1. Spreadsheet Modeling & Decision Analysis
A Practical Introduction to Management Science
5th edition
Cliff T. Ragsdale

2. © 2007 South-Western College Publishing

3. Chapter 5
Network Modeling

4. 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

5. 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.

6. A Transshipment Problem: The Bavarian Motor Company
+100
Boston
$30
$50 2
Newark
-200 1
Columbus
+60 3
$40
$40
$35
$30
Richmond
+80
Atlanta 4
+170 5
$25
$45
$50
$35
Mobile
+70
J'ville 6
-300
$50 7

7. 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!

8. Defining the Objective Function
Minimize total shipping costs.
MIN: 30X X X X35
+40X X X X65
+ 50X X X76

9. Constraints for Network Flow Problems: The Balance-of-Flow Rules
For Minimum Cost Network Apply This Balance-of-Flow
Flow Problems Where: 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

10. Defining the Constraints
In the BMC problem:
Total Supply = 500 cars
Total Demand = 480 cars
For each node we need a constraint like this:
Inflow - Outflow >= Supply or Demand
Constraint for node 1:
–X12 – X14 >= – (Note: there is no inflow for node 1!)
This is equivalent to:
+X12 + X14 <= 200
(Supply >= Demand)

11. Defining the Constraints
Flow constraints
–X12 – X14 >= – } node 1
+X12 – X23 >= } node 2
+X23 + X53 – X35 >= } node 3
+ X14 + X54 + X74 >= } node 4
+ X35 + X65 + X75 – X53 – X54 – X56 >= +170 } node 5
+ X56 + X76 – X65 >= } node 6
–X74 – X75 – X76 >= –300 } node 7
Nonnegativity conditions
Xij >= 0 for all ij

12. Implementing the Model
See file Fig5-2.xls

13. Optimal Solution to the BMC Problem
+100
Boston
$30
$50 2
Newark
-200
120 20 1
Columbus
+60 80 3
$40
$40 40
Richmond
+80
Atlanta 4
+170 5
$45
210 70
+70
Mobile
J'ville 6
-300
$50 7

14. 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

15. The American Car Association+0
3.3 hrs +1
L'burg
5 pts 9
Va Bch 11
5.0 hrs
9 pts
2.0 hrs
4 pts
4.7 hrs +0
2.7 hrs
9 pts +0
4 pts
1.1 hrs
K'ville
3 pts 5
G'boro
2.0 hrs
Raliegh
3.0 hrs
9 pts 8
4 pts 10 +0
1.7 hrs
5 pts
A'ville
1.5 hrs 6 +0
3 pts
2.3 hrs +0
3 pts
Chatt.
2.8 hrs 3
7 pts
2.0 hrs
Charl.
8 pts 7 +0
3.0 hrs
1.7 hrs
4 pts
4 pts
1.5 hrs
G'ville
2 pts 4
Atlanta +0
B'ham 2
2.5 hrs 1
3 pts
2.5 hrs
3 pts +0 -1

16. 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

17. 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…

18. 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

19. Network for Contract 1 +1 -1 +0 1 3 5 2 4
$28,520
$60,363
$30,231
$63,985
$32,045
$67,824
$33,968
Cost of trading after 1 year: *$62, *$62,000 = $28,520
Cost of trading after 2 years: *$62, *$62,000 = $60,363
etc, etc….

20. Solving the Problem
See file Fig5-12.xls

21. Transportation & Assignment Problems
Some network flow problems don’t have trans-shipment nodes; only supply and demand nodes.
Mt. Dora 1
Eustis 2
Clermont 3
Ocala 4
Orlando 5
Leesburg 6
Distances (in miles)
Capacity
Supply
275,000
400,000
300,000
225,000
600,000
200,000
Groves
Processing
Plants 21 50 40 35 30 22 55 25 20
These problems are implemented more effectively using the technique described in Chapter 3.

22. 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.

23. Coal Bank Hollow Recycling
Process 1 Process 2
Material Cost Yield Cost Yield Supply
Newspaper $13 90% $12 85% 70 tons
Mixed Paper $11 80% $13 85% 50 tons
White Office Paper $9 95% $10 90% 30 tons
Cardboard $13 75% $14 85% 40 tons
Newsprint Packaging Paper Print Stock
Pulp Source Cost Yield Cost Yield Cost Yield
Recycling Process 1 $5 95% $6 90% $8 90%
Recycling Process 2 $6 90% $8 95% $7 95%
Demand 60 tons 40 tons 50 tons

24. Network for Recycling Problem
-70
Newspaper
$13 1
$12 +0
95%
Newsprint
90%
+60
pulp 7
80% $5
$11
Mixed
Recycling
90%
-50
paper
Process 1
95% $6 2 5
$13 $8
75%
90%
Packing
paper
+40
pulp $9 8
85%
95%
White
85%
-30
office $6
paper 3
$10
Recycling
90%
90% $8
Process 2 $7
Print 6
stock
+50
85%
$13
95%
pulp +0 9
-40
Cardboard
$14 4

25. Defining the Objective Function
Minimize total cost.
MIN: 13X X X X26
+ 9X35+ 10X X X46 + 5X X58 + 8X59 + 6X67 + 8X68 + 7X69

26. Defining the Constraints-I
Raw Materials
-X15 -X16 >= -70 } node 1
-X25 -X26 >= -50 } node 2
-X35 -X36 >= -30 } node 3
-X45 -X46 >= -40 } node 4

27. Defining the Constraints-II
Recycling Processes
+0.9X15+0.8X X X45- X57- X58-X59 >= 0 } node 5
+0.85X X26+0.9X X46-X67-X68-X69 >= 0 } node 6

28. Defining the Constraints-III
Paper Pulp
+0.95X X67 >= 60 } node 7
+0.90X X67 >= 40 } node 8
+0.90X X67 >= 50 } node 9

29. Implementing the Model
See file Fig5-17.xls

30. 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.

31. 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?

32. The Northwest Petroleum Company
Pumping
Pumping
Station 1
Station 3 3 4 2 6 2 6
Oil Field 1
Refinery 6 2 4 4 3 5 5
Pumping
Pumping
Station 2
Station 4

33. The Northwest Petroleum Company
Pumping
Pumping
Station 1
Station 3 3 4 2 6 2 6
Oil Field 1
Refinery 6 2 4 4 3 5 5
Pumping
Pumping
Station 2
Station 4

34. Formulation of the Max Flow Problem
MAX: X61
Subject to: +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

35. Implementing the Model
See file Fig5-24.xls

36. Optimal Solution 3 4 2 6 2 6 1 6 2 4 4 3 5 5 3 5 5 2 2 4 4 2 Pumping
Station 1
Station 3 3 3 4 2 5 5 6 2 2 6
Oil Field 1
Refinery 6 2 4 2 4 4 4 3 5 5 2
Pumping
Pumping
Station 2
Station 4

37. Special Modeling Considerations: Flow Aggregation+0 $3 $5
-100 1 3 5
+75 $3 $4 $5 $4 2 4 6
+50
-100 $5 $6 +0
Suppose the total flow into nodes 3 & 4 must be at least 50 and 60, respectively. How would you model this?

38. Special Modeling Considerations: Flow Aggregation+0 +0 $3 $5
-100 1 30 3 5
+75
L.B.=50 $4 $3 $4 $5 2 40 4 6
+50
-100 $5
L.B.=60 $6 +0 +0
Nodes 30 & 40 aggregate the total flow into nodes 3 & 4, respectively.

39. Special Modeling Considerations: Multiple Arcs Between Nodes$8
-75 1 2 $6
+50
U.B. = 35
Two two (or more) arcs cannot share the same beginning and ending nodes. Instead, try... +0 10 $0 $8 $6 1 2
+50
-75
U.B. = 35

40. Special Modeling Considerations: Capacity Restrictions on Total Supply1
-100 2 3
+75 4
+80
$5, UB=40
$3, UB=35
$6, UB=35
$4, UB=30
Supply exceeds demand, but the upper bounds prevent the demand from being met.

41. Special Modeling Considerations: Capacity Restrictions on Total Supply
+75
-100
$5, UB=40 1 3
$999, UB=100
+200
$4, UB=30
$6, UB=35 2 4
$999, UB=100
$3, UB=35
+80
-100
Now demand exceeds supply. As much “real” demand as possible will be met in the least costly way.

42. 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.

43. Minimal Spanning Tree Example: Windstar Aerospace Company
$150 2 4
$100
$85
$75
$40 1
$80 5
$85
$90 3
$50 6
$65
Nodes represent computers in a local area network.

44. 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.

45. Solving the Example Problem - 12 4
$100
$85 1
$80 5
$85
$90 3 6

46. Solving the Example Problem - 24
$100
$85
$75 1
$80 5
$85
$90 3
$50 6

47. Solving the Example Problem - 32 4
$100
$85
$75 1
$80 5
$85 3
$50 6
$65

48. Solving the Example Problem - 42 4
$100
$85
$75
$40 1
$80 5 3
$50 6
$65

49. Solving the Example Problem - 5
$150 2 4
$85
$75
$40 1
$80 5 3
$50 6
$65

50. Solving the Example Problem - 62 4
$75
$40 1
$80 5 3
$50 6
$65

51. End of Chapter 5