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Kramer’s (a.k.a Cramer’s) RulePowerPoint Presentation

Kramer’s (a.k.a Cramer’s) Rule

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Kramer’s (a.k.a Cramer’s) Rule

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- Component j of
x = A-1b is

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

- 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.

- 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.

General Form of the MCNF Problem

2

1

3

Flow Balance Constraint Matrix

Capacity Constraints

Constraints in Standard Form

- 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

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

- Optimally Select Non-Overlapping Bids for 10 periods

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

- 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

- 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

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.

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)

- Let Arc Cost = 0 for all Arcs
- Add an infinite capacity arc from t to s
- Give this arc a cost of -1

2

D

C

5

4

SF

4

NY

6

7

5

H

A

D

C

5

4

SF

4

NY

6

5

H

A

- 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.