How to find the maximum of this using Nmaximise?

Question

I have a function

 f[p_, q_, c_, x_] := 
     Log[c]*(2*x + 1 - p - q) + Log[1 - c]*(p + q - 2*x) + 
      q*((x/q)*Log[x/q] - (1 - x/q)*Log[1 - x/q]) + (1 - 
         q)*((1 - (p - x)/(1 - q))*
          Log[1 - (p - x)/(1 - q)] - ((p - x)/(1 - q))*
          Log[(p - x)/(1 - q)])

This is subject to certain conditions. $0<p<1, 0<q<1,0<c<1, 0<x<p$, I tried to maximise this f subject to these conditions.

NMaximize[{f[p, q, c, x], 0 < p < 1 && 0 < q < 1 && 0 < c < 1 && 0 < x < p}, {p, q, c, x}]

It gives me an error saying "0.2`" is not a valid variable". How can I do this?

Answer 1

This is an extremely difficult optimization problem. First, the maximum value of a continuous function on a noncompact set may not be achieved. Second, the arguments of the logarithms must be positive. Here is my best done with NMaximize:

NMaximize[{Log[c]*(2*x + 1 - p - q) + Log[1 - c]*(p + q - 2*x) + q*((x/q)*Log[x/q] - 
(1 - x/q)*Log[1 - x/q]) + (1 - q)*((1 - (p - x)/(1 - q))*Log[1 - (p - x)/(1 - q)] - 
((p - x)/(1 - q))*Log[(p - x)/(1 - q)]), 
0 <= p <= 1 && 0 <= q <= 1 && 0 <= c <= 1 && 0 <= x <= p && 
 1 - (p - x)/(1 - q) > 0 && (p - x)/(1 - q) > 0 && 1 - x/q > 0}, {p, q, c, x}, 
Method -> {"RandomSearch", "SearchPoints" -> 6000}, 
 AccuracyGoal -> 8, PrecisionGoal -> 8, WorkingPrecision -> 100]

{-0.040235724512206291627660187339000988341300360257660957486419128212
42038286521895562117758792508277991, {p -> 0.43170814406681159730265583187083824112441622120650523314960727729
69897992977574812748455238551242614, q -> 0.4301340679090147534723708508594863670430177236954453316714468
492587270496156672563111251203220147869, c -> 0.9569090295389766009595904849664351705051956310898170047261569
646576489145425430011107039749579302614, x -> 0.4298933184280109540416959995873164308236214711970976650767745
555271997494320105427232650983623018400}}

I think the supremum equals 0 , considering $p=q$, $c$ being close to $1$, and $x<p$ being close to $p$. The result of the command

MaxLimit[Log[c]*(2*x + 1 - p - q) + Log[1 - c]*(p + q - 2*x) + 
   q*((x/q)*Log[x/q] - (1 - x/q)*Log[1 - x/q]) + (1 - 
  q)*((1 - (p - x)/(1 - q))*
   Log[1 - (p - x)/(1 - q)] - ((p - x)/(1 - q))*
   Log[(p - x)/(1 - q)]) /. {p -> 1/2, q -> 1/2}, {c, x}->{1, 1/2},Direction -> "FromBelow"]

0

and Maple's results

Matrix(3, 3, [[-0.526452049410^(-8), [c = 0.9999999943976654, p = 0.24140178813121238, q = 0.2414017880807429, x = 0.2414017880467214], 579], [-0.228576294210^(-7), [c = 0.9999999871956152, p = 0.9546206287899156, q = 0.954620639904515, x = 0.9546206287884551], 491], [-0.2347957064*10^(-7), [c = 0.9999999729593567, p = 0.15817527515336488, q = 0.15817527345368837, x = 0.15817527263676995], 335]])

confirm it.

Answer 2

In this approach I changed the constraints slightly to force "real f"

NMaximize[{f[p, q, c, x], 
{0 < c < 1,1 - (p - x)/(1 - q) > 0
, (p - x)/(1 - q) > 0, 0 < x < q}}
, {p, q,c, x}] 
(*{-0.721159, {p -> 0.160626, q -> 0.614775, c ->0.340624,x -> 0.0928292}}*)