JNTU B.Tech IV OPTIMIZATION TECHNIQUES IN ENGINEERING Previous Question Paper (July/August - 2022 - R16)

To find the minimum of the function f(λ) = λ / log(λ) using quadratic interpolation, we define the objective function f(λ) = λ / ln(λ). Given the initial trial step length Δλ = 0.1, we start at an initial point λ₀ = 2 (since ln(λ) must be positive and λ > 1 for the function to be defined).

Step 1: Evaluate the function at three points to perform the first interpolation.

Step 2: Use the quadratic interpolation formula to find the stationary point λ*:

Substituting the values:

Step 3: Verify the minimum. The derivative f'(λ) = (ln(λ) - 1) / (ln(λ))². Setting f'(λ) = 0 gives ln(λ) = 1, which implies λ = e ≈ 2.718. The quadratic interpolation method converges iteratively toward this value. Repeating the process with the new point λ* will refine the estimate toward the global minimum at λ = e.

To minimize the function f(λ) = λ⁵ - 5λ³ - 20λ + 5 using the quadratic interpolation method, we iteratively approximate the function with a quadratic polynomial P(λ) = a₀ + a₁λ + a₂λ² and find its minimum at λ* = -a₁ / (2a₂).

Let us choose initial points λ₀ = 0, λ₁ = 1, and λ₂ = 2. The function values are:

Using the quadratic interpolation formula for three points, the minimum λ* is given by:

Substituting the values:

Since the denominator is zero, the points are collinear, indicating the quadratic approximation is not suitable for these initial points. Choosing new points λ₀ = 1, λ₁ = 2, λ₂ = 3:

Applying the formula again:

The process is repeated by replacing the point with the highest function value with λ* until the desired convergence is achieved. For this function, the analytical minimum occurs where f'(λ) = 5λ⁴ - 15λ² - 20 = 0. Letting x = λ², we get x² - 3x - 4 = 0, giving x = 4 (λ = 2) or x = -1 (imaginary). The quadratic interpolation method converges to this value of λ = 2.