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Diffstat (limited to 'code/intfmincon')
| -rw-r--r-- | code/intfmincon/ProcessSel.sce | 148 |
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diff --git a/code/intfmincon/ProcessSel.sce b/code/intfmincon/ProcessSel.sce new file mode 100644 index 0000000..d2efefb --- /dev/null +++ b/code/intfmincon/ProcessSel.sce @@ -0,0 +1,148 @@ +//This is an MINLP example.The solution to the problem is Y1=1,Y2=0;Y3=1;C1=1;B1=1.111;B2=0;B3=1.111;BP=0;A2=0;A3=1.52; +//Netprofit = 1.923 (10^3$/hr)) +//Ref:Optimization of chemical processes, second edition. By Thomas F. Edgar, David M. Himmelblau, and Leon S. Lasdon, McGraw Hill, New York, 2001, Chapter 9 +//The manufacture of a chemical C in process 1 that uses raw material B.B can either be purchased or +//manufactured via two processes, 2 or 3, both of which use chemical A as a raw mate-rial.Data and +//specifications for this example problem, involving several nonlinear input4utput relations (mass balances), +//are shown in Table.We want to deter-mine which processes to use and their production levels in order to +//maximize profit.The processes represent design alternatives that have not yet been built. Their fixed +//costs include amortized design and construction costs over their anticipated lifetime,which are incurred +//only if the process is used +// +// A2------Process2-------------B2 +// A1------Process3-------------B3 +// B2+B3+BP---------------------B1 +// B1------Process1-------------C1 +// +//Problem Data +//Conversions Process 1 C = 0.9B +// Process 2 B = ln(1+A) +// Process 3 B = 1.2ln(1+A) (A,B,C in tons) +// +//Maximum Capacity Process 1 2 ton/h of C +// Process 1 4 ton/h of B +// Process 1 5 ton/h of B +// +//Price A: $1800/ton +// B: $7000/ton +// C: $13000/ton +// +//Demand of C: 1 ton/h maximum +//Costs: +//--------------------------------------------------------------------------------- +// Fixed(10^3 $/hr) variable (10^3 $/ton of product) +//Process 1 3.5 2 +//Process 1 1 1 +//Process 1 1.5 1.2 +//--------------------------------------------------------------------------------- +////====================================================================== +// 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; +//Objective function to calculate profit +function costval = ProcessSel(x) + Y1 = x(1);Y2 = x(2);Y3 = x(3); + C1 = x(4);B1 = x(5);B2 = x(6); + B3 = x(7);BP = x(8);A2 = x(9);A3 = x(10); + + Income = 13*C1; + Purchase = 7 * BP; + Chemical = 1.8*A2 + 1.8*A3; + Invest = 3.5*Y1 + 2*C1 + Y2 + B2 + 1.5*Y3 + 1.2*B3; + + costval = -(Income - (Purchase + Chemical + Invest)); +endfunction + +//Nonlinear constraint function +function [C,Ceq] = Nonlincon(x) + Y1 = x(1);Y2 = x(2);Y3 = x(3); + C1 = x(4);B1 = x(5);B2 = x(6); + B3 = x(7);BP = x(8);A2 = x(9);A3 = x(10); + //No Non-linear inequality constraint + C = []; + Ceq(1) = C1-0.9*B1; + Ceq(2) = B2-log(1+A2); + Ceq(3) = B3 - 1.2*log(1+A3); + +endfunction + +// Decision vector structure +// Y1 = x(1);Y2 = x(2);Y3 = x(3);C1 = x(4);B1 = x(5);B2 = x(6);B3 = x(7);BP = x(8);A2 = x(9);A3 = x(10); + +//Inequality constraints +A = [0 -4 0 0 0 1 0 0 0 0 +0 0 -5 0 0 0 1 0 0 0 +-2 0 0 1 0 0 0 0 0 0]; + +b = zeros(3,1); + +//Equality Constraint +Aeq = [0 0 0 0 1 -1 -1 -1 0 0;]; +beq = 0; +//Number of variables +Var = 10; +//Capacity Constraints are used for deciding upper bounds of variables +C1max = 1; +B2max = 4; +B3max = 5; + +A2max = exp(B2max)-1; +A3max = exp(B3max/1.2)-1; +B1max = C1max/0.9; +BPmax = B1max; + +//Bounds on variables +lb = zeros(1,Var); +// Important to give proper UB as otherwise solution might become sub optimal +//users can try ub = [1 1 1 1 %inf %inf %inf %inf %inf %inf]; + +ub = [1 1 1 C1max B1max B2max B3max BPmax A2max A3max]; +//Initial guess +x0 = lb; +//Integer ariables +intcon = [1 2 3]; +mprintf('The decision variables and their bounds are") + +var = ['Y1','Y2','Y3','C1','B1','B2','B3','BP','A2','A3']; +Table = [['Variables','lower Bound','Upper Bound'];[var' string(lb') string(ub')]]; +disp(Table) + +input("Press enter to solve the problem"); +clc +//Using intfmin for solving MINLP +mprintf("Scilab is solving the problem") +[xopt,fopt,exitflag,gradient,hessian]= intfmincon(ProcessSel,x0,intcon,A,b,Aeq,beq,lb,ub,Nonlincon) +clc; +Y1 = xopt(1);Y2 = xopt(2);Y3 = xopt(3);C1 = xopt(4);B1 = xopt(5);B2 = xopt(6); +B3 = xopt(7);BP = xopt(8);A2 = xopt(9);A3 = xopt(10); + +select exitflag +case 0 + mprintf("Optimal Solution Found"); + mprintf("\n Net profit is %f $ \n Net production is : %f ton/hr",-fopt*1000,C1) + mprintf("\n The following production process should be followed \n") + mprintf("\n Amount of raw material A2: %f ton/hr \n Amount of raw material A3: %f ton/hr \n amount of B1 purchased : %f ton/hr",A2,A3,B1) + +case 1 + mprintf("InFeasible Solution."); +case 2 + mprintf("Objective Function is Continuous Unbounded."); +case 3 + mprintf("Limit Exceeded."); +case 4 + mprintf("User Interrupt"); +case 5 + mprintf("MINLP Error") +end + + + |