Re: [Help-glpk] Stock Cutting problem

[Jeffrey Kantor]

Niitin,

The simple MathProg model I developed for your small problem will never work for your full scale application.  Producing 200 product pieces from 500 stock pieces would correspond to 200 x 500 = 10,000 binary variables. Given the large number of symmetries in the solution to the problem, that's not a feasible approach.

Column generation is a standard approach to these problems. Rather than directly assign product pieces to stock pieces, in column generation you consider a set of cutting patterns.  Each cutting pattern breaks a stock piece into known product pieces. The optimization problem is to then determine an optimal number of times to use each pattern.  Each pattern corresponds to a column in an integer LP.  The trick is to start with a base set of patterns then generate new patterns on the fly  (hence 'column generation') to solve your specific problem.

Unfortunately, MathProg doesn't provide a means to add columns on the fly.  Therefore you'll need to look into use of the glpk api.  It's not as bad as it sounds, and you can into the cspsol for a head start.  Also google 'cutting stock column generation' for some excellent resources.

Jeff

On Sun, Mar 10, 2013 at 5:34 AM, Nitin Patel <address@hidden> wrote:

Thank you Mr Jeff

 

Thanks for quick replay. Actual problem has more stocks (10 to 15 different lengths with 500 quantities) and demands (10 to 15 different lengths with 200 quantities).

I am new to GLPK-Mathprog . It will take some time to learn things. I know VB / Excel VBA.

 

I do not understand matter you have mention in second paragraph. It is my fault as I am new to this subject. (OR & GLPK)

 

AS per you “Larger problems would require more care. In particular, stock cutting is often solved using a column generation.” Can you provide some more information for the same?

For below small data it took 413 Seconds. Can we reduce time using “column generation”?

 

 

param : PRODUCTS : pLength demand :=

    '110m'   110   1

    '107m'   107   6

    '105m'   105   4

    '103m'   103   3

    '100m'   100   2

    '96m'     96   4

    '94m'     94   1

    '91m'     91   3

    '86m'     86   3

    '78m'     78   2

    '76m'     76   2

    '69m'     69   5;

 

 

param : RAW : rLength avail :=

    '280m'  280 14;

 

 

Thanks

 

Niitn Patel

 

 

 

 

From: Jeffrey Kantor [mailto:address@hidden]
Sent: Sunday, March 10, 2013 1:57 AM
To: address@hidden
Cc: GLPK
Subject: Re: [Help-glpk] Stock Cutting problem

 

This is a relatively small scale problem so it can be solved as the assignment of product pieces to stock pieces.  Appended below is a MathProg solution which you can cut and paste into http://MathP.org or http://www.nd.edu/~jeff/mathprog/mathprog.html for testing.  This solution uses indexed sets to enumerate the individual product and stock pieces of materials.

 

For the given data is no waste piece could be less than 1 meter, so it isn't necessary to consider the issue of minimizing small scrap. That could be handled with an additional binary variable for each piece of raw material, but wasn't necessary given the problem data. Another aspect is that minimizing the number of pieces cut leaves a lot of solution symmetries. The computation is faster by introducing weights, which has the nice side effect of producing a 'no-waste' solution for the given problem data.

 

Larger problems would require more care. In particular, stock cutting is often solved using a column generation.

 

Jeff

 

 

# Stock Cutting Problem

 

# Product catalog

set PRODUCTS;

param pLength{PRODUCTS};

param demand{PRODUCTS};

 

# Raw Materials

set RAW;

param rLength{RAW};

param avail{RAW};

 

# Set of production pieces indexed by products

set Q{p in PRODUCTS} := 1..demand[p] ;

 

# Set of stock pieces indexed by raw materials

set S{r in RAW} := 1..avail[r];

 

# Cutting assignments

var y{p in PRODUCTS, q in Q[p], r in RAW, s in S[r]} binary;

 

# Indicator if an item of raw material is used

var u{r in RAW, s in S[r]} binary;

 

# Length of waste from each piece of raw material

var w{r in RAW, s in S[r]} >= 0;

 

# Cut each product piece only once

s.t. A{p in PRODUCTS, q in Q[p]} : sum{r in RAW, s in S[r]} y[p,q,r,s] = 1;

 

# For each product, cut enough pieces to exactly meet demand

s.t. B{p in PRODUCTS} : sum{q in Q[p], r in RAW, s in S[r]} y[p,q,r,s] = demand[p];

 

# For each piece of raw material, do not exceed length

s.t. C{r in RAW, s in S[r]} : 

    sum{p in PRODUCTS, q in Q[p]} pLength[p]*y[p,q,r,s] + w[r,s] = rLength[r];

    

# Determine if a piece of raw material is used.

s.t. D{r in RAW, s in S[r]} : 15*u[r,s] >= sum{p in PRODUCTS, q in Q[p]} y[p,q,r,s];

 

# Minimize number of bars, with weights to favoring long pieces

minimize NumberOfBars: sum{r in RAW, s in S[r]} rLength[r]*u[r,s];

 

solve;

 

printf "Cutting Plan\n";

for {r in RAW} : {

    printf "    Raw Material Type %s \n", r;

    for {s in S[r]} : {

        printf "        Piece %g : Remainder = %2g : Cut products ", s, w[r,s];

        for {p in PRODUCTS} : {

            for {q in Q[p] : y[p,q,r,s]} : {

                printf "%s ", p;

            }

        }

        printf "\n";

    }

    printf "\n";

}

 

printf "Production Plan\n";

for {p in PRODUCTS} : {

    printf "    Product %s \n", p;

    for {q in Q[p]} : {

        printf "        Piece %g : Cut from stock ", q;

        for {r in RAW} : {

            for {s in S[r] : y[p,q,r,s]} : {

                printf "%s ", r;

            }

        }

        printf "\n";

    }

    printf "\n";

}

 

data;

 

param : PRODUCTS : pLength demand :=

    '7m'   7   3

    '6m'   6   2

    '4m'   4   6

    '3m'   3   1 ;

  

param : RAW : rLength avail :=

    '15m'  15  3

    '10m'  10  3;

  

end;

 

 

On Sat, Mar 9, 2013 at 12:07 PM, Nitin Patel <address@hidden> wrote:

How to solve stock cutting problem with multiple length stock available and demand is under.

 

Stock length available---

15 m – 3 Numbers

10 m – 3 Numbers

 

Demand---

7m – 3 Numbers

6m – 2 Numbers

4m – 6 Numbers

3m – 1 Number

 

How to solve it so that minimize wastage

 

We should utilize minimum number of stocks

If unused length is more than 0.5 m then we can utilize it for next cutting schedule and it is not wastage.

 

 

Thanks

 

Nitin patel

 

 

 

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