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Process Optimization

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Tier I: Mathematical Methods of Optimization

Section 3:

Nonlinear Programming


Introduction to Nonlinear Programming

• We already talked about a basic aspect of nonlinear programming (NLP) in the Introduction Chapter when we considered unconstrained optimization.


Introduction to Nonlinear Programming

• We optimized one-variable nonlinear functions using the 1st and 2nd derivatives.

• We will use the same concept here extended to functions with more than one variable.


Multivariable Unconstrained Optimization

• For functions with one variable, we use the 1st and 2nd derivatives.

• For functions with multiple variables, we use identical information that is the gradient and the Hessian.

• The gradient is the first derivative with respect to all variables whereas the Hessian is the equivalent of the second derivative


The Gradient

• Review of the gradient ():

For a function “f ”, of variables x1, x2, …, xn:

nx

f

x

f

x

ff

21

Example: 231

321 )(3)(215 xxxxf

312

22

3 6)(6)(315 xxxxf


The Hessian

• The Hessian (2) of f(x1, x2, …, xn) is:

2

2

2

2

1

2

2

2

22

2

12

21

2

21

2

21

2

2

nnn

n

n

x

f

xx

f

xx

f

xx

f

x

f

xx

fxx

f

xx

f

x

f

f


Hessian Example

• Example (from previously):2

313

21 )(3)(215 xxxxf

13

2

32

606

0120

600

xx

x

x

f

312

22

3 6)(6)(315 xxxxf


Unconstrained Optimization

The optimization procedure for multivariable functions is:

1. Solve for the gradient of the function equal to zero to obtain candidate points.

2. Obtain the Hessian of the function and evaluate it at each of the candidate points

• If the result is “positive definite” (defined later) then the point is a local minimum.

• If the result is “negative definite” (defined later) then the point is a local maximum.


Positive/Negative Definite

• A matrix is “positive definite” if all of the eigenvalues of the matrix are positive (> 0)

• A matrix is “negative definite” if all of the eigenvalues of the matrix are negative (< 0)


Positive/Negative Semi-definite

• A matrix is “positive semi-definite” if all of the eigenvalues are non-negative (≥ 0)

• A matrix is “negative semi-definite” if all of the eigenvalues are non-positive (≤ 0)


Example Matrix

Given the matrix A:

211

175

542

A

The eigenvalues of A are:22 702.31 702.23

This matrix is negative definite


Unconstrained NLP Example

Consider the problem:

Minimize f(x1, x2, x3) = (x1)2 + x1(1 – x2) + (x2)2 – x2x3 + (x3)2 + x3

First, we find the gradient with respect to xi:

1 2

1 2 3

2 3

2 1

2

2 1

x x

f x x x

x x



Next, we set the gradient equal to zero:

0f

0

0

0

12

2

12

32

321

21

xx

xx

xx

So, we have a system of 3 equations and 3 unknowns. When we solve, we get:

1

1

1

3

2

1

x

x

x

x



So we have only one candidate point to check.

Find the Hessian:

210

121

0122 f



The eigenvalues of this matrix are:

414.31 586.02 23

All of the eigenvalues are > 0, so the Hessian is positive definite.

So, the point is a minimum

1

1

1

3

2

1

x

x

x

x



Unlike in Linear Programming, unless we know the shape of the function being minimized or can determine whether it is convex, we cannot tell whether this point is the global minimum or if there are function values smaller than it.


Method of Solution

• In the previous example, when we set the gradient equal to zero, we had a system of 3 linear equations & 3 unknowns.

• For other problems, these equations could be nonlinear.

• Thus, the problem can become trying to solve a system of nonlinear equations, which can be very difficult.


Method of Solution

• To avoid this difficulty, NLP problems are usually solved numerically.

• We will now look at examples of numerical methods used to find the optimum point for single-variable NLP problems. These and other methods may be found in any numerical methods reference.


Newton’s Method

When solving the equation f (x) = 0 to find a minimum or maximum, one can use the iteration step:

)(

)(''

'1

k

kkk

xf

xfxx

where k is the current iteration.

Iteration is continued until |xk+1 – xk| < where is some specified tolerance.


Newton’s Method Diagram

Newton’s Method approximates f (x) as a straight line at xk and obtains a new point (xk+1), which is used to approximate the function at the next iteration. This is carried on until the new point is sufficiently close to x*.

xx* xk+1 xk

Tangent of f (x) at xk

f (x)


Newton’s Method Comments

• One must ensure that f (xk+1) < f (xk) for finding a minimum and f (xk+1) > f (xk) for finding a maximum.

• Disadvantages:– Both the first and second derivatives must be

calculated– The initial guess is very important – if it is not

close enough to the solution, the method may not converge


Regula-Falsi Method

This method requires two points, xa & xb that bracket the solution to the equation f (x) = 0.

)()(

)()(''

'

ab

abbbc

xfxf

xxxfxx

where xc will be between xa & xb. The next interval will be xc and either xa or xb, whichever has the sign opposite of xc.


Regula-Falsi Diagram

The Regula-Falsi method approximates the function f (x) as a straight line and interpolates to find the root.

xxa

xb

xc

x*

f (x)


Regula-Falsi Comments

• This method requires initial knowledge of two points bounding the solution

• However, it does not require the calculation of the second derivative

• The Regula-Falsi Method requires slightly more iterations to converge than the Newton’s Method


Multivariable Optimization

• Now we will consider unconstrained multivariable optimization

• Nearly all multivariable optimization methods do the following:

1. Choose a search direction dk

2. Minimize along that direction to find a new point:

where k is the current iteration number and k is a positive scalar called the step size.

kkkk dxx 1


The Step Size

• The step size, k, is calculated in the following way:

• We want to minimize the function f(xk+1) = f(xk +kdk) where the only variable is k

because xk & dk are known.

• We set and solve for k

using a single-variable solution method

such as the ones shown previously.

0

d

d

k

kkkf

dx


Steepest Descent Method

• This method is very simple – it uses the gradient (for maximization) or the negative gradient (for minimization) as the search direction:

)( kk f xd

for

min

max

So, )(1 kkkk f xxx


Steepest Descent Method

• Because the gradient is the rate of change of the function at that point, using the gradient (or negative gradient) as the search direction helps reduce the number of iterations needed

x1

x2 f(x) = 5

f(x) = 20

f(x) = 25

xk

f(xk)

-f(xk)


Steepest Descent Method Steps

So the steps of the Steepest Descent Method are:

1. Choose an initial point x0

2. Calculate the gradient f(xk) where k is the iteration number

3. Calculate the search vector:

4. Calculate the next x:Use a single-variable optimization method to determine k.

)( kk f xd kkkk dxx 1


Steepest Descent Method Steps

5. To determine convergence, either use some given tolerance 1 and evaluate:

for convergence

Or, use another tolerance 2 and evaluate:

for convergence

11 )()( kk ff xx

2)( kf x


Convergence

• These two criteria can be used for any of the multivariable optimization methods discussed here

Recall: The norm of a vector, ||x|| is given by:22

22

1 )()()( nT xxx xxx


Steepest Descent Example

Let’s solve the earlier problem with the Steepest Descent Method:


Let’s pick

0

0

00x



10000001)0(2)( 00 xd f

122)1(2)( 3232121 xxxxxxxf x

101101

101000 01 x

Now, we need to determine 0



2)(4)( 0

0

1

d

df x

)()(00)1)(()()( 0200201 xf

)(2)(2 020

Now, set equal to zero and solve:

21

420 2)(4 0



So,

101000 01 x

2

10

2

1000

2

10

2

11x



110

2

10

2

1011)( 11 xd f

0101 d

Take the negative gradient to find the next search direction:



0102

10

2

1 12

x

2

1

2

1 1

Update the iteration formula:



1)(2)( 1

1

1

d

df x

2

1

4

1

2

1)()(1

2

1

4

1)( 12112

xf

21)( 121

Insert into the original function & take the derivative so that we can find 1:



Now we can set the derivative equal to zero and solve for 1:

211 1)(2 1



Now, calculate x2:

0102

10

2

1 12

x

0

2

10

2

10

2

1

2

1

2

1

2

12x



So,

11

2

1

2

11

2

1

2

111)( 22 xd f

2

10

2

12d

2

10

2

1

2

1

2

1

2

1 23 x

)1(

2

1

2

1)1(

2

1 22



2

3)1(

)( 22

3

d

df x

4

1)1(

2

3)1(

2

1)( 2223 xf

Set the derivative equal to zero and solve:

212

2

3)1( 2

Find 2:



Calculate x3:

2

10

2

1

2

1

2

1

2

1 23 x

4

10

4

1

2

1

2

1

2

1

4

3

2

1

4

33x



0

2

100

2

10)( 33 xd f

0

2

10

4

3

2

1

4

3 34 x

4

3)1(

2

1

4

3 3

Find the next search direction:



08

9)1(

2

1)( 33

4

d

df x

2

3)(

2

3)1(

4

1)( 3234 xfFind 3:

4

53



So, x4 becomes:

0

8

50

4

3

2

1

4

34x

4

3

8

9

4

34x



8

5

4

3

8

5

8

5

4

3

8

5)( 44 xd f

8

5

4

3

8

5

4

3

8

9

4

3 45 x

)

2

53(

4

1)

2

3(

4

3)

2

53(

4

1 444

The next search direction:



032

43

16

73)( 44

5

d

df x

64

51

32

43)(

32

73)( 4245 xf

146434

Find 4:



Update x5:

8

5

4

3

8

5

146

43

4

3

8

9

4

35x

1168

1091

73

66

1168

10915x



Let’s check to see if the convergence criteria is satisfied

Evaluate ||f(x5)||:

584

21

584

35

584

21)( 5xf

0786.058421

58435

58421)(

2225 xf



So, ||f(x5)|| = 0.0786, which is very small and we can take it to be close enough to zero for our example

Notice that the answer of

1168

1091

73

66

1168

1091x

is very close to the value of that we obtained analytically

111* x


Quadratic Functions

• Quadratic functions are important for the next method we will look at

• A quadratic function can be written in the form: xTQx where x is the vector of variables and Q is a matrix of coefficients

Example: )2(x x2x )(x2

210

121

0122

2212

1

3

2

1

321

x

x

x

xxx

)2(x x2x– 2332


Conjugate Gradient Method

• The Conjugate Gradient Method has the property that if f(x) is quadratic, it will take exactly n iterations to converge, where n is the number of variables in the x vector

• Although it works especially well with quadratic functions, this method will work with non-quadratic functions also


Conjugate Gradient Steps

1. Choose a starting point x0 and calculate f(x0). Let d0 = -f(x0)

2. Calculate x1 using:Find 0 by performing a single-variable optimization on f(x0 +0d0) using the methods discussed earlier. (See illustration after algorithm explanation)

0001 dxx



3. Calculate f(x1) and f(x1). The new search direction is calculated using the equation:

)()(

)()()(

00

11011

xx

xxdxd

ff

fff

T

T

This can be generalized for the kth iteration:

)()(

)()()(

1111

kkT

kkTkkk

ff

fff

xx

xxdxd



4. Use either of the two methods discussed earlier to determine tolerance:

11 )()( kk ff xx

2)( kf x

Or,


Number of Iterations

• For quadratic functions, this method will converge in n iterations (k = n)

• For non-quadratic functions, after n iterations, the algorithm cycles again with dn+1 becoming d0.


Step Size for Quadratic Functions

• When optimizing the step size, we can approximate the function to be optimized in the following manner:

)()()()()()( 2

2

1 kkTkkkTkkk ffff dxddxxdx

• For a quadratic function, this is not an approximation – it is exact



We take the derivative of that function with respect to and set it equal to zero:

0)()()()( 2

kkTkkkTkk

ffd

dfdxddx

dx

The solution to this equation is:

kkTk

kkToptk

f

f

dxd

dx

)()(

)(2

,



• So, for the problem of optimizing a quadratic function,

kkTk

kkToptk

f

f

dxd

dx

)()(

)(2

,

is the optimum step size.

• For a non-quadratic function, this is an approximation of the optimum step size.


Multivariable Newton’s Method

We can approximate the gradient of f at a point x0 by:

)()()()( 0020 xxxxx fff

We can set the right-hand side equal to zero and rearrange to give:

)()( 01020 xxxx ff


Multivariable Newton’s Method

We can generalize this equation to give an iterative expression for the Newton’s Method:

)()(121 kkkk ff xxxx

where k is the iteration number


Newton’s Method Steps

1. Choose a starting point, x0

2. Calculate f(xk) and 2f(xk)

3. Calculate the next x using the equation

4. Use either of the convergence criteria discussed earlier to determine convergence. If it hasn’t converged, return to step 2.

)()(121 kkkk ff xxxx


Comments on Newton’s Method

• We can see that unlike the previous two methods, Newton’s Method uses both the gradient and the Hessian

• This usually reduces the number of iterations needed, but increases the computation needed for each iteration

• So, for very complex functions, a simpler method is usually faster


Newton’s Method Example

For an example, we will use the same problem as before:


12212)( 3232121 xxxxxxxf x



The Hessian is:

210

121

012

)(2 xf

And we will need the inverse of the Hessian:

43

21

41

2112

14

12

14

3

210

121

012

)(

1

12 xf



So, pick

0

0

00x

Calculate the gradient for the 1st iteration:

100000100)( 0 xf

101)( 0 xf


Newton’s Method ExampleSo, the new x is:

11 0 2 0 0( ) ( )

3 1 14 2 40 1

1 1 0 1 02 20 131 1

4 2 4

f f

x x x x

1

1

11x



Now calculate the new gradient:

000121121112)( 1 xf

Since the gradient is zero, the method has converged


Comments on Example

• Because it uses the 2nd derivative, Newton’s Method models quadratic functions exactly and can find the optimum point in one iteration.

• If the function had been a higher order, the Hessian would not have been constant and it would have been much more work to calculate the Hessian and take the inverse for each iteration.


Constrained Nonlinear Optimization

• Previously in this chapter, we solved NLP problems that only had objective functions, with no constraints.

• Now we will look at methods on how to solve problems that include constraints.


NLP with Equality Constraints

• First, we will look at problems that only contain equality constraints:

Minimize f(x) x = [x1 x2 … xn]

Subject to: hi(x) = bi i = 1, 2, …, m


Illustration

Consider the problem:

Minimize x1 + x2

Subject to: (x1)2 + (x2)2 – 1 = 0

The feasible region is a circle with a radius of one. The possible objective function curves are lines with a slope of -1. The minimum will be the point where the lowest line still touches the circle.


Graph of Illustration

f(x) = 1

f(x) = 0

f(x) = -1.414

Feasible region

707.0

707.0*x

)(xf

The gradient of f points in the direction of

increasing f


More on the Graph

• Since the objective function lines are straight parallel lines, the gradient of f is a straight line pointing toward the direction of increasing f, which is to the upper right

• The gradient of h will be pointing out from the circle and so its direction will depend on the point at which the gradient is evaluated.


Further Details

f(x) = 1

f(x) = 0

f(x) = -1.414

Feasible region

707.0

707.0*x

)( *xh

)( *xf

0

11x

)( 1xf

Tangent Plane

)( 1xhx1

x2


Conclusions

• At the optimum point, f(x) is perpendicular to h(x)

• As we can see at point x1, f(x) is not perpendicular to h(x) and we can move (down) to improve the objective function

• We can say that at a max or min, f(x) must be perpendicular to h(x) – Otherwise, we could improve the objective

function by changing position


First Order Necessary Conditions

So, in order for a point to be a minimum (or maximum), it must satisfy the following equation:

This equation means that f(x*) and h(x*) must be in exactly opposite directions at a minimum or maximum point

0)()( *** xx hf


The Lagrangian Function

To help in using this fact, we introduce the Lagrangian Function, L(x,):

)()(),( xxx hfL

Review: The notation x f(x,y) means the gradient of f with respect to x.

So, )()(),( xxx xxx hfL



• So, using the new notation to express the First Order Necessary Conditions (FONC), if x* is a minimum (or maximum) then

0),( ** ,

xx xL

and to ensure feasibility.0)( * xh



• Another way to think about it is that the one Lagrangian function includes all information about our problem

• So, we can treat the Lagrangian as an unconstrained optimization problem with variables x1, x2, …, xn and 1, 2, …, m.

We can solve it by solving the equations

0x

L

& 0λ

L


Using the FONC

Using the FONC for the previous example,

)()(),( xxx hfL 12

22

121 xxxx

0

0),(

2

1

xL

xL

Lx x

And the first FONC equation is:


FONC Example

This becomes:

&

021 11

xx

L

021 22

xx

L

The feasibility equation is:

or,

0122

21 xx

0122

21

xxL


FONC Example

So, we have three equations and three unknowns.

When they are solved simultaneously, we obtain

707.021 xx & 707.0

We can see from the graph that positive x1 & x2 corresponds to a maximum while negative x1 & x2 corresponds to the minimum.


FONC Observations

• If you go back to the LP Chapter and look at the mathematical definition of the KKT conditions, you may notice that they look just like our FONC that we just used

• This is because it is the same concept

• We simply used a slightly different derivation this time but obtained the same result


Limitations of FONC

• The FONC do not guarantee that the solution(s) will be minimums/maximums.

• As in the case of unconstrained optimization, they only provide us with candidate points that need to be verified by the second order conditions.

• Only if the problem is convex do the FONC guarantee the solutions will be extreme points.


Second Order Necessary Conditions (SONC)

For where

and for y where

)()(),( 222 xxx hfLx ),(2 xLx

0y

x

xyxJ

x

)(

1

*

*

)(m

hh

h

If x* is a local minimum, then 0),( *2 yxy LxT


Second Order Sufficient Conditions (SOSC)

• y can be thought of as being a tangent plane as in the graphical example shown previously– Jh is just the gradients of each h(x) equation

and we saw in the example that the tangent plane must be perpendicular to h(x) and that is why

0),( *2 yxy LxT

0yJ )(xh


The y Vector

x1

x3

x2

Tangent Plane(all possible y vectors))(xh

0)( xh

*x

The tangent plane is the location of all y vectors and intersects with x*

It must be orthogonal (perpendicular) to h(x)


Maximization Problems

• The previous definitions of the SONC & SOSC were for minimization problems

• For maximization problems, the sense of the inequality sign will be reversed

For maximization problems:

SONC:

SOSC:

0),( *2 yxy LxT

0),( *2 yxy LxT


Necessary & Sufficient

• The necessary conditions are required for a point to be an extremum but even if they are satisfied, they do not guarantee that the point is an extremum.

• If the sufficient conditions are true, then the point is guaranteed to be an extremum. But if they are not satisfied, this does not mean that the point is not an extremum.


Procedure

1. Solve the FONC to obtain candidate points.

2. Test the candidate points with the SONC– Eliminate any points that do not satisfy the

SONC

3. Test the remaining points with the SOSC– The points that satisfy them are min/max’s – For the points that do not satisfy, we cannot

say whether they are extreme points or not


Problems with Inequality Constraints

We will consider problems such as:

Minimize f(x)

Subject to: hi(x) = 0 i = 1, …, m

& gj(x) ≤ 0 j = 1, …, p

An inequality constraint, gj(x) ≤ 0 is called “active” at x* if gj(x*) = 0.Let the set I(x*) contain all the indices of the active constraints at x*:

0)( * xjg for all j in set I(x*)


Lagrangian for Equality & Inequality Constraint

ProblemsThe Lagrangian is written:

We use ’s for the equalities & ’s for the inequalities.

p

jjji

m

ii ghfL

11

)()()(),,( xxxμλx


FONC for Equality & Inequality Constraints

For the general Lagrangian, the FONC become

p

jjji

m

ii ghfL

1

*

1

* *)(*)(*)(*)*,*,( 0xxxμλx

and the complementary slackness condition:

,0)( ** xjj g,0* j pj ,1


SONC for Equality & Inequality Constraints

The SONC (for a minimization problem) are:

where as before.

0yxy x ),,( ***2 LT

0yxJ )( *

This time, J(x*) is the matrix of the gradients of all the equality constraints and only the inequality constraints that are active at x*.


SOSC for Equality & Inequality Constraints

• The SOSC for a minimization problem with equality & inequality constraints are:

0yxy ),,( ***2 LxT


Generalized Lagrangian Example

• Solve the problem:

Minimize f(x) = (x1 – 1)2 + (x2)2

Subject to: h(x) = (x1)2 + (x2)2 + x1 + x2 = 0

g(x) = x1 – (x2)2 ≤ 0

The Lagrangian for this problem is: 21

221

22

21 2)1(),,( xxxxxxL x

221 xx



• The first order necessary conditions:

0212 111

xxx

L

0222 2222

xxxx

L

0212

22

1

xxxxL

0221 xx



Solving the 4 FONC equations, we get 2 solutions:

4534.0

2056.0)1(x 9537.045.0 &

and

0

0)2(x 20 &2)

1)



Now try the SONC at the 1st solution:

Both h(x) & g(x) are active at this point (they both equal zero). So, the Jacobian is the gradient of both functions evaluated at x(1):

9068.01

0932.0411.1

21

1212

)1(2

211

x

xJx

xx



The only solution to the equation:

0yxJ )( )1(

is:

00

00y

And the Hessian of the Lagrangian is:

993.00

09.2

2220

022

)1(

2

x

x

L



So, the SONC equation is:

0

00

00

00

00

993.00

09.2

00

00

This inequality is true, so the SONC is satisfied for x(1) and it is still a candidate point.



The SOSC equation is:

And we just calculated the left-hand side of the equation to be the zero matrix. So, in our case for x2:

0yxy x ),,( ***2 LT

0yyxx

00

00)1(

2LT

So, the SOSC are not satisfied.



For the second solution:

Again, both h(x) & g(x) are active at this point. So, the Jacobian is:

01

11

21

1212

)2(2

21)2(

x

xJx

xx



The only solution to the equation:

0yxJ )( )2(

is:

00

00y

And the Hessian of the Lagrangian is:

20

02

2220

022

)2(

2

x

x

L



So, the SONC equation is:

0

00

00

00

00

20

02

00

00

This inequality is true, so the SONC is satisfied for x(2) and it is still a candidate point



The SOSC equation is:

And we just calculated the left-hand side of the equation to be the zero matrix. So, in our case for x2:

0yxy x ),,( ***2 LT

0yyxx

00

00)2(

2LT

So, the SOSC are not satisfied.


Example Conclusions

• So, we can say that both x(1) & x(2) may be local minimums, but we cannot be sure because the SOSC are not satisfied for either point.


Numerical Methods

• As you can see from this example, the most difficult step is to solve a system of nonlinear equations to obtain the candidate points.

• Instead of taking gradients of functions, automated NLP solvers use various methods to change a general NLP into an easier optimization problem.


Excel Example

Let’s solve the previous example with Excel:

Minimize f(x) = (x1 – 1)2 + (x2)2

Subject to: h(x) = (x1)2 + (x2)2 + x1 + x2 = 0

g(x) = x1 – (x2)2 ≤ 0


Excel Example

We enter the objective function and constraint equations into the spreadsheet:

Variables:x1 x20 0

Value LimitObjective Function: =(A3-1) 2̂+B3 2̂

Constraint 1: =A3 2̂+B3 2̂+A3+B3 0Constraint 2: =A3-B3 2̂ 0


Excel Example

Now, open the solver dialog box under the Tools menu and specify the objective function value as the target cell and choose the Min option. As it is written, A3 & B3 are the variable cells. And the constraints should be added – the equality constraint and the ≤ constraint.


Excel Example

The solver box should look like the following:


Excel Example

• This is a nonlinear model, so unlike the examples in the last chapter, we won’t choose the “Assume Linear Model” in the options menu

• Also, x1 & x2 are not specified to be positive, so we don’t check the “Assume Non-negative” box

• If desired, the tolerance may be decreased to 0.1%


Excel Example

• When we solve the problem, the spreadsheet doesn’t change because our initial guess of x1 = 0 & x2 = 0 is an optimum solution, as we found when we solved the problem analytically.


Excel Example

However, if we choose initial values of both x1 & x2 as -1, we get the following solution


Conclusions

• So, by varying the initial values, we can get both of the candidate points we found previously

• However, the NLP solver tells us that they are both local minimum points


References

Material for this chapter has been taken from:

• Optimization of Chemical Processes 2nd Ed.; Edgar, Thomas; David Himmelblau; & Leon Lasdon.

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