Network Optimization Models:Maximum Flow Problems

In this handout:

• The problem statement

• Solving by linear programming

• Augmenting path algorithm

Maximum Flow Problem• Given: Directed graph G=(V, E),

Supply (source) node O, demand (sink) node T

Capacity function u: E R .

• Goal: Given the arc capacities,

send as much flow as possible

from supply node O to demand node T

through the network.

• Example:

4

4

56

445

5

O

A

DB

C

T

Characteristics of a feasible flow

• Let xij denote the flow through arc i j .• Capacity kij of arc i j is the upper bound on the flow shipped

through arc i j . Thus, we have the following constraints:

0 xij kij , for any arc i j • Every node i, except the source and the sink,

should satisfy the conservation-of-flow constraint, i.e.,flow into node i = flow out of node i

In terms of xij the constraint is

• For any flow that satisfies the conservation-of-flow constraints,flow out of the source = flow into the sink

This is the amount we want to maximize.

TO,i nodeany for ,xxkiarcs

ikijarcs

ji

Linear Program for the Maximum Flow Problem

• Summarizing, we have the following linear program:

maxxmaximize

TO,i nodeany for ,xx s.t.kiarcs

ikijarcs

ji

maxTiarcs

iTTjarcs

Oj xxx

0 xij kij , for any arc i j

This linear program can be solved by a Simplex method or Excel Solver add-in.

x12

4 4

5 64

45

5

1

2

53

4

6

x13

x14

x35

x34

x56

x46

x25

Objective function: maximize xmax

Capacity constraints:x12 ≤ 4, x13 ≤ 5, x14 ≤ 4, x25 ≤ 4, x34 ≤ 4, x35 ≤ 6, x45 ≤ 5, x56 ≤ 5Conservation-of-flow constraint:x12 = x25, x13 = x34+x35, x14+x34=x46, x24+x35=x56

Constraint for the sourse and sink node:x12+x13+x14=x46+x56=xmax

Non-negativity constraints: xij≥0

• There is also a method, which can be applied directly in the graph:

Augmenting Path Algorithm.

Augmenting Path Algorithm• Idea: Find a path from the source to the sink,

and use it to send as much flow as possible.

• In our example,

5 units of flow can be sent through the path O B D T ;

Then use the path O C T to send 4 units of flow.

The total flow is 5 + 4 = 9 at this point.

• Can we send more?

O

A

DB

C

T

4

4

5 644 5

5

5 5

4 4

5

Towards the Augmenting Path Algorithm

• If we redirect 1 unit of flow

from path O B D T to path O B C T,

then the freed capacity of arc D T could be used

to send 1 more unit of flow through path O A D T,

making the total flow equal to 9+1=10 .

• To realize the idea of redirecting the flow in a systematic way,

we need the concept of residual capacities.

O

A

DB

C

T

44

5 644 5

5

5 5 5

4 4

44

1 5

1 1

Residual capacities• Suppose we have an arc with capacity 6 and current flow 5:

• Then there is a residual capacity of 6-5=1

for any additional flow through B D .

• On the other hand,

at most 5 units of flow can be sent back from D to B, i.e.,

5 units of previously assigned flow can be canceled.

In that sense, 5 can be considered as

the residual capacity of the reverse arc D B .

• To record the residual capacities in the network,

we will replace the original directed arcs with undirected arcs:

B D6

5

B D1 5 The number at B is the residual capacity of BD;the number at D is the residual capacity of DB.

Residual Network• The network given by the undirected arcs and residual capacities

is called residual network.• In our example,

the residual network before sending any flow:

Note that the sum of the residual capacities on both ends of an arc is equal to the original capacity of the arc.

• How to increase the flow in the network based on the values of residual capacities?

O

A

DB

C

T

4

4

5 644 5

5

0

0

0

000

0 0

Augmenting paths• An augmenting path is a directed path

from the source to the sink in the residual network

such that

every arc on this path has positive residual capacity.• The minimum of these residual capacities

is called the residual capacity of the augmenting path.

This is the amount

that can be feasibly added to the entire path.• The flow in the network can be increased

by finding an augmenting path

and sending flow through it.

Updating the residual network by sending flow through augmenting pathsContinuing with the example, • Iteration 1: O B D T is an augmenting path

with residual capacity 5 = min{5, 6, 5}.• After sending 5 units of flow

through the path O B D T,

the new residual network is:

O

A

DB

C

T

4

4

44

5

0

0

000

0 1 0 55 55 6 5 00 0

Updating the residual network by sending flow through augmenting paths• Iteration 2:

O C T is an augmenting path

with residual capacity 4 = min{4, 5}.• After sending 4 units of flow

through the path O C T,

the new residual network is:

O

A

DB

C

T

4

4

0

0

0

4

50

0

0 1 0 55 5

0

14

4

4

Updating the residual network by sending flow through augmenting paths• Iteration 3:

O A D B C T is an augmenting path

with residual capacity 1 = min{4, 4, 5, 4, 1}.• After sending 1 units of flow

through the path O A D B C T ,

the new residual network is:

O

A

DB

C

T

00 5

50

4

4

4

0

0

0

1 5

1

4

4

3

3

1

1

1

2 4

0

5

3

Terminating the Algorithm:Returning an Optimal Flow

• There are no augmenting paths in the last residual network.

So the flow from the source to the sink cannot be increased further, and the current flow is optimal.

• Thus, the current residual network is optimal.

The optimal flow on each directed arc of the original network

is the residual capacity of its reverse arc:

flow(OA)=1, flow(OB)=5, flow(OC)=4,

flow(AD)=1, flow(BD)=4, flow(BC)=1,

flow(DT)=5, flow(CT)=5.

The amount of maximum flow through the network is

5 + 4 + 1 = 10

(the sum of path flows of all iterations).

The Summary of the Augmenting Path Algorithm

• Initialization: Set up the initial residual network.• Repeat

– Find an augmenting path.– Identify the residual capacity c* of the path; increase the flow

in this path by c*.– Update the residual network: decrease by c* the residual

capacity of each arc on the augmenting path; increase by c* the residual capacity of each arc in the opposite direction on the augmenting path.

Until no augmenting path is left• Return the flow corresponding to

the current optimal residual network