Azzera filtri
Azzera filtri

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how can i change this code

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praveen thakur
praveen thakur il 19 Feb 2020
Chiuso: MATLAB Answer Bot il 20 Ago 2021
clear all
close all
clc
clear workspace;
fun=[1,-1,-1];
coeff=[0,2.5,1; 0,1.5,2; 0,-1,5];
const=[70;100;0];
eqs=[1;1;1];
sign=[1,1,1];
slack=eye(3);
table=[zeros(1,1),fun,zeros(1,4);zeros(3,1),coeff,slack,const;zeros(1,8)];
disp('table=');
disp(table);
for i=2:4
tmp=-table(1,i);
for j=2:4
tmp = tmp + (table(j,i)*table(j,1));
end
table(5,i)=tmp;
end
minim=min(table(5,:));
while minim<0
index = find(table(5,2:7)==min(table(5,2:7)));
index = index+1;
mini = inf;
in = 0;
for i=2:4
tmp = table(i, 8)/table(i, index);
if tmp<mini && tmp>0
mini = tmp;
in = i;
end
end
pivot = table(in,index);
for j=2:8
table(in, j) = table(in, j)/pivot;
end
for i=2:5
if i~=in
tmp=table(i, index);
for j=2:8
table(i,j) = table(i,j)-((tmp*table(in,j))/pivot);
end
end
end
table(in, 1) = table(1,index);
minim = min(table(5,:));
end
disp(table);
disp('Maximum value of z=');
disp(table(5,8));
  3 Commenti
praveen thakur
praveen thakur il 19 Feb 2020
i am getting an error in line 36 that si array indices should be positive or logical value
madhan ravi
madhan ravi il 19 Feb 2020
Don’t use table as a variable name, there’s an inbuilt function table()

Risposte (1)

KSSV
KSSV il 19 Feb 2020
in = 0 ;
pivot = table(in,index);
In the above line you have indexed in to zero. There is no zero and negaitve indices in MATLAB. You should change it to 1.
in = 1 ;
pivot = table(in,index);
Also there are some warnings in the code. There is divison with zero. Read about matlab indexing.

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R2019b

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