INTRODUCTION TO OPERATIONS
RESEARCH
Maximum Flow Problem
MAXIMUM FLOW PROBLEM

Maximum flow problem description




All flow through a directed and connected network originates
at one node (source) and terminates at one another node
(sink)
All the remaining nodes are transshipment nodes
Flow through an arc is allowed only in the direction indicated
by the arrowhead, where the maximum amount of flow is
given by the capacity of that arc. At the source, all arcs point
away from the node. At the sink, all arcs point into the node
The objective is to maximize the total amount of flow from the
source to the sink (measured as the amount leaving the
source or the amount entering the sink)
MAXIMUM FLOW PROBLEM

Typical applications
Maximize the flow through a company’s distribution
network from its factories to its customers
 Maximize the flow through a company’s supply network
from its vendors to its factories
 Maximize the flow of oil through a system of pipelines
 Maximize the flow of water through a system of
aqueducts
 Maximize the flow of vehicles through a transportation
network

MAXIMUM FLOW ALGORITHM

Some Terminology
 The
residual network shows the remaining arc
capacities for assigning additional flows after
some flows have been assigned to the arcs
The residual capacity for assigning some flow from node B to node O
O
2
The residual capacity for flow from node O to Node B
5
B
MAXIMUM FLOW ALGORITHM



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 strictly positive residual capacity
The residual capacity of the augmenting path is the
minimum of these residual capacities (the amount of flow
that can feasibly be added to the entire path)
Basic idea

Repeatedly select some augmenting path and add a flow
equal to its residual capacity to that path in the original
network. This process continues until there are no more
augmenting paths, so that the flow from the source to the
sink cannot be increased further
MAXIMUM FLOW ALGORITHM

The Augmenting Path Algorithm

Assume that the arc capacities are either integers or
rational numbers
1.
Identify an augmenting path by finding some directed
path from the source to the sink in the residual network
such that every arc on this path has strictly positive
residual capacity. If no such path exists, the net flows
already assigned constitute an optimal flow pattern
2.
Identify the residual capacity c* of this augmenting
path by finding the minimum of the residual capacities
of the arcs on this path. Increase the flow in this path
by c*
MAXIMUM FLOW EXAMPLE

During the peak season the park management of the
Seervada park would like to determine how to route the
various tram trips from the park entrance (Station O) to the
scenic (Station T) to maximize the number of trips per day.

Each tram will return by the same route it took on the
outgoing trip so the analysis focuses on outgoing trips only.

To avoid unduly disturbing the ecology and wildlife of the
region, strict upper limits have been imposed on the number
of outgoing trips allowed per day in the outbound direction
on each individual road.

For each road the direction of travel for outgoing trips is
indicated by an arrow in the next slide.

The number at the base of the arrow gives the upper limit on
the number of outgoing trips allowed per day.
MAXIMUM FLOW EXAMPLE

Consider the problem of sending as many units from node O
to node T for the following network (current flow, capacity):
(0,3)
A
(0,9)
D
(0,5)
(0,1)
(0,4)
T
(0,1)
(0,7)
(0,6)
B
O
(0,5)
(0,4)
(0,2)
(0,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 1: one of the several augmenting paths is OBET, which has a
residual capacity of min{7, 5, 6} = 5. By assigning the flow of 5 to this path, the
resulting network is shown above
(0,3)
A
(0,9)
D
(0,5)
(0,1)
(0,4)
T
(0,1)
(5,7)
(5,6)
B
O
(5,5)
(0,4)
(0,2)
(0,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 2: Assign a flow of 3 to the augmenting path OADT. The
resulting residual network is
(3,3)
A
(3,9)
D
(3,5)
(0,1)
(0,4)
T
(0,1)
(5,7)
(5,6)
B
O
(5,5)
(0,4)
(0,2)
(0,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 3: Assign a flow of 1 to the augmenting path OABDT. The
resulting residual network is
(3,3)
A
(4,9)
D
(4,5)
(1,1)
(1,4)
T
(0,1)
(5,7)
(5,6)
B
O
(5,5)
(0,4)
(0,2)
(0,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 4: Assign a flow of 2 to the augmenting path OBDT. The
resulting residual network is
(3,3)
A
(6,9)
D
(4,5)
(1,1)
(3,4)
T
(0,1)
(7,7)
(5,6)
B
O
(5,5)
(0,4)
(0,2)
(0,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 5: Assign a flow of 1 to the augmenting path OCEDT. The
resulting residual network is
(3,3)
A
(7,9)
D
(4,5)
(1,1)
(3,4)
T
(1,1)
(7,7)
(5,6)
B
O
(5,5)
(1,4)
(0,2)
(1,4)
C
E
MAXIMUM FLOW EXAMPLE

Iteration 6: Assign a flow of 1 to the augmenting path OCET. The
resulting residual network is
(3,3)
A
(7,9)
D
(4,5)
(1,1)
(3,4)
T
(1,1)
(7,7)
(6,6)
B
O
(5,5)
(2,4)
(0,2)
(2,4)
C
E
MAXIMUM FLOW EXAMPLE

There are no more flow augmenting paths, so the
current flow pattern is optimal
3
A
7
D
1
4
T
13
3
O
1
B
6
7
5
2
C
13
2
E
MAXIMUM FLOW EXAMPLE





Recognizing optimality
Max-flow min-cut theorem can be useful
A cut is defined as any set of directed arcs
containing at least one arc from every directed
path from the source to the sink
For any particular cut, the cut value is the sum of
the arc capacities of the arcs of the cut
The theorem states that, for any network with a
single source and sink, the maximum feasible
flow from the source to the sink equals the
minimum cut value for all cuts of the network
MAXIMUM FLOW PROBLEM

Many situations can be modeled by a network
in which the arcs may be thought of as having a
capacity that limits the quantity of a product
that may be shipped through the arc. In these
situations, it is often desired to transport the
maximum amount of flow from a starting point
(called the source) to a terminal point (called
the sink). Such problems are called maximum
flow problems.
MAXIMUM FLOW PROBLEM

Sunco Oil wants to ship the maximum possible amount of
oil (per hour) via pipeline from node so to node si as shown
in the figure below.
Arc
Capacity
a0
(1)3
so
(2)2
1
(1)3
(1)4
2
3
(2)2
(1)1
si
(so,1)
2
(so,2)
3
(1,2)
3
(1,3)
4
(3,si)
1
(2,si)
2
The various arcs represent pipelines of different diameters. The
maximum number of barrels of oil that can be pumped through
each arc is shown in the table above (also called arc capacity).
MAXIMUM FLOW PROBLEM

Xij = Millions of barrels of oil per
hour that will pass through arc(i,j) of
pipeline.

For a flow to be feasible it needs to be in the
following range:
 0 <= flow through each arc <= arc capacity
and
 Flow into node i = Flow out from node i
MAXIMUM FLOW PROBLEM

Let X0 be the flow
through the artificial
arc, the conservation
of flow implies that
X0 = total amount of
oil entering the sink.
Thus, Sunco’s goal is
to maximize X0.
Max Z= X0
S.t. Xso,1<=2
Xso,2<=3
X12<=3
X2,si<=2
X13<=4
X3,si<=1
X0=Xso,1+Xso,2
Xso,1=X12+X13
Xso,2+X12=X2,si
X13=X3,si
X3,si+X2,si=X0
Xij>=0
(Arc Capacity constraints)
(Node so flow constraints)
(Node 1 flow constraints)
(Node 2 flow constraints)
(Node 3 flow constraints)
(Node si flow constraints)
One optimal solution to this LP is Z=3,
Xso,1=2, X13=1, X12=1, Xso,2=1, X3,si=1, X2,si=2, Xo=3.
MINIMUM SPANNING TREE



A spanning tree of a graph is just a sub-graph that
contains all the vertices. A graph may have many
spanning trees
The minimum spanning tree of a graph defines the
cheapest subset of edges that keeps the graph in
one connected component.
Telephone companies are particularly interested in
minimum spanning trees, because the minimum
spanning tree of a set of sites defines the wiring
scheme that connects the sites using as little wire
as possible.
MINIMUM SPANNING TREE

the complete graph on four vertices
has sixteen spanning trees:
MINIMUM SPANNING TREE

Minimum spanning trees prove important for several
reasons:




They can be computed quickly and easily, and they create a
sparse sub-graph that reflects a lot about the original graph.
They provide a way to identify clusters in sets of points.
Deleting the long edges from a minimum spanning tree
leaves connected components that define natural clusters in
the data set
They can be used to give approximate solutions to hard
problems
As an educational tool, minimum spanning tree algorithms
provide graphic evidence that greedy algorithms can give
provably optimal solutions.
MINIMUM SPANNING TREE PROBLEM

Suppose that each arc (i,j) in a network has a length
associated with it and that arc (i,j) represents a way of
connecting node i to node j.


For example, if each node in a network represents a computer in a
computer network, arc(i,j) might represent an underground cable
that connects computer i to computer j.
In many applications, we want to determine the set of arcs
in a network that connect all nodes such that the sum of
the length of the arcs is minimized.
APPLICATIONS

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Design of telecommunication networks (fiber-optic,
computer, telephone, cable television)
Design of lightly used transportation network to minimize
the total cost of providing the links (rail lines, roads, etc.)
Design of a network of high-voltage electrical transmission
lines
Design of a network of wiring on electrical equipment (e.g.,
a digital computer system) to minimize the total length of
the wire
Design of a network of pipelines to connect a number of
locations
DESCRIPTION

For a network with n nodes, a spanning tree is a group
of n-1 arcs that connects all nodes of the network and
contains no loops.
12
1
2
(1,2)-(2,3)-(3,1) is a loop
4
7
(1,3)-(3,2) is the minimum spanning tree
3
MINIMUM SPANNING TREE



You are given the nodes of the network but not the
links. Instead you are given the potential links and
the positive length for each if it is inserted into the
network (alternative measures for length of a link
include distance, cost, and time)
You wish to design the network by inserting enough
links to satisfy the requirement that there be a
path between every pair of nodes
The objective is to satisfy this requirement in a way
that minimizes the total length of links inserted
into the network
MINIMUM SPANNING TREE ALGORITHM

Greedy Algorithm
1)
2)
3)
Select any node arbitrarily, and then connect (i.e., add a
link) to the nearest distinct node
Identify the unconnected node that is closest to a
connected node, and then connect these two nodes
(i.e., add a link between them). Repeat the step until all
nodes have been connected
Tie breaking: Ties for the nearest distinct node (step 1)
or the closest unconnected node (step 2) may be broken
arbitrarily, and the algorithm will still yield an optimal
solution.
MINIMUM SPANNING TREE EXAMPLE

The State University campus has five computers. The
distances between computers are given in the figure
below. What is the minimum length of cable required
to interconnect the computers?
1
1
2
2
6
4
5
3
2
4
4
5
2
3
MINIMUM SPANNING TREE EXAMPLE

Iteration 1: Following the MST algorithm discussed before, we arbitrarily
choose node 1 to begin. The closest node is node 2. Now C={1,2}, Ć={3,4,5},
and arc(1,2) will be in the minimum spanning tree.
1
1
2
2
6
4
5
3
2
4
4
5
2
3
MINIMUM SPANNING TREE EXAMPLE

Iteration 2: Node 5 is closest to C since node 5 is two blocks from node 1
and node 2, we may include either arc(2,5) or arc(1,5) in the minimum
spanning tree. We arbitrarily choose to include arc(2,5).
Then C={1,2,5} and Ć={3,4}.
1
1
2
2
6
4
5
3
2
4
4
5
2
3
MINIMUM SPANNING TREE EXAMPLE

Iteration 3: Since node 3 is two blocks from node 5, we may include
arc(5,3) in the minimum spanning tree. Now C={1,2,5,3} and Ć={4}.
1
1
2
2
6
4
5
3
2
4
4
5
2
3
MINIMUM SPANNING TREE EXAMPLE

Iteration 4: Node 5 is the closest node to node 4. Thus, we add arc(5,4) to

the minimum spanning tree. We now have a minimum spanning tree
consisting of arcs(1,2), (2,5), (5,3), and (5,4).
The length of the minimum spanning tree is 1+2+2+4 = 9 blocks.
1
1
2
2
6
4
5
3
2
4
4
5
2
3
NOTES
The number of links in a spanning tree are one
less than the number of nodes.
 Each node is directly connected by a single link
to at least one other node.
 It is possible to have multiple minimum
spanning trees.
