Help me to run the code
Show older comments
% file modified by sujogya %dated 9.1.2021
% min z=x1-3x2+2x3 % max z'=-x1+3x2-2x3 % s.t. %3x1-x2+2x3<=7, %-2x1+4x2+0x3<=12, %-4x1+3x2+8x3<=10
NVar = 3; C = [-1 3 2]; CMat = [3 -1 2;-2 4 0;-4 3 8]; b = [7; 12; 10;];
s = eye(size(CMat,1)); A = [CMat s b]; cost = zeros(1,size(A,2)); cost(1:NVar) = C;
%% Constraint
BV = NVar+1:1:size(A,2)-1; zjcj = cost(BV)*A - cost;
%% print the Table
zcj = [zjcj;A]; simpletable = array2table(zcj);
simpletable.Properties.VariableNames(1:size(zcj,2)) = {'x_1','x_2','x_3','s_1','s_2','s_3','sol'}
if any(zjcj<0);
fprintf('the current solution is not optimal');
fprintf('\n--------------next iteration----------')
disp('old basic variable(BV)=')
else('optimal solution reached');
end
disp(BV);
%% find entering variable
zc = zjcj(1:end-1); [entercol,pvt_col] = min(zc);
fprintf('the most negative element in zj-cj row is %d corresponding to column%d\n',entercol,pvt_col)
fprintf('entering variable is %d\n',pvt_col);
%% finding leaving variable
sol = A(:,end)
column = A(:,pvt_col);
if all(column <= 0)
printf(' solution is unbdd');
for i = 1:size(column,1)
if column(i)>0
ratio(i) = sol(i)./column(i);
else
ratio(i) = inf
end
end
[MinRatio, pvt_row] = min(ratio);
fprintf('min ratio correspond to pivot row is %d\n',pvt_row);
fprintf('leaving variable %d\n',BV(pvt_row));
end
%% Finding the minimum
BV(pvt_row) = pvt_col;
disp('New basic variables(BV)=');
disp(BV);
%% Pivot key
pivot_key = A(pvt_row,pvt_col);
%%% update the table
A(pvt_row,:) = A(pvt_row,:)./pvt_key;
for i = 1:size(A,1)
if i ~= pvt_row
A(i,:) = A(i,:) - A(i,pvt_col).*A(pvt_row,:);
end
zjcj = zjcj-zjcj(pvt_col).*A(pvt_row,:);
end
%%% for printing purpose
zcj = [zjcj;A]; simptable = array2table(zcj);
simptable.properties.variable(1:size(zcj,2)) = {'x_1','x_2','x_3','s_1','s_2','s_3','sol'};
%%%% simplex method start
run = true
while run
if any(zjcj < 0);
fprintf(' the current solution is not optimal\n')
fprintf(' ----next iteration start----\n')
disp('the old basic variable (BV)=');
disp(BV);
%% print the optimal solution
BFS = zeros(1,size(A,2)); BFS(BV) = A(:,end); BFS(end) = sum(BFS.*cost);
current_BFS = array2table(BFS);
current_BFS.properties.variablenames(1:size(current_BFS,2)) = {'x_1','x_2','x_3','s_1','s_2','s_3','sol'};
else
run = false
fprintf('-------**********--------\n')
fprintf('the solution is optimal\n')
disp('optimal solution')
end
end
Accepted Answer
More Answers (1)
Sulaymon Eshkabilov
on 30 Oct 2021
0 votes
Some of the assigned variables don't get their values from the second [if .. end] and [for end] operations, e.g.: pvt_row
Thus, its default value has to be assigned. Moreover, one variable name is misspelled: pivot_key vs. pivt_key
There is no need to rename simptable variable names, since it is already done.
In addition, to rename table variable names (e.g. current_BFS), it is better to use renamevars().
1 Comment
MINATI PATRA
on 11 Nov 2021
Categories
Find more on Solver Outputs and Iterative Display in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!