Constrained Optimization doesn't converge?

Question

Given a $n\times J$ matrix $T$ with strictly positive elements, I'm trying to obtain the values ($\lambda_i$) that minimize $\sum T_{ij}\lambda_i^{\alpha_i}$ with the constraints $\lambda_i \geq 0$ and $\sum \lambda_i = 1$ for each of the $J$ columns. Provided $\alpha_i \geq 1$, the problem has a unique solution such that $\lambda_i > 0$, so only constraint $\sum \lambda_i = 1$ will be binding. I have used

Alpha=4/3
T = {{2, 3}, {3, 2}, {1, 4}}
tecniques = Dimensions[T][[1]]
factors = Dimensions[T][[2]]
AlphaVec = ConstantArray[Alpha, tecniques]
LambdaVec = Array[Lambda, tecniques]
onesVec = ConstantArray[1, tecniques]
maxmin = Table[LambdaVec /. Minimize[{(Transpose[T].LambdaVec^(AlphaVec))[[i]],
     onesVec.LambdaVec == 1}, LambdaVec][[2]], {i, 1, factors}]

And this works fine when $\alpha=2$. However, when I set $\alpha=\frac{4}{3}$, Mathematica fails to return an answer, just keeps running. I'm sure there is some problem in my code, but I cannot see it (also, it looks kind of awkard to me). I've computed the solutions manually and it did not take long. Of course, this is an example and I'd like the code to work for arbitrary $n>J,\alpha_i$ and $T$.

Answer 1

Replacing Minimize by NMinimize . You will see that in the process of computation at $\alpha =4/3$ complex numbers are encountered. It is necessary to add the restrictions of $\lambda\ge 0$

T = {{2, 3}, {3, 2}, {1, 4}};
Alpha = 4/3;
tecniques = Dimensions[T][[1]];
factors = Dimensions[T][[2]];
AlphaVec = ConstantArray[Alpha, tecniques];
LambdaVec = Array[Lambda, tecniques];
onesVec = ConstantArray[1, tecniques];
h = (Transpose[T].LambdaVec^(AlphaVec));
maxmin = Table[
   LambdaVec /. 
    NMinimize[{h[[i]], {onesVec.LambdaVec == 1, 
        Table[Lambda[i] >= 0, {i, 1, tecniques}]}}, LambdaVec][[
     2]], {i, 1, factors}] // Quiet
{{0.201207, 0.0314277, 0.767365}, {0.239077, 0.683654, 0.077269}}

This task has an exact solution, which the author indicated and which can easily be found using the Mathematica. But it does not coincide with the numerical solution that we found using NMinimize. We consider the function h={2 Lambda[1]^(4/3) + 3 Lambda[2]^(4/3) + Lambda[3]^(4/3), 3 Lambda[1]^(4/3) + 2 Lambda[2]^(4/3) + 4 Lambda[3]^(4/3)}, find its extrema, using the standard analysis. Put h[[1]]=q, h[[2]]=q1, Lambda[1]=x,Lambda[2]=y, using constraints, we find

q = 2*x^(4/3) + 3*y^(4/3) + (1 - x - y)^(4/3); q1 = 
 3*x^(4/3) + 2*y^(4/3) + 4*(1 - x - y)^(4/3);

Necessary conditions for an extremum

eq = {D[q, x] == 0, D[q, y] == 0} // FullSimplify
Out[]= {9 x + y == 1, x + 28 y == 1}

 Solve[eq, {x, y}]

Out[]={{x -> 27/251, y -> 8/251}}
eq1 = {D[q1, x] == 0, D[q1, y] == 0} // FullSimplify

Out[]= {91 x + 64 y == 64, 8 x + 9 y == 8}

 Solve[eq1, {x, y}]

Out[]= {{x -> 64/307, y -> 216/307}}

The third value is found as Lambda[3]=1-x-y. And so we got the exact solution, which the author indicated in the comments. Now we need to get a numerical solution that would not be as rude as we indicated above using NMinimize. I will indicate a simple solution to the problem. We use a special method

maxmin2 = 
 Table[LambdaVec /. 
    NMinimize[{h[[i]], {onesVec.LambdaVec == 1, 
        Table[Lambda[i] >= 0, {i, 1, tecniques}]}}, LambdaVec, 
      WorkingPrecision -> 30, MaxIterations -> 100, 
      Method -> "RandomSearch"][[2]], {i, 1, factors}] 
{{0.10756972111553840466091432374, 0.0318725099601588814911179025068, 
  0.860557768924302713847967773754}, {0.20846905537459283387622149837,
 0.703583061889250814332247557003, 0.0879478827361563517915309446257}}

We compare it with the exact solution

{{27/251, 8/251, 216/251}, {64/307, 216/307, 27/307}}*1.`30

{{0.107569721115537848605577689243, 0.0318725099601593625498007968127, 
  0.860557768924302788844621513944},{0.208469055374592833876221498371,
 0.703583061889250814332247557003, 0.0879478827361563517915309446254}}

And so, we numerically reproduced the exact solution with an error of $10^{-15}$.