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diff --git a/code/Symphonymat/Factory planing.sce b/code/Symphonymat/Factory planing.sce new file mode 100644 index 0000000..e140f42 --- /dev/null +++ b/code/Symphonymat/Factory planing.sce @@ -0,0 +1,143 @@ +//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) + + |