//Design a circular tank, closed at both ends, with a volume of 200 m3.The cost is proportional to the surface area of material, which is priced
//at $400/m2. The tank is contained within a shed with a sloping roof,thus the height of the tank h is limited by h ≤ 12 − d/2, where d is
//the tank diameter. Formulate the minimum cost problem and solve the design problem.
//Ref:Diwekar, Urmila,Introduction to Applied Optimization, Introduction to Applied Optimization, Editor:Ding-Zhu Du, Springer Optimization and Its Applications Springer Optimization and Its Applications, VOL 22, Chapter 3

// 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;

function cost = Tankprob(x)
    r = x(1);h = x(2);
    cost = (2*%pi*r*h+2*%pi*r^2)*400;
endfunction

function [ceq,c] = Nonlinearcon(x)
    r = x(1);h = x(2);
    c = [];
    ceq = 200-%pi*r^2*h;
endfunction

function y=Gradobj(x)
    y= [800*%pi*x(2) + 1600*%pi*x(1),800*%pi*x(1)];
endfunction

mprintf('\nDesign a circular tank, closed at both ends, with a volume of 200 m3 with minimum cost.\n The tank is contained within a shed with a sloping roof,thus the height of the tank h is limited')
mprintf('\nCost of material is: %f',400);
mprintf('The design variables are radius and height of the tank')

A = [1 1];b = 12;

x0 = input('Enter initial guess as vector:');
if  (sum(x0<=0) | (length(x0)~=2))
    x0 = [1 2];
    mprintf('Incorrect initial guess...\n changing initial guess to r = %d and h = %d',x0(1),x0(2));
end


lb = [0 0];
input('press enter to continue')
options=list("MaxIter",1000,"GradObj", Gradobj);

[xopt,fopt,exitflag,output1] = fmincon(Tankprob,x0,A,b,[],[],lb,[],Nonlinearcon,options);

[xopt1,fopt1,exitflag1,output2] = fmincon(Tankprob,x0,A,b,[],[],lb,[],Nonlinearcon);

clc
select exitflag
case 0
    mprintf('Optimal Solution Found');
    mprintf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
    mprintf('\nTime taken to solve the problem with gradient information is %f s and without gradient information is %f s',output1.Cpu_Time,output2.Cpu_Time);
case 1
    mprintf('Maximum Number of Iterations Exceeded. Output may not be optimal');
    input('press enter to view results');
    printf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
case 2
    mprintf('Maximum amount of CPU Time exceeded. Output may not be optimal.');
    input('press enter to view results');
    printf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
case 3
    mprintf('Stop at Tiny Step');
    input('press enter to view results');
    printf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
case 4
    mprintf('Solved To Acceptable Level');
    input('press enter to view results');
    printf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
case 5
    mprintf('Converged to a point of local infeasibility.');
    input('press enter to view results');
    printf('\nThe optimal radius and height of the tank are %f m and %f m',xopt(1),xopt(2));
    mprintf('\nThe volume of the tank is %f m^3',%pi*xopt(1)^2*xopt(2));
    mprintf('\nThe total surface area and cost of the tank are %f m^2 and %f $',fopt/400,fopt)
end