ORF 411 20 Substitutable resources - I - Princeton …...Linear programming review Notes: » If we...

© 2013 W.B. Powell 21

Lecture outline

Substitutable resources Linear programming review

Linear programming review

Linear programming formulation

» You can use any of a number of solvers (matlab, Cplex, Gurobi, MP-Express, …) to solve this problem.

» Our goal is to derive insights from the solution.

© 2013 W.B. Powell 22

min

subject to 0

ij iji j

c x

Ax bx ux


The linear programming formulation

© 2013 W.B. Powell 23

We begin by writing out the constraint matrix for our network problem. Genericflow conservation constraint is written as

Net supply (>0)

or d

ij ki ij k

x x S

14 15 16

24 25 26 27

14 24

15 25

emand (<0)

For the graph at the right: 12

18 ... 8 19

x x xx x x x

x xx x

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6


The constraint matrix

© 2013 W.B. Powell 24

1 1 1 1 0 0 0 0 0 02 0 0 0 1 1 1 1 0 03 0 0 0 0 0 0 0 1 14 1 0 0 1 0 0 0 0 05 0 1 0 0 1 0 0 0 06 0 0 1 0 0 1 0 1 07 0 0 0 1 0 0 1 0 1

A

Nodes

1 4 1 5 1 6 2 4 2 5 2 6 2 7 3 6 3 7 Links


Notes:» If we add the rows of the matrix A, they sum to zero.

This means that one of the rows is redundant – if we have n nodes, we only need n-1 constraints.

» To illustrate: consider a two node network:

» If we guarantee flow conservation at node 1, then we automatically guarantee flow conservation at node 2.

» In the larger network, pick one node (any node) and drop it. We call this node the root node.

» Let’s pick node 1.

© 2013 W.B. Powell 25

1 2 66


Dropping node 1, our constraint matrix now looks like:

© 2013 W.B. Powell 26

2 0 0 0 1 1 1 1 0 03 0 0 0 0 0 0 0 1 14 1 0 0 1 0 0 0 0 05 0 1 0 0 1 0 0 0 06 0 0 1 0 0 1 0 1 07 0 0 0 0 0 0 1 0 1

A

1 4 1 5 1 6 2 4 2 5 2 6 2 7 3 6 3 7


Simplex review:» Remember – we pick a set of columns that form a

basis. We are going to then divide our decision vector into basic variables and nonbasic variables.

» How do we do this?» We are going to find an initial solution using a simple

rule called the “northwest corner rule.”

© 2013 W.B. Powell 27

Linear programming review» Start with node 1 (the

“highest” node on the left)» Assign as much flow to node

4 (the highest node on the right).

© 2013 W.B. Powell 28

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8




» When one node is saturated, “pivot” to the next lower node on the list.

© 2013 W.B. Powell 29

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

84





© 2013 W.B. Powell 30

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8415





» We use this logic to assign the remaining flow.

© 2013 W.B. Powell 31

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8415396

Linear programming review» The links with flow (wide

solid links) represent the basis.

» The links without flow are nonbasic.

© 2013 W.B. Powell 32

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8415396

1 41 52 52 63 63 7

Bx

1 62 42 7

Nx


This allows us to partition the A matrix into basic and non-basic columns:

© 2013 W.B. Powell 33

2 0 0 1 1 0 03 0 0 0 0 1 14 1 0 0 0 0 05 0 1 1 0 0 06 0 0 0 1 1 07 0 0 0 0 0 1

BA

1 4 1 5 2 5 2 6 3 6 3 7

2 0 1 13 0 0 04 0 1 05 0 0 06 1 0 07 0 0 1

NA

1 6 2 4 2 7


Let’s restate our constraints:

» … becomes

» …. or

»

© 2013 W.B. Powell 34

Ax b

BB N

N

xA A b

x

B B N NA x A x b

We can now solve for as a function of B Nx x1B B N Nx A b A x


»

» We are going to do this by inspection! You can view this as magic, or a cheap parlor trick.

© 2013 W.B. Powell 35

By construction, 0, but we can use this equation to understandthe effect of increasing a nonbasic variable by adjusting the basic variables so that we maintain conservation of flow. But it means

Nx

we have to find the inverse of BA

1B B N Nx A b A x


The inverse

» Each column traces a path from a node to the root node.

© 2013 W.B. Powell 36

1

1 4 0 0 1 0 0 01 5 1 1 0 1 1 12 5 1 1 0 0 1 12 6 0 1 0 0 1 13 6 0 1 0 0 0 13 7 0 0 0 0 0 1

BA

2 1 3 1 4 1 5 1 6 1 7 1

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8415396

LinksPaths


Verify that we have an inverse:

© 2013 W.B. Powell 37

1BA

0 0 1 0 0 01 1 0 1 1 1

1 1 0 0 1 10 1 0 0 1 10 1 0 0 0 10 0 0 0 0 1

0 0 1 1 0 00 0 0 0 1 11 0 0 0 0 0

0 1 1 0 0 00 0 0 1 1 00 0 0 0 0 1

BA

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

I


Notes:»

» We can rewrite our objective function using

© 2013 W.B. Powell 38

1

We refer to as a "node-arc" incide matrix (rows are nodes,

columns are arcs (links). The matrix is an "arc-chain"

incidence matrix. Rows are arcs (links), while columns correspondto paths

B

B

A

A

from each node in the network to the root node (node 1).

1

1 1

1

1

min

=

=

=

where

"reduced costs"

T TT B B N N

T TB B N N N N

T T TB B B B N N N N

TB B N N

T TN N B B N

c x c x c x

c A b A x c x

c A b c A A x c x

c A b c x

c c c A A

Linear programming review Reduced costs tell us if we should increase flow on a

nonbasic link, while adjusting flows on all basic links.

© 2013 W.B. Powell 39

1

16 24 27 14 15 25 26 36 37

The "reduced cost"

0 0 1 0 0 0 0 1 11 1 0 1 1 1 0 0 0

1 1 0 0 1 1 0 1 00 1 0 0 1 1 0 0 00 1 0 0 0 1 1 0 00 0 0 0 0 1 0 0 1

T TN B B Nc c c A A

c c c c c c c c c

2 3 4 5 6 7 1 6 2 4 2 7

“paths”Nonbasic

links

Linear programming review Reduced costs (cont’d)

© 2013 W.B. Powell 40

16 24 27 14 15 25 26 36 37

16 24 27 2 3 4 5 6 7

16

0 0 1 0 0 0 0 1 11 1 0 1 1 1 0 0 0

1 1 0 0 1 1 0 1 00 1 0 0 1 1 0 0 00 1 0 0 0 1 1 0 00 0 0 0 0 1 0 0 1

0 1 10 0 00 1 00 0 01 0 0

0 0 1

c c c c c c c c c c

c c c v v v v v v

c c

24 27 6 2 4 2 7

16 1 6 24 2 4 27 2 7

c v v v v v

c v v c v v c v v

1TB Bc A


Computing “reduced costs”» Assume that we have the following costs:» The path costs for each node are:

» The reduced costs are

© 2013 W.B. Powell 41

1

2

3

4

5

6

7

148

915 16

417

85

16 24 27

16 6 24 2 4 27 2 7

9 4 15 8 ( 14) 17 8 7

13 7 16

c c c c

c v c v v c v v

2

3

4

5

6

7

16 8 88 4 16 8 12

1484 16 8 45 8 4 16 8 7

vvvvvv

So, should we increase flow on any of the nonbasic links? If so, which one(s)?

1TB Bc A


Computing “reduced costs”» Consider what happens if we

want to increase flow on the nonbasic link from 2 to 4

• To maintain flow conservation, we need to adjust flows on the basic links.

• We do this by “pushing” flow from the root node to node 2, and then from 4 back to the root node. This creates a cycle.

© 2013 W.B. Powell 42

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

8415396


Pushing flow around the cycle» Imagine that we want to increase the flow from 2 to 4.

How does this affect the flows on the rest of the network? This is called the “flow augmenting cycle”

© 2013 W.B. Powell 43

1

2

3

4i

j

5

6

7

12

18

15

8

19

12

6

84

15

39

6

» How much flow can we move around the cycle produced by adding the nonbasic link from 2 to 4?

» As we move flow around the cycle, can we guarantee that we maintain flow conservation at all times?

0


All of these calculations depend on our choice of node 1 as the root node.» What if we picked a different node?

» Notice that we do not need to do any fancy linear algebra to calculate the path costs.

© 2013 W.B. Powell 44

1

2

3

4

5

6

7

1 0v

2 8v

3 12v

4 14v

5 8v

7 7v

148

915 16

417

85

Node 1 = root node

1

2

3

4

5

6

7

1 12v

2 4v

3 0v

4 26v

5 20v

6 8v

7 5v

148

915 16

417

85

Node 3 = root node

6 4v

Linear programming review Changing the root node (cont’d)

» How did changing the root node affect our reduced costs?» Root node = node 1

» Root node = node 3

» They are the same! Choice of root node does not change the reduced costs.

© 2013 W.B. Powell 45

16 24 27

16 1 6 24 2 4 27 2 7

9 0 4 15 8 ( 14) 17 8 7

13 7 16

c c c c

c v v c v v c v v

16 24 27

16 1 6 24 2 4 27 2 7

9 ( 12) ( 8) 15 ( 4) ( 26) 17 ( 4) ( 5)

13 7 16

c c c c

c v v c v v c v v


Dual variables» Recall that our constraints look like:

» What is the meaning of a dual variable? » How can we calculate it for this problem?

© 2013 W.B. Powell 46

14

15

16

24

25

26

27

36

37

0 0 0 1 1 1 1 0 0 180 0 0 0 0 0 0 1 1 151 0 0 1 0 0 0 0 0 8

0 1 0 0 1 0 0 0 0 190 0 1 0 0 1 0 1 0 120 0 0 1 0 0 1 0 1 6

xxxxxxxxx

2

3

4

5

6

7

vvvvvv

Dualvariables

Linear programming review The dual variable is often described as the marginal value

of a resource (the right hand side constraint). » What happens if we increase the flow into node 2 from 18 to 19?» What happens if we “increase” the right hand side at node 6 from

-12 to -11? (Recall how the flow conservation constraint is written).

© 2013 W.B. Powell 47

1

2

3

4

5

6

7

1 0v

2 8v

3 5v

4 14v

5 8v

7 7v

148

915 16

417

85

12

18

15

8

19

12

6

6 4v


Dual variables (cont’d)» The root node serves as a slack node.» It allows us to change the supply/demand at any node,

and perturb the solution without violating conservation of flow (remembering that we no longer have a conservation of flow constraint at the root node).

» If we change the root node, do the dual variables change? Do we care?

» So – dual variables can be arbitrary, but differences between dual variables are meaningful.

© 2013 W.B. Powell 48

ORF 411 20 Substitutable resources - I - Princeton …...Linear programming review Notes: » If we...

Documents

Transcript of ORF 411 20 Substitutable resources - I - Princeton …...Linear programming review Notes: » If we...