Flow models, optimal routing, and topological network design

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Flow models, optimal routing, and topological
network design

• Single commodity network flows
• Max-flow/min cut
• The Ford-Fulkerson algorithm
• Minimum cost flows
• Constrained MST algorithms
• The Essau-William algorithm
• (Sharma algorithm)

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Flows in a network

• In a directed graph, if node i and node j are
  joined by an arc (link), over this arc some
  quantity f(i,j) (link flow) in data units/sec can
  flow.
• Associated with the arc connecting node i and
  node j is a capacity c(i,j), so that f(i,j)
  cannot exceed c(i,j).
• There is a source node S, which has an external
  supply of flow v. There is a sink node T, which
  has a demand of flow v.

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Flows in a network example

• A network with the arc capacities as indicated

• conservation of flow (flow_out-flow_in0)
• node s f(S,1) f(S,2) v 0
• node 1 f(1,3) f(1,2) f(S,1) 0
• node 2 f(2,3) f(2,4)- f(S,2) f(1,2) 0
• node 3 f(3,T) f(3,4) f(2,3) f(1,3) 0
• node 4 f(4,T) - f(3,4) - f(2,4) 0
• v f(3,T) f(4,T) 0

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Flows in a network example (cont.)

• Arc capacity constraints f(S,1) ? 6, f(S,2) ?
  8, f(1,2) ? 3, f(1,3) ? 3, f(2,3) ? 4, f(2,4) ?
  2, f(3,4) ? 2, f(3,T) ? 8, f(4,T) ? 6

• A (path) flow is a collection of the link flows
  f(i,j) . A feasible flow is a flow that
  satisfies the flow conservation and arc
  capacities constraints.
• The objective is to maximize the total flow v
  under the flow conservation and capacity
  constraints.

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8
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S
T
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v
v
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Flow-augmenting paths

• Consider the following feasible flow example

• The arcs in a flow network can be put into two
  categories
• I, the set of arcs whose flow can be increased,
  these are those arcs whose flow is smaller than
  its capacity, e.g. arcs S-1, 1-2, 2-6, 3-S, 4-2,
  4-3
• R, the set of arcs whose flow can be reduced,
  these are those arcs whose flow is non-zero, e.g.
  all arcs in I and 3-2, T-4

capacity
Link flow
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Flow-augmenting paths (cont.)

• To increase the net flow from S to T, we can
• find a directed path from S to T which only
  consists of arcs belonging to I then determine
  the maximum increase in flow for these arcs, e.g.
  the path S-1-2-T is such a path and the maximum
  increase in flow without exceeding the arc
  capacities is 1. This path may be called a flow
  augmenting path.
• find a directed path from T to S which only
  consists of arcs belonging to R then determine
  the maximum decrease in flow for these edges,
  e.g. the path T-4-3-S is such a path and the
  maximum decrease in flow is 1. This is also a
  flow augmenting path.
• combine the two approaches above i.e., find a
  path P from S to T with the following properties
• if arc(x,y) is on P and is directed from S to T
  (a forward arc), then (x,y) is a member of I.
• if arc(x,y) is on P and is directed from T to S
  (a backward arc), then (x,y) is a member of R.

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Flow-augmenting paths (cont.)

• For the combined approach, the maximum increase
  in flow is the minimum of the possible increase
  in flow for those forward arcs, or the minimum
  decrease in flow for those backward arcs,
  whichever smaller.
• e.g., in the previous example S-1-3-4-T is such a
  path. The maximum increase in flow is 2 units.

Illustrating flow augmentation on S-1-3-4-T
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Residual network

• The residual network for any feasible flow is
  constructed as follows.
• If arc(x,y) is a member of I, then construct an
  arc in the residual network from x to y and label
  it with a residual capacity r(x,y) c(x,y)
  f(x,y).
• If arc(x,y) is a member of R, then construct an
  arc in the residual network from y to x and label
  it with a residual capacity r(y,x) f(x,y).

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Residual network example
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Flow-augmenting algorithm

• Every flow augmenting path from S to T in the
  original network is a directed path from S to T
  in the residual network.
• A flow augmenting algorithm can be designed as
  follows.
• Step 1 Construct the residual network
  corresponding to the current feasible flow. label
  node S with e(s) ? and p(S) 0. All other
  nodes are initially unlabeled. All arcs are
  unmarked.
• Step 2 Select a labeled node x which has not yet
  been considered. If none exits, then stop no
  flow-augmenting path from S to T exits. Otherwise
  go to step 3.
• Step 3 If y is currently unlabeled and if (x,y)
  is a member of the forward star (outgoing arc) of
  x, then label node y with e(y) min e(x),
  r(x,y) and p(y) x. Mark arc(x,y). If T is
  labeled, then stop since a flow-augmenting path
  from S to T has been found Otherwise return to
  step 2.

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Flow-augmenting algorithm example
e extra flow that can be sent down the
pathp predecessor
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Cut

• If a set of arcs are removed from a connected
  graph, the number of components is increased.
  This set of arcs is called a disconnecting set.
  
• A cut is a set of arcs in a directed graph such
  that each path from S to T includes an arc in
  this set. Thus a cut in a network is an S-T
  disconnecting set (one component containing S,
  the other T) in the corresponding directed graph.

a disconnecting set
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Cut (cont.)

• The capacity of a cut is the sum of the
  capacities of the forward arcs in the cut.
• When the cut capacity is the smallest, the cut is
  called a minimum cut.
• Note that there may be more than one minimum cut
  in a network.

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Cut examples
cut 2
cut 1
Cut3
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a
c
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T
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Cap.2
Cap.2
Cap.3

• The minimum cut is (S,a), (b,T) and the
  capacity is equal to 2

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Ford and Fulkerson algorithm (Maximum-flow
algorithm)

• Max-Flow/Min-Cut theorem The value of the
  maximum feasible flow from S to T is equal to the
  capacity of the minimum S-T cut.
• This algorithm makes use of the concepts of
  residual network and flow-augmenting path
  discussed previously.

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Ford and Fulkerson algorithm (cont.)

• Step 1 Let S denote the source node, and let T
  denote the sink node.Select any initial flow
  from S to T, i.e., any set of values for f(x,y)
  that satisfies conservation of flow and capacity
  constraints. If no such initial flow is known,
  use as the initial flow f(x,y)0 for all (x,y).
• Step 2 Construct the residual network relative
  to the current flow.
• Step 3 Perform the flow-augmenting algorithm to
  determine a flow-augmenting path if no path is
  found, stop the current flow is maximum
  otherwise make the maximum possible flow
  augmentation along the flow-augmenting path
  discovered by the flow-augmenting algorithm.
  Return to Step 2.

An example in flash movie
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Minimum cost flow

• A given cost per unit C(i,j) is associated with
  the arc (i,j).
• Given s-t flow, find the minimum cost of the s-t
  flow, where the cost of a flow is defined as the
  sum of the flow over all arcs times the cost per
  unit for that arc.
• Find the maximum flow with a minimum cost.

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Optimal routing

• In a network, there is a set of
  origin-destination (O-D) pairs. The traffic for
  each pair w(i,j) is denoted rw.
• The transmission capacity of link (i,j) is
  denoted c(i,j) in data units/sec. The link flow
  for link (i,j) is f(i,j).
• Optimal routing given traffic and link
  capacities, find path flows to minimize
  subject to flow conservation, where Dij
  is a cost function.
• Example use the average number of waiting data
  units in the network as cost function. On link
  (i,j), are waiting. Optimal
  routing finding path flows to minimize
  subject to flow
  conservation.

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Subnet design and capacity assignment

• Subnet design problem given the location of
  network nodes and traffic flow for each pair of
  nodes, select capacity and flow of each link to
  meet some delay and/or reliability constraints
  while minimizing costs.
• Capacity assignment problem choose the capacity
  c(i,j) of each link (i,j) so as to minimize the
  linear cost where p(i,j)
  is a known positive price per unit capacity,
  subject to average delay constraint.

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Iterative heuristic method for capacity
assignment problem

• At the start of each iteration, there is
  available a current best topology and a trial
  topology.
• Current bets topology a topology found so far
  that satisfies the constraints.
• Trial topology the topology that will be
  evaluated in the current iteration.
• If the cost of the trial topology is less than
  that of the current best topology, replace the
  current best topology with the trial topology.
• Heuristic rules for generating trial topologies
• Lower the capacity of a link with low
  f(i,j)/c(i,j)
• Increase the capacity of links with high
  f(i,j)/c(i,j) when the delay constraint is not
  met.
• Branch exchange heuristic one link is deleted
  and another link is added.

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Centralized network topology

• Multipoint line topology Selection of the links
  connecting terminals to the center.
• Terminal assignment The association of terminals
  with specific concentrators.
• Concentrator location Deciding where to place
  concentrators, and whether or not to use them at
  all.

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Multipoint line layout formalated

• Given N sites, site 1 to site N, and a center
  site denoted by site 0.
• The traffic from site i (i1..N) to the center is
  denoted ti0 and t0i denotes the traffic from the
  center to site i.
• The cost of connecting site i and and site j is
  denoted cij.
• Let xij 1 if (i,j) is included in the topology
  xij 0 otherwise.

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Problem statement

• Find zmin cij xij subject
  to a b c d

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Constrained MST algorithm

• A matrix of input traffic tij from every node to
  every other node is given.
• An MST is to be designed suject to the constraint
  that the flow on each arc will not exceed a given
  upper bound.
• Kruskal or Prim algorithms can be modified for
  finding constrained MST at each iteration, a
  check is made on the flow constraint.

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The Essau-Williams algorithm

• Start with a spanning tree where the central node
  is directly connected with each of the N nodes.
  At each successive itearation, a link (i,0) is
  deleted from the current spanning tree and a link
  (i,j) is added. The link (i,j) is so chosen
• No cycle is formed
• The capacity constraints of all links of the new
  spanning tree are satisfied.
• The saving wi0 - wij in link weight obtained by
  exchanging (i,0) with (i,j) is positive and is
  maximized over all nodes i and j for which the
  previous two conditions are satisfied.

example