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I am trying to formulate the Lagrangian function for the following problem and constraints: $$f(x,y,z)= \sum_{n=1}^N x_{nk}\log(1+y_{nk})+\sum_{m=1}^Mz_{mk}C,$$ and maximize $\sum_{k=1}^Kf(x,y,z)$ subject to$$f_k(x,y,z)\geq b_k,\forall k,\ \sum_{k=1}^K\sum_{n=1}^N y_{nk}\leq Q,\ \sum_{k=1}^K z_{mk}\leq1,\forall m,\ z_{mk}\geq0, \forall k,m. $$

Is the Lagrangian function for the above problem given below correct? \begin{align*} L(x,y,z,\lambda,\nu,\mu_1,\mu_2)&=\sum_{k=1}^Kf(x,y,z)+\sum_{k=1}^K\lambda_k(f_{k}(x,y,z)-b_k)\\ &\mathrel{\phantom{=}} +\nu\left(Q-\sum_{k=1}^K\sum_{n=1}^Ny_{nk}\right)+\sum_{m=1}^M\mu_m\left(1-\sum_{k=1}^Kz_{mk}\right)+\sum_{k=1}^K\sum_{m=1}^M\mu_{mk}z_{mk}. \end{align*}

Fawad
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  • You seem to have two uses of the subscript $k$ that are throwing you off. The function $f$ is a sum over the $K$ entries of $x$, $y$, and $z$. Presumably you mean for all $K$ values of $z$ to be nonnegative, but you have no summation for those. And why is there only one value of $b_k$ as well? – Michael Grant Feb 11 '18 at 21:59
  • So I'd say that you need to fix your constraints so that you understand how many you actually have---I'm guessing $2K+2$---and then I suspect your Lagrangian will follow. – Michael Grant Feb 11 '18 at 22:01
  • Yes you are right this $z_k \geq 0,\forall k$. $b_k$ is same for all $k=1....K$ – Fawad Feb 12 '18 at 01:31
  • Then you need a separate Lagrange multiplier for each nonnegativity constraint. – Michael Grant Feb 12 '18 at 01:41
  • @MichaelGrant I edit the problem. Can you please check it. I am really obliged. – Fawad Feb 12 '18 at 17:03
  • I'm still not convinced you're doing what you intend with $f(x,y,z)\geq b_k$. There's no need for $K$ separate constraints. What you have written is equivalent to $f(x,y,z)\geq \max_k b_k$. – Michael Grant Feb 12 '18 at 18:03
  • Or do you have $k$ separate functions $f_k$? If so you haven't indicated that properly. – Michael Grant Feb 12 '18 at 18:05
  • This is the first constraint $ \sum_{n=1}^N x_{nk}\log(1+y_{nk})+\sum_{m=1}^Mz_{mk}C\geq b_k, \forall k$. Yes I have $f_k$ separate functions. – Fawad Feb 12 '18 at 18:38
  • That's still not notationally correct. $f(x,y,z)$ is a single function. If you have $K$ functions you have to have $K$ names. – Michael Grant Feb 12 '18 at 18:39
  • Then what should be the correct format – Fawad Feb 12 '18 at 18:41
  • Just call the $k$th function $f_k$. – Michael Grant Feb 12 '18 at 18:45
  • you mean the second term will become like this $\sum_{k=1}^K \lambda_{k}(f(x,y,z){k}-b{k})$ – Fawad Feb 12 '18 at 18:53
  • @MichaelGrant $f(x,y,z)_k \geq b_k, \forall k$ should be the constraint. This is what you mean? – Fawad Feb 12 '18 at 19:02
  • You really need to learn some basic notation here. The functions should be $f_k(x,y,z)$, not $f(x,y,z)_k$. – Michael Grant Feb 12 '18 at 19:56
  • ohh I am so sorry. you are right. Thank you so much – Fawad Feb 12 '18 at 20:05

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