1. Lecture 12
Network Models

2. Lecture 5 – Integration of Network Flow Programming Models
Topics
Min-cost flow problem (general model)
Mathematical formulation and problem characteristics
Pure vs. generalized networks

3. Distribution Problem 2 6 1 3 4 5 8 7 [supply / demand]
arc lower bounds = 0
(shipping cost)
arc upper bounds = 200
[700]
[–200]
(6)
CHIC 2 NY 6
[–250]
PHOE 1
(4)
(7)
(6)
(4)
(3)
(3)
(5)
(5)
(2)
[–200] LA 3
DAL 4
ATL 5
[–150]
(7)
(2)
[–300]
(4)
(2)
(7)
(6)
(5)
GAINS 8
[200]
AUS 7
[200]

4. Example: Distribution problem
Min-Cost Flow Problem
Example: Distribution problem
Warehouses store a particular commodity in Phoenix, Austin and Gainesville.
Customers - Chicago, LA, Dallas, Atlanta, & New York
Supply [ si ] at each warehouse i
Demand [ dj ] of each customer j
Shipping links depicted by arcs, flow on each arc is limited to 200 units.
Dallas and Atlanta - transshipment hubs
Per unit transportation cost (cij ) for each arc
Problem: Determine optimal shipping plan that minimizes transportation costs

5. Notation for Min-Cost Flow Problem
In general: [supply/demand] on nodes
(shipping cost per unit) on arcs
In example: all arcs have an upper bound of 200
nodes labeled with a number 1,...,8
Must indicate notation that is included in model:
(cij ) unit flow cost on arc (i, j )
(uij ) capacity (or simple upper bound) on arc (i, j )
(gij ) gain or loss on arc (i, j )
All 3 could be included: (cij , uij , gij )

6. Spreadsheet Input Data
arc
name
origin
node
termination
node
lower
bound
upper
bound
cost
gain
xij i j
lij
uij
cij
gij i j
external
flow
si or -di
The origin node is the arc’s tail
The termination node is called the head
Supplies are positive and demands are negative
External flow balance: total supply = total demand

7. Data Entry Using Jensen Network Solver
And here is the solution ...

8. Solution to Distribution Problem
[supply / demand]
(flow)
[-200]
[-250]
(200)
[700]
CHIC NY
PHOE
(50)
(200)
(100)
(200)
[-150] LA
ATL
(200)
[-300]
DAL
(50)
[-200]
(200)
GAINS
[200]
AUS
[200]

9. Sensitivity Report for Distribution Problem

10. Characteristics of Network Flow Problems
Conservation of flow at nodes. At each node
flow in = flow out.
At supply nodes there is an external inflow
(positive)
At demand nodes there is an external outflow
(negative).
Flows on arcs must obey the arc bounds; i.e.,
lower bound & upper bound (capacity)
Each arc has a per unit cost & the goal is to minimize total cost.

11. Distribution Network Used in Formulation
[external flow]
(cost)
lower = 0, upper = 200
[-200]
[-250]
(6) 2 6
[700] 1
(4)
(6)
(7)
(4)
(5)
(3)
(3)
(7)
(5)
(2)
[-200]
[-150] 3 4 5
[-300]
(2)
(4)
(6)
(7)
(2)
(5) 8
[200] 7
[200]
 Notation

12. Pure Minimum Cost Flow Problem
G = (N, A)  network with node set N and arc set A
Indices i, j Î N denote nodes and (i, j ) Î A denote arcs
Originating set of arcs for node i (tails are i ) is the forward star of i
FS(i ) = { (i, j ) : (i, j ) Î A }
Terminating set of arcs for node i is the reverse star of i
RS(i ) = { (j,i ) : (j,i ) Î A }.

13. å xij – å xji = bi In our example:
FS(1) = { (1,2), (1,3), (1,4), (1, 5) }
RS(1) = Ø
FS(4) = { (4,2), (4,3), (4,5), (4,6) }
RS(4) = { (1,4), (5, 4), (7,4), (8,4) }
å xij – å xji = bi
(i, j )ÎFS(i )
where bi = positive for supply node i
= negative for demand node i
= 0 otherwise
(j,i )ÎRS(i )
Flow balance equation for node i :

14. Pure Min-Cost Flow Model
Indices/sets
i, j Î N
nodes
arcs
forward star of i
reverse star of i
(i, j ) Î A
FS(i )
RS(i )
Data
cij
unit cost of flow on (i, j )
lower bound on flow (i, j )
upper bound on flow (i, j )
external flow at node i
lij
uij bi
Total supply = total demand: i bi = 0

15. å cijxij å xij - å xji = bi, " i Î N Decision variables
xij = flow on arc (i, j )
Formulation for pure min-cost flow model
Min
å cijxij
(i, j )ÎA
s.t.
å xij å xji = bi, " i Î N
(i, j )ÎFS(i )
(j, i )ÎRS(i )
lij £ xij £ uij, " (i, j ) Î A

16. ( or in the other order if i > j )
Decision variables are the flow variables xij i j
By examining the flow balance constraints we see that xij appears in exactly two of them:
xij . . . +1
node i
( or in the other order if i > j ) . . . - 1
node j . . .
This structure is called total unimodularity and guarantees integer solutions

17. Observations from LP Model
If we add the constraints we obtain zero on the left-hand side so the right-hand side must also be zero for feasibility.
In particular, this means
sum of supplies = sum of demands.
Mathematically, we have one redundant constraint.
Must be careful in interpreting shadow prices on the flow balance constraints.
Cannot change only a supply or demand and have model make sense.

18. Generalized Minimum Cost Network Flow Model
Only one modification to “pure” formulation
 a possible gain (or loss) on each arc, denoted by gij
If gij = 0.95 then 100 units of flow leaves node i and 95 units arrive at node j

19. Generalized Formulation
å cijxij
Min
(i, j )ÎA
s.t.
å xij å gjixji = bi, " i Î N
(i, j )ÎFS(i )
(j, i )ÎRS(i )
lij £ xij £ uij, " (i, j )ÎA
Note that if gij =1 " (i, j ) Î A, then we obtain the “pure” model

20. gij = 0.95 for the associated arc (i,j).
Gains and Losses
Might experience 5% spoilage of a perishable good during transportation on a particular arc.
gij = 0.95 for the associated arc (i,j).
In production of manufacturing formulations we might incur losses due to production defects.
In financial examples we can have gains due to currency exchange or gains due to returns on investments. US $
Swiss
francs
Year 1
Year 2
Currency
exchange
15% return
on investment
Gain = 1.78
Gain = 1.15

21. Pure Network Problems vs. General Network Problems
If bi, lij and uij are integer-valued then all extreme points of the feasible region for a pure network flow problem give integer values for xij.
Fact:
(Same cannot be said for generalized network models.)
This integer property means that if we use the simplex method to solve a pure network flow problem then we are guaranteed that xij will be integer at optimality.

22. This is critical when we formulate the assignment, shortest path problems, and other network problems.
Special cases of the pure min-cost flow model:
Transportation problem
Assignment problem
Shortest path problem
Maximum flow problem

23. Checking for Arbitrage Opportunities
US $
Yen(100)
CHF
D-Mark
Brit £ 1
US $ 1
1.05
1.45
1.72
.68 2
Yen(100)
.95 1
1.41
1.64
.64 3
CHF
.69
.71 1
1.14
.48 4
D-Mark
.58
.61
0.88 1
.39 5
Brit £
1.50
1.56
2.08
2.08 1
The table is to be read as follows:
The 1.45 in row 1 column 3 means that $1 US will purchase 1.45 Swiss Francs (CHF).
In addition, there is a 1% fee that is charged on each exchange.

24. Arbitrage Network: Generalized Min-Cost Flow Problem
Arc costs:
cij = $ equivalent
(first column of table)
For example:
c12 = 1.05, c35 = 0.48
[-1]
US $ 1
Yen 2 5
Brit £ 4 3
D-Mark
CHF
Each arc has a gain of gij. For example,
g12 = (1.05)(0.99)
g35 = (0.48)(0.99)

25. Solution to Arbitrage Network
US $
Arc gains in
optimal cycle: 1
0.674
g54 = 2.535
g43 = 0.871
g35 = 0.475
Brit £ 5
30.473
13.801
CHF 3
Total cycle gain: =
= 4.88% 4
34.986
D-Mark
Note (£  $):
g51 = 1.485
Start with £  D-Mark  CHF  £
Remove £  $1 leaving £

26. What You Should Known About General Network Flow Problems
How to view flow on an arch with a gain or loss.
How to formulate a general network flow problem as a linear program.
What the implications are for a network flow problem with gains.
How to solve general network flow problems using the Excel add-ins.