lec6_column1

lec6_column1 - Column Generation Teo Chung-Piaw (NUS) 25th...

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Column Generation Teo Chung-Piaw (NUS) 25 th February 2003, Singapore
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1L e c t u r e 1.1 Outline Cutting Stock Problem Classical Integer Programming Formulation Set Covering Formulation Column Generation Approach Connection with Lagrangian Relaxation Computational issues 2 Cutting Stock Problem 2.1 Introduction 2.1.1 Example Slide 1 Slide 2 A paper company has a supply of large rolls of paper, each of width W . Slide 3 Customers demand n i rolls of width w i ( i =1 ,...,m ). ( w i W ) Example: Quantity Ordered n i Order Width (inches) w i 97 45 610 36 395 31 211 14 Slide 4 The demand can be met by slicing a large roll in a certain way, called a pattern . For example, a large roll of width 100 can be cut into 1
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± ± ± 4 rolls each of width 25, or 2 rolls each of width 35, with a waste of 30. 3 Solution Approach I 3.1 L. V. Kantorovich 3.1.1 Formulation Slide 5 (1939 Russian, 1960 English) “Mathematical Methods of Planning and Organ- ising Production” Management Science , 6 , 366-422. • K : Set of available rolls. y k : 1 if roll k is cut, 0 otherwise. x k : number of times item i is cut on roll k . i Objective: To minimize the number of rolls used to meet all the demand k min y k ∈K Slide 6 Constraints Total number of times item i is cut is not less than the demand. k x i n i k ∈K The width of a roll is at most W k w i x i Wy k i 2
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Slide 7 k min k ∈K y k s.t. k ∈K x i n i for a fixed item i, k Wy k i w i x i for fixed roll k, x k 0 , 0 y k 1 i Integrality constraints on all variabes 3.1.2 Quality of Solution Slide 8 Scenario I : N1 n i : uniform, between 1 and 100 (rand(100)+1); w i : uniform, between 1 and 30 (rand(30)+1); Width of Roll, W = 3000; Rolls Items constr variables CPU (s) 30 60 90 50 100 150 100 200 300 200 400 600 1830 2.8 5050 14.33 20100 179 80200 3048 Note 1 Code The OPL code for the problem: 3
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± ± ± Slide 9 Scenario II : Change width of the roll from 3000 to 150. Rolls Items constr variables CPU (s) 70 10 80 770 3.18 - never 140 20 160 2940 58.28 - never 210 30 240 6510 Out of memory The performance has deteriorated. Why? Slide 10 How good is the LP relaxation? w i n i Observation : Z LP = i W . N2 This bound is trivial: The objective is to k min y k ∈K the optimal solution will satisfy: Choose y k as small as possible. Therefore k w i x i = Wy k i for all k Slide 11 k Choose x i as small as possible. Therefore k x i = n i k 4
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± ± ± ± ± ± ± ± for all i . ± y k = ± i w i x k i W = ± ± w i W x k i k ∈K k ∈K i k ∈K = ± w i W ± x k i = ± w i n i W i k ∈K i Note 2 Proof We have the constraints k w i x i Wy k . i In the LP, since the objective is to minimize k y k , the optimal LP solution will be such that k i w i x i = y k . W k y k will be small if the x i values are small. At the same time, because k x i n i , k ∈K k to make x i small, at the optimal solution, we must have k x i = n i .
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This note was uploaded on 12/04/2011 for the course ESD 15.094 taught by Professor Jiesun during the Spring '04 term at MIT.

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lec6_column1 - Column Generation Teo Chung-Piaw (NUS) 25th...

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