Proof of Theorem 4.1, p. 97 — the Lagrangian is minimized at φ_j = a_j + √(a_j/(y f_j))
ProvedKellyReversibility.Allocation.lagrangian_minimizercapacity-allocationlagrange-multiplierp2o-batch-pfp2bp2o-gran-per-chapterp2o-plan-bookp2o-v1stochastic-networks
Let and for , let and let be a Lagrange multiplier. Over capacity vectors with for all , the Lagrangian
is minimized by the choice
which satisfies , and every other vector with for all has .
This is the unconstrained step of the Lagrangian method used to prove Theorem 4.1.
Formalization Note The book says " is minimized by the choice"; the strict inequality for every other point (uniqueness of the minimizer) is a slight strengthening, true because is strictly convex on the region . The minimization is over the open region , where every term of is finite.
Preamble
import Mathlib import Definitions.Def_KellyReversibility_Allocation_CapacityAllocation
Formal statement
namespace KellyReversibility.Allocation
theorem lagrangian_minimizer {J : ℕ} (a f : Fin J → ℝ) (F y : ℝ)
(ha : ∀ j, 0 < a j) (hf : ∀ j, 0 < f j) (hy : 0 < y) :
let φstar : Fin J → ℝ := fun j => a j + Real.sqrt (a j / (y * f j))
(∀ j, a j < φstar j) ∧
∀ φ : Fin J → ℝ, (∀ j, a j < φ j) → φ ≠ φstar →
lagrangian a f F y φstar < lagrangian a f F y φ := by sorry
end KellyReversibility.Allocation
Source
Kelly, Reversibility and Stochastic Networks, Wiley 1979, p. 97, proof of Theorem 4.1 (the Lagrangian L and the choice φ_j = a_j + √(a_j/(y f_j)))
Human review
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.