Please go through the code I have tried to modify the neccessary:
function simplex2p(type, c, A, rel, b)
% The Two-Phase Method for solving the LP problem
% min(or max) z = c*x
% Subject to Ax rel b
% x >= 0
% The input parameter type holds information about type of the LP
% problem to be solved. For the minimization problem type = ‘min’ and
% for the maximization problem type = ‘max’.
% The input parameter rel is a string holding the relation signs.
% For instance, if rel = ‘‘, then the constraint system consists
% of one inequality =.
clc
if (type == ‘min’)
mm = 0;
else
mm = 1;
c = -c;
end
b = b(:);
c = c(:)’;
[m, n] = size(A);
n1 = n;
les = 0;
neq = 0;
red = 0;
if length(c) ‘)
A = [A -vr(m,i)];
else
neq = neq + 1;
end
end
ncol = length(A);
if les == m
c = [c zeros(1,ncol-length(c))];
A = [A;c];
A = [A [b;0]];
[subs, A, z, p1] = loop(A, n1+1:ncol, mm, 1, 1);
disp(‘ End of Phase 1’)
disp(‘ *********************************’)
else
A = [A eye(m) b];
if m > 1
w = -sum(A(1:m,1:ncol));
else
w = -A(1,1:ncol);
end
c = [c zeros(1,length(A)-length(c))];
A = [A;c];
A = [A;[w zeros(1,m) -sum(b)]];
subs = ncol+1:ncol+m;
av = subs;
[subs, A, z, p1] = loop(A, subs, mm, 2, 1);
if p1 == ‘y’
disp(‘ End of Phase 1’)
disp(‘ *********************************’)
end
nc = ncol + m + 1;
x = zeros(nc-1,1);
x(subs) = A(1:m,nc);
xa = x(av);
com = intersect(subs,av);
if (any(xa) ~= 0)
disp(sprintf(‘nn Empty feasible regionn’))
return
else
if ~isempty(com)
red = 1;
end
end
A = A(1:m+1,1:nc);
A =[A(1:m+1,1:ncol) A(1:m+1,nc)];
[subs, A, z, p1] = loop(A, subs, mm, 1, 2);
if p1 == ‘y’
disp(‘ End of Phase 2’)
disp(‘ *********************************’)
end
end
if (z == inf | z == -inf)
return
end
[m, n] = size(A);
x = zeros(n,1);
x(subs) = A(1:m-1,n);
x = x(1:n1);
if mm == 0
z = -A(m,n);
else
z = A(m,n);
end
disp(sprintf(‘nn Problem has a finite optimal solutionn’))
disp(sprintf(‘n Values of the legitimate variables:n’))
for i=1:n1
disp(sprintf(‘ x(%d)= %f ‘,i,x(i)))
end
disp(sprintf(‘n Objective value at the optimal point:n’))
disp(sprintf(‘ z= %f’,z))
t = find(A(m,1:n-1) == 0);
if length(t) > m-1
str = ‘Problem has infinitely many solutions’;
msgbox(str,’Warning Window’,’warn’)
end
if red == 1
disp(sprintf(‘n Constraint system is redundantnn’))
end
function [subs, A, z, p1]= loop(A, subs, mm, k, ph)
% Main loop of the simplex primal algorithm.
% Bland’s rule to prevente cycling is used.
tbn = 0;
str1 = ‘Would you like to monitor the progress of Phase 1?’;
str2 = ‘Would you like to monitor the progress of Phase 2?’;
if ph == 1
str = str1;
else
str = str2;
end
question_ans = questdlg(str,’Make a choice Window’,’Yes’,’No’,’No’);
if strcmp(question_ans,’Yes’)
p1 = ‘y’;
end
if p1 == ‘y’ & ph == 1
disp(sprintf(‘nn Initial tableau’))
A
disp(sprintf(‘ Press any key to continue …nn’))
pause
end
if p1 == ‘y’ & ph == 2
tbn = 1;
disp(sprintf(‘nn Tableau %g’,tbn))
A
disp(sprintf(‘ Press any key to continue …nn’))
pause
end
[m, n] = size(A);
[mi, col] = Br(A(m,1:n-1));
while ~isempty(mi) & mi eps
t = A(1:m-k,col);
if all(t %g pivot column->%g’,…row,col))
end
A(row,:)= A(row,:)/A(row,col);
subs(row) = col;
for i = 1:m
if i ~= row
A(i,:)= A(i,:)-A(i,col)*A(row,:);
end
end
[mi, col] = Br(A(m,1:n-1));
end
tbn = tbn + 1;
if p1 == ‘y’
disp(sprintf(‘nn Tableau %g’,tbn))
A
disp(sprintf(‘ Press any key to continue …nn’))
pause
end
end
z = A(m,n);
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