% MATH 4/5794: Optimization Modeling

% EXAMPLES: H. Greenberg - Example 1, Capital Investment

% Example 2, Production

% Example 3, Network Flow

% Example 4, Transportation

% Spring Semester 2005

function [x,y,z] = classdemo1(exnum)

switch exnum

case 1

% Example 1 - H. Greenberg Capital Investment

A = zeros(1,5);

% Matrix A

A(1,:) = [1 1 1 1 1];

% Left and right endpoint of the extremes of fuzzy right-hand sides

b(1) = 100000;

% Objective function coefficients - since MATLAB minimizes, need negative to maximize

c = [-1;-2;-3;-4;-5];

for i=1:5

% lower bound, upper bound

vlb(i,1) = 0;

vub(i,1) = 100000;

end

plusminus = -1;

Aeq = [];

beq = [];

%

case 2

% Example 2 - H. Greenberg Production

A = zeros(7,4);

% Matrix A - MATLAB's inequalities are all <= for for >= use negative

A(1,:) = [-1 0 0 0];

A(2,:) = [-1 -1 0 0];

A(3,:) = [-1 -1 -1 0];

A(4,:) = [-1 -1 -1 -1];

A(5,:) = [ 1 -1 0 0];

A(6,:) = [ 0 1 -1 0];

A(7,:) = [ 0 0 1 -1];

% Left and right endpoint of the extremes of fuzzy right-hand sides

b(1,1) = -5;

b(2,1) = -11;

b(3,1) = -19;

b(4,1) = -25;

b(5,1) = 0;

b(6,1) = 0;

b(7,1) = 0;

% Cost coefficients, since minimizing with matlab, need no sign change

c = [3;7;4;7];

for i=1:4

% starting point

vlb(i,1) = 0;

end

plusminus = 1;

vub = [];

Aeq = [];

beq = [];

%

case 3

% Example 3 - H. Greenberg Network Flow

Aeq = zeros(3,6);

% Matrix Aeq - MATLAB's equalities

Aeq(1,:) = [ 1 0 -1 -1 0 0];

Aeq(2,:) = [ 0 1 1 0 -1 0];

Aeq(3,:) = [ 0 0 0 1 1 -1];

% Left and right endpoint of the extremes of fuzzy right-hand sides

beq(1,1) = 0;

beq(2,1) = 0;

beq(3,1) = 0;

% Cost coefficients, since minimizing with matlab, need no sign change

c = [0;0;0;0;0;-1];

for i=1:6

% starting point

vlb(i,1) = 0;

end

% Capacities along the arcs

vub(1,1) = 4;

vub(2,1) = 3;

vub(3,1) = 1;

vub(4,1) = 2;

vub(5,1) = 3;

vub(6,1) = 100;

plusminus = -1;

A = [];

b = [];

case 4

% Example 4 - H. Greenberg Tranportation with integer variables

case 5

% Example 5 - H. Greenberg Diet Problem

A = zeros(4,3);

% Matrix A - MATLAB's inequalities are all <= for for >= use negative

A(1,:) = [ -2 -3 -7];

A(2,:) = [ -1 -1 0];

A(3,:) = [ -5 -3 0];

A(4,:) = [-0.6 -0.25 -1];

% Left and right endpoint of the extremes of fuzzy right-hand sides

b(1,1) = -1250;

b(2,1) = -250;

b(3,1) = -900;

b(4,1) = -232.5;

% Cost coefficients, since minimizing with matlab, need no sign change

c = [41;35;96];

for i=1:3

% starting point

vlb(i,1) = 0;

end

% Capacities along the arcs

plusminus = 1;

Aeq = [];

beq = [];

vub = [];

%

otherwise

out = 'exnum not valid'

exnum

end % end case of exnum

% Call the lp solver

x = linprog(c,A,b,Aeq,beq,vlb,vub);

soln = 'The solution is: ';

soln

x

% Objective function value

obj = 'Objective function value is: ';

obj

z = plusminus*(c'*x)

% Constraint values

if exnum == 3

A = Aeq;

b = beq;

end %end if

cons = 'The excess on the constraints at the solution are: ';

cons

y = b - A*x

out = 'solution is:';