diff options
Diffstat (limited to 'code/linprog/Blending_problem.sce')
| -rw-r--r-- | code/linprog/Blending_problem.sce | 135 |
1 files changed, 135 insertions, 0 deletions
diff --git a/code/linprog/Blending_problem.sce b/code/linprog/Blending_problem.sce new file mode 100644 index 0000000..24b80ad --- /dev/null +++ b/code/linprog/Blending_problem.sce @@ -0,0 +1,135 @@ +// A practcal problem of blending of different products to acheive designed performance levels. +// This is an example from the class of allocation problems. +// Ref:H.A. Eiselt and C.-L.SAndblom,"Linear Programming and its Applications",Springer-Verlag Berlin Heidelberg 2007,chapter 2.7 +//Example: +//A firm faces the problem of blending three raw materials into two final +//products. The required numerical information is provided in Table. +// +//----------------------------------------------------------------------- +// Raw Products Amount available Unit Cost +//Materials 1 2 (in tons) ($) +//----------------------------------------------------------------------- +// 1 [0.4;0.6][0.5;0.6] 2000 1 +// 2 [0.1;0.2][0.1;0.4] 1000 1.5 +// 3 [0.2;0.5][0.2;0.3] 500 3 +//----------------------------------------------------------------------- +//quantity required 600 700 +//(in tons) +//----------------------------------------------------------------------- +//unit selling price 10 8 +//($) +//----------------------------------------------------------------------- +//we assume that raw materials blend linearly, meaning that taking, +//say, α units of raw material A and β units of raw material B, then +//the resulting blend C has features that are proportional to the +//quantities of A and B that C is made of. As an example, take 3 gallons +//of 80º water and 2 gallons of 100º water,then the result would be 5 +//gallons of water, whose temperature is [3(80) + 2(100)]/5 = 88º. +//The parameters max and min indicate the smallest and largest proportion +//of rawmaterial that is allowed in the final product.Determine the +//optimal transportation plan to maximize profit. + +// 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; +Nprod = 2; +Nraw = 3; +demand = [600 700]; +cost = [10 8]; +con(1,:,:) = [0.4 0.6;0.1 0.2;0.2 0.5]; +con(2,:,:) = [0.5 0.6;0.1 0.4;;0.2 0.3]; +raw = [2000 1000 500]; +RawC = [1 1.5 3]; +mprintf('Received Data') +mprintf('Number of products %d',Nprod); +mprintf('number of raw materials %d',Nraw); +disp(demand,'Demand of products'); +disp(cost,'Unit cost of products'); + +product = [];Min_Max = []; +for i = 1:Nprod + product = [product,['Min-Max',string(i)]]; + for j = 1:Nraw + MM(j,:) = con(i,j,:); + end + Min_Max = [Min_Max string(MM)]; +end +Row1 = ['Raw Material',product,'Amount_avail','unit_Cost'] +RawMaterial = string([1:Nraw]'); +Amount_avail = string(raw(:)); +unit_C = string(RawC(:)); +Table = [Row1;[RawMaterial Min_Max Amount_avail unit_C]]; +disp(Table); +input('Press Enter to proceed ') +A1 = zeros(Nraw,Nraw*Nprod); +for i = 1:Nraw + A1(i,i:Nraw:Nraw*Nprod)=1; + b1(i) = raw(i); +end +A2 = zeros(Nprod*Nraw*2,Nprod*Nraw);b2 = zeros(Nprod*Nraw*2,1); +A3 = zeros(Nprod,Nprod*Nraw); +for i = 1:Nprod + Areq = ones(2*Nraw,Nraw); + Areq(Nraw+1:2*Nraw,:) = -1; + for j =1:Nraw + Areq(j,:) = con(i,j,1)* Areq(j,:); + Areq(j,j) = Areq(j,j)-1; + Areq(Nraw+j,:) = Areq(Nraw+j,:)*con(i,j,2); + Areq(Nraw+j,j) = 1 + Areq(Nraw+j,j); + end + A2((i-1)*2*Nraw+1:i*2*Nraw,(i-1)*Nraw+1:i*Nraw) = Areq; + A3(i,(i-1)*Nraw+1:i*Nraw)= ones(1,Nraw); + b3(i) = demand(i); +end + +A = [A1;A2];b = [b1;b2]; +Aeq = A3;beq = b3; +C = []; +for i = 1:Nprod + C1 = -cost(i)*ones(1,Nraw) + RawC; + C = [C ;C1']; +end + +lb = zeros(1,Nprod*Nraw); +intcon = 1:Nprod*Nraw; +[xopt,fopt,exitflag,output,lambda] = linprog(C,A,b,Aeq,beq,lb,[]); +clc +select exitflag +case 0 then + mprintf('Optimal Solution Found') + input('Press enter to view results') + //Display results + mprintf('Total profit is %d\n',-fopt); + + A1 = [];product = []; + for i=1:Nprod + Raw_M(:,i) = xopt((i-1)*Nraw+1:i*Nraw) ; + product(i) = string(sum(Raw_M(:,i))); + A1 = [A1 strcat(['Product' string(i)])]; + end + + table = [['RawMaterial', A1];[RawMaterial string(Raw_M)];['Total Product' product']]; + disp(table); +case 1 then + mprintf('Primal Infeasible') +case 2 then + mprintf('Dual Infeasible') +case 3 + mprintf('Maximum Number of Iterations Exceeded. Output may not be optimal') +case 4 + mprintf('Solution Abandoned') +case 5 + mprintf('Primal objective limit reached') +else + mprintf('Dual objective limit reached') +end + + |