MIE376 Lec8 Column Generation

MIE376 Lec8 Column Generation - MIE376 Mathematical...

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MIE376 Mathematical Programming Lecture Notes Daniel Frances 2011 1 Lecture 3: Cutting Stock Problem with Column Generation This is a classical problem that leads to a totally new expansion of the capabilities of linear programming. Suppose you sell boards of lumber to the retail market in lengths of 3 m., 5 m. and 9 m. to meet a demand of 25 units of 3 m. boards 20 units of 5 m. boards 15 units of 9 m. Length Suppose also that you can only buy boards in the wholesale market in 17 m lengths. How many 17 m lengths do we need and how should we cut them to meet the demand at minimum waste. Unfortunately we cannot simply start to define decision variables to formulate this problem into a mathematical model. We need some preliminary analysis. It turns out that one of the issues we have to deal with, is to define in advance all the possible ways that we can cut a 17 m board into one or more of the target lengths. This is best done methodically by using a table of all possible cuts: Cut Type 3 m 5 m 9 m waste 1 5 0 0 2 2 4 1 0 0 3 2 2 0 1 4 2 0 1 2 5 1 1 1 0 6 0 3 0 2 Note that we have omitted any cut which results in a waste ≥ 3 m. Now the problem can be formulated as follows: Let x i be the number of boards cut according to cut type i. and z the overall waste. Then we could formulate the following LP problem: Min z = 2x 1 + x 3 + 2x 4 + 2x 6 subject to 5x 1 + 4x 2 + 2x 3 + 2x 4 + x 5 ≥ 25 x 2 +2x 3 +x 5 + 3x 6 ≥ 20
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MIE376 Mathematical Programming Lecture Notes Daniel Frances 2011 2 x 4 + x 5 ≥ 15 All x i ≥ 0 While it would seem that we would need to restrict the variables to integers, in practice rounding off works quite well for this type of problem, we are done, or are we? Two remaining problems: Is the waste expression correct? If we exceed demand is that not also waste? What if instead of 17 m boards, we were dealing with 107 m wire lengths to be cut into 3, 5 and 9 m lengths – the number of possible cuts would render this method impractical. A 2 nd look at the objective In addition to the waste that we cannot sell because there is no demand for the “wrong” size, we also need to account for the waste that we cannot sell because we are producing too many units of the “right” size. The new objective which includes both types of waste becomes z = 2x 1 + x 3 + 2x 4 + 2x 6 + 3*(5x 1 + 4x 2 + 2x 3 + 2x 4 + x 5 – 25) + 5*(x 2 + 2x 3 + x 5 + 3x 6 – 20) + 9*(x 4 + x 5 – 15) = 17x 1 + 17x 2 + 17x 3 + 17x 4 + 17x 5 + 17x 6 – 310 Clearly this objective can be replaced by the simpler z = x 1 + x 2 + x 3 + x 4 + x 5 + x 6 minimize the total number of 17 foot lengths cut. Thus the new LP becomes Min z = x 1 + x 2 + x 3 + x 4 + x 5 + x 6 subject to 5x 1 + 4x 2 + 2x 3 + 2x 4 + x 5 ≥ 25 x 2 +2x 3 +x 5 + 3x 6 ≥ 20 x 4 + x 5 ≥ 15 All x i ≥ 0
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MIE376 Mathematical Programming Lecture Notes Daniel Frances 2011 3 What to do when the number of cut types is too large to be listed This is where it gets interesting! Thus suppose that the material comes in 107 m lengths
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This note was uploaded on 04/20/2011 for the course MIE 376 taught by Professor Daniel during the Spring '11 term at University of Toronto.

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MIE376 Lec8 Column Generation - MIE376 Mathematical...

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