%MATLAB code to pivot on a dictionary  (thanks to Lauren Baker)
 
%Usage:
%Set up:
%Specify...
%   an mxn matrix, A, of constraint coeffs.
%   an mx1 matrix, b, of RHS values (assumed >=0)
%   a 1xn matrix, c, of objective fn coeffs.
%   and an initial value z (the objective value), usually zero.
 
%   There are three MATLAB functions to be used with these definitions
%   dict() prints the current A, b, c, z in dictionary format
%   ratios(j) determines all upper bounds for entering variable xj.
%   lppivot(i,j) modifies A, b, c, z by pivoting on eqn i, variable j
 
%   Define A, b, c, z in the space indicated, then run this script. The
%   three functions dict, ratios, and lppivot can then be used.
 
global A b c m n z nonbas bas myeps
 
%Define A, b, c, z here.
 
[m,n]=size(A);
nonbas = [1:n-m];
%vars = horzcat([1:n-m],zeros(1,n-n+m));
bas = [n-m+1:n];
myeps = .1^8;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Contents of dict.m
 
function dict()
%dict is a function to print out the current A,b,c,z in dictionary format.
global A b c m n z bas nonbas myeps
 
fprintf('zeta = %8.2f',z)
sortnonbas=sort(nonbas);
for i=1:n-m
    var = sortnonbas(i);
    %pos = find(vars==var);
    if c(1,var) < 0
        fprintf('  - %6.2f x%d',-c(1,var),var)
    else    
        fprintf('  + %6.2f x%d',c(1,var),var)
    end
end
fprintf('\n')
fprintf('---------------')
for i=1:n-m
    fprintf('-------------')
end
fprintf('\n')
for i=1:m
   fprintf('x%d   = %8.2f',bas(1,i),b(i,1))
for j=1:n-m
    var = sortnonbas(j);
    if A(i,var) < -myeps
        fprintf('  + %6.2f x%d',-A(i,var),var)
    elseif A(i,var) > myeps
        fprintf('  - %6.2f x%d',A(i,var),var) 
    else
        fprintf('             ')
    end
end
fprintf('\n')
end
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Contents of ratios.m
 
function ratios(enter)
%ratios(j) determines all upper bounds for entering variable xj.
global A m bas b myeps 
 
for i=1:m
    if A(i,enter) > myeps
        fprintf('Row %d (x%d basic): Upper bound = %.2f\n',i,bas(1,i),b(i,1)/A(i,enter))
    else
        fprintf('Row %d (x%d basic): No upper bound\n',i,bas(1,i))
    end
end
end
 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Contents of lppivot.m

function lppivot(leave, enter)
%lppivot(i, j) modifies A, b, c, z by pivoting on eqn i, variable j
global A b c m n z bas nonbas
 
if isempty(find(bas==leave))
    error('Error (lppivot): The chosen exiting variable x%d is NOT basic.\n',leave)
end
 
pivcol = find(nonbas==enter);
pivrow = find(bas==leave);
if A(pivrow,enter) == 0
    error('Error (lppivot): Division by zero.')
end
b(pivrow,1)=b(pivrow,1)/A(pivrow,enter);
for i=1:n
    if ~isempty(find(nonbas==i)) && i~=enter
    A(pivrow,i)=A(pivrow,i)/A(pivrow,enter);
    end
end
tempcol = A(1:m,leave);
%temp = A(pivrow, leave);
A(pivrow, leave) = 1/A(pivrow, enter);
%A(pivrow, enter) = temp;
 
for i=1:m
    if i~=pivrow
    b(i,1)=b(i,1)-A(i,enter)*b(pivrow,1);
    for j=1:n
        if ~isempty(nonbas==j) && j~=enter
        A(i,j)=A(i,j)-A(i,enter)*A(pivrow,j);
        end
    end
    A(i,leave)=-A(pivrow,leave)*A(i,enter);
    end
end
z=z+c(1,enter)*b(pivrow,1);
for i=1:n
    if ~isempty(nonbas==i) && i ~= enter
        c(1,i)=c(1,i)-A(pivrow, i)*c(1,enter);
    end
end
c(1,leave)=-c(1,enter)*A(pivrow,leave);
c(1,enter)=0;
temp = bas(1,pivrow);
bas(1,pivrow)=nonbas(1,pivcol);
nonbas(1,pivcol)=temp;
for i=1:m
    A(i,enter) = tempcol(i,1);
end
dict()
end