1. MATS-30004 Quantitative Methods Dr Huw Owens

2. Network Models Many managerial problems in areas such as
transportation design,
information system design, and
project scheduling
have been successfully solved with the aid of network models and network analysis techniques.
In Network Models, we will discuss three typical network problems, which are
the shortest route problem
the minimal spanning tree problem
the maximal flow problem.
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3. Before we start, let us get familiar with the network definitions.
A network consists of nodes linked by arcs or branches.
The nodes are represented by circles, and the arcs by lines. The standard notation for describing a network G is G = (N, A), where N is the set of nodes and A is the set of arcs. For example, if network G = (N, A), and
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4. N = {1, 2, 3, 4, 5}
A = {(1, 3), (1, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}
then the equivalent network is
Arc or Branch 1 3 5 2 4
Node
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5. The Shortest Route Problem
The primary objective of solving this problem is to determine the shortest route or path between the source node and any other node in the network.
Let us discuss this through the following example.
Example:
The Manchester Construction Company (MCC) has six construction sites, the geographical relations of these six sites to MCC central office being described in Figure 1. Everyday, a number of trips need to be arranged to each site many times carrying personnel, equipment, and supplies. Therefore, it is important for the MCC manager to work out the shortest route to every site so that such trips are most economical. We are asked to help the MCC manager to determine the routes that will minimise the total distance from the MCC office to each site.
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6. Example 17 15 10 6 5 3 2 4 7 1
MCC office
Road distance in miles
Figure 1
The algorithm used in this type of problems employs a labelling procedure to find the shortest distance from node 1 to each of the other nodes.
A label is placed about a node in the form of two figures within a pair of square brackets.
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7. A node may be either permanently or tentatively labelled
A node may be either permanently or tentatively labelled. That is, whenever the algorithm has determined the shortest distance for a node, that node is permanently labelled.
If the shortest distance has not been determined for a node, the node is then tentatively labelled.
A permanently labelled node is marked by shading it. Node 1, the starting node, is permanently labelled as [0, S], indicating that the distance occurred so far is 0 and this is the starting node.
[19, 5] n
Distance from node 1 to this node is 19
Preceding-node on the route from node 1 to this node is 5
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8. Accordingly, we permanently label node 3, and re-label node 2.
[15, 1]
[0, S]
[10, 1] 17 15 10 6 5 3 2 4 7 1
To start, we give a permanent label to node 1. Node 2 and 3 can be reached from the permanent node, then tentatively label nodes 2 and 3.
Examining the distances between node 1 and nodes 2 & 3 respectively, we find that node 3 has been labelled by the shortest distance from the starting node, and node 2 has not.
Accordingly, we permanently label node 3, and re-label node 2.
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9. Thus , we permanently label node 2.
After re-labelling node 2, we examine the distance again, and find that the node has been labelled by the shortest distance from node 1.
Thus , we permanently label node 2.
[13, 3]
[15, 1]
[0, S]
[10, 1] 17 15 10 6 5 3 2 4 7 1
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10. The procedure is illustrated by the following diagrams.
[13, 3]
[0, S]
[10, 1] 17 15 10 6 5 3 2 4 7 1
Using the same principle, we can label the rest of the nodes to make them permanently labelled.
The procedure is illustrated by the following diagrams.
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11. [13, 3]
[30, 2]
[19, 2]
[0, S]
[10, 1]
[14, 3] 17 15 10 6 5 3 2 4 7 1
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12. Example 1 03/12/2018 [13, 3] [30, 2] [19, 2] [18, 5] [0, S] [10, 1]
[14, 3]
[16, 5] 17 15 10 6 5 3 2 4 7 1
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13. [13, 3]
[22, 6]
[30, 2]
[18, 5]
[0, S]
[10, 1]
[16, 5]
[14, 3] 17 15 10 6 5 3 2 4 7 1
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14. [13, 3]
[22, 6]
[18, 5]
[0, S]
[10, 1]
[16, 5]
[14, 3] 17 15 10 6 5 3 2 4 7 1
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15. [13, 3]
[22, 6]
[18, 5]
[0, S]
[10, 1]
[16, 5]
[14, 3] 17 15 10 6 5 3 2 4 7 1
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16. Using this approach, the following shortest routes are identified for the MCC transportation network:
For simple cases such as the MCC problem, we may determine the shortest routes just as fast, if not faster, by inspection.
However, as problems grow larger, e. g. having more than 15 nodes, it becomes very time-consuming and mistake-prone to find the shortest routes by inspection.
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17. Summary of the Shortest Route Algorithm
The labelling procedure describes will then be required. For even larger problems, it becomes necessary to implement the algorithm on a computer.
Summary of the shortest-route algorithm:
Step 1. Permanently label node 1;
Step 2. Label tentatively all other nodes which are directly linked to permanent node(s);
Step 3. Examine all the tentative nodes to see whether they have shortest routes from node 1;
if YES, change label to permanent;
if NO, re-label the node using the shortest distance.
Repeat 2 and 3 until all the nodes are permanently labelled.
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18. L2 The Minimum Spanning Tree Problem
In network terminology, the minimal spanning tree problem involves using the arcs of the network to reach all nodes of the network in such a fashion that the total length of all the arcs used is minimum.
The objective of using the minimum spanning tree technique is to connect all nodes in the network such that the sum of connecting arc lengths is minimal.
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19. The network algorithm that can be used to solve the minimal spanning tree problem can be explained in the following steps:
Step 1: Arbitrarily start at any node and connect it to the closest node in terms of the criterion being used (e.g. time, cost, distance);
Step 2: Identify the unconnected node that is closest to one of the connected nodes. Add this new node to the set of connected nodes;
Step 3: Repeat Step 2 until all the nodes are connected.
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20. Example
UMIST computer centre is to lay network cables to 5 academic departments. The locations of these 5 departments relative to the computer centre are as shown in the diagram. To reduce costs, you are asked to help the computer centre to determine the minimal spanning tree.
400
500
200
100
300 5 2 3 4 1 6
UMIST Computer Centre
Distance in metres
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21. Let us choose node 1 as our starting point by shading it (step 1)
Let us choose node 1 as our starting point by shading it (step 1). Then, identify the closest unconnected node that is node 2 (step 2). Shade node 2. Next, in the neighbourhood of nodes 1 and 2, identify the closest unconnected node. We find node 4 (step 2). Repeat the procedure, we then find node 3, node 6, and node 5.
400
500
200
100
300 5 2 3 4 1 6
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22. To visualise the determined minimal spanning tree, we highlight the arcs involved in the diagram below.
This spanning tree connects the Manchester computer centre and the 5 departments most economically (using minimal length of cable).
It can be easily worked out that the total length of cable to be used in the project is 1100 metres.
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23. 200
100
300 5 2 3 4 1 6
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24. L3 The Maximum Flow Problem
Consider a network with one input or source node and one output or sink node.
The maximum flow problem asks:
What is the maximum amount of flow (vehicles, messages, fluid, etc.) that can go through the network in a given period of time?
In this problem, the amount of flow is limited due to capacity restrictions on the various arcs of the network.
For example, pipe sizes limit oil flow in a oil distribution system.
In the network for maximum flow problem, while flow capacity applies to the arcs, we assume that the flow out of a node is equal to the flow into the node.
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25. Example
Consider the South-North traffic flow problem on the road system through Manchester. The South-North traffic flow reaches a level of 15,000 vehicles per hour at peak times. Due to planned road maintenance, a program calling for lane closures and slower speed limits has been created involving a network of alternate roads. The proposed network is shown below, with flow capacities on each arc of the network.
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26. Can the proposed network support a flow of 15,000 vehicles per hour?
What is the maximum flow of the network?
How much flow should go over each road? 6 8 1 3 2 5 7
Entering M/cr 4
Flow capacity (,000) from node 4 to node 6
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27. The following algorithm is to used to solve the maximal flow problem:
Step 1: Find any path from the source node to the sink node that has capacities in the direction of the flow greater than zero for all arcs on the path. If no path is available, the optimal solution has been reached;
Step 2: Find the smallest arc capacity, Pf, on the path selected in step 1. Increase the flow through the network by sending an amount Pf over the path elected in step 1;
Step 3: For the path selected in step 1, reduce all arc flow capacities in the direction of the flow by Pf, and increase all arc flow capacities in the reverse direction by Pf. Go to Step 1.
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28. Notation for increasing the capacity of an arc:
From node 3 to node 6, it permits 7, 000 vehicles to go through; none in the reverse direction;
If we choose to let 6,000 vehicles to pass in the 3-6 direction, the network is then revised as shown in the revised network. It does not mean that 6,000 vehicles will be able to pass in the direction of 6-3, since the directional rule has be set in the original network.
Now, let’s get on with the problem by choosing the path (Step 1). The smallest capacity on this path is Pf = 6 (Step 2). 7 6 3 6 1 3
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29. By using step 3, the revised network becomes6 8 1 3 2 5 7 4
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30. 6 8 1 3 2 5 7 4
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31. Example
Then, choose path The smallest capacity is 3. The network is then revised to 6 5 3 1 2 7 4
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32. Objective function coefficients
Choose path The smallest capacity is 2. The revised network is as follows: 6 3 1 2 5 4 7
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33. Choose path 1-4-6-7; and the smallest capacity is 1
Choose path ; and the smallest capacity is 1. The network is revised as follows: 6 3 1 2 5 4 7
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34. The next available path is 1-4-6-5-7; and the smallest capacity is 1
The next available path is ; and the smallest capacity is 1. The revised network is now 6 2 3 4 1 7 5
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35. The last available path is 1-4-6-3-5-7, with the smallest capacity 1
The last available path is , with the smallest capacity 1. We revise the network to the following. 6 1 3 2 7 4 5
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36. Note that a flow of 1,000 vehicles per hour was permitted in the 6-3 direction. From the initial network, however, we know that the flow capacity in the 6-3 direction is zero; thus, the 1,000 units of flow in the 6-3 direction represent a fictitious flow. The real effect of this flow is to divert 1,000 units of flow, originally committed to the 3-6 arc, along the 3-5 arc to enable us to get 1,000 units more of flow through the network.
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37. Now, we will determine the maximal flow through the network.
This can be done by comparing the final network to the initial.
Summing up the increases in the second figures on arcs 1-2, 1-3, and 1-4 produces the maximal entering flow, which is 14(,000) vehicles per hour.
Adding up the increases in the second figures on arcs 5-7 and 6-7 generates the maximal leaving flow, which is 14(,000). The fact that the entering flow and leaving flow are both 14(,000) shows in this case that the maximal flow though this network is 14,000 vehicles per hour.
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38. The following diagram shows maximal flow and its direction of each arc.14 5 6 3 1 7 2 4
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39. The answers to the questions asked:
The proposed network cannot accommodate the maximal flow of 15,000 vehicles per hour;
The maximal flow is 14,000 vehicles per hour;
The amount of flow that each arc can bear is as show in the above diagram.
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