i want to declare that x will be variable for binary 0,1
i and k represent facilities in flow matrix, j and q represent location in distance matrix ..
x(i,j) mean that x will be equal to 1 if i (facility) is assign in j (location) ..
x(i, j) = 1 if facility i is assigned to location j and if otherwise, xij = 0,
so otherwise mean that if x(k,q) =1 , x(i,J) will be 0...
example of the manual calculation
Min =(f i1,k1 * d j1,q1 * x i1,j1 * x k1,q1) + (f i1,k1 * d j1,q2 * x i1,j1 * x k1,q2) + (f i1,k1 * d j1,q3 * x i1,j1 * x k1,q3)....
( 0*0* 1*1 ) + ( 0* 6* 1*0 ) + ( 0*8 * 1 *0 ).....
I want to *xi1,j1 * xk1,q1 to be 0 or 1.. if i choose i1, j1=1 the other will be 0.. for example i2,j1 will be equal to 0
below is the coding
clc;
clear;
%sum sum sum sum(fik*djq*xij*xkq)
%i,k= facilities
%j,q= location
%f(i,k)= flow between facilities i and k
%d(j,q)= distance between locations j and q
%xij = 1 if facility i is assigned to location j and if otherwise, xij = 0
% Flow matrix: flow assigning facility i (column) to facility k (row)
f = [0 5 7 9;
5 0 4 6;
7 4 0 3;
9 6 3 0];
%Distance matrix: distance assigning location j (column) to location q (row)
d = [0 6 8 9;
6 0 5 1;
8 5 0 2;
9 1 2 0];
z= 0;
nf= 4;
nd= 4;
for i=1:nf
for j=1:nf
for k=1:nd
for q=1:nd
z = min('z','f(i,k)*d(j,q)*x(i,j)*x(k,q)');
end
end
end
end
%Constraints
%The first set of constraints requires that each facility gets exactly one
%location, that is for each facility, the sum of the location values
%corresponding to that facility is exactly one
Constraints.constr1 = sum(x,2) == 1;
%The second set of constraints are inequalities. These constraints specify
%that each office has no more than one facility in it.
Constraints.constr2 = sum(x,1) == 1;
disp (z);
by using recperm
this recperm is used to fine permuation 0,1 in the coding
Related
I have created a function to add one to an element as follows;
function xz = addOne(x)
nrow = size(x,1);
ncol = size(x,1);
for K = 1:nrow
for J = 1:ncol
xz = x(:) + 1;
end
end
Example: given 1 the function results in 2:
addOne(1) [2]
I have tried using a matrix as an argument for the function...
x = [1 2 3; 0 0 0; 4 5 6];
x =
1 2 3
0 0 0
4 5 6
addOneWithFors(x)
ans =
2
1
5
3
1
6
4
1
7
How would I update this function to accept a matrix with multiple rows and columns and output it as such instead of just 1 number or a list of elements. Any help would be greatly appreciated.
In Matlab, you don't need a special function to do this. Matlab natively supports adding a scalar to a matrix. For example:
x = [1 2 3; 0 0 0; 4 5 6];
y = x + 1
will produce:
y =
2 3 4
1 1 1
5 6 7
However, if you specifically want to write this out explicitly using for loops, then your addOne() function only needs minor modifications. For example:
function xz = addOne(x)
nrow = size(x,1);
ncol = size(x,2);
xz = zeros(nrow,ncol);
for K = 1:nrow
for J = 1:ncol
xz(K,J) = x(K,J) + 1;
end
end
Note that ncol = size(x,2); has been defined correctly.
I'm trying to generate an n x n matrix like
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
where n = 5 or n 50. I'm at an impasse and can only generate a portion of the matrix. It is Problem 2.14 from Numerical Methods using MATLAB 3rd Edition by Penny and Lindfield. This is the best I have so far:
n = 5;
m = n;
A = zeros(m,n);
for i = 1:m
for j = 1:n
A(i,j) = m;
end
m = m - 1;
end
Any feedback is appreciated.
That was a nice brain-teaser, here’s my solution:
[x,y] = meshgrid(5:-1:1);
out = min(x,y)
Output:
ans =
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Here's one loop-based approach:
n = 5;
m = n;
A = zeros(m, n);
for r = 1:m
for c = 1:n
A(r, c) = n+1-max(r, c);
end
end
And here's a vectorized approach (probably not faster, just for fun):
n = 5;
A = repmat(n:-1:1, n, 1);
A = min(A, A.');
That's one of the matrices in Matlab's gallery, except that it needs a 180-degree rotation, which you can achieve with rot90:
n = 5;
A = rot90(gallery('minij', n), 2);
Given this objective function:
Minimize:
f = (Ax + By)' * G * (Ax + By)
subject to some equalities and inequalities.
where x and y are real-valued vectors (decision variables) with p and q elements, respectively. A of size m * p, B of size m * q, G is a symmetric matrix of size m * m.
My question is how to write f in the form v' * G * v, such that it can be easily be used in quadprog. In other words, how to mix A, B and G?
This looks incompletely specified!
It seems, for whatever reason, you want to model in terms of two variable components. Now you did not specify how they interact with each other.
As most optimizers work on a single variable-vector, you need to concatenate yours.
As you did not show G, i'm assuming you got one G for x and one for y, let's call it H.
(Remark: not a matlab user; don't take example-syntax for granted!)
z = [x y]
P = blkdiag(G,H)
assuming x and y independent in regards to quadratic-term
e.g. no x0*y1 like terms
Solve: for z` P z
Example:
x = [x0 x1 x2]
y = [y0 y1]
G = [6 2 1; 2 5 2; 1 2 4]
H = [8 2; 2 10]
# G
6 2 1
2 5 2
1 2 4
# H
8 2
2 8
z = [x0 x1 x2 y0 y1]
P = [6 2 1 0 0; 2 5 2 0 0; 1 2 4 0 0; 0 0 0 8 2; 0 0 0 2 8]
# P
6 2 1 0 0
2 5 2 0 0
1 2 4 0 0
0 0 0 8 2
0 0 0 2 8
I am trying to use the convhull function in a loop and for that I need to split matrices into submatrices of different sizes. Here is the code I am using:
x1=data(:,5); % x centre location
y1=data(:,16); % y centre location
z1=phi*90; % phi angle value
n=300;
%Create regular grid across data space
[X,Y] = meshgrid(linspace(min(x1),max(x1),n), linspace(min(y1),max(y1),n));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% PLOT USING SCATTER - TRYING TO ISOLATE SOME REGIONS %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
c=z1>10 & z1 < 20;
c=c.*1;
j=1;
for i=1:length(z1)
if z1(i)< 20 && z1(i)> 10
c(i) = 1;
else
c(i)= 0;
end
end
C=[c c c];
C = ~C;
elementalLengthA = cellfun('length',regexp(sprintf('%i',all(C,2)),'1+','match'));
elementalStartA = regexp(sprintf('%i',all(C,2)),'1+','start');
result = cell(length(elementalLengthA),1);
for i = 1:length(elementalLengthA)
result(i) = {C(elementalStartA(i):elementalStartA(i)+elementalLengthA(i)-1,:)};
length(x1(i))=length(cell2mat(result(i)));
length(y1(i))=length(cell2mat(result(i)));
end
My for loop doens't work properly and I get this error: ??? Subscript indices must either be real positive integers or
logicals.
My matrix C is an nx3 matrix made of lines of 1 and 0. With the result(i) line I am splitting the C matrix into submatrices of 1. Let's say
c = [1 1 1;
0 0 0;
0 0 0;
1 1 1;
1 1 1;
1 1 1;
0 0 0;
1 1 1;
1 1 1;]
Then
>> cell2mat(result(1))
ans =
1 1 1
>> cell2mat(result(2))
ans =
1 1 1
1 1 1
1 1 1
>> cell2mat(result(3))
ans =
1 1 1
1 1 1
Now x1 and y1 are two vector column nx1. And I want to split them according to the length of C submatrices. so length(x1(1)) should be 1, length(x1(2))=3, length(x1(3))=2 and same for the y vector.
Is it possible to do that?
EDIT:
Just to make it more clear
For instance
x1 =
1
2
3
4
5
6
7
8
9
and
y1 =
2
4
6
8
10
12
14
16
18
I want to get this as an output:
x1(1)=[1], x1(2)=[4 5 6]' and x1(3)=[8 9]'
y1(1)=[2], y1(2)[8 10 12]' and y1(3)=[16 18]'
Thanks
Dorian
In Matlab, I need to find all possible combinations of a vector that satisfies some constraints that I find quite irregular.
The vector, x, has 12 entries: x_0,x_1,...,x_11
Local constraints:
x_0 in {1,...,6}
x_i in {0,...,6}, i = 1,...,6
x_j in {0,...,12-j}, j = 7,...,11
Global constraints:
sum(x) = 12
for any k, x_k = y ==> x_{k+j} = 0 for j = 1,...,y-1
I have thought about this for quite a while now and can't seem to solve it myself - any ideas anyone?
Use recursion and intermediate constraint checking. Something like this does the trick:
function dispConstraintSatisfyingVectors()
global lb ub sumVal
lb = [1 0 0 0 0 0 0 0 0 0 0 0];
ub = [6 6 6 6 6 6 6 5 4 3 2 1];
sumVal = 12;
x = zeros(1,12);
addValueRecursive(x,1);
end
function addValueRecursive(x,idx)
global lb ub sumVal
for val = lb(idx):ub(idx)
x(idx) = val;
if checkZerosConstraint(x) == 0 && sum(x) <= sumVal
if idx < 12
addValueRecursive(x,idx+1);
else
if sum(x) == sumVal
disp(num2str(x))
end
end
end
end
end
function c = checkZerosConstraint(x)
c = 0;
for j=1:11
c = c + sum(x(j+1:j+x(j)-1));
end
end