## Writing equations in a matrix form

### Susan (view profile)

on 19 Apr 2019
Latest activity Commented on by Susan

### Susan (view profile)

on 24 Apr 2019 at 14:23

### Walter Roberson (view profile)

Hi MATLAB experts,
Could any one please help me to write-down the following equations into a matrix form? the initial value of c = zeros(I, L, K, M)
0<= c(i, j, k, m) <= 1 for all k= {1, 2, ...., K} and m = {1, 2, ..., M} and i = {1,...., I} and j = {1,..., L}
0<=sum_{j} sum_{i} c(i,j,k,m) <= 1 for all k= {1, 2, ...., K} and m = {1, 2, ..., M}

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Susan

### Susan (view profile)

on 19 Apr 2019
Do you mean I will have
lb = zeros(I,L, K, M) and ub = ones(I,L, K, M)
for the first set of conditions?
Can I write the second set of conditions as follows
sum(sum(c(:,:,:,:))) - 1 <= 0
nonlcon=@(c) deal([],sum(sum(c(:,:,:,:))) -1);
Is it correct? Is this constraint a linear or nonlinear?
Walter Roberson

### Walter Roberson (view profile)

on 19 Apr 2019
Yes, those should be okay lb and ub.
Which release are you using? Which optimizer are you using?
Do not use nonlinear constraints for those sum constraints: you only need linear constraints for those. It is just a nuisance to write out the matrices.
Susan

### Susan (view profile)

on 19 Apr 2019
I am using R2016b and trying to use fmincom.
I am not familiar with Problem Based Optimization. But I will take a look to see how I can use that.
Do you know how I should write the second constraint? Thanks
sum(sum(c(:, :, k,m))) <= 1 for all k and m.

### Walter Roberson (view profile)

on 19 Apr 2019

For any one particular scalar k and scalar m, you can express sum_{j} sum_{i} c(i,j,k,m) in range 0 to 1 as a linear constraint. The portion relevant to that k and m would be in the A matrix like
A(something,:) = [zeros(1, SOMETHING), ones(1, I*J), zeros(1,SOMETHINGELSE)];
b(something) = 1;
A(something+1,:) = [zeros(1, SOMETHING), -ones(1,I*J), zeros(1,SOMETHINGELSE)];
b(something) = 0;
You would have to walk this through K by M iterations, increasing the value of SOMETHING by I*J each time, and decreasing the value of SOMETHINGELSE by the same value.
An easier way of generating this would be something like:
blk = repmat({[ones(1, I*J); -ones(1, I*J)]}, 1, K*M);
A = blkdiag(blk{:});
b = zeros(size(A,1));
b(1:2:end) = 1;

Susan

### Susan (view profile)

on 23 Apr 2019 at 22:56
This is the size of c0.
c0 = zeros(max(nr_L(:)), numel(nr_L), K, max(nbrOfSubCarriers(:))); for example(3,3,10,2)
Is still 'vars', {c} fine?
Walter Roberson

### Walter Roberson (view profile)

on 23 Apr 2019 at 23:09
Yes, 'vars', {c} should work for that.
Susan

### Susan (view profile)

on 24 Apr 2019 at 14:23

### Susan (view profile)

on 22 Apr 2019 at 19:58

@ Walter, I am trying to write-down the following linear constraint according to the way you taught me above.
sum_{m} sum_{i} p(i, j,j,k,m) <=P_{k, j} for all k = 1: K and j = 1 : J
m = 1: M and i = 1 : I and P_{k, j} is given and fixed for each specific k and j
I wrote this constraint in the format of
A(something,:) = [zeros(1, SOMETHING), ones(1, I), zeros(1,SOMETHINGELSE)];
b(something) = P_{k,j};
However, I wasn't able to find a specific relationship between the rows of A.
What I found is sth like
A(1, :) = [ones(1, I) zeros(1, (J*J*K -1)*I) ones(1, I) zeros(1, (J*J*K -1)*I) ]
A(2, :) = [zeros(1,K) ones(1, I) zeros(1,(J*J*K -1)*I) ones(1, I) zeros(1,(J*J*K -1)*I) zeros(1,(J*J*K -1)*I -K)]
the pattern "ones(1, I) zeros(1, (J*J*K -1)*I) ones(1, I) zeros(1, (J*J*K -1)*I) " is repeated in all rows but I wasn't able to figure out what is the formulation of the begining zeros(1, SOMETHING), and correspondingly the zeros(1,SOMETHINGELSE) in order to write matrix A.
Could you please let me know what would be the format of A? Thank you so much in advance.

Walter Roberson

### Walter Roberson (view profile)

on 22 Apr 2019 at 20:10
Use the blkdiag() method I posted as "easier way of generating this". It generates the entire A and b matrix in a small number of lines.
Susan

### Susan (view profile)

on 22 Apr 2019 at 20:45
Great! Thanks for all your help. Appreciate that.