1. Chapter 5
Network Modeling

2. Introduction
A number of business problems can be represented graphically as networks.
This chapter focuses on several such problems:
Transshipment Problems (see section of your text)
Shortest Path Problems (see section 5.2 of text)
Maximal Flow Problems (see section 5.6 of text)
Above sections are the only required readings from this chapter.

3. 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 supply nodes and positive numbers to represent demand nodes.

4. 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

5. 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!

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

7. 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

8. 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)

9. 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
Non-negativity conditions
Xij >= 0 for all ij

10. 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

11. Implementing the Model
Before understanding how to set up this
model in an Excel spreadsheet, it will be a
good idea to view the Sumif Function and
the Vlookup Function video clips.
See Excel file Fig5-2.xls for the
spreadsheet set up of the Bavarian Motor
Company (MBC) example problem.

12. 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

13. 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

14. Formulating the Shortest Route Problem
Note that this problem is formulated exactly like the transshipment problem, where every node must obey the balance of flow rule.
Since supply = demand, all constraints will have the equality sign.
The beginning or starting node will always have a supply value of -1.

15. Formulation (continued)
The ending node will always have a demand value of +1.
Note that in the previous network diagram (i.e., The American Car Association), the objective is to find the shortest path from node 1 to node 11.

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 to understand how this
problem may be modeled in an Excel
spreadsheet.

17. 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?

18. 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

19. 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

20. 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 <= infinity

21. Formulation (continued)
To summarize, when you formulate the max flow problem always create a fictitious or dummy arc that will connect the ending node to the starting node (and not the other way!).
The objective will then be to maximize the flow along this dummy arc (e.g., max X61). Note that we theoretically set an “infinity” value (∞) as an upper limit on the capacity of this arc.

22. Formulation (continued)
In terms of constraints, you will always have two sets of constraints.
The first set is just like the transshipment and the shortest path problems (i.e., every node must obey the balance of flow rule)

23. Formulation (continued)
The second set of constraints will put an upper limit on how many units can flow along each arc.
Thus, each arc will have its own upper boundary constraint to ensure that we do not ship more than the capacity of each arc.

24. Formulation (continued)
For example, if say an arc represents a highway with 3 lanes, then that would constitute the upper limit. In other words, we can not have 4 cars simultaneously traveling along that highway.
Note that your author uses a very large value such as “9999” as the upper limit for the fictitious arc (i.e., X61). Any other large value will also work.

25. 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

26. Implementing the Model
See file Fig5-24.xls to understand how this
problem may be modeled in an Excel
spreadsheet.

27. End of Chapter