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Lecture 8 – Nonlinear Programming ModelsPowerPoint Presentation

Lecture 8 – Nonlinear Programming Models

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Lecture 8 – Nonlinear Programming Models

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Topics

- General formulations
- Local vs. global solutions
- Solution characteristics
- Convexity and convex programming
- Examples

In LP ... the objective function & constraints are linear and the problems are “easy”to solve.

Many real-world engineering and business problems have nonlinear elements and are hard to solve.

Minimize f(x)

s.t. gi(x) (, , =) bi, i = 1,…,m

x = (x1,…,xn) is the n-dimensional vector of decision variables

f(x) is the objective function

gi(x) are the constraint functions

bi are fixed known constants

…

c

x

c

x

c

x

Example 2

Max f(x) = e

e

e

1

1

2

2

n

n

s.t. Ax = b, x³0

n

å

fj (xj)

Example 3

Min

Problems with

“decreasing efficiencies”

j =1

s.t. Ax = b, x³0

fj(xj)

where each fj(xj)is of the form

xj

Examples of NLPs

4

Example 1

Max f(x) =

3x1 + 2x2

2

s.t. x1 + x2£ 1, x1³ 0, x2 unrestricted

Examples 2 and 3 can be reformulated as LPs

Max f(x1, x2) = x1x2

s.t.4x1 + x2£ 8

x1 ³ 0, x2 ³ 0

x2

8

f(x) = 2

f(x) = 1

x1

2

Optimal solution will lie on the line g(x) = 4x1 + x2 – 8 = 0.

Solution Characteristics

Gradient of f(x) = f(x1, x2) (f/x1, f/x2)T

This gives f/x1 = x2, f/x2 = x1

and g/x1 = 4, g/x2 = 1

At optimality we have f(x1, x2) = g(x1, x2)

or x1* = 1 and x2* = 4

- Solution is not a vertex of feasible region. For this particular problem the solution is on the boundary of the feasible region. This is not always the case.
- In a more general case, f(x1, x2) = g(x1, x2) with 0. (In this case, = 1.)

Nonconvex Function

global

max

stationary

point

f(x)

local

max

local

min

local

min

x

Let Sn be the set of feasible solutions to an NLP.

Definition: A global minimum is any x0S such than

f(x0) f(x)

for all feasible x not equal to x0.

Function with Unique Global Minimum at x = (1, –3)

If g1 = x1³ 0 and g2 = x2³ 0, what is the optimum ?

At (1, 0), f(x1, x2) = 1g1(x1, x2) + 2g1(x1, x2)

or (0, 6) = 1(1, 0) + 2(0, 1), 1³ 0, 2³ 0

so 1 = 0 and 2 = 6

Function with Multiple Maxima and Minima

Min { f(x)= sin(x) : 0 x 5p}

Constrained Function with Unique Global Maximum and Unique Global Minimum

Convexity condition for

univariate f :

d2 f(x)

≥ 0 for all x

dx2

Convex function:If you draw a straight line between any two points on f(x) the line will be above or on the line.

Concave function:If f(x) is convex than –f(x) is concave.

Linear functions are both convex and concave.

Let x1 and x2 be two points in Sn. A function f(x) is convex if and only if

f(lx1 + (1–l)x2) ≤ lf(x1) + (1–l)f(x2)

for all 0 < l < 1. It is strictly convex if the inequality sign ≤ is replaced with the sign <.

1-dimensional example

Nonconvex -- Nonconave Function

f(x)

x

2

2

2

d

f

d

f

d

f

Hessian of f at x : s2f(x) =

. . .

dx

dx

dx

dx

2

dx

1

2

1

n

1

.

.

2

d

f

.

.

.

dx

dx

.

2

1

.

.

.

2

2

d

f

d

f

. . .

dx

dx

2

dx

n

1

n

A positively weighted sum of convex functions is convex:

if fk(x) k =1,…,m are convex and 1,…,m³ 0,

then f(x) = å ak fk(x) is convex.

m

k =1

f(x)

x1 x2

One-Dimensional Functions:

A function f(x) ÎC1 is convex if and only if it is underestimated by linear extrapolation; i.e.,

f(x2) ≥ f(x1) + (df(x1)/dx)(x2 – x1) for all x1 and x2.

A function f(x) C2 is convex if and only if its second derivative is nonnegative.

d2f(x)/dx2≥ 0 for all x

If the inequality is strict (>), the function is strictly convex.

Example: f(x) = 3(x1)2 + 4(x2)3 – 5x1x2 + 4x1

f(x) is convex if only if f(x2) ≥ f(x1) + Tf(x1)(x2 – x1) for all x1 and x2.

Definition: The Hessian matrix H(x) associated with f(x) is the nn symmetric matrix of second partial derivatives of f(x) with respect to the components of x.

When f(x) is quadratic, H(x) has only constant terms; when f(x) is linear, H(x) does not exist.

How can we use Hessian to determine whether or not f(x) is convex?

- H(x) is positive definite if and only if xTHx> 0 for all x0.
- H(x) is positive semi-definite if and only if xTHx≥ 0 for all x and there exists and x 0 such that xTHx = 0.
- H(x) is indefinite if and only if xTHx> 0 for some x, and xTHx< 0 for some other x.

- f(x) is strictly convex (convex) if its associated Hessian matrix H(x) is positive definite (semi- definite) for all x.
- f(x) is neither convex nor concave if its associated Hessian matrix H(x) is indefinite

The terms negativedefiniteandnegative-semi definiteare also appropriate for the Hessian and provide symmetric results for concave functions. Recall that a function f(x) is concave if –f(x) is convex.

Definition: The ith leading principal submatrix of H is the matrix formed taking the intersection of its first i rows and i columns. Let Hi be the value of the corresponding determinant:

, where hij= 2f(x)/xixj

LetHessian, H =

- H is positive definite if and only if the determinants of all the leading principal submatrices are positive; i.e., Hi> 0 for i = 1,…,n.
- His negative definite if and only if H1 < 0 and the remaining leading principal determinants alternate in sign:
- H2 > 0, H3 < 0, H4 > 0, . . .

Positive-semidefinite and negative semi-definiteness require that all principal submatrices satisfy the above conditions for the particular case.

Example 1: f(x) = 3x1x2 + x12 + 3x22

so H1 = 2 and H2 = 12 – 9 = 3

Conclusion f(x) is convex because H(x) is positive definite.

Example 2: f(x) = 24x1x2 + 9x12 + 16x22

- so H1 = 18 and H2 = 576 – 576 = 0
- Thus H is positive semi-definite (determinants of all submatrices are nonnegative) so f(x) is convex.

- Note, xTHx = 2(3x1 + 4x2)2≥ 0. For x1 = -4, x2 = 3, we get xTHx = 0.

Example 3: f(x) = (x2 – x12)2 + (1 – x1)2

Thus the Hessian depends on the point under consideration:

At x = (1, 1), which is positive definite.

At x = (0, 1), which is indefinite.

Thus f(x) is not convex although it is strictly convex near (1, 1).

- How to identify the decision variables.
- How to write constraints.
- How to identify a convex function.
- The difference between a local and global solution.