EMIS 8373: Integer Programming

EMIS 8373: Integer Programming

Column Generationupdated 18 April 2005

Example 1: Adapted from

“Optimal Placement of ADD/DROP Multiplexers: Heuristic and Exact

Algorithms” by

Alain Sutter, Francis Vanderbeck, Laurence Wolsey

Operations Research, Vol. 46 (5), pp. 719-728, 1998

3

Formulation• Sets

– N is the set of nodes in the demand graph– E N N is the set of edges in the demand graph– M is a set of candidate SONET rings to may be used

to construct the network

• Parameters– de is weight of edge e E – b is the ring capacity– aej = 1 if demand e can be routed on ring j– cj is cost of ring j (the number of ADMs required)

4

Formulation Continued

• Decision Variables – zj = 1 if ring j is select, and 0, otherwise.

• BIP Formulation:

• Constraint set (1) ensures that each demand is assigned to exactly one of the candidate rings.

– Views the problem as edge partitioning

)2(.1,0

)1(,,1s.t.

min

Mj

Ee

zza

zc

j

Mjjej

Mjjj

5

Example Demand Graph and Candidate Set with b = 18

3

21

4 5

3 2

3

34

2

8 2

1. {(1, 2)} c1 = 22. {(1, 3)} c2 = 23. {(1, 4)} c3 = 24. {(2, 3)} c4 = 25. {(2, 5)} c5 = 26. {(3, 4)} c6 = 27. {(3, 5)} c7= 28. {(4, 5)} c8 = 29. {(1, 2), (1, 3), (2, 3)}

c9 = 310. {(1, 2), (1, 3), (1, 4),

(2, 3), (3, 5), (4, 5)} c10 = 5

Optimal Solution:z5 = z6 = z10 = 1cost = 9 ADMs

6

Comments• The best solution for the given demand graph uses two

rings and eight ADMs:1. {(1, 2), (2, 3), (2, 5), (3, 5)} 2. {(1, 3), (1, 4), (3, 4), (4, 5)}

• The edge-partitioning formulation cannot find this solution unless these candidate rings are given as part of the input.

• Let BIP(M) be the edge-partitioning formulation for a given candidate set M.

– The exact formulation is BIP(M*) where M* is the set of all feasible candidate rings.

– For a given demand graph |M*| = O(2 |E|)

7

Matrix Representation of BIP(M)

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

1010000000

1001000000

0000100000

0000010000

1100001000

1000000100

1100000010

11000000015322222222

A

c

BIP(M*) could have up to 28 –1 = 255 columns depending on b.

8

LP Relaxation of BIP(M)

.

,,1s.t.

min LP(M)

0 Mj

Ee

zza

zc

j

Mjjej

Mjjj

.,s.t.

max

MjEe

jeje

Eee

ca

Dual Problem

9

Additional Notation

• For a given basic feasible solution (BFS) for LP(M) – Let B denote the basis matrix– Let Aj denote column j of the constraint matrix A– Let cB denote the vector of objective coefficients of the

basic variables– Let denote the vector of corresponding dual variables – Recall that the reduced cost for variable zj is given by the

formula cj- cB B-1 Aj = cj- Aj.- B is optimal if all variables have a non-negative reduced

cost.

10

Reduced Cost Example 1

Suppose z1,z2, …, z8 are the basic variables.

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

10000000

01000000

00100000

00010000

00001000

00000100

00000010

0000000122222222

B

cB

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

0

0

0

0

1

0

1

1

3

9

9

A

c

3639

1

9

ABcc B

The current BFS uses 16 ADMs.Bringing ring 9 into the basis reduces this to 16 – 3 = 13

11


• Suppose z1,z2, …, z8 are the basic variables• Each demand assigned to its own ring• Using ring 9 to cover three demands saves 3 ADMs.

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

0

0

0

0

1

0

1

1

3

9

9

A

c


3639

1

9

ABcc B

12


Suppose z1,z2, …, z8 are the basic variables.

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

10000000

01000000

00100000

00010000

00001000

00000100

00000010

0000000122222222

B

cB

)5,4(

)5,3(

)4,3(

)5,2(

)3,2(

)4,1(

)3,1(

)2,1(

1

1

0

0

1

1

1

1

5

10

10

A

c

510510

1

10

ABcc B


13

A Column-Generation Heuristic

1. Solve restricted LP master problem LP(M) and let B be the optimal basis matrix

• LP(M*) is referred to as the linear programming master problem.

2. Look for a ring (column) j that is not in M, but would have a negative reduced cost if it were added to M

• This is referred as as solving the pricing problem.

3. If j is found then add j to M and goto step 1.4. Solve BIP(M)

• At this point an optimal solution to LP(M) is also an optimal solution to LP(M*).

14

Ni

ik xc

yyyyyyyy

A k

45

35

34

25

23

14

13

12

Feasible Rings with Negative Reduced Costs

Ejiij

Niikk yxAc ij

),(

Reduced cost for the new ring:

Ni Eji

iji yx ij),(

0

We want to find a ring with negative reduced cost

15

Generating Feasible Rings

• Let yij = 1 if the new ring contains edge (i,j), and 0, otherwise

• Let xi = 1 if the new ring requires an ADM at node i

• For the x’s and y’s to represent a feasible ring we need

.),(

,),(

,),(

Eji

Eji

xyxy

byd

jij

iij

Ejiijij

16

Ring Generation BIP

.),(

,),(

,s.t.

min

),(

),(

Eji

Eji

b

ij

xyxy

yd

yx

jij

iij

Ejiijij

Ni Ejiiji

• This problem is NP-hard, but in practical terms is much easier to solve than BIP(M*). • If the optimal value of the objective function is zero then the optimal basis for LP(M) is optimal for L(M*).

17

Column Generation Flow Chart

Solve LP(M)

Restore Integrality Constraints and Solve BIP(M)

New column found ?

Add column to M

Yes

No

Solve the Pricing Problem

18

Restricted LP Master Problem (LP(M)) for Iteration 1

min2z1+2z2+2z3+2z4+2z5+2z6+2z7+2z8+3z9+5z10s.tz1 + z9 + z10 = 1 (demand between 1 and 2)z2 + z9 + z10 = 1 (demand between 1 and 3)z3 + z10 = 1 (demand between 1 and 4)z4 + z9 + z10 = 1 (demand between 2 and 3)z5 = 1 (demand between 2 and 5)z6 = 1 (demand between 3 and 4)z7 + z10 = 1 (demand between 3 and 5)z8 + z10 = 1 (demand between 4 and 5)0 · zi · 1

19

Optimal Solution for LP(M): Iteration 1

910001100002

2

2

2

2

2

2

5

1010000000

1001000000

0000100000

0000010000

1100001000

1000000100

1100000010

11000000015322222222

45

35

34

25

23

14

13

12

czz

A

c

20

Pricing Problem for Iteration 1

1822832433

22222225

4535342523141312

4535342523141312

5

1

s.t.

min

yyyyyyyy

yyyyyyyyxi

i

,),(

,),(

Eji

Eji

xyxy

jij

iij

,1,0yij

.1,0xi

Optimal Solution:

Add ring 11 = {(1,3), (1, 4), (2, 3), (2, 5), (3, 5), (4,5)}

1453525231413

yyyyyy1

54321 xxxxx

Objective function value = -7

21


min2z1+2z2+2z3+2z4+2z5+2z6+2z7+2z8+3z9+5z10+5z11s.tz1 + z9 + z10 = 1 (demand between 1 and 2)z2 + z9 + z10 + z11 = 1 (demand between 1 and 3)z3 + z10 + z11 = 1 (demand between 1 and 4)z4 + z9 + z10 + z11 = 1 (demand between 2 and 3)z5 + z11 = 1 (demand between 2 and 5)z6 = 1 (demand between 3 and 4)z7 + z10 + z11 = 1 (demand between 3 and 5)z8 + z10 + z11 = 1 (demand between 4 and 5)0 · zi · 1

22


9010001100002

2

2

2

2

2

5

2

11010000000

11001000000

00000100000

10000010000

11100001000

11000000100

11100000010

0110000000155322222222

45

35

34

25

23

14

13

12

czz

A

c

23


1822832433

22222252

4535342523141312

4535342523141312

5

1

s.t.

min

yyyyyyyy

yyyyyyyyxi

i

,),(

,),(

Eji

Eji

xyxy

jij

iij

,1,0yij

.1,0xi

Optimal Solution:

Add ring 12 = {(1,2), (1, 4), (2, 3), (2, 5), (3, 5), (4,5)}

1453525231412

yyyyyy1

54321 xxxxx


24


min2z1+2z2+2z3+2z4+2z5+2z6+2z7+2z8+3z9+5z10+5z11+5z12s.tz1 + z9 + z10 + x12 = 1 (demandbetween1and2)z2 + z9 + z10 + z11 = 1 (demandbetween1and3)z3 + z10 + z11 + x12 = 1 (demandbetween1and4)z4 + z9 + z10 + z11 + x12 = 1 (demandbetween2and3)z5 + z11 + x12 = 1 (demandbetween2and5)z6 = 1 (demandbetween3and4)z7 + z10 + z11 + x12 = 1 (demandbetween3and5)z8 + z10 + z11 + x12 = 1 (demandbetween4and5)0· zi · 1

25


90010001100002

2

2

2

1

2

2

2

111010000000

111001000000

000000100000

110000010000

111100001000

111000000100

011100000010

101100000001555322222222

45

35

34

25

23

14

13

12

czz

A

c

26


1822832433

2222222

4535342523141312

4535342523141312

5

1

s.t.

min

yyyyyyyy

yyyyyyyyxi

i

,),(

,),(

Eji

Eji

xyxy

jij

iij

,1,0yij

.1,0xi

Optimal Solution:

14535342513

yyyyy1

54321 xxxxx

Add ring 13 = {(1, 3), (2, 5), (3, 4), (3, 5), (4,5)}


27


min2z1+2z2+2z3+2z4+2z5+2z6+2z7+2z8+3z9+5z10+5z11+5z12+5z13s.tz1 + z9 + z10 + x12 = 1 (demand between 1 and 2)z2 + z9 + z10 + z11 + x13 = 1 (demand between 1 and 3)z3 + z10 + z11 + x12 = 1 (demand between 1 and 4)z4 + z9 + z10 + z11 + x12 = 1 (demand between 2 and 3)z5 + z11 + x12 + x13 = 1 (demand between 2 and 5)z6 + x13 = 1 (demand between 3 and 4)z7 + z10 + z11 + x12 + x13 = 1 (demand between 3 and 5)z8 + z10 + z11 + x12 + x13 = 1 (demand between 4 and 5)0 · zi · 1

28


5.82

1

2

100

2

100

2

100

2

100

2

5.2

2

2

0

2

5.1

5.1

1111010000000

1111001000000

1000000100000

1110000010000

0111100001000

0111000000100

1011100000010

01011000000015555322222222

45

35

34

25

23

14

13

12

czz

A

c

29

Comments

• Repeat until optimal solution to the pricing problem has objective function value zero.

• In many cases, the pricing problem is NP-hard.• There is no guarantee that the column-generation

procedure will generate all of the columns that are selected in the optimal solution to BIP(M*)– A Branch-and-Price procedure does column

generation at each node of the branch-and-bound tree• The extra constraints added by branching usually complicate

the pricing the problem.

EMIS 8373: Integer Programming

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