Network Optimization - PowerPoint PPT Presentation

1 / 25
About This Presentation
Title:

Network Optimization

Description:

We wish to ship goods (a single commodity) from. m sources ... CHIC. 2. AUS. 7. LA. 3. PHOE. 1 (6) (3) (5) (7) (4) (2) (4) (5) (5) (6) (4) (7) (6) (3) [ 150] ... – PowerPoint PPT presentation

Number of Views:1360
Avg rating:3.0/5.0
Slides: 26
Provided by: nt3
Category:

less

Transcript and Presenter's Notes

Title: Network Optimization


1
Network Optimization
Network optimization models
Special cases of linear programming models
Important to identify problems that can be
modeled as networks because

(1)
network representations make optimization
models easier to visualize and explain

(2)
very efficient algorithms are available
2
Example of (Distribution) Network
3
Terminology
  • Nodes and arcs
  • Arc flow
  • Upper and lower bounds
  • Cost
  • Gains and losses
  • External flow
  • Optimal flow

4
Network Flow Problems
5
Transportation Problem
We wish to ship goods (a single commodity) from
m sources to n destinations at minimum
cost.
Warehouse i has si units available i 1, . . .
,m and destination

j has a demand of dj, j 1, . . . ,n.
Goal - Ship the goods from sources to
destinations at minimum cost.
Plants
Supply
Markets
Demand
Example
San Francisco
350
New York
325
Los Angeles
Chicago
600
300
Austin
275
From/To
NY
Chi
Aus
Unit Shipping Costs
SF
2.5
1.7
1.8
LA
--
1.8
1.4
6
Total supply 950, total demand 900
Transportation problem is defined on a bipartite
network
Arcs only go from supply nodes to destination
nodes To handle excess supply
create a dummy destination with a
demand of 50.
The min-cost flow network for this transportation
problem is given by
NY
-325
(2.5)
(1.7)
SF
350
(1.8)
CHI
-300
(0)
(M)
(1.8)
AUS
-275
(1.4)
LA
600
(0)
DUM
-50
7

Costs on arcs to dummy destination 0
(In some settings it would be necessary
to include a nonzero warehousing cost.)


The objective coefficient on the LA
NY arc is M.
This denotes a large value and effectively
prohibits
use of this arc (could eliminate arc).

We are assured of integer solutions
because technological matrix A is totally
unimodular.
(important in some applications)
8
The LP formulation of the transportation problem
with m
sources and n destinations is given by
m
n
å
å
Min



c
x
ij
ij
j1
i1
n
å
s.t.

x
s
i 1,,m
ij
i
j1
m
å

x
d
j 1,,n
ij
j
i1
x
³
0
i 1,,m j 1,,n
ij
9
Shortest Path Problem
  • Given a network with distances on the arcs our
    goal is to find the shortest path from the origin
    to the destination.
  • These distances might be length, time, cost, etc,
    and the values can be positive or negative. (A
    negative cij can arise if we earn revenue by
    traversing an arc.)
  • The shortest path problem may be formulated as a
    special case of the pure min-cost flow problem.

10
Example
(cij)
cost or length
(2)
2
4
(3)
(4)
(1)
(1)
(2)
-1
6
1
1
(6)
(7)
(2)
3
5
  • We wish to find the shortest path from node 1 to
    node 6.
  • To do so we place one unit of supply at node 1
    and push it through the network to node 6 where
    there is one unit of demand.
  • All other nodes in the network have external
    flows of zero.

11
Network Notation
A set of Arcs, N set of nodes Forward Star
for node i FS(i) (i,j) (i,j) Î A
Reverse Star for node i RS(i) (j,i)
(j,i) Î A
RS(i)
FS(i)
i
i
12
In general, if node s is the source node and node
t is the termination node then the shortest path
problem may be written as follows.
å
x
c
Min
ij
ij
(i,j)ÎA


1, i s 1, i t 0, i Î N \ s,t
å
å
x
x
s
.t.

-
ij


ji
(i,j)ÎFS(i)
(j,i)ÎRS(i)

xij ³ 0, " (i,j)ÎA
13
Maximum Flow Problem
  • In the maximum flow problem our goal is to send
    the largest amount of flow possible from a
    specified source node to a specified destination
    node subject to arc capacities.
  • This is a pure network flow problem (i.e., gij
    1) in which all the (real) arc costs are zero
    (cij 0) and at least some of the arc capacities
    are finite.

Example
(2)
2
4
(3)
(4)
(uij) arc capacities
(1)
(1)
6
(2)
1
(7)
(6)
(2)
3
5
1
14
Goal for Max Flow Problem Send as much flow as
possible from node 1 to node 6
Solution
2 (2)
2 (3)
3 (4)
2
4
xij (uij) flow capacities
1 (1)
1
3 (7)
2 (6)
3
5
2 (2)
Maximum flow 5
15
Cut A partition of the nodes into two sets S
and T. The origin node must be in S and the
destination node must be in T.
Examples of cuts in the network above are
2,3,4,5,6
S1
1
T1

4,5,6
T2
S2
1,2,3

S3
2,4,6
T3
1,3,5

The value of a cut V(S,T) is the sum of all the
arc capacities that have their tails in S and
their heads in T.
V(S2,T2) 5
V(S3,T3) 12
V(S1,T1) 10
16
Max-Flow Min-Cut Theorem
The value of the maximum flow is equal to the
value of the minimum cut.
  • In our problem, S 1,2,3 / T 4,5,6 is a
    minimum cut.
  • The arcs that go from S to T are (2,4), (2,5) and
    (3,5).
  • Note that the flow on each of these arcs is at
    its capacity. As such, they may be viewed as the
    bottlenecks of the system.

17
Max Flow Problem Formulation
  • There are several different linear programming
    formulations.
  • We will use one based on the idea of a
    circulation.
  • We suppose an artificial return arc from the
    destination to the origin with uts and cts
    1.
  • External flows (supplies and demands) are zero at
    all nodes.

s
t
18
Maximum Flow Model
Max xts

å
xij
å
xji 0, " iÎN
s.t.
-




(i,j)ÎFS(i)
(j,i)ÎRS(i)
0 xij uij
" (i,j)ÎA


Identify minimum cut from sensitivity report
  • If the reduced cost for xij has value 1 then arc
    (i,j) has its tail (i) in S and its head (j) in
    T.
  • Reduced costs are the shadow prices on the simple
    bound constraint xij uij.
  • Value of another unit of capacity is 1 or 0
    depending on whether or not the arc is part of
    the bottleneck


Note that the sum of the arc capacities with
reduced costs of 1 equals the max flow value.
19
Sensitivity Report for Max Flow Problem
Adjustable Cells
Final
Reduced
Objective
Allowable
Allowable
Cell
Name
Value
Cost
Coefficient
Increase
Decrease
E9
Arc1 Flow
3
0
0
1E30
0
E10
Arc2 Flow
2
0
0
0
1
E11
Arc3 Flow
0
0
0
0
1E30
E12
Arc4 Flow
2
1
0
1E30
1
E13
Arc5 Flow
1
1
0
1E30
1
E14
Arc6 Flow
2
1
0
1E30
1
E15
Arc7 Flow
0
0
0
0
1E30
E16
Arc8 Flow
2
0
0
0
1
E17
Arc9 Flow
3
0
0
1E30
0
E18
Arc10 Flow
5
0
1
1E30
1
Constraints
Final
Shadow
Constraint
Allowable
Allowable
Cell
Name
Value
Price
R.H. Side
Increase
Decrease
N9
Node1 Balance
0
0
0
0
3
N10
Node2 Balance
0
0
0
1E30
0
N11
Node3 Balance
0
0
0
0
3
N12
Node4 Balance
0
1
0
0
2
N13
Node5 Balance
0
1
0
0
3
N14
Node6 Balance
0
1
0
0
3
20
Minimum 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
21
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

22
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

23
This network flow problem is based on


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 arcs bounds.
lower bound upper bound (capacity)

Each arc has a per unit cost
the goal is to minimize total cost.
24
Distribution Network
-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

25
Linear Program Model for Distribution Problem
Minimize z 6x12 3x13 3x14 7x15
7x86 Subject to conservation of flow constraints
at each node Node 1 x12 x13 x14 x15
700 Node 2 x12 x62 x52 200 Node 3
x13 x43 x73 200 Node 4 x42 x43
x45 x46 x14 x54 x74
300 . . . . . . Node 8 x84 x85 x86
200 Bounds 0 xij 200 for all (i,j) ? A
Write a Comment
User Comments (0)
About PowerShow.com