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multiple selections from an iteration

1 Ansicht (letzte 30 Tage)
summyia qamar
summyia qamar am 19 Dez. 2016
Geschlossen: MATLAB Answer Bot am 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 Kommentare
José-Luis
José-Luis am 19 Dez. 2016
Bearbeitet: José-Luis am 19 Dez. 2016
You are not providing all the code necessary to answer this question. What is Part_cell? One could of course browse all your previous questions but it is better (as in easier for us) if your questions were self-contained.
summyia qamar
summyia qamar am 19 Dez. 2016
Bearbeitet: summyia qamar am 19 Dez. 2016
I mean that part_cell is generated as
part_machine* rand_cell)
where rand_cell should satisfy
all(sum(rand_cell,2)==1)

Antworten (1)

Jan
Jan am 19 Dez. 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 Kommentare
summyia qamar
summyia qamar am 19 Dez. 2016
now the result of all iterations is same value
Jan
Jan am 20 Dez. 2016
Then please the relevant part of the code which reproduces the problem.

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