Transportation networks
Lecture notes in Transportation Systems Engineering
4 August 2009
The conservation equation for a network can be written as:
where,
denote the flow value, and
is the flow on link i,j.
The net flow across any cut-set
separating the origin and destination is equal to the flow value.
The conservation equation can be conveniently represented in a matrix form.
The node-link incidence matrix in an
matrix
where the rows
corresponds to nodes (
)and the column corresponds to links
and each
cell denoted by
defined as:
For example ...
The flow conservation equations () can now be written as:
 |
(1) |
where
is the vector of link flows,
is the O-D vector.
In a connected weighted graph, all spanning trees have n-1 edges and will have minimum or maximum sum of the weights.
Tow algorithms are proposed: Prims algorithma dn Kruskal's algorithm.
`
Prims algorithm grows a spanning tree from a given vertex of a connected
weighted graph G, iteratively adding the cheapest edge from a vertex already
reached to a vertex not yet reached, finishing when all the vertices of G have
been reached.
Break tie arbitrarily.
Input: A weighted, connected graph G(N,L)
Initilisation: T(G)=
Iteration: Add cheapest edge that incorporate a new vertex
Find the minimum spanning tree of the network given in figure 1
Figure 1:
Example netowrk for minimum spanning tree problem 1
 |
- Start with node 1, N={1}, T={
}
- Node 1 can be connected by nodes 2, 3, and 4, having weights 12, 17, and 10
- Minimim weight is 10, for node 4. So N={1,4}, T={(1,4)}
- Node 1 can be connected by nodes 2 and 3, having weights 12 and 17
- Node 4 can be connected by nodes 3 and 5, having weights 14 and 19
- Minimim weight is 12, for node 2. So N={1,4,2}, T={(1,4),(1,2)}
- Node 1 can be connected by node 3, having weight 17
- Node 4 can be connected by nodes 3 and 5, having weights 14 and 19
- Node 2 can be connected by nodes 3 and 5, having weights 18 and 16
- Minimim weight is 14, for node 3. So N={1,4,2,3}, T={(1,4),(1,2),(3,4)}
- Node 1 can be connected by no nodes without loop
- Node 4 can be connected by node 5, having weight 19
- Node 2 can be connected by node 5, having weight 16
- Node 3 can be connected by node 5, having weight 11
- Minimim weight is 11, for node 3. So N={1,4,2,3}, T={(1,4),(1,2),(3,4),(3,5)} W=49.
Figure 2:
Solution for the example netowrk for minimum spanning tree problem 1
 |
The main concept of this algorithm is to maintain an acyclic spanning sub-graph H, enlargin it by edges with low weight to form a spanning tree.
Consider edges in non-descending order of weight, breaking ties arbitrarily.
Input: A weighted connected graph G(N,L)
Initilization: Set T=
a sub-graph of G
Iteration: If the next cheapest edge joins two components of T, then include it, otherwise, discard it.
Terminate when T is connected.
Find the minimum spanning tree of the network given in figure 1
- Sort the links in the descending order of weights reslts in
- 19 (4,5); 18 (2,3); 17 (1,3); 16 (2,5); 14 (3,4); 12 (1,2); 11 (3,5); and 10 (1,4)
- T=(1,4) W=10
- T=(1,4),(3,5) W=10+11
- T=(1,4),(3,5),(1,2), W=10+11+12
- T=(1,4),(3,5),(1,2),(3,4), W=10+11+12+14=47
Input: A graph with nonnegative edge weights and a starting vertex
.
The weight of
is
; let
if
is not an edge.
Initilization: Set
for
Iteration: Select a vertex
outside
such that
Add
to
.
Explore edges from
to update tentative distances: for each edge
with
, update
to minimize
Iteration continues until
or until
for every
.
At the end, set
Table 1:
Solution to the shortest path algorithm
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u |
u |
ua |
0+1=1 |
1 |
a |
ua |
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ub |
0+3=3 |
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ua |
u |
ub |
0+3=3 |
3 |
b |
uab |
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ad |
1+5=6 |
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ac |
1+4=4 |
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uab |
a |
ad |
1+5=6 |
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ac |
1+4=5 |
5 |
c |
uabc |
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3+4=7 |
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3+5=8 |
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ad |
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6 |
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uabcd |
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- 1
Renfrey B Potts and Robert M Oliver.
Flows in transportation networks.
Academic Press, New York, 1972.
- 2
Douglas B West.
Introduction to graph theory.
Pearson education Asia, New Delhi, India, 2001.
Prof. Tom V. Mathew
2009-08-04