Mathematical models of networks give us algorithms so computationally efficient that we can employ them to evaluate problems too big to be solved any other way. - PowerPoint PPT Presentation

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Mathematical models of networks give us algorithms so computationally efficient that we can employ them to evaluate problems too big to be solved any other way.

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The Minimal Spanning Tree A tree is a set of arcs connecting nodes in such a way that only one route involving those arcs connects any two nodes. – PowerPoint PPT presentation

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Title: Mathematical models of networks give us algorithms so computationally efficient that we can employ them to evaluate problems too big to be solved any other way.


1
Chapter 13
  • Mathematical models of networks give us
    algorithms so computationally efficient that we
    can employ them to evaluate problems too big to
    be solved any other way.
  • Network Models

2
The Structure of Network Problems
  • Networks have features of the following arcs
    (e.g., roads), nodes (e.g., cities), and arc
    values representing distance or flow.

3
Types of Networks and Applications
  • Networks have been used in the following.
  • There are four basic network models.
  • Shortest route problems.
  • Minimal spanning trees.
  • Maximum flow.
  • Minimum-cost maximum flow.

4
The Shortest Route Problem
  • This problem uses the network as prop. We find
    the shortest route from A to G.
  • The START node is evaluated first (shaded). Then
    all direct links between evaluated and
    unevaluated nodes are identified and the
    distances back to START are computed along
    connecting arcs.
  • The node with smallest distance, written above
    it, becomes the next evaluated one and is shaded.
  • An arrow points from there along the connecting
    arc.

5
The Shortest Route Problem
  • Node C joins the evaluated set. An arrow is
    added pointing back to A, and the cumulative
    distance, 3, back to START from C is entered
    above.
  • The process continues.
  • Node B is next to join the evaluated set, at a
    distance of 4 back to START.

6
The Shortest Route Problem
  • The process continues.
  • Node D is next to join the evaluated set with a
    distance of 5 back to START.

7
The Shortest Route Problem
  • The process continues.
  • Node E is next to join the evaluated set with a
    distance of 7 back to START.

8
The Shortest Route Problem
  • The process continues.
  • Two nodes, F and G, are next to join the
    evaluated set, each with with a distance of 9
    back to START.

9
The Shortest Route Problem
  • All nodes are evaluated. The shortest route is
    found by tracing back from FINISH following the
    arrows.
  • The shortest route from A to G is
  • A-B-D-E-G for a distance of C 9.

10
The Minimal Spanning Tree
  • A tree is a set of arcs connecting nodes in such
    a way that only one route involving those arcs
    connects any two nodes.
  • Imagine an ant on a real tree. It has just one
    way to walk from any leaf to another.
  • A spanning tree connects with all nodes.
  • It is like railroad tracks connecting all cities,
    but with only one routing between any two.
  • A minimal spanning tree has the smallest sum of
    its arc distances C (tree size).
  • In connecting all circuit-board solder points
    with gold wire, it would use the least gold.
  • It would have the least tracks for a railway.

11
Finding the Minimal Spanning Tree
  • As first connected node pick any (here A). Find
    all arcs directly joining connected to
    unconnected. Join the shortest arc to tree.
  • Connected nodes are shaded. A-C joins the tree.

12
Finding the Minimal Spanning Tree
  • As new arcs join the tree, more nodes become
    connected. We consider only arcs directly
    joining connected to unconnected nodes.
  • Arc A-B joins the tree.

13
Finding the Minimal Spanning Tree
  • The process continues.
  • Arc C-F joins the tree.

14
Finding the Minimal Spanning Tree
  • The process continues.
  • Arcs B-E, D-E, F-J, H-I, and I-J join tree.

15
Finding the Minimal Spanning Tree
  • The process continues.
  • Arc G-H joins the tree. Since all nodes are
    connected, the tree has been found. The sum of
    the arc lengths gives its size C.

16
Maximizing Flow
  • Arcs in a maximum flow problem are directed and
    have upper bounds. Flow moves one way.
  • A node is designated as the SOURCE and another as
    the SINK.
  • Flow-augmenting paths from SOURCE to SINK are
    found and flows sent over the arcs. If no path
    can be found, flow is maximized.
  • A flow-augmenting path ordinarily involves arcs
    directed away from the SOURCE toward the SINK.
  • But an arc can point in the opposite direction if
    some of its current flow would be reduced and be
    redirected to another arc.
  • It doesnt matter which path is used. The
    possibilities shrink as more are found.
  • Flow into an interior node must equal the flow
    out.

17
Maximizing Flow
  • The bottleneck arc on a path has the lowest
    remaining capacity. Here it is H-J.

18
Maximizing Flow
  • Arc H-J is saturated. Flow over saturated arcs
    may be decreased only.
  • The next paths bottleneck arc is E-I.

19
Maximizing Flow
  • The next path has two bottleneck arcs C-F and
    K-L.

20
Maximizing Flow
  • The next path has two bottleneck arcs D-F and
    I-J.

21
Maximizing Flow
  • This path goes against the direction of J-K
    flow. Some J-K flow is redirected over J-L.
  • The bottlenecks are B-D and I-K.

22
Maximizing Flow
  • There are no more flow-augmenting paths. The
    optimal solution has been found with maximum flow
    (sum into SINK) of C 14.

23
Minimum-Cost Maximum Flow
  • Transportation problems are special cases of
    minimum-cost maximum flow problems.
  • The general problem has bounded arcs (routes) and
    is represented as a network.
  • It may be solved by an elaborate procedure, the
    out-of-kilter algorithm, involving shortest
    routes and maximum flows.
  • However, it is best solved on the computer.
  • QuickQuant may be used for this purpose.
  • It can perform the out-of-kilter algorithm.
  • The problem can also be solved as a general
    linear program (with QuickQuant or Excel).

24
Solving with QuickQuant
  • The following first iteration involves a smaller
    version of the problem in the text.

25
Solving with QuickQuant
  • The initial solution is infeasible. A series of
    iterations yields the optimal solution.

26
NetworkTemplates
  • shortest route
  • maximum flow
  • minimum cost maximum flow

27
Shortest Route for Yellow Jacket Freightways
(Figure 13-11)
This is the upper portion of Figure 13-11. The
lower portion is shown next.
1. Enter the problem name in B3.
3. If 1000 is not large enough to denote the
impossibility of going between two cities, use a
larger number.
2. Enter the distances above the diagonal in the
table B9H15. They will automatically be entered
below the diagonal.
28
Shortest Route for Yellow Jacket Freightways
(Figure 13-11)
This is the lower portion of Figure 13-11.
The length of the shortest route is in cell E9.
Here it is 9.
The shortest route is found from the table in
cells A20H27. Here it is A-B-D-E-G.
29
Shortest Route for Yellow Jacket Freightways
(Figure 13-11)
This is the lower portion of Figure 13-11.
4. Click on Tools and use Solver to find the
shortest route. The Solver Parameters dialog box
is shown on the next slide.
5. The starting point is assumed to be A and the
ending point G. If this is different adjust the
required flow in cells B31H31 accordingly.
6. For problems with more than 7 cities, expand
the distance and path tables and check to make
sure that all the formulas have the proper ranges.
30
Solver Parameters Dialog Box (Figure 13-12)
1. Enter the value of the objective function,
E17, in the Target Cell line, either with or
without the signs.
NOTE Normally all these entries appear in the
Solver Parameter dialog box so you only need to
click on the Solve button. However, you should
always check to make sure the entries are correct
for the problem you are solving.
2. The Target Cell is to be minimized so click
on Min in the Equal To line.
3. Enter the decision variables in the By
Changing Cells line, B21H27.
4. The constraints are entered in the Subject to
Constraints box by using the Add Constraints
dialog box shown next (obtained by clicking on
the Add button). If a constraint needs to be
changed, click on the Change button. The Change
and Add Constraint dialog box function in the
same manner.
31
The Add Constraint Dialog Box
Normally, all these entries already appear. You
will need to use this dialog box only if you need
to add a constraint.
3. Enter the required flow B31H31 in the
Constraint line (or B31H31).
1. Enter the net flows B30H30 (or B30H30)
in the Cell Reference line.
4. Click the OK button.
If you need to change a constraint, the Change
Constraint dialog box functions just like this
one.
2. Enter as the sign because the net flow must
be equal to the required flow, given next in Step
3.
32
Maximum Flow for Lulliput Telephone Company
(Figure 13-31)
This is the upper portion of Figure 13-31. The
lower portion is shown next.
1. Enter the problem name in B3.
2. (a) Enter the capacities in the table B9M20.
2. (b) A big number is entered for the upper
limit on the return flow from L to A.
33
Maximum Flow for Lulliput Telephone Company
(Figure 13-31)
This is the lower portion of Figure 13-31.
The maximum flow is in cell E22. Here it is 14.
The flow along each arc is found from the table
in cells A20H27. For example, cell C26 has a 1
in it. This means one unit of flow goes from A
to B.
34
Maximum Flow for Lulliput Telephone Company
(Figure 13-31)
This is the lower portion of Figure 13-31.
3.Click on Tools and use Solver to find the
maximum flow. The Solver Parameters dialog box is
shown on the next slide.
4. The starting point is the first node and the
ending point the last one.
5. For problems with more than 12 nodes, expand
the capacities and flows tables and check to make
sure that all the formulas have the proper ranges.
35
Solver Parameters Dialog Box (Figure 13-32)
1. Enter the value of the objective function,
E22, in the Target Cell line, either with or
without the signs.
NOTE Normally all these entries appear in the
Solver Parameter dialog box so you only need to
click on the Solve button. However, you should
always check to make sure the entries are correct
for the problem you are solving.
2. The Target Cell is to be maximized so click
on Max in the Equal To line.
3. Enter the decision variables in the By
Changing Cells line, B26M37.
4. The constraints are entered in the Subject to
Constraints box by using the Add Constraints
dialog box (obtained by clicking on the Add
button) as was done for the shortest route
template. If a constraint needs to be changed,
click on the Change button. The Change and Add
Constraint dialog box function in the same manner.
36
Minimum Cost Maximum Flow for BigCo (Figure
13-40)
This is the upper portion of Figure 13-40. The
lower portion is shown next.
1. Enter the problem name in B3.
2. Enter the costs and capacities in the table
B8G10 and the corresponding From and To names in
cells A8A9 and B7F7.
3. Enter the minimum quantities in the table
B15F16.
4. Enter the maximum quantities in the table
B21F22.
37
Minimum Cost Maximum Flow for BigCo (Figure
13-40)
This is the lower portion of Figure 13-40.
The minimum cost is in cell E24. Here it is
12,110.
The optimal shipping schedule is given in the
table in cells A28G29.
38
Minimum Cost Maximum Flow for BigCo (Figure
13-40)
This is the lower portion of Figure 13-40.
5.Click on Tools and use Solver to find the
optimal solution. The Solver Parameters dialog
box is shown on the next slide.
6. For other problems insert (or delete) the
appropriate number of rows or columns and check
to make sure that all the formulas have the
proper ranges.
39
Solver Parameters Dialog Box (Figure 13-41)
1. Enter the value of the objective function,
E24, in the Target Cell line, either with or
without the signs.
NOTE Normally all these entries appear in the
Solver Parameter dialog box so you only need to
click on the Solve button. However, you should
always check to make sure the entries are correct
for the problem you are solving.
2. The Target Cell is to be minimized so click
on Min in the Equal To line.
3. Enter the decision variables in the By
Changing Cells line, B27F28.
4. The constraints are entered in the Subject to
Constraints box by using the Add Constraints
dialog box (obtained by clicking on the Add
button) as was done for the shortest route
template. If a constraint needs to be changed,
click on the Change button. The Change and Add
Constraint dialog box function in the same manner.
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