
 j: i , j 
 bi   0
iN
Like feasible flow in application 6.1.
The maximum flow problem has a feasible solution iff the circulation has a
feasible flow.
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Can show theorem 6.11
Theorem 6.11 (Circulation Feasibility Conditions). A circulation
problem with nonnegative lower bounds on arc flows is feasible if and only
if for every set S of nodes
l  u

  
   
ij
i , j  s , s
i , j  s ,s
ij
Theorem 6.12. The feasible flow problem stated in (6.2) has a feasible
solution if and only if for every subset
s  N , bs   us, s   0 , where bs   is bi 
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So, an algorithm to check feasibility works like this:
Can start with zero flow, look at infeasible arcs (xij < lij)
Increase xij to lij by finding an augmenting path from j to i.
Continue until
1. All xij  lij
2. Can't find the augmenting path you need which implies the
problem is infeasible because theorem 6.11 is violated.
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