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//A practical problem of factory planning which determines the optimum product
// mix subject to production capacity and marketing limitation.
//This example shows how to use spreadsheet data directly in scilab.
//Ref: H. Paul Williams ,Model Building in Mathematical Programming ,
//A John Wiley & Sons, Ltd., Publication, Fifth Ed, Chapter 12.
//
//Example: An engineering factory makes seven products (PROD 1 to PROD 7) on the
//following machines: four grinders, two vertical drills, three horizontal drills, one
//borer and one planer. Each product yields a certain contribution to profit (defined
//as £/unit selling price minus cost of raw materials). These quantities (in £/unit)
//together with the unit production times (hours) required on each process are given
//below. A dash indicates that a product does not require a process.
//---------------------------------------------------------------------------------------
// Prod 1 Prod 2 Prod 3 Prod 4 Prod 5 Prod 6 Prod 7
//
//Contribution to profit 10 6 8 4 11 9 3
//Grinding 0.5 0.7 - - 0.3 0.2 0.5
//Vertical drilling 0.1 0.2 - 0.3 - 0.6 -
//Horizontal drilling 0.2 - 0.8 - - - 0.6
//Boring 0.05 0.03 - 0.07 0.1 - 0.08
//Planing - - 0.01 - 0.05 - 0.05
//--------------------------------------------------------------------------------------
//In the present month (January) and the five subsequent months, certain
//machines will be down for maintenance. These machines will be as follows:
//--------------------------------------------
//January 1 Grinder
//February 2 Horizontal drills
//March 1 Borer
//April 1 Vertical drill
//May 1 Grinder and 1 Vertical drill
//June 1 Planer and 1 Horizontal drill
//--------------------------------------------
//
//There are marketing limitations on each product in each month. These are
//given in the following table:
//-------------------------------------------
// 1 2 3 4 5 6 7
//-------------------------------------------
//January 500 1000 300 300 800 200 100
//February 600 500 200 0 400 300 150
//March 300 600 0 0 500 400 100
//April 200 300 400 500 200 0 100
//May 0 100 500 100 1000 300 0
//June 500 500 100 300 1100 500 60
//-------------------------------------------
//
//The factory works at six days a week with two shifts of 8 h each day.
//When and what should the factory make in order to maximise the total profit?
//This problem considers single period only with no storage
// Copyright (C) 2018 - IIT Bombay - FOSSEE
// This file must be used under the terms of the CeCILL.
// This source file is licensed as described in the file COPYING, which
// you should have received as part of this distribution. The terms
// are also available at
// http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt
// Author:Debasis Maharana
// Organization: FOSSEE, IIT Bombay
// Email: toolbox@scilab.in
//=================================================================================
clc
filepath = 'C:\Users\Iball\Desktop\scilab problems\Date 21-05-2018-Debasis\factory.xls';
[fd,SST,Sheetnames,Sheetpos] = xls_open(filepath);
S = readxls(filepath);
[Contributation,TextInd] = xls_read(fd,Sheetpos(Sheetnames == 'Contributation')); //Sheetpos gives the position of all the sheets
[Available,TextInd] = xls_read(fd,Sheetpos(Sheetnames == 'Available'));
[Failure,TextInd] = xls_read(fd,Sheetpos(Sheetnames == 'Machine Failure'));
[Limitation,TextInd] = xls_read(fd,Sheetpos(Sheetnames == 'Limitation'));
mclose(fd) //close the file
Sheetdata = readxls(filepath);
mprintf('Problem Data received\n')
for i = 1:4
disp(Sheetnames(i))
disp(Sheetdata(i))
end
input('press enter to continue')
clc
mprintf('Scilab is solving your problem')
Shift_T = 8;N_shift = 2;N_Workingdys = 24;//Data is fixed according to example. However it can be made as user defined parameter
Wrk_hrs = Shift_T*N_shift*N_Workingdys;
[Nmonth,Nprod] = size(Limitation);
Nprod = Nprod - 1;
Nmonth = Nmonth - 1;
Ntype = size(Available,'r');
Available = Available(:,2);
Contributation = Contributation(2:Ntype+2,2:Nprod+1);// Extracting the data matrix only
Limitation = Limitation(2:Nmonth+1,2:Nprod+1);
Failure = Failure(2:Nmonth+1,2:Ntype+1);
Failure(isnan(Failure))=0;
ub = [];
for i = 1:Ntype
Amonth(i,:) = Contributation(i+1,:);
breq(i) = Available(i)*Wrk_hrs;
end
b = [];A = zeros(Nmonth*Ntype,Nmonth*Nprod);C = [];lb = zeros(1,Nmonth*Nprod);
// Instead of seperate loops, we can use a single loop for determining all the possible variables
for i = 1:Nmonth
A((i-1)*Ntype+1:i*Ntype,(i-1)*Nprod+1:i*Nprod) = Amonth;
B_month = breq - (Failure(i,:))'*Wrk_hrs;
b = [b;B_month];
C = [C;-Contributation(1,:)'];
ub = [ub Limitation(i,:)];
end
intcon = 1:Nmonth*Nprod; // all production units are integers
options = list("time_limit", 2500);
[xopt,fopt,status,output] = symphonymat(C,intcon,A,b,[],[],lb,ub,options);
clc
select status
case 227 then
mprintf('Optimal Solution Found')
case 228 then
mprintf('Maximum CPU Time exceeded')
case 229 then
mprintf('Maximum Number of Node Limit Exceeded')
else
mprintf('Maximum Number of Iterations Limit Exceeded')
end
input('Press enter to view results')
// Solution representation
A2 = ['January','Februry','March','April','May','June']';// users can modify it to accept from the excel sheet
A1 = [" ", 'Prod1','Prod2','Prod3','Prod4','Prod5','Prod6','Prod7'];
for i = 1:Nmonth
solution(i,:) = string(xopt((i-1)*Nprod+1:i*Nprod)');
end
mprintf('Production schedule for all the months')
table = [A1;[A2 solution]];
disp(table)
mprintf('The profit is %d',-fopt)
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