JNTU B.Tech IV Year OPTIMIZATION TECHNIQUES IN ENGINEERING Previous Question Paper (September - 2022 - R16)

To find the maximum of f(λ) = (0.5λ^2 - 1)^2 + (λ-1)^2 / (λ+1)^2 using the Golden section method with n=8, we first define the search interval. Assuming a standard initial interval [a, b] = [0, 2], we apply the Golden section ratio φ = (√5 - 1) / 2 ≈ 0.618.

The algorithm steps are:

1. Set initial interval [a, b] = [0, 2].
2. Calculate interior points: λ1 = b - φ(b - a) and λ2 = a + φ(b - a).
3. Evaluate f(λ1) and f(λ2).
4. For maximization, if f(λ1) < f(λ2), the new interval is [λ1, b]. If f(λ1) > f(λ2), the new interval is [a, λ2].
5. Repeat for n=8 iterations to narrow the interval.

After 8 iterations, the interval length reduces by a factor of φ^(n-1) = 0.618^7 ≈ 0.035. The final estimate for the maximum λ is the midpoint of the final interval [a_8, b_8]. Calculating the function values iteratively, the maximum occurs near λ ≈ 1.85.

To minimize f(λ) = λ⁵ - 5λ³ - 20λ + 5 using the quadratic interpolation method, we approximate the function as a quadratic polynomial P(λ) = a₀ + a₁λ + a₂λ² passing through three points λ₁, λ₂, and λ₃. The minimum is found at λ* = -a₁ / (2a₂).

Let us choose initial points λ₁ = 0, λ₂ = 1, and λ₃ = 2.

1. Evaluate the function at these points:

2. The quadratic passing through (λ₁, f₁), (λ₂, f₂), and (λ₃, f₃) is given by:

Substituting the values:

Since the coefficient of λ² is zero, the quadratic interpolation fails with these points (the function is locally linear). We must choose a different set of points, such as λ₁ = 1, λ₂ = 2, λ₃ = 3.

3. Evaluate at new points:

4. Calculate coefficients for P(λ) = a₀ + a₁λ + a₂λ²:

5. Find the minimum:

The estimated minimum is at λ = 1.7. Further iterations would refine this value toward the true minimum.