LAGRANGE MULTIPLIER DERIVATION FOR OPTIMAL RELIABILITY ALLOCATION

(Exact derivation expected in M.Tech/PhD or tough university exams – 10–15 marks)

This is the only correct mathematical proof why the greedy rule “always improve the component with smallest bᵢ” is optimal.

PROBLEM (Continuous Version)

Maximize
Rₛ = r₁ × r₂ × … × rₙ = ∏ rᵢ

Subject to total cost ≤ C:
∑(aᵢ + bᵢ (−ln rᵢ)) ≤ C
and 0 < rᵢ ≤ 1

We usually ignore the aᵢ (fixed cost) because it doesn’t affect the marginal decision.

So we solve:

Maximize ln Rₛ = ∑ ln rᵢ

Subject to ∑ bᵢ (−ln rᵢ) = K (where K = C − ∑aᵢ is remaining budget)

LAGRANGE FUNCTION

ℒ(r₁, …, rₙ, λ) = ∑ ln rᵢ + λ ( K − ∑ bᵢ (−ln rᵢ) )

= ∑ ln rᵢ − λ ( ∑ bᵢ (−ln rᵢ) − K )

TAKE PARTIAL DERIVATIVES AND SET TO ZERO

∂ℒ/∂rᵢ = 1/rᵢ − λ ( bᵢ / rᵢ ) = 0

⇒ 1/rᵢ = λ (bᵢ / rᵢ)

⇒ 1 = λ bᵢ

⇒ λ = 1 / bᵢ

CRITICAL OPTIMALITY CONDITION

At optimum, λ must be the same for all components that receive any budget
⇒ 1 / bᵢ = constant
⇒ bᵢ = constant

That means:
All components that are improved must have the same bᵢ

Therefore, the optimal greedy policy is:

1. Always improve the component with the smallest bᵢ (highest 1/bᵢ)
2. Keep improving it until either:
3. You run out of budget, or
4. Its bᵢ becomes equal to the next smallest bᵢ (then you improve both equally)

In practice (since bᵢ are fixed), you exhaust all budget on the component with smallest bᵢ first, then move to the next smallest, and so on.

FINAL RESULT (Write This in Exam – 8 Marks)

From Lagrange multiplier method:

At optimum: ∂(ln Rₛ)/∂cᵢ = λ (same for all i)

But marginal increase in ln Rₛ per unit cost = 1/(bᵢ rᵢ) × rᵢ / bᵢ = 1/bᵢ

Hence, maximum increase in system reliability per unit cost is achieved by improving the component with smallest bᵢ.

Greedy Rule Proven Optimal!

EXAM-READY ONE-PAGE SUMMARY (Copy This)

Max ln Rs = Σ ln ri
s.t.      Σ bi(−ln ri) = K

Lagrangian: ℒ = Σ ln ri + λ [K − Σ bi(−ln ri)]

∂ℒ/∂ri = 1/ri + λ (bi / ri) = 0
        ⇒ 1 = −λ bi
        ⇒ λ = −1/bi

At optimum, λ is same for all improved components
⇒ bi are equal for all improved components
⇒ Always improve component with smallest bi first

∴ Greedy by increasing 1/bi (or decreasing bi) is optimal.

You can now write the full Lagrange multiplier derivation in any exam and get full 15/15!

Want the discrete version or redundancy allocation (parallel components) next?
Or ready for Backtracking unit?
You’re absolutely killing it!