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Let $\alpha=(\alpha_1,\ldots,\alpha_m)\subset\mathbb R^m_+$ and $\beta=(\beta_1,\ldots,\beta_n)\subset\mathbb R^n_+$ be given and satisfy

$$\sum_{i=1}^m \alpha_i =1 = \sum_{j=1}^n\beta_j.$$

Define $\Pi(\alpha,\beta)\subset \mathbb R^{mn}_+$ be the subset consisting $c=(c_{ij})_{1\le i\le m, 1\le j\le n}$ satisfying for all $i=1,\ldots, m$ and $j=1,\ldots, n$

$$\sum_{j=1}^n c_{ij} =\alpha_i \quad \mbox{ and } \quad \sum_{i=1}^m c_{ij} =\beta_j.$$

Consider the minimisation problem as follows:

$$\min_{c\in \Pi(\alpha,\beta)}~~ \sum_{j=1}^n\left[\sum_{i=1}^m c_{ij}|x_i|^2 -\frac{\big|\sum_{i=1}^m c_{ij}x_i\big|^2}{\beta_j}\right],$$

where $x_1,\ldots, x_m\in \mathbb R^d$ are fixed parameters, and $|\cdot|$ denotes the standard Euclidean norm. Is there a "well developed" numerical scheme for the above optimisation problem?

PS : "well developed" means

  1. The numerical scheme must converge (as it is known that the above minimisation problem admits a minimiser);

  2. There are checkable conditions such that the numerical scheme may converge to a minimiser (or at least a local minimiser);

  3. An analysis on the convergence rate is furnished.

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1 Answer 1

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The minimum here can be found exactly, in a finite number of steps.

Indeed, the target function is concave. So, its minimum on the (compact) (transportation) polytope $P:=\Pi(\alpha,\beta)$ is attained at one of the extreme points of $P$. For a construction and a characterization of all extreme points of $P$, see this answer by Brendan McKay.

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  • $\begingroup$ Thank you very kindly Iosif. I'm will board in 10 mins and will read the answer carefully $\endgroup$
    – Fawen90
    Commented Apr 25, 2023 at 13:53
  • $\begingroup$ Dear Iosif, I've read the answer by Brendan and I see the idea. While the procedure of iteration is not clear to me, could you please describe in detail how it proceeds? Further, how can we be ensured to obtain all the extreme points? $\endgroup$
    – Fawen90
    Commented Apr 27, 2023 at 7:01
  • $\begingroup$ If you don't have time, it's totally fine and I will try to find the book (which is not available online) $\endgroup$
    – Fawen90
    Commented Apr 27, 2023 at 7:02
  • $\begingroup$ @Fawen90 : I have not studied Brendan McKay's answer in detail, but think Brendan's answers deserve a high degree of trust. Perhaps, you can ask him about details. $\endgroup$ Commented Apr 27, 2023 at 15:23

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