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

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Network optimization

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


Example of distribution network

Example of (Distribution) Network


Terminology

Terminology

  • Nodes and arcs

  • Arc flow

  • Upper and lower bounds

  • Cost

  • Gains and losses

  • External flow

  • Optimal flow


Network flow problems

Network Flow Problems


Network optimization

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


Network optimization

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]


Network optimization

·

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)


Network optimization

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


Network optimization

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.


Network optimization

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


Network optimization

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


Network optimization

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


Network optimization

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


Network optimization

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


Network optimization

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.


Network optimization

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


Network optimization

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.


Network optimization

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


Network optimization

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


Network optimization

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]


Network optimization

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]


Network optimization

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]


Network optimization

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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