JNTU B.Tech IV OPTIMIZATION TECHNIQUES IN ENGINEERING Previous Question Paper (July - 2021 - R16)

To minimize f(λ) = λ⁵ - 5λ³ - 20λ + 5 in the interval [0, 5] using Fibonacci search, we first determine the number of iterations n. For a reduction ratio of 1/100 (assuming a typical precision requirement), F_n ≥ (b-a)/ε = 5/0.05 = 100. Thus, n = 11 (F₁₁ = 144). The search points are determined by L_i = (F_n-i+1/F_n-i+2) * L_i-1. In the first step, λ₁ = a + (F_n-2/F_n)(b-a) = 0 + (55/144)*5 ≈ 1.91 and λ₂ = a + (F_n-1/F_n)(b-a) = 0 + (89/144)*5 ≈ 3.09. Evaluating f(λ₁) ≈ -33.5 and f(λ₂) ≈ -17.6. Since f(λ₁) < f(λ₂), the new interval is [0, 3.09]. This process repeats until the interval is sufficiently small, converging to the minimum near λ ≈ 2.0.

Using the Golden Section method with the ratio φ = (√5 - 1)/2 ≈ 0.618, we define the search points λ₁ = b - φ(b-a) and λ₂ = a + φ(b-a). For the interval [0, 5]:
λ₁ = 5 - 0.618(5) = 1.91
λ₂ = 0 + 0.618(5) = 3.09
Calculating the function values: f(1.91) ≈ -33.51 and f(3.09) ≈ -17.65. Since f(λ₁) < f(λ₂), the new interval is [0, 3.09]. In the next iteration, the new λ₂ becomes the previous λ₁ (1.91), and a new λ₁ is calculated as 3.09 - 0.618(3.09) ≈ 1.18. The interval continues to shrink by a factor of 0.618 per iteration, converging to the local minimum at λ = 2.