Kramer s a k a cramer s rule
This presentation is the property of its rightful owner.
Sponsored Links
1 / 15

Kramer’s (a.k.a Cramer’s) Rule PowerPoint PPT Presentation


  • 45 Views
  • Uploaded on
  • Presentation posted in: General

Kramer’s (a.k.a Cramer’s) Rule. Component j of x = A -1 b is Form B j by replacing column j of A with b. Total Unimodularity. A square, integer matrix B is unimodular (UM) if its determinant is 1 or -1.

Download Presentation

Kramer’s (a.k.a Cramer’s) Rule

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Kramer s a k a cramer s rule

Kramer’s (a.k.a Cramer’s) Rule

  • Component j of

    x = A-1b is

  • Form Bj by replacing column j of A with b.


Total unimodularity

Total Unimodularity

  • A square, integer matrix B is unimodular (UM) if its determinant is 1 or -1.

  • An integer matrix A is called totally unimodular (TUM) if every square, nonsingular submatrix of A is UM.

  • From Cramer’s rule, it follows that if A is TUM and b is an integer vector, then every BFS of the constraint system Ax = b is integer.


Tum theorem

TUM Theorem

  • An integer matrix A is TUM if

    • All entries are -1, 0 or 1

    • At most two non-zero entries appear in any column

    • The rows of A can be partitioned into two disjoint sets such that

      • If a column has two entries of the same sign, their rows are in different sets.

      • If a column has two entries of different signs, their rows are in the same set.

  • The MCNFP constraint matrices are TUM.


Kramer s a k a cramer s rule

General Form of the MCNF Problem


Kramer s a k a cramer s rule

2

1

3

Flow Balance Constraint Matrix

Capacity Constraints

Constraints in Standard Form


Shortest path problems

Shortest Path Problems

  • Defined on a Network

    • Nodes, Arcs and Arc Costs

    • Two Special Nodes

      • Origin Node s

      • Destination Node t

  • A path from s to t is an alternating sequence of nodes and arcs starting at s and ending at t:

    s,(s,v1),v1,(v1,v2),…,(vi,vj),vj,(vj,t),t


Kramer s a k a cramer s rule

s=1, t=3

We Want a Minimum Length Path From s to t.

5

10

1

2

3

7

1

7

4

1,(1,2),2,(2,3),3Length = 15

1,(1,2),2,(2,4),4,(4,3)Length = 13

1,(1,4),4,(4,3),3Length = 14


Maximizing rent example

Maximizing Rent Example

  • Optimally Select Non-Overlapping Bids for 10 periods


Kramer s a k a cramer s rule

Shortest Path Formulation

t

d10

-7

-2

d9

-3

-7

-5

-2

0

0

-1

0

-4

d1

d2

d3

d4

d5

d6

d7

d8

-3

-6

s

-1

-11


Mcnf formulation of shortest path problems

MCNF Formulation of Shortest Path Problems

  • Origin Node s has a supply of 1

  • Destination Node t has a demand of 1

  • All other Nodes are Transshipment Nodes

  • Each Arc has Capacity 1

  • Tracing A Unit of Flow from s to t gives a Path from s to t


Maximum flow problems

Maximum Flow Problems

  • Defined on a Network

    • Source Node s

    • Sink Node t

    • All Other Nodes are Transshipment Nodes

    • Arcs have Capacities, but no Costs

  • Maximize the Flow from s to t


Example rerouting airline passengers

Example: Rerouting Airline Passengers

Due to a mechanical problem, Fly-By-Night Airlines had to cancel flight 162 - its only non-stop flight from San Francisco to New York. The table below shows the number of seats available on Fly-By-Night's other flights.


Kramer s a k a cramer s rule

Max Flow from SF to NY

= 2+2+5=9

2

D

C

5

4

SF

4

NY

6

7

5

H

A

Formulate a maximum flow problem that will tell Fly-By-Night

how to reroute as many passengers from San Francisco to

New York as possible.

(flow, capacity)

(2,2)

D

C

(4,5)

(2,4)

SF

(2,4)

NY

(5,6)

H

A

(7,7)

(5,5)


Mcnf formulation of maximum flow problems

MCNF Formulation of Maximum Flow Problems

  • Let Arc Cost = 0 for all Arcs

  • Add an infinite capacity arc from t to s

    • Give this arc a cost of -1


Maximum flow minimum cut theorem

2

D

C

5

4

SF

4

NY

6

7

5

H

A

D

C

5

4

SF

4

NY

6

5

H

A

Maximum-Flow Minimum-Cut Theorem

  • Removing arcs (D,C) and (A,NY) cuts off SF from NY.

  • The set of arcs{(D,C), (A,NY)} is an s-t cut with capacity 2+7=9.

  • The value of a maximum s-t flow = the capacity of a minimum s-t cut.


  • Login