Dantzig-Wolfe Decomposition

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Presentation transcript:

Dantzig-Wolfe Decomposition 1

Outline block structure of a problem representing a point by extreme points Revised Simplex to the extreme point representation an example 2

General Problem common in network-based problems min s.t. B1x1 = b1, B2x2 = b2, … BKxK = bK, 0  xk, k = 1, 2, …, K.   common in network-based problems distribution of K types of products Bkxk = bk: constraints related to the flow of the kth type of products constraints of common resources for the K products: 3

General Problem min s.t. B1x1 = b1, B2x2 = b2, … BKxK = bK, 0  xk, k = 1, 2, …, K.   4

Possible to avoid solving a large problem? General Problem min s.t. B1x1 = b1, B2x2 = b2, … BKxK = bK, 0  xk, k = 1, 2, …, K.   Possible to avoid solving a large problem? 5

A Numerical Example Problem P2: min –3x1 – 2x2 – 2x3 – 4x4, s.t. x1 + x2 + 2x3 + x4  10, x1 + 2x2  8, x2  3, x3 + 3x4  6, x3  4, xi  0. 6

A Numerical Example feasible region: {(x1, x2): x1 + 2x2  8, x2  3} extreme points: a feasible point: convex combination of the extreme points 7

A Numerical Example feasible region: {(x3, x4): x3 + 3x4  6, x3  4} extreme points: a feasible point: convex combination of the extreme points 8

a subspace Bkxk = bk is represented by a single constraint n kn = 1 A Numerical Example Problem P3: problem in terms of extreme points min -3(812+213) – 2(313+314) – 2(422+423) – 4( 23+224), s.t. (812+213) + (313+314) + 2(422+423) + ( 23+224)  10, 11 + 12 + 13 + 14 = 1, 21 + 22 + 23 + 24 = 1, ij  0, i = 1, 2; j = 1, 2, 3, 4. a subspace Bkxk = bk is represented by a single constraint n kn = 1 9

The previous representation suits DW Decomposition more. A Numerical Example Problem P3: problem in terms of extreme points min –2412 – 1213 – 614 – 822 – (32/3)23 – 824, s.t. 812 + 513 + 314 + 822 +8 23 +224  10, 11 + 12 + 13 + 14 = 1, 21 + 22 + 23 + 24 = 1, ij  0, i = 1, 2; j = 1, 2, 3, 4. The previous representation suits DW Decomposition more. 10

Practical? impractical approach impossible to generate all the extreme points of a block Bkxk = bk Dantzig-Wolfe Decomposition: check all extremely points without explicitly generating them 11

General Problem K blocks, Bkxk = bk, k = 1, …, K Nk extreme points in the kth block dual variable 0  dual variable 1  dual variable 2  … dual variable K  12

Reduced Cost for a Non-Basic Variable kn reduced cost of kn 13

Most Negative Reduced Cost where Check each extreme point for each block, which is equivalent to solving a linear program. Result: Solving K+1 small linear programs. 14

General Approach 1 Form the master program (MP) by the representation 2 Get a feasible solution of the MP; find the corresponding  3 Solve the subproblems to check the reduced costs of kn 3.1 stop if the MP is optimal; 3.2 else carry a standard revised simplex iteration (i.e., identifying the entering and leaving variables, stopping for an unbounded problem , and determining B-1 otherwise) 3.3 go back to 2 if the problem is not unbounded 15

An Example of the Dantzig-Wolfe Approach min –3y1 – 2y2 – 2y3 – 4y4, s.t. y1 + y2 + 2y3 + y4  10, y1 + 2y2  8, y2  3, y3 + 3y4  6, y3  4, yi  0. 16

An Example of the Dantzig-Wolfe Approach problem in general form with 17

An Example of the Dantzig-Wolfe Approach let be the extreme points of B1x1 = b1 any point in B1x1 = b1: similarly, any point in B2x2 = b2: 18

An Example of the Dantzig-Wolfe Approach in terms of the extreme points 19

An Example of the Dantzig-Wolfe Approach to solve the problem, introduce the slack variable x5 and artificial variables x6 and x7 20

An Example of the Dantzig-Wolfe Approach initial basic variable xB = (x5, x6, x7)T, B = I, (cB)T = (0, M, M), b = (10, 1, 1)T  = (cB)TB = (0, M, M) for the kth subproblem, the reduced costs: 21