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
1
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.
2
Types of Networks and
Applications
 Networks have been used in the following.
 There are four basic network models.




3
Shortest route problems.
Minimal spanning trees.
Maximum flow.
Minimum-cost maximum flow.
The Shortest Route Problem
 This problem uses the network as prop. We find
the shortest route from A to G.
4
 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.
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.
5
 Node B is next to join the evaluated set, at a
distance of 4 back to START.
The Shortest Route Problem
 The process continues.
 Node D is next to join the evaluated set with
a distance of 5 back to START.
6
The Shortest Route Problem
 The process continues.
 Node E is next to join the evaluated set with
a distance of 7 back to START.
7
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.
8
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.
9
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).
10
 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.
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.
11
 Connected nodes are shaded. A-C joins the tree.
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.
12
Finding the Minimal
Spanning Tree
 The process continues.
 Arc C-F joins the tree.
13
Finding the Minimal
Spanning Tree
 The process continues.
 Arcs B-E, D-E, F-J, H-I, and I-J join tree.
14
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.
15
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.
16
 It doesn’t matter which path is used. The
possibilities shrink as more are found.
 Flow into an interior node must equal the flow out.
Maximizing Flow
 The bottleneck arc on a path has the lowest
remaining capacity. Here it is H-J.
17
Maximizing Flow
 Arc H-J is saturated. Flow over saturated
arcs may be decreased only.
 The next path’s bottleneck arc is E-I.
18
Maximizing Flow
 The next path has two bottleneck arcs: C-F
and K-L.
19
Maximizing Flow
 The next path has two bottleneck arcs: D-F
and I-J.
20
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.
21
Maximizing Flow
 There are no more flow-augmenting paths.
The optimal solution has been found with
maximum flow (sum into SINK) of C = 14.
22
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).
23
Solving with QuickQuant
 The following first iteration involves a
smaller version of the problem in the text.
24
Solving with QuickQuant
 The initial solution is infeasible. A series of
iterations yields the optimal solution.
25
NetworkTemplates
 shortest route
 maximum flow
 minimum cost maximum flow
26
Shortest Route for Yellow Jacket
Freightways (Figure 13-11)
This is the upper portion of Figure 13-11.
The lower portion is shown next.
A
3. If 1000 is not
large enough to
denote the
impossibility of
going between
two cities, use a
larger number.
27
1
2
3 PROBLEM:
4
5
6
7
8
From Node
9
A
10
B
11
C
12
D
13
E
14
F
15
G
16
17
18
19
B
C
D
E
F
G
H
I
J
K
SHORTEST ROUTE SOLUTION
1. Enter the
problem name
in B3.
Yellow Jacket Freightways
Distances
A
1000
4
3
6
1000
1000
1000
B
4
1000
1000
1
4
1000
1000
To Node
C
D
E
3
6 1000
1000 1
4
1000 3 1000
3 1000 2
1000 2 1000
6
4 1000
1000 5
2
Solution C =
9
To Node
17
2. Enter the
distances above
the diagonal in
the table
B9:H15. They
will
automatically be
E
entered below
=SUMPRODUCT(B9:H15,B21:H27)
the diagonal.
F
1000
1000
6
4
1000
1000
2
G
1000
1000
1000
5
2
2
1000
Total
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 A20:H27.
Here it is A-B-DE-G.
28
A
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
B
C
D
Solution C =
From Node
A
B
C
D
E
F
G
Total Inflow
Total Outflow
Net Flow
Required Flow
28
29
30
31
F
9
G
H
I
J
K
E
17 =SUMPRODUCT(B9:H15,B21:H27)
Total
To Node
I
D
E
F
G Outflow
0
0
0
0
1 21 =SUM(B21:H21)
1
0
0
0
1 22 =SUM(B22:H22)
0
0
0
0
0 23 =SUM(B23:H23)
0
1
0
0
1 24 =SUM(B24:H24)
0
0
0
1
1 25 =SUM(B25:H25)
0
0
0
0
0 26 =SUM(B26:H26)
0
0
0
0
0 27 =SUM(B27:H27)
1
1
0
1
1
1
0
0
0
0
0
-1
0
0
0
-1
C
D
E
=SUM(C21:C27) =SUM(D21:D27) =SUM(E21:E27)
=I22
=I23
=I24
=C29-C28
=D29-D28
=E29-E28
0
0
0
F
G
H
=SUM(F21:F27) =SUM(G21:G27) =SUM(H21:H27)
=I25
=I26
=I27
=F29-F28
=G29-G28
=H29-H28
0
0
-1
A
B
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
1
0
B
=SUM(B21:B27)
=I21
=B29-B28
1
28
29
30
31
E
C
0
0
0
0
0
0
0
0
0
0
0
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
B31:H31
accordingly.
29
A
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
B
C
D
Solution C =
From Node
A
B
C
D
E
F
G
Total Inflow
Total Outflow
Net Flow
Required Flow
28
29
30
31
F
9
G
H
I
J
K
E
17 =SUMPRODUCT(B9:H15,B21:H27)
Total
To Node
I
D
E
F
G Outflow
21
=SUM(B21:H21)
0
0
0
0
1
1
0
0
0
1 22 =SUM(B22:H22)
0
0
0
0
0 23 =SUM(B23:H23)
0
1
0
0
1 24 =SUM(B24:H24)
0
0
0
1
1 25 =SUM(B25:H25)
0
0
0
0
0 26 =SUM(B26:H26)
0
0
0
0
0 27 =SUM(B27:H27)
1
1
0
1
6. For problems
1
1
0
0
with more than 7
0
0
0
-1
0
0
0
-1
cities, expand
C
D
E
the
distance and
=SUM(C21:C27) =SUM(D21:D27) =SUM(E21:E27)
=I22
=I23
=I24 path tables and
=C29-C28
=D29-D28
=E29-E28
check to make
0
0
0
sure that all the
F
G
H
=SUM(F21:F27) =SUM(G21:G27) =SUM(H21:H27) formulas have
=I25
=I26
=I27
the proper
=F29-F28
=G29-G28
=H29-H28
ranges.
0
0
-1
A
B
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
1
0
B
=SUM(B21:B27)
=I21
=B29-B28
1
28
29
30
31
E
C
0
0
0
0
0
0
0
0
0
0
0
1. Enter the
value of the
objective
function, E17, in
the Target Cell
line, either with
or without the $
signs.
Solver Parameters Dialog Box
(Figure 13-12)
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,
B21:H27.
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
30
Change and Add Constraint dialog box function in the same manner.
The Add Constraint Dialog Box
1. Enter the net flows B30:H30
(or $B$30:$H$30) in the Cell
Reference line.
Normally, all
these entries
already appear.
You will need to
use this dialog
box only if you
need to add a
constraint.
3. Enter the
required flow
B31:H31 in the
Constraint line (or
=$B$31:$H$31).
4. Click the OK
button.
If you need to
change a
constraint, the
Change
Constraint
dialog box
functions just
31
like this one.
2. Enter = as the sign because the net flow must be
equal to the required flow, given next in Step 3.
Maximum Flow for Lulliput
Telephone Company (Figure 13-31)
This is the upper portion of Figure 13-31.
The lower portion is shown next.
A
B
C
D
E
F
G
H
I
J
1
MAXIMUM FLOW
2
3 PROBLEM: Lilliput Telephone Company
4
5
Capacities
6
7
To Node
8 From Node
A
B
C
D
E
F
G
H
I
9
A
5
5
4
6
10
B
4
11
C
5
4
12
D
3
6
13
E
2
3
14
F
5
6
15
G
4
16
H
3
17
I
18
J
19
K
20
L
20
32
K
L
M
N
O
P
SOLUTION
1. Enter the problem
name in B3.
J
K
L
2. (a) Enter the
capacities in the
table B9:M20.
2
3
10
3
7
9
2. (b) A big number is entered for the upper
limit on the return flow from L to A.
Maximum Flow for Lulliput
Telephone Company (Figure 13-31)
This is the lower portion of Figure 13-31.
A
The
maximum
flow is in cell
E22. Here it
is 14.
The flow
along each
arc is found
from the
table in cells
A20:H27. For
example, cell
C26 has a 1
in it. This
means one
unit of flow
goes from A
to B.
33
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
B
C
D
Solution C =
From Node
A
B
C
D
E
F
G
H
I
J
K
L
Total Inflow
Total Outflow
Net Flow
Required Flow
38
39
40
41
A
0
0
0
0
0
0
0
0
0
0
0
14
14
14
0
0
B
1
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
E
14
C
4
0
0
0
0
0
0
0
0
0
0
0
4
4
0
0
D
4
1
0
0
0
0
0
0
0
0
0
0
5
5
0
0
F
G
H
I
J
K
L
M
N
O
P
E
22 =B37
E
5
0
0
0
0
0
0
0
0
0
0
0
5
5
0
0
To Node
F
G
0
0
0
0
4
0
3
0
0
2
0
1
0
0
0
0
0
0
0
0
0
0
0
0
7
3
7
3
0
0
0
0
H
0
0
0
2
0
0
0
0
0
0
0
0
2
2
0
0
I
0
0
0
0
3
6
3
0
0
0
0
0
12
12
0
0
J
0
0
0
0
0
0
0
2
3
0
0
0
5
5
0
0
K
0
0
0
0
0
0
0
0
9
0
0
0
9
9
0
0
L
0
0
0
0
0
0
0
0
0
5
9
0
14
14
0
0
Total
Outflow
14
1
4
5
5
7
3
2
12
5
9
14
26
27
28
29
30
31
32
33
34
35
36
37
N
=SUM(B26:M26)
=SUM(B27:M27)
=SUM(B28:M28)
=SUM(B29:M29)
=SUM(B30:M30)
=SUM(B31:M31)
=SUM(B32:M32)
=SUM(B33:M33)
=SUM(B34:M34)
=SUM(B35:M35)
=SUM(B36:M36)
=SUM(B37:M37)
B
C
D
E
F
G
H
=SUM(B26:B37) =SUM(C26:C37)=SUM(D26:D37)=SUM(E26:E37) =SUM(F26:F37) =SUM(G26:G37)=SUM(H26:H37)
=N26
=N27
=N28
=N29
=N30
=N31
=N32
=B38-B39
=C38-C39
=D38-D39
=E38-E39
=F38-F39
=G38-G39
=H38-H39
0
0
0
0
0
0
0
38
39
40
41
I
=SUM(I26:I37)
=N33
=I38-I39
0
J
=SUM(J26:J37)
=N34
=J38-J39
0
K
L
=SUM(K26:K37) =SUM(L26:L37)
=N35
=N36
=K38-K39
=L38-L39
0
0
M
=SUM(M26:M37)
=N37
=M38-M39
0
Maximum Flow for Lulliput
Telephone Company (Figure 13-31)
This is the lower portion of Figure 13-31.
A
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.
34
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
B
C
D
Solution C =
From Node
A
B
C
D
E
F
G
H
I
J
K
L
Total Inflow
Total Outflow
Net Flow
Required Flow
38
39
40
41
A
0
0
0
0
0
0
0
0
0
0
0
14
14
14
0
0
B
1
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
E
14
C
4
0
0
0
0
0
0
0
0
0
0
0
4
4
0
0
D
4
1
0
0
0
0
0
0
0
0
0
0
5
5
0
0
F
G
H
I
J
K
L
M
To Node
F
G
0
0
0
0
4
0
3
0
0
2
0
1
0
0
0
0
0
0
0
0
0
0
0
0
7
3
7
3
0
0
0
0
H
0
0
0
2
0
0
0
0
0
0
0
0
2
2
0
0
I
0
0
0
0
3
6
3
0
0
0
0
0
12
12
0
0
J
0
0
0
0
0
0
0
2
3
0
0
0
5
5
0
0
K
0
0
0
0
0
0
0
0
9
0
0
0
9
9
0
0
L
0
0
0
0
0
0
0
0
0
5
9
0
14
14
0
0
N
O
P
E
22 =B37
E
5
0
0
0
0
0
0
0
0
0
0
0
5
5
0
0
Total
Outflow
14
1
4
5
5
7
3
2
12
5
9
14
26
27
28
29
30
31
32
33
34
35
36
37
N
=SUM(B26:M26)
=SUM(B27:M27)
=SUM(B28:M28)
=SUM(B29:M29)
=SUM(B30:M30)
=SUM(B31:M31)
=SUM(B32:M32)
=SUM(B33:M33)
=SUM(B34:M34)
=SUM(B35:M35)
=SUM(B36:M36)
=SUM(B37:M37)
B
C
D
E
F
G
H
=SUM(B26:B37) =SUM(C26:C37)=SUM(D26:D37)=SUM(E26:E37) =SUM(F26:F37) =SUM(G26:G37)=SUM(H26:H37)
=N26
=N27
=N28
=N29
=N30
=N31
=N32
=B38-B39
=C38-C39
=D38-D39
=E38-E39
=F38-F39
=G38-G39
=H38-H39
0
0
0
0
0
0
0
38
39
40
41
I
=SUM(I26:I37)
=N33
=I38-I39
0
J
=SUM(J26:J37)
=N34
=J38-J39
0
K
L
=SUM(K26:K37) =SUM(L26:L37)
=N35
=N36
=K38-K39
=L38-L39
0
0
M
=SUM(M26:M37)
=N37
=M38-M39
0
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.
1. Enter the
value of the
objective
function, E22, in
the Target Cell
line, either with
or without the $
signs.
Solver Parameters Dialog Box
(Figure 13-32)
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,
B26:M37.
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
35
button. The Change and Add Constraint dialog box function in the same manner.
Minimum Cost Maximum Flow
for BigCo (Figure 13-40)
This is the upper portion of Figure 13-40.
The lower portion is shown next.
A
2. Enter the costs
and capacities in
the table B8:G10
and the
corresponding
From and To
names in cells
A8:A9 and B7:F7.
36
1
2
3 PROBLEM:
4
5
6
7 From Node
8
Centralia
9
Gotham
10
Demand
11
12
13
14 From Node
15
Centralia
16
Gotham
17
18
19
20 From Node
21
Centralia
22
Gotham
B
C
D
E
F
G
H
MINIMUM COST MAXIMUM FLOW SOLUTION
BigCo Discount Retail Chain
1. Enter the problem
name in B3.
Costs, Capacities, and Demands
To Node
Jay Street 7th Avenue River Cross Downtown Exurbia
10
12
11
13
13
15
12
9
14
9
200
250
350
150
150
Minimum Quantities (Lower Bounds)
To Node
Jay Street 7th Avenue River Cross Downtown Exurbia
40
60
40
0
60
0
40
80
40
0
Maximum Quantities (Upper Bounds)
To Node
Jay Street 7th Avenue River Cross Downtown Exurbia
300
200
200
200
100
200
200
200
200
200
Capacity
500
600
3. Enter the
minimum
quantities in the
table B15:F16.
4. Enter the
maximum
quantities in the
table B21:F22.
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.
A
B
C
D
E
F
G
H
23
E
24
Solution Quantities, C = $ 12,110 24 =SUMPRO DUCT(B8:F9,B27:F28)
25
To Node
26 From Node Jaystreet 7th Avenue River Cross Downtown Exurbia Total Quantity
G
27
Centralia
200
60
150
30
60
500
28
Gotham
0
190
200
120
90
600 27 =SUM(B27:F27)
28 =SUM(B28:F28)
29 Total Quantity
200
250
350
150
150
30
B
C
D
E
F
31 29 =SUM(B27:B28) =SUM(C27:C28)=SUM(D27:D28)=SUM(E27:E28) =SUM(F27:F28)
The optimal shipping schedule is
given in the table in cells A28:G29.
37
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.
A
B
C
D
E
F
G
H
23
E
24
=SUMPRO
DUCT(B8:F9,B27:F28)
24
Solution Quantities, C = $ 12,110
25
To Node
26 From Node Jaystreet 7th Avenue River Cross Downtown Exurbia Total Quantity
G
27
Centralia
200
60
150
30
60
500
28
Gotham
0
190
200
120
90
600 27 =SUM(B27:F27)
28 =SUM(B28:F28)
29 Total Quantity
200
250
350
150
150
30
B
C
D
E
F
31 29 =SUM(B27:B28) =SUM(C27:C28)=SUM(D27:D28)=SUM(E27:E28) =SUM(F27:F28)
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.
38
1. Enter the
value of the
objective
function, E24, in
the Target Cell
line, either with
or without the $
signs.
Solver Parameters Dialog Box
(Figure 13-41)
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,
B27:F28.
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
39
button. The Change and Add Constraint dialog box function in the same manner.