Network Optimization
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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).

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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



Terminology
Terminology

  • Nodes and arcs

  • Arc flow

  • Upper and lower bounds

  • Cost

  • Gains and losses

  • External flow

  • Optimal flow



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

New York

350

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


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)

[600]

LA

(0)

DUM

[-50]


·

Costs on arcs to dummydestination = 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)


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

j=1

i=1

n

å

s.t.

x

= s

i = 1,…,m

ij

i

j=1

m

å

x

= d

j = 1,…,n

ij

j

i=1

x

³

0

i = 1,…,m; j = 1,…,n

ij


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.


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.


Network notation
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


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)

xij³ 0, " (i,j)ÎA


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


6

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


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}

S2

T2

= {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


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.


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


Maximum Flow Model

Max xts

å

xij

å

xji = 0, "iÎN

s.t.

-

(i,j)ÎFS(i)

(j,i)ÎRS(i)

"(i,j)ÎA

0 £xij£ uij

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.


Sensitivity Report for Max Flow Problem

Adjustable Cells

Final

Reduced

Objective

Allowable

Allowable

Cell

Name

Value

Cost

Coefficient

Increase

Decrease

$E$9

Arc1 Flow

3

0

0

1E+30

0

$E$10

Arc2 Flow

2

0

0

0

1

$E$11

Arc3 Flow

0

0

0

0

1E+30

$E$12

Arc4 Flow

2

1

0

1E+30

1

$E$13

Arc5 Flow

1

1

0

1E+30

1

$E$14

Arc6 Flow

2

1

0

1E+30

1

$E$15

Arc7 Flow

0

0

0

0

1E+30

$E$16

Arc8 Flow

2

0

0

0

1

$E$17

Arc9 Flow

3

0

0

1E+30

0

$E$18

Arc10 Flow

5

0

1

1E+30

1

Constraints

Final

Shadow

Constraint

Allowable

Allowable

Cell

Name

Value

Price

R.H. Side

Increase

Decrease

$N$9

Node1 Balance

0

0

0

0

3

$N$10

Node2 Balance

0

0

0

1E+30

0

$N$11

Node3 Balance

0

0

0

0

3

$N$12

Node4 Balance

0

1

0

0

2

$N$13

Node5 Balance

0

1

0

0

3

$N$14

Node6 Balance

0

1

0

0

3


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


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

[–150]

ATL

5

[–200]

(7)

(2)

[–300]

(4)

(2)

(7)

(6)

(5)

GAINS

8

[200]

AUS

7

[200]


Solution to Distribution Problem

[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]


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 arc’s bounds.

lower bound & upper bound (capacity)

·

Each arc has a per unit cost &

the goal is to minimize total cost.


Distribution network

[external flow]

(cost)

lower = 0, upper = 200

Distribution Network

[-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)

[200]

8

7

[200]


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


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