Lecture 5

1
Lecture 5 Integration of Network Flow
Programming Models

• Topics
• Min-cost flow problem (general model)
• Mathematical formulation and problem
  characteristics
• Pure vs. generalized networks

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

3
Min-Cost Flow Problem
Example Distribution problem

• Warehouses store a particular commodity in
  Phoenix, Austin and Gainesville.

• 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
4
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 )

5
Spreadsheet Input Data
arc name
termination node
origin node
lower bound
upper bound

cost
gain
xij
i
j
lij
uij
cij
gij
external flow si or -di
j
i
The origin node is the arcs tail The termination
node is called the head Supplies are positive and
demands are negative
6
Data Entry Using Math Programming/Network Add-in.
And here is the solution ...
7
Solution to Distribution Problem
supply / demand
(flow)
-200
-250
(200)
NY
CHIC
700
(50)
PHOE
(100)
(200)
(200)
-150
(200)
ATL
LA
DAL
-300
(50)
-200
(200)
200
GAINS
AUS
200

8
Sensitivity Report for Max Flow Problem
9
Characteristics of Network Flow Problems
10
Distribution Network Used in Formulation
-200
-250
(6)
2
6
700
1
(4)
(6)
(7)
(4)
(5)
(3)
(3)
(7)
(2)
(5)
-150
-200
4
5
3
(2)
-300
(4)
(6)
(2)
(5)
(7)
200
8
7
200

? Notation
11
LP for Distribution Problem
Pure network ? flow at each node is conserved ?
flow across an arc is conserved ? no gains or
losses can occur on arcs
¼
6x
7x
Min


3x





4x

3x
7x
86
16
12
14
13
85


x

x
x
700
Node 1 Node 2 . . .
x
s.t.

x


x

x

x
sit.
13
14
21
23
24
12
15
-
- 200


x
x
-
-x

-
-
x


x
x
43
73
23
43
63
13




x

x
-
-
x
- 300
x
x
x
-
-

x

-x

x


x

x

x

x


x
42
43
54
84
41
45
46
74
41
43
46
47
24
54
84
. . .
.
Flow balance constraints for each of the 8 nodes
.
.
x
x

x

85
84
86
0 xij 200, for all (i, j ) combinations
which are arcs
12
Decision variables are the flow variables xij

j
i
By examining the flow balance constraints we see
that xij appears in exactly two of them
xij

0
.

.


.
0

1

node i
0


(or in the other order if i gt j)
.

.


.
0

-
1

node j
0

.

.


.
0

13
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.

14
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 .
15
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)
16
Pure Min-Cost Flow Problem
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
17
Decision Variables
xij flow on arc (i, j )
Formulation
å cijxij
Min
(i, j )ÎA
å xij - å xji bi, " i Î N
s.t.
(i, j )ÎFS(i )
(j, i )ÎRS(i )
lij xij uij, " (i, j ) Î A
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
å xij - å gjixji bi, " i Î N
s.t.
(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
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.

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
US
1
1
1.05
1.45
1.72
.68
Yen(100)
2
.95
1
1.41
1.64
.64
3
CHF
.69
.71
1
1.14
.48
4
D-Mark
.58
.61
1
.39
0.88
Brit
5
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 c14 1, c35 0.69
-1
US
1
Yen
5
2
Brit
4
D-Mark
3
CHF
g12 (1.05)(0.99) g51 (1.50)(0.99)
Each arc has a gain of gij. For example,
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
3
CHF
Total cycle gain 1.0488 4.88
4
34.986
D-Mark
Note ( ? ) g51 1.485
Start with 13.801 ? 34.986 D-Mark ? 30.473 CHF
? 14.475 Remove 0.674 ? 1 leaving 13.801
26
What You Should Known About General Network Flow
Problems

• How to formulate a general network flow problem
  as a linear program.
• What the relationship is between the maximum flow
  and the minimum cut in a network.
• What the implications are for a network flow
  problem with gains.
• How to solve general network flow problems using
  the Excel add-ins.