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Re: [Help-glpk] Stock Cutting problem

From: Jeffrey Kantor
Subject: Re: [Help-glpk] Stock Cutting problem
Date: Sat, 9 Mar 2013 15:26:52 -0500

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 or 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.


# Stock Cutting Problem

# Product catalog
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];


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";


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;

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



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.





Nitin patel




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