Spreadsheet Modeling
& Decision Analysis
A Practical Introduction to
Management Science
5th edition
Cliff T. Ragsdale
Chapter 5
Network Modeling
Introduction
 A number of business problems can be
represented graphically as networks.
 This chapter focuses on several such problems:
–
–
–
–
–
–
Transshipment Problems
Shortest Path Problems
Maximal Flow Problems
Transportation/Assignment Problems
Generalized Network Flow Problems
The Minimum Spanning Tree Problem
Network Flow Problem
Characteristics
 Network flow problems can be represented as a
collection of nodes connected by arcs.
 There are three types of nodes:
– Supply
– Demand
– Transshipment
 We’ll use negative numbers to represent supplies
and positive numbers to represent demand.
A Transshipment Problem:
The Bavarian Motor Company
+100
Boston
2
$50
+60
$30
Newark
1
-200
Columbus
$40
3
$40
$35
+170
$30
Atlanta
5
Richmond
+80
4
$25
$45
$35
+70
Mobile
6
$50
$50
J'ville
7
-300
Defining the Decision Variables
For each arc in a network flow model
we define a decision variable as:
Xij = the amount being shipped (or flowing) from node i to node j
For example…
X12 = the # of cars shipped from node 1 (Newark) to node 2 (Boston)
X56 = the # of cars shipped from node 5 (Atlanta) to node 6 (Mobile)
Note: The number of arcs determines
the number of variables!
Defining the Objective Function
Minimize total shipping costs.
MIN: 30X12 + 40X14 + 50X23 + 35X35
+40X53 + 30X54 + 35X56 + 25X65
+ 50X74 + 45X75 + 50X76
Constraints for Network Flow Problems:
The Balance-of-Flow Rules
For Minimum Cost Network
Flow Problems Where:
Total Supply > Total Demand
Apply This Balance-of-Flow
Rule At Each Node:
Inflow-Outflow >= Supply or Demand
Total Supply < Total Demand
Inflow-Outflow <=Supply or Demand
Total Supply = Total Demand
Inflow-Outflow = Supply or Demand
Defining the Constraints
 In the BMC problem:
Total Supply = 500 cars
Total Demand = 480 cars
(Supply >= Demand)
 For each node we need a constraint like this:
Inflow - Outflow >= Supply or Demand
 Constraint for node 1:
–X12 – X14 >= – 200
 This is equivalent to:
+X12 + X14 <= 200
(Note: there is no inflow for node 1!)
Defining the Constraints
 Flow constraints
–X12 – X14 >= –200
+X12 – X23 >= +100
+X23 + X53 – X35 >= +60
+ X14 + X54 + X74 >= +80
+ X35 + X65 + X75 – X53 – X54 – X56 >= +170
+ X56 + X76 – X65 >= +70
–X74 – X75 – X76 >= –300
 Nonnegativity conditions
Xij >= 0 for all ij
} node 1
} node 2
} node 3
} node 4
} node 5
} node 6
} node 7
Implementing the Model
See file Fig5-2.xls
Optimal Solution to the BMC Problem
+100
Boston
2
$50
Newark
1
120
20
+60
$30
Columbus
80
3
-200
$40
$40
40
+170
Richmond
+80
4
Atlanta
5
$45
+70
Mobile
6
210
70
$50
J'ville
7
-300
The Shortest Path Problem
 Many decision problems boil down to
determining the shortest (or least costly) route
or path through a network.
– Ex. Emergency Vehicle Routing
 This is a special case of a transshipment
problem where:
– There is one supply node with a supply of -1
– There is one demand node with a demand of +1
– All other nodes have supply/demand of +0
The American Car Association
+0
L'burg
9
11
2.0 hrs
9 pts
1.7 hrs
5 pts
+0
2.0 hrs
4 pts
4.7 hrs
9 pts
K'ville
5
+0
3.0 hrs
4 pts
A'ville
6
Chatt.
3
2.8 hrs
7 pts
-1
Atlanta
2.5 hrs
3 pts
+0
10
2.3 hrs
3 pts
+0
2.5 hrs
3 pts
2
Raliegh
Charl.
7
2.0 hrs
8 pts
1.5 hrs
2 pts
+0
2.7 hrs
4 pts
1.1 hrs
3 pts
8
1.5 hrs
3 pts
+0
G'ville
4
B'ham
1
G'boro
1.7 hrs
4 pts
3.0 hrs
4 pts
+1
Va Bch
5.0 hrs
9 pts
+0
3.3 hrs
5 pts
+0
Solving the Problem
 There are two possible objectives for this
problem
– Finding the quickest route (minimizing travel
time)
– Finding the most scenic route (maximizing the
scenic rating points)
See file Fig5-7.xls
The Equipment
Replacement Problem
 The problem of determining when to
replace equipment is another common
business problem.
 It can also be modeled as a shortest
path problem…
The Compu-Train Company
 Compu-Train provides hands-on software training.
 Computers must be replaced at least every two years.
 Two lease contracts are being considered:
– Each requires $62,000 initially
– Contract 1:
Prices increase 6% per year
60% trade-in for 1 year old equipment
15% trade-in for 2 year old equipment
– Contract 2:
Prices increase 2% per year
30% trade-in for 1 year old equipment
10% trade-in for 2 year old equipment
Network for Contract 1
+0
+0
$63,985
4
2
$30,231
$28,520
$33,968
$32,045
-1
1
$60,363
3
$67,824
5
+1
+0
Cost of trading after 1 year:
1.06*$62,000 - 0.6*$62,000 = $28,520
Cost of trading after 2 years: 1.062*$62,000 - 0.15*$62,000 = $60,363
etc, etc….
Solving the Problem
See file Fig5-12.xls
Transportation
& Assignment Problems
 Some network flow problems don’t have transshipment nodes; only supply and demand nodes.
Supply
275,000
400,000
Groves
Distances (in miles)
21
Mt. Dora
1
50
Processing
Plants
Capacity
Ocala
4
200,000
40
These problems
are implemented
more effectively
using the technique
35
30
Eustis described
in Chapter 3. Orlando 600,000
2
5
22
55
300,000
20
Clermont
3
25
Leesburg
6
225,000
Generalized
Network Flow Problems
 In some problems, a gain or loss occurs
in flows over arcs.
– Examples
Oil or gas shipped through a leaky pipeline
Imperfections in raw materials entering a
production process
Spoilage of food items during transit
Theft during transit
Interest or dividends on investments
 These problems require some modeling
changes.
Coal Bank Hollow Recycling
Process 1
Material
Cost
Newspaper
$13
Mixed Paper
$11
White Office Paper $9
Cardboard
$13
Yield
90%
80%
95%
75%
Newsprint
Pulp Source
Recycling Process 1
Recycling Process 2
Demand
Cost Yield
$5 95%
$6 90%
60 tons
Process 2
Cost
$12
$13
$10
$14
Yield
85%
85%
90%
85%
Supply
70 tons
50 tons
30 tons
40 tons
Packaging Paper
Print Stock
Cost Yield
$6 90%
$8 95%
40 tons
Cost Yield
$8 90%
$7 95%
50 tons
Network for Recycling Problem
-70
Newspaper
$13
1
$12
-50
Mixed
paper
2
-30
White
office
paper
3
$11
80%
5
75%
$9
85%
4
+60
7
90%
$8
90%
95%
Packing
paper
pulp
+40
8
$6
$10
$13
Cardboard
Newsprint
pulp
$6
85%
90%
Recycling
Process 2
6
-40
$5
Recycling
Process 1
95%
$13
95%
+0
90%
$14
85%
+0
$8
90%
$7
95%
Print
stock
pulp
9
+50
Defining the Objective Function
Minimize total cost.
MIN: 13X15 + 12X16 + 11X25 + 13X26
+ 9X35+ 10X36 + 13X45 + 14X46 + 5X57
+ 6X58 + 8X59 + 6X67 + 8X68 + 7X69
Defining the Constraints-I
 Raw Materials
-X15 -X16 >= -70
-X25 -X26 >= -50
-X35 -X36 >= -30
-X45 -X46 >= -40
} node 1
} node 2
} node 3
} node 4
Defining the Constraints-II
 Recycling Processes
+0.9X15+0.8X25+0.95X35+0.75X45- X57- X58-X59 >= 0
} node 5
+0.85X16+0.85X26+0.9X36+0.85X46-X67-X68-X69 >= 0
} node 6
Defining the Constraints-III

Paper Pulp
+0.95X57 + 0.90X67 >= 60 } node 7
+0.90X57 + 0.95X67 >= 40 } node 8
+0.90X57 + 0.95X67 >= 50 } node 9
Implementing the Model
See file Fig5-17.xls
Important Modeling Point
 In generalized network flow problems,
gains and/or losses associated with flows
across each arc effectively increase and/or
decrease the available supply.
 This can make it difficult to tell if the total
supply is adequate to meet the total
demand.
 When in doubt, it is best to assume the
total supply is capable of satisfying the
total demand and use Solver to prove (or
refute) this assumption.
The Maximal Flow Problem
 In some network problems, the objective is to
determine the maximum amount of flow that can
occur through a network.
 The arcs in these problems have upper and
lower flow limits.
 Examples
– How much water can flow through a network
of pipes?
– How many cars can travel through a network
of streets?
The Northwest Petroleum Company
Pumping
Station 3
Pumping
Station 1
3
2
6
2
6
1
4
Oil Field
Refinery
2
4
4
3
Pumping
Station 2
5
5
Pumping
Station 4
6
The Northwest Petroleum Company
Pumping
Station 3
Pumping
Station 1
3
2
6
2
6
1
4
Oil Field
Refinery
2
4
4
3
Pumping
Station 2
5
5
Pumping
Station 4
6
Formulation of the Max Flow Problem
MAX:
Subject to:
X61
+X61 - X12 - X13 = 0
+X12 - X24 - X25 = 0
+X13 - X34 - X35 = 0
+X24 + X34 - X46 = 0
+X25 + X35 - X56 = 0
+X46 + X56 - X61 = 0
with the following bounds on the decision variables:
0 <= X12 <= 6
0 <= X25 <= 2 0 <= X46 <= 6
0 <= X13 <= 4
0 <= X34 <= 2 0 <= X56 <= 4
0 <= X24 <= 3
0 <= X35 <= 5 0 <= X61 <= inf
Implementing the Model
See file Fig5-24.xls
Optimal Solution
Pumping
Station 1
3
3
2
5
Pumping
Station 3
4
6
2 2
6
Oil Field
1
5
6
Refinery
2
4
2
4
4
3
5
5
2
Pumping
Station 2
Pumping
Station 4
4
Special Modeling Considerations:
Flow Aggregation
+0
$3
-100
-100
1
2
3
$5
$4
$3
$4
$5
4
$5
5
+75
6
+50
$6
+0
Suppose the total flow into nodes 3 & 4 must be at least 50
and 60, respectively. How would you model this?
Special Modeling Considerations:
Flow Aggregation
+0
+0
$3
-100
-100
1
30
L.B.=50
3
$5
$4
$3
$4
$5
2
40
4
L.B.=60
$5
+0
5
+75
6
+50
$6
+0
Nodes 30 & 40 aggregate the total flow into nodes
3 & 4, respectively.
Special Modeling Considerations:
Multiple Arcs Between Nodes
$8
-75
1
$6
2
+50
U.B. = 35
Two two (or more) arcs cannot share the same
beginning and ending nodes. Instead, try...
+0
10
$0
-75
1
$8
$6
U.B. = 35
2
+50
Special Modeling Considerations:
Capacity Restrictions on Total Supply
-100
+75
$5, UB=40
1
3
$4, UB=30
$6, UB=35
2
-100
$3, UB=35
4
+80
Supply exceeds demand, but the upper
bounds prevent the demand from being met.
Special Modeling Considerations:
Capacity Restrictions on Total Supply
-100
$999, UB=100
+200
+75
$5, UB=40
1
3
$4, UB=30
0
$6, UB=35
$999, UB=100
-100
2
$3, UB=35
4
+80
Now demand exceeds supply. As much “real” demand
as possible will be met in the least costly way.
The Minimal
Spanning Tree Problem
 For a network with n nodes, a spanning
tree is a set of n-1 arcs that connects all
the nodes and contains no loops.
 The minimal spanning tree problem
involves determining the set of arcs that
connects all the nodes at minimum cost.
Minimal Spanning Tree Example:
Windstar Aerospace Company
$150
2
$100
4
$85
$75
$40
$80
1
$85
5
$90
3
$50
$65
6
Nodes represent computers in a local area network.
The Minimal Spanning Tree Algorithm
1.
Select any node. Call this the current subnetwork.
2.
Add to the current subnetwork the cheapest arc that
connects any node within the current subnetwork to
any node not in the current subnetwork. (Ties for
the cheapest arc can be broken arbitrarily.) Call this
the current subnetwork.
3.
If all the nodes are in the subnetwork, stop; this is
the optimal solution. Otherwise, return to step 2.
Solving the Example Problem - 1
4
2
$100
$85
$80
1
$85
5
$90
3
6
Solving the Example Problem - 2
4
2
$100
$85
$80
1
$85
$75
5
$90
3
$50
6
Solving the Example Problem - 3
4
2
$100
$85
1
$75
$80
5
$85
3
$50
$65
6
Solving the Example Problem - 4
4
2
$100
$85
$75
$40
1
$80
5
3
$50
$65
6
Solving the Example Problem - 5
$150
2
4
$85
$75
$40
1
$80
5
3
$50
$65
6
Solving the Example Problem - 6
4
2
$75
$40
1
$80
5
3
$50
$65
6
End of Chapter 5