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multiple selections from an iteration

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summyia qamar
summyia qamar on 19 Dec 2016
Closed: MATLAB Answer Bot on 20 Aug 2021
I'm running this code.
part_machine=[1 0 0 1 0 1 1
0 1 1 1 0 0 1
1 0 0 1 1 0 0
1 0 0 0 1 0 1
1 1 0 0 0 1 0
0 1 0 0 0 1 1];
demand=[600;550;620;500;590;600];
numIterations=100;
for k=1:numIterations
rand_cell=randi([0,1],7,3);
if all(sum(rand_cell,2)==1);
Part_cell(k)=part_machine*rand_cell;
end
Part_cell(Part_cell>=1)=1
no_of_movements=sum(Part_cell,2)-1;
movement_cost(k)=sum(bsxfun(@times,no_of_movements,demand));
end
minval=min(movement_cost);
[rand_cell minval(ones(7,1))]
the idea behind this code is * generate rand_cell * select that rand_cell in which sum of rows is 1 * multiply Part_machine matrix with that rand_cell * then apply the other functions on part_machine matrix * then at the end select the minval from all iterations * result is displayed with that rand_cell matrix which has generated the minimum value
now the result I'm getting is like this
ans =
1 0 1 6920
1 0 0 6920
0 1 0 6920
1 1 0 6920
1 1 1 6920
1 1 1 6920
1 0 1 6920
it is not the rand_cell matrix according to condition
all(sum(rand_cell,2)==1)
where is the problem?
  2 Comments
summyia qamar
summyia qamar on 19 Dec 2016
I mean that part_cell is generated as
part_machine* rand_cell)
where rand_cell should satisfy
all(sum(rand_cell,2)==1)

Answers (1)

Jan
Jan on 19 Dec 2016
After
rand_cell = randi([0,1],7,3);
the condition
if all(sum(rand_cell,2)==1)
Part_cell(k)=part_machine*rand_cell;
end
is very unlikely. The body of this IF-Block is most likely not entered.
But if it is satisfied
no_of_movements = sum(Part_cell, 2) - 1
replies zeros only.
To get a random matrix with one 1 per row:
index = randi([1,3], 1, 7);
Part_cell = zeros(7, 3);
Part_cell(sub2ind([7,3], 1:7, index)) = 1;
  2 Comments
Jan
Jan on 20 Dec 2016
Then please the relevant part of the code which reproduces the problem.

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