linear equality constraint writing

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I am facing difficulty to write linear inequality constraint given below. Here Φ, 48 element long vector, is decision variable where as P and μ are vectors of 48 elements each and λ is a constant. I shall appreciate any help.
λ * P μ * Φ

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Walter Roberson
Walter Roberson 2017-6-21
lambda .* P <= mu * phi so
lambda .* P / mu <= phi
With lambda, P, and mu all being constants, this does not need to be written as a linear constraint. Instead, it can be written as lower bound: phi >= lambda .* P / mu so
x0 = .... %starting point
A = []; b = [];
Aeq = []; beq = [];
LB = lamda .* P ./ mu; UB = inf(1,48);
fmincon( fun, x0, A, b, Aeq, beq, LB, UB )
If you prefer to implement it as a linear constraint (perhaps because you are using a different minimizer) then
A = -eye(48); B = lambda .* P ./ mu;
  6 个评论
Saifullah Khalid
Saifullah Khalid 2017-6-21
I am grateful for prompt support, it solved my problem. I have used the the last suggested approach of UB and LB given above.
Saifullah Khalid
Saifullah Khalid 2017-6-21
Once more a little query, after running optimization, the value of phi (optimized) is set to all 1s, meaning 100% resource utilization which is not practically true. For example, if we have a cloud with 48 servers and 30 processing tasks to be executed, they would be assigned to any of the 30 servers whose phi may be between 0 to 1 but for rest of the servers, phi should not be 1. If you can kindly comments on this? Is my understanding correct?

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更多回答(1 个)

KSSV
KSSV 2017-6-21
编辑:KSSV 2017-6-21
% take some random data for test
lambda = rand ;
P = rand(48,1) ;
mu = rand(48,1) ;
phi = rand(48,1) ;
% inequality
idx = lambda*P<=mu.*phi ;
%%check result
R = [lambda*P(idx) mu(idx).*phi(idx)]
  1 个评论
Saifullah Khalid
Saifullah Khalid 2017-6-21
sir, thanks for response, I am actually trying to optimize using fmincon, the guide says constraints should be Ax-b < 0. I am not sure I can use this as it is?

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