EASTERN MEDITERRANEAN UNIVERSITYDepartment of Industrial Engineering

Non linear Optimization Spring 2014-15

Instructor: Prof.Dr.Sahand DaneshvarSubmited by: AAKASH AHMEDStudent number: 145322

APPLICATIONS OF JACOBIAN OF JACOBIAN

METHODMETHOD

APPLICATIONS OF JACOBIAN OF JACOBIAN

METHODMETHOD

Constrained Derivatives (Jacobian) Method

Minimize z = f(X)subject to g(X) = 0where

The functions f(X) and g(X), i = 1,2, ... , m, are twice continuously differentiable.The idea of using constrained derivatives is to develop a closed-form expressionfor the first partial derivatives of f(X) at all points that satisfy the constraints g(X) = O.This corresponding stationary points are identified as the points at which these partialderivatives vanish. The sufficiency conditions introduced in Section can then beused to check the identity of stationary points.

To clarify the proposed concept, consider f(xl , X2) illustrated in Figure 18.4. This

function is to be minimized subject to the constraint

where b is a constant. From Figure , the curve designated by the three points A, B, and C represents the values of f(x1, X2) for which the given constraint is always satisfied. The constrained derivatives method defines the gradient of f(Xl, X2) at any point on the curve ABC. Point B at which the constrained derivative vanishes is a stationary point for the constrained problem.The method is now developed mathematically. By Taylor's theorem, for in the feasible neighborhood of X, we have and

Demonstration of the idea of the Jacobian method

Demonstration of the idea of the Jacobian method

For feasibility, we must have , and it follows that

This gives (m + 1) equations in (n + 1) unknowns, and . Note that

is a dependent variable, and hence is determined as soon as is known.

This means that, in effect, we have m. equations in n unknowns.

If m > n, at least (m - n) equations are redundant. Eliminating redundancy, the system reduces to m < n. If m = n, the solution is , and X has no feasible

neighborhood, which means that the solution space consists of one point only. The remaining case, where m < n, requires further elaboration.

Define,

x = (Y, Z)

such that,

The vectors Y and Z are called the dependent and independent variables, respectively. Rewriting the gradient vectors of f and g in terms of Y and Z, we get,

Define,

J(m*m) is called the Jacobian matrix and C(m*n- m) the control matrix. The Jacobian

J is assumed non-singular. This is always possible because the given m equations are independent by definition. The components of the vector Y must thus be selected from

among those of X such that J is nonsingular.

The original set of equations in partial df(x) and partial df(x) may be written as

Because J is nonsingular, its inverse J-1 exists. Hence,

Substituting for partial d(Y) in the equation for partial df(x) gives partial d f as a function of partial d ( Z ) -that is,

From this equation, the constrained derivative with respect to the independent vector Z is given by

The sufficiency conditions are similar to those developed in Section . The

Hessian matrix will correspond to the independent vector Z, and the elements of the

Hessian matrix must be the constrained second derivatives. To show how this is obtained,

Let

It thus follows that the “i” th row of the (constrained) Hessian matrix is a Notice that W is a function of Y and Y is a function of Z. Thus, the partial derivative of with respect to Zi is based on the following chain rule:

Example: 1 Consider the following problem:

Hence, the incremental value of constrained f is given as

Example: 2 Application of the Jacobian Method to an LP Problem : Consider the linear program

Maximize z = 2x1 + 3x2

subject to

Xl + X2 + X3 = 5

Xl – X2 + X4 = 3

Xl , X2, X3, X4 > 0

To account for the nonnegativity constraints , substitute . With this substitution, the nonnegativity conditions become implicit and the original problem becomes

Subject to

To apply the Jacobian method, let

(In the terminology of linear programming, Y and Z correspond to the basic and non basic variables, respectively.) Thus

So that,

The corresponding dual objective value is 5UI + 3U2 = 15, which equals the optimal

primal objective value. The given solution also satisfies the dual constraints and hence

is optimal and feasible. This shows that the sensitivity coefficients are the same as the

dual variables. In fact, both have the same interpretation.

Figure: Extreme points of the solution space of the linear program