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Susan
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Writing equations in a matrix form

Asked by Susan
on 19 Apr 2019
Latest activity Commented on by Susan
on 24 Apr 2019
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
on 19 Apr 2019
Thanks for your reply.
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?
Yes, those should be okay lb and ub.
Which release are you using? Which optimizer are you using?
Your task might be easier to express with Problem Based Optimization.
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
on 19 Apr 2019
Thanks for your reply.
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.

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2 Answers

Answer by Walter Roberson
on 19 Apr 2019
 Accepted Answer

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;

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Susan
on 23 Apr 2019
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?
Thanks in advance
Yes, 'vars', {c} should work for that.
Susan
on 24 Apr 2019
Thanks for your reply.

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Answer by Susan
on 22 Apr 2019

@ 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.

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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
on 22 Apr 2019
Great! Thanks for all your help. Appreciate that.

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