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Lecture 5 – Integration of Network Flow Programming Models. Topics Mincost flow problem (general model) Mathematical formulation and problem characteristics Pure vs. generalized networks. Distribution Problem. [supply / demand]. arc lower bounds = 0 . arc upper bounds = 200.
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Topics
[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
[–150]
ATL
5
[–200]
(7)
(2)
[–300]
(4)
(2)
(7)
(6)
(5)
GAINS
8
[200]
AUS
7
[200]
Customers  Chicago, LA, Dallas, Atlanta, & New York
Supply [ si ] at each warehouse i
Demand [ dj] of each customer j
MinCost Flow ProblemExample: Distribution problem
Problem: Determine optimal shipping plan that minimizes transportation costs
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
arc
name
origin
node
termination
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 arc’s tail
The termination node is called the head
Supplies are positive and demands are negative
External flow balance: total supply = total demand
And here is the solution ...
[supply / demand]
(flow)
[200]
[250]
(200)
NY
CHIC
[700]
(50)
PHOE
(100)
(200)
(200)
[150]
(200)
LA
ATL
DAL
[300]
(50)
[200]
(200)
[200]
GAINS
AUS
[200]
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.
Characteristics of Network Flow Problems(cost)
lower = 0, upper = 200
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
3
5
(2)
[300]
(4)
(6)
(2)
(5)
(7)
8
[200]
7
[200]
Notation
G = (N, A) network with node set N and arc set A
Indices i, jÎN denote nodes and (i, j) ÎA denote arcs
Originating set of arcs for node i (tails are i) is the forward starof 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) ÎA}.
Flow balance equation for node i :
åxij – åxji = bi
(i,j)ÎFS(i)
(j,i)ÎRS(i)
where bi = positive for supply node i
= negative for demand node i
= 0 otherwise
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) }
Indices/sets
i, jÎN
nodes
arcs
forward star of i
reverse star of i
(i, j) ÎA
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: ibi = 0
xij = flow on arc (i,j)
Formulation for pure mincost flow model
åcijxij
Min
(i,j)ÎA
åxij å xji = bi, "i Î N
s.t.
(i,j)ÎFS(i)
(j,i)ÎRS(i)
lij £ xij£ uij, " (i,j) Î A
Decision variables are the flow variables xij
j
i
By examining the flow balance constraints we see that xijappears in exactly two of them:
xij
0
.
.
.
0
+1
node i
0
( or in the other order if i >j )
.
.
.
0

1
node j
0
.
.
.
This structure is called total unimodularity and guarantees integer solutions
0
åcijxij
Min
(i,j)ÎA
åxij ågjixji = bi, "iÎ N
s.t.
(i,j)ÎFS(i)
(j,i)ÎRS(i)
lij £ xij£ uij, " (i,j)ÎA
Note that if gij =1 " (i,j) ÎA, then we obtain the “pure” model
exchange
Gain = 1.78
Swiss
US
francs
$
15% return
on investment
Gain = 1.15
Year 2
Year 1
Gains and LossesIf bi, lij and uijare integervalued 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.
This is critical when we formulate the assignment, shortest path problems, and other network problems.
Special cases of the pure mincost flow model:
US $
Yen(100)
CHF
DMark
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
DMark
.58
.61
1
.39
0.88
Brit £
5
1.50
1.56
2.08
2.08
1
Arc costs:
cij = $ equivalent
(first column of table)
For example:
c12 = 1.05, c35 = 0.48
[1]
US $
1
Yen
5
Brit £
2
4
DMark
3
CHF
g12 = (1.05)(0.99)
g35 = (0.48)(0.99)
Each arc has a gain of gij. For example,
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
DMark
Note (£ $):
g51 = 1.485
Start with 13.801 £ 34.986 DMark 30.473 CHF 14.475 £
Remove 0.674 £ $1 leaving 13.801 £