2

Function to maximize

$R(P) = \sum\limits_{i=1}^4 (-\frac{1}{2}p_i^2 + 1000p_i)$

Parameters

$P = (p_1, p_2, p_3, p_4)$

Constraints

  • $p_1, p_2, p_3, p_4 > 0$

  • $\sum\limits_{i=1}^4 (-\frac{1}{2}p_i + 1000) \leq 2000$

m_ocean
  • 129
  • Convex optimization. – David G. Stork Dec 22 '20 at 20:56
  • @DavidG.Stork Thanks for the hint! A quick google-search suggests that it should probably be done programmatically, right? – m_ocean Dec 22 '20 at 20:59
  • Because the $p_i$ are interchangeable, try setting them all equal so you have a one-dimensional optimization problem. $p_i = 1000$ – David G. Stork Dec 22 '20 at 21:57
  • @DavidG.Stork Thank you for suggestion! That approach, as I understand, stands on the assumption that the prices for each period are going to be the same at the optimum. It makes intuitive sense to me but I haven't figured it out formally yet. Just documenting my thoughts... – m_ocean Dec 24 '20 at 17:26
  • I think you can argue as follows: the objective function is strictly concave, so there can only be one maximizer. Due to the symmetry in the problem, if $p_1, \ldots, p_4$ is a maximizer then any permutation of these four values must also be a maximizer. Since there can't be more than one maximizer, a maximizer must satisfy $p_1 = \cdots = p_4$. – littleO Dec 29 '20 at 20:05
  • An alternative technique from Convex Optimization is "Geometric Programing"
    https://en.wikipedia.org/wiki/Geometric_programming I quite like it but it is basically the same. In any case, a big clue is the symmetry under permutations of the variables.     – rrogers Dec 29 '20 at 21:30

1 Answers1

1

This is the first time I have answered a question, so please bear with me.

Maximizing $R(P)=\sum_1^4(−1/2p^2_i+1000p_i)$ is the same as minimizing its negative. Doubling it and adding 1000000 to each summand doesn't change the minimum and we get $R(P)=\sum_1^4(p^2_i-2000p_i+1000000)$ $ =\sum_1^4(p_i-1000)^2$. Setting each $p_1=1000$ satisfies the constraints and makes $R(P)=0$ and, since it is a sum of squares this must be the minimum.

Jack Fearnley
  • 11