1.On P. 196, Fig. E6.2, the Contours of F at (0,0.5) Appear Not to Be Convex. Show

ChE 356

Exam #2

In-Class

1.On p. 196, Fig. E6.2, the contours of f at (0,0.5) appear not to be convex. Show

(25)that the Hessian matrix is not convex at that point.

The function is given by

The Hessian is defined as

Calculating the partial derivatives gives

Evaluating the Hessian at the point (0, 0.5) gives

Because this matrix has a zero on the off diagonal, the eigenvalues are given by the values on the diagonal.

With one positive eigenvalue and one negative eigenvalues, the matrix is indefinite, therefore the function is not convex at the point (0, 0.5).

2.Show that is normal to for a starting point of (1, 1) using the

(25)steepest descent method. You will need to calculate the optimum step size for , then find a new point x2. The quadratic objective function is

Note: this is the same optimization problemas in 6.14 in your 3/30/07 homework.

To show that the two directions are orthogonal we take

As the inner product is zero, the vectors are orthogonal.

Take-home

3.A company has two alkylate plants, A1 and A2, from which a given product is

(30)distributed to customers C1, C2, and C3. The transportation costs are given as follows:

RefineryA1A1A1A2A2A2

CustomerC1C2C3C1C2C3

Cost ($/ton)256075205085

The maximum refinery production rates and minimum customer demand rates are fixed and known to be as follows:

Customer or refineryA1A2C1C2C3

Rate (tons/day)1.60.80.90.70.3

The cost of production for A1 is $30/ton for production levels less than or equal to 0.5 ton/day: for production levels greater than 0.5 ton/day, the production cost is $40/ton. A2’s production cost is uniform at $35/ton.

Find the optimum distribution policy to minimize the company’s total costs.

We may define the problem with a simple block diagram

Let Alkylate from A1 to C1 (ton/day)

Alkylate from A1 to C2 (ton/day)

Alkylate from A1 to C3 (ton/day)

Alkylate from A2 to C1 (ton/day)

Alkylate from A2 to C2 (ton/day)

Alkylate from A2 to C3 (ton/day)

Total cost = transportation cost + production cost

Transportation cost = ($/day)

Production cost = ($/day)

Objective function ($/day):

Inequality Constraints:

Production Capacity

Refinery A1 (ton/day)

Refinery A2 (ton/day)

Consumer Demand

Consumer C1 (ton/day)

Consumer C2 (ton/day)

Consumer C3 (ton/day)

Non-negativity

Let’s assume that . We modify the objective function (but not the constraint) and plug it into Excel to get the answer:

However, we can see that , so our assumption was incorrect. If we use the other constraint and change the objective function accordingly, then we get

4.In Marquardt’s method, the Hessian matrix is augmented with a positive diagonal

(20)matrix so that the eigenvalues of the resulting matrix, , are positive. When  is large, show this modification is equivalent to the gradient method when Newton’s method is employed.

The algorithms are essentially identical, except when the new search direction is calculated. In Marquardt’s method the new search direction is calculated as

While, for the gradient method the new search direction is simply

If is sufficiently large, we may say

Therefore, we have

Notice that and are the same direction, with different magnitudes. As the magnitude of the direction does not matter in the algorithm (one method will have a larger optimal step size than the other, but both methods will reach the same successive point), we can say that the two algorithms are essentially identical.