Jawaharlal Nehru Technological University Hyderabad B.Tech III OPTIMIZATION TECHNIQUES Previous Question Paper (February/March - 2022 - R16)

Optimization involves finding the best solution from a set of available alternatives by maximizing or minimizing an objective function. The classification depends on the number of decision variables and the presence of constraints.

Single Variable Optimization: This involves a function of a single variable, f(x). To find the optimum (maximum or minimum), we determine the stationary points where the first derivative is zero. For a function f(x), the necessary condition is f'(x) = 0. The sufficient condition for a minimum is f''(x) > 0 and for a maximum is f''(x) < 0. For example, consider f(x) = x² - 4x + 5. The first derivative is f'(x) = 2x - 4. Setting 2x - 4 = 0 gives x = 2. Since the second derivative f''(x) = 2 is positive, x = 2 is a point of global minimum.

Multi-Variable Optimization (Unconstrained): This involves a function of several variables, f(x₁, x₂, ..., x_n). The necessary condition for an extremum is that the gradient of the function must be zero, meaning all partial derivatives must vanish simultaneously: ∂f/∂x_i = 0 for all i = 1, 2, ..., n. To classify these points, we examine the Hessian matrix, H, which is the square matrix of second-order partial derivatives. If H is positive definite, the point is a local minimum; if H is negative definite, it is a local maximum; if H is indefinite, it is a saddle point.

Example: Consider f(x, y) = x² + y² - 2x - 4y + 10. The partial derivatives are ∂f/∂x = 2x - 2 and ∂f/∂y = 2y - 4. Setting these to zero yields the stationary point (1, 2). The Hessian matrix is:

Since the principal minors are 2 and 4 (both positive), the Hessian is positive definite, confirming that (1, 2) is a local minimum.

To find the value of β for which the point (x₁*, x₂*) = (1, 2) is optimal, we apply the Kuhn-Tucker conditions for the maximization problem: Maximize Z = 2x₁ + βx₂ subject to g₁(x₁, x₂) = x₁² + x₂² - 5 ≤ 0 and g₂(x₁, x₂) = x₁ - x₂ - 2 ≤ 0.

First, we evaluate the constraints at (1, 2):

Since g₁ = 0, the first constraint is active (binding). Since g₂ < 0, the second constraint is inactive, implying its Lagrange multiplier λ₂ = 0 by complementary slackness.

The Lagrangian function is L = 2x₁ + βx₂ - λ₁(x₁² + x₂² - 5) - λ₂(x₁ - x₂ - 2). The stationarity conditions are:

Substituting x₁ = 1, x₂ = 2, and λ₂ = 0 into these equations:

Substituting λ₁ = 1 into the second equation:

For the point to be optimal, we must satisfy the non-negativity condition λ₁ ≥ 0. Since λ₁ = 1 ≥ 0, the condition is satisfied. Thus, the value of β for which (1, 2) is optimal is 4.