JNTU B.Tech III OPTIMIZATION TECHNIQUES Previous Question Paper (November/December - 2020 - R16)

An objective function is a mathematical expression that quantifies the goal of an optimization problem. It represents the quantity to be either maximized (such as profit, efficiency, or utility) or minimized (such as cost, time, error, or energy consumption) subject to a set of constraints. Mathematically, it is expressed as f(x), where x is a vector of decision variables.

Objective function surfaces represent the geometric visualization of the function's values over the domain of decision variables. In a two-variable problem, the surface is a three-dimensional landscape where the height at any point (x₁, x₂) corresponds to the value of f(x₁, x₂). Key aspects of these surfaces include:

• Contours (Level Sets): These are projections of the surface onto the decision variable plane, defined by f(x₁, x₂) = c, where c is a constant. For linear objective functions, these are parallel lines. For non-linear functions, they can be complex curves such as ellipses or hyperbolas.
• Gradient Vector: The gradient, denoted by ∇f, points in the direction of the steepest ascent of the surface. In optimization, moving in the direction of the negative gradient helps in finding the minimum of the function.
• Stationary Points: These are locations on the surface where the gradient is zero. They include local minima, local maxima, and saddle points.
• Convexity: If the surface is convex (like a bowl), any local minimum is also the global minimum, which is a highly desirable property for optimization algorithms. If the surface is non-convex, it may contain multiple local optima, making it more difficult to find the global optimum.

Understanding the topology of these surfaces is essential for selecting appropriate optimization algorithms, such as gradient descent, which relies on the local slope of the surface to navigate toward the optimal solution.

The optimization problem is to minimize f(X) = x₁² + x₂² + x₃² subject to g₁(X) = 5 - x₁ - x₂ - x₃ ≤ 0, g₂(X) = 2 - x₂x₃ ≤ 0, and non-negativity constraints x₁ ≥ 0, x₂ ≥ 0, x₃ ≥ 2. The Kuhn-Tucker (KT) conditions require:

1. ∇f(X) + λ₁∇g₁(X) + λ₂∇g₂(X) = 0
2. λ₁g₁(X) = 0, λ₂g₂(X) = 0
3. λ₁, λ₂ ≥ 0

The gradients are ∇f = (2x₁, 2x₂, 2x₃), ∇g₁ = (-1, -1, -1), and ∇g₂ = (0, -x₃, -x₂).

For X₁ = (3/2, 3/2, 2): g₁ = 5 - (3/2 + 3/2 + 2) = 0 (active), g₂ = 2 - (3/2)(2) = -1 < 0 (inactive, so λ₂ = 0). The stationarity conditions are: 2(3/2) - λ₁ = 0 ⇒ λ₁ = 3; 2(3/2) - λ₁ = 0 ⇒ λ₁ = 3; 2(2) - λ₁ = 0 ⇒ λ₁ = 4. Since λ₁ is inconsistent (3 ≠ 4), the KT conditions are not satisfied.

For X₂ = (4/3, 2/3, 3): g₁ = 5 - (4/3 + 2/3 + 3) = 0 (active), g₂ = 2 - (2/3)(3) = 0 (active). Stationarity: 2(4/3) - λ₁ = 0 ⇒ λ₁ = 8/3; 2(2/3) - λ₁ - 3λ₂ = 0 ⇒ 4/3 - 8/3 - 3λ₂ = 0 ⇒ λ₂ = -4/9. Since λ₂ < 0, the KT conditions are not satisfied.

For X₃ = (2, 1, 2): g₁ = 5 - (2 + 1 + 2) = 0 (active), g₂ = 2 - (1)(2) = 0 (active). Stationarity: 2(2) - λ₁ = 0 ⇒ λ₁ = 4; 2(1) - λ₁ - 2λ₂ = 0 ⇒ 2 - 4 - 2λ₂ = 0 ⇒ λ₂ = -1. Since λ₂ < 0, the KT conditions are not satisfied.