Introduction to
Mathematical Programming
OR/MA 504
Chapter 4
Network Modeling
4-1
Introduction
• A large variety of linear programming applications
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
4-2
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.
4-3
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
4-4
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!
4-5
Defining the Objective Function
Minimize total shipping costs.
MIN: 30X12 + 40X14 + 50X23 + 35X35
+40X53 + 30X54 + 35X56 + 25X65
+ 50X74 + 45X75 + 50X76
4-6
Constraints for Network Flow Problems:
The Balance-of-Flow Rules
For Minimum Cost Network
Flow Problems Where:
Apply This Balance-of-Flow
Rule At Each Node:
Total Supply > Total Demand
Inflow-Outflow >= Supply or Demand
Total Supply < Total Demand
Inflow-Outflow <=Supply or Demand
Total Supply = Total Demand
Inflow-Outflow = Supply or Demand
4-7
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
(Note: there is no inflow for node 1!)
• This is equivalent to:
+X12 + X14 <= 200
4-8
Defining the Constraints
• Flow constraints
–X12 – X14 >= –200
} node 1
+X12 – X23 >= +100
} node 2
+X23 + X53 – X35 >= +60
} node 3
+ X14 + X54 + X74 >= +80
} node 4
+ X35 + X65 + X75 – X53 – X54 – X56 >= +170} node 5
+ X56 + X76 – X65 >= +70
} node 6
–X74 – X75 – X76 >= –300
} node 7
• Nonnegativity conditions
Xij >= 0 for all ij
4-9
Implementing the Model
See file Fig4-1.xls
4-10
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
4-11
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
4-12
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
2.5 hrs
3 pts
+0
10
+0
2.3 hrs
3 pts
+0
2.5 hrs
3 pts
2
Raleigh
Charl.
7
2.0 hrs
8 pts
Atlanta
2.7 hrs
4 pts
1.1 hrs
3 pts
8
1.5 hrs
3 pts
+0
1.5 hrs
2 pts
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
4-13
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 Fig4-2.xls
4-14
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…
4-15
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
4-16
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
Cost of trading in year 2 after trading in year 1:
1.062*$62,000 – 0.6(1.06*$62,000) = $69,663 - $39,432 = $30,231
4-17
Solving the Problem
See file Fig4-3.xls
4-18
Transportation & Assignment Problems
• Some network flow problems don’t have transshipment nodes; only supply and demand nodes.
Supply
275,000
Groves
Distances (in miles)
21
Mt. Dora
1
50
Processing
Plants
Capacity
Ocala
4
200,000
40
400,000
These problems are implemented
35
more
effectively
using the technique
30
Orlando
Eustis
600,000
described
in
Chapter
2.
2
5
22
55
300,000
20
Clermont
3
25
Leesburg
6
225,000
4-19
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.
4-20
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
4-21
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%
85%
+0
$8
90%
$7
95%
Print
stock
pulp
+50
9
$14
4-22
Defining the Objective Function
Minimize total cost.
MIN: 13X15 + 12X16 + 11X25 + 13X26
+ 9X35+ 10X36 + 13X45 + 14X46 + 5X57
+ 6X58 + 8X59 + 6X67 + 8X68 + 7X69
4-23
Defining the Constraints-I
• Raw Materials
-X15 -X16 >= -70
-X25 -X26 >= -50
-X35 -X36 >= -30
-X45 -X46 >= -40
}
}
}
}
node
node
node
node
1
2
3
4
4-24
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
4-25
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
4-26
Implementing the Model
See file Fig4-4.xls
4-27
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.
4-28
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?
4-29
The Northwest Petroleum Company
Pumping
Station 3
Pumping
Station 1
3
2
6
2
6
1
4
Oil Field
Refinery
6
2
4
4
3
Pumping
Station 2
5
5
Pumping
Station 4
4-30
Max Flow Problem Set-Up
• Solve as transshipment problem:
– Add return arc from ending node to the
starting node
– Assign demand of 0 to all nodes in network
– Maximize flow over the return arc
4-31
The Northwest Petroleum
Company
Pumping
Pumping
Station 1
+0
1
3
2
+0
4
6
2
6
+0
Station 3
+0
Oil Field
Refinery
6
2
4
4
+0
3
Pumping
Station 2
5
5
+0
Pumping
Station 4
4-32
Formulation of the Max Flow Problem
MAX:
Subject to:
X61
+X61
+X12
+X13
+X24
+X25
+X46
- X12 - X13 = 0
- X24 - X25 = 0
- X34 - X35 = 0
+ X34 - X46 = 0
+ X35 - X56 = 0
+ 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
4-33
Implementing the Model
See file Fig4-5.xls
4-34
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
4
5
2
Pumping
Station 2
Pumping
Station 4
4-35
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?
4-36
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.
4-37
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
4-38
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.
4-39
Special Modeling Considerations:
Capacity Restrictions on Total Supply
-100
$999, UB=100
+75
$5, UB=40
1
3
+200
$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.
4-40
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.
4-41
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.
4-42
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.
4-43
Solving the Example Problem - 1
4
2
$100
$85
$80
1
$85
5
$90
3
6
4-44
Solving the Example Problem - 2
4
2
$100
$85
$80
1
$85
$75
5
$90
3
$50
6
4-45
Solving the Example Problem - 3
4
2
$100
$85
1
$75
$80
5
$85
3
$50
$65
6
4-46
Solving the Example Problem - 4
4
2
$100
$85
$75
$40
1
$80
5
3
$50
$65
6
4-47
Solving the Example Problem - 5
$150
2
4
$85
$75
$40
1
$80
5
3
$50
$65
6
4-48
Solving the Example Problem - 6
4
2
$75
$40
1
$80
5
3
$50
$65
6
4-49
End of Chapter 4
4-50