constrained non linear programming

7/27/2019 constrained non linear programming

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CONSTRAINED NONLINEAR PROGRAMMING

We now turn to methods for general constrained nonlinear programming. These may

be broadly classified into two categories:

1. TRANSFORMATION METHODS: In this approach the constrained nonlinear

program is transformed into an unconstrained problem (or a series of unconstrained

problems) and solved using one (or some variant) of the algorithms we have already

seen.

Transformation methods may further be classified as

(1) Exterior Penalty Function Methods

(2) Interior Penalty (or Barrier) Function Methods

(3) Augmented Lagrangian (or Multiplier) Methods

2. PRIMAL METHODS: These are methods that work directly on the original

problem.

We shall look at each of these in turn; we first start with transformation methods.


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Exterior Penalty Function Methods

These methods generate a sequence of infeasible points whose limit is an optimal

solution to the original problem. The original constrained problem is transformed into

an unconstrained one (or a series of unconstrained problems).The constraints are

incorporated into the objective by means of a "penalty parameter" which penalizes

any constraint violation. Consider the problem

Minf(x)

st gj(x) 0,j=1,2,,m, hj(x) = 0,j=1,2,,p, xR

n

Define a penalty functionp as follows:

Max0,

where is some positive integer. It is easy to see that

Ifgj(x) 0 then Max{0, gj(x)} = 0

Ifgj(x) > 0 then Max{0, gj(x)} = gj(x) (violation)

Now define the auxiliary function as

P(x)=f(x) + (x),where >> 0.Thus if a constraint is violated,

(x)>0 and a "penalty" of

(x) is incurred.


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The (UNCONSTRAINED) problem now is:

Minimize P(x)=f(x) + (x)

st xRn

EXAMPLE:

Minimizef(x)=x, st g(x)=2-x 0,xR (Note that minimum is atx* =2)

Let (x)= Max[{0,g(x)}]2 ,

i.e., 2 if 20if 2

If(x)=0, then the optimal solution to Minimize is atx*= - and this isinfeasible; so 2

Minimize hasoptimumsolution 2 12.

:As , 2

1

2 2

Penalty Function Auxiliary Function

3 2 1 0 1 2 1 0 1 2 3

1p

2p

p

f+ 2p

f+ 1p

1=0.5

2=1.5


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In general:

1. We convert the constrained problem to an unconstrained one by using thepenalty function.

2. The solution to the unconstrained problem can be made arbitrarily close to thatof the original one by choosing sufficiently large .

In practice, ifis very large, too much emphasis is placed on feasibility. Often,

algorithms for unconstrained optimization would stop prematurely because the step

sizes required for improvement become very small.

Usually, we solve a sequence of problems with successively increasing

values; the optimal point from one iteration becomes the starting point for the next

problem.

PROCEDURE: Choose the following initially:

1. a tolerance ,2. an increase factor,3. a starting pointx1, and4. an initial 1.

At Iteration i

1. Solve the problem Minimizef(x)+ip(x); stxX (usually Rn). Usexias thestarting point and let the optimal solution bex

i+1

2. Ifip(xi+1) < then STOP; else let i+1 =(i)and start iteration (i+1)


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EXAMPLE: Minimizef(x) =x12+ 2x2

2

st g(x) = 1x1 x2 0;xR2

Define the penalty function

1 if 00if 0

The unconstrained problem is

Minimizex12+ 2x2

2+ (x)

If(x)=0, then the optimal solution isx* =(0,0) INFEASIBLE!(x) = (1x1 x2)2, f(x) =x12 + 2x22+ [(1x1 x2)2], and the necessaryconditions for the optimal solution (f(x)=0) yield the following: 2 21 1 0,

4 21 1 0

Thus

2

2 3 and

2 3

Starting with =0.1, =10 andx1=(0,0) and using a tolerance of 0.005 (say), we have

the following:

Iter. (i) i xi+1 g(x

i+1) ip(x

i+1)

1

2

34

5

0.1

1.0

10100

1000

(0.087, 0.043)

(0.4, 0.2)

(0.625, 0.3125)(0.6622, 0.3311)

(0.666, 0.333)

0.87

0.40

0.06250.0067

0.001

0.0757

0.16

0.0390.00449

0.001

Thus the optimal solution isx*= (2/3, 1/3).


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INTERIOR PENALTY (BARRIER) FUNCTION METHODS

These methods also transform the original problem into an unconstrained one;

however the barrier functions prevent the current solution from ever leaving the

feasible region. These require that the interior of the feasible sets be nonempty, which

is impossible if equality constraints are present. Therefore, they are used with

problems having only inequality constraints.

A barrier functionB is one that is continuous and nonnegative over the interior

of {x | g(x)0}, i.e., over the set {x | g(x)


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Q. Why should be SMALL?

A. Ideally we would likeB(x)=0 ifgj(x)


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EXAMPLE: Minimizef(x) =x12+ 2x2

2

st g(x) = 1x1 x2 0;xR2

Define the barrier function

1The unconstrained problem is

Minimizex12+ 2x2

2+ (x) =x12 + 2x22 - 1

The necessary conditions for the optimal solution (f(x)=0) yield the following:

2 / 1 0, 4 / 1 0

Solving,weget 1 1 33 and 1 1 3

6

Since the negative signs lead to infeasibility, we and

Starting with =1, =0.1 andx

1

=(0,0) and using a tolerance of 0.005 (say), we have

the following:

Iter. (i) i xi+1 g(x

i+1) iB(x

i+1)

1

2

3

4

5

1

0.1

0.01

0.001

0.0001

(1.0, 0.5)

(0.714, 0.357)

(0.672, 0.336)

(0.6672, 0.3336)

(0.6666, 0.3333)

-0.5

-0.071

-0.008

-0.0008

-0.0001

0.693

0.265

0.048

0.0070

0.0009

Thus the optimal solution isx*= (2/3, 1/3).


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Penalty Function Methods and Lagrange Multipliers

Consider the penalty function approach to the problem below

Minimize f(x)

st gj(x) 0;j = 1,2,,m

hj(x) = 0;j = m+1,m+2,,l xkRn.

Suppose we use the usual interior penalty function described earlier, withp=2.

The auxiliary function that we minimize is then given by

P(x) =f(x) + p(x) =f(x) +{j[Max(0, gj(x))]2

+ j[hj(x)]2

}

=f(x) +{j[Max(0, gj(x))]2

+ j[hj(x)]2}

The necessary condition for this to have a minimum is that

P(x) = f(x) + p(x) = 0, i.e.,

f(x) + j2[Max(0, gj(x)]gj(x) + j2[hj(x)]hj(x) = 0 (1)

Suppose that the solution to (1) for a fixed (say k>0) is given byxk.

Let us also designate

2k[Max{0, gj(xk

)}] = j(k); j=1,2,,m (2)

2k[hj(xk)] = j(k); j=m+1,,l (3)

so that for=kwe may rewrite (1) as


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f(x) + jj(k) gj(x) + jj(k) hj(x) = 0 (4)

Now consider the Lagrangian for the original problem:

L(x,) =f(x) + jjgj(x) + jjhj(x)

The usual K-K-T necessary conditions yield

f(x) + j=1..m {jgj(x)} + j=m+1..l {jhj(x)} = 0 (5)

plus the original constraints, complementary slackness for the inequality constraints

and j0 forj=1,,m.

Comparing (4) and (5) we can see that when we minimize the auxiliary function

using =k, thej(k) values given by (2) and (3) estimate the Lagrange multipliers

in (5). In fact it may be shown that as the penalty function method proceeds and

kand thexkx*, the optimum solution, the values ofj(k) j*, the optimum

Lagrange multiplier value for constraintj.

Consider our example on page 153 once again. For this problem the Lagrangian is

given byL(x,)=x12

+ 2x22+ (1-x1-x2).

The K-K-T conditions yield

L/x1 = 2x1-= 0; L/x2 = 4x2-= 0; (1-x1-x2)=0

Solving these results inx1*=2/3;x2

*=1/3;

*=4/3; (=0 yields an infeasible solution).


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Recall that for fixed kthe optimum value ofxk was [2k/(2+3k) k/(2+3k)]T. As

we saw, when k, these converge to the optimum solution ofx*= (2/3,1/3).

Now, suppose we use (2) to define () = 2[Max(0, gj(x

k

)]

= 2[1 - {2/(2+3)} - {/(2+3)}] (since gj(xk)>0 if>0)

= 2[1 - {3/(2+3)}] = 4/(2+3)

Then it is readily seen that Lim() = Lim4/(2+3) = 4/3 = *

Similar statements can also be made for the barrier (interior penalty) function

approach, e.g., if we use the log-barrier function we have

P(x) =f(x) + p(x) =f(x) +j-log(-gj(x)),

so that

P(x) = f(x) - j{/gj(x)}gj(x) = 0 (6)

If (like before) for a fixed kwe denote the solution byxkand define -k/gj(x

k) =

j(k), then from (6) and (5) we see that j(k) approximates j. Furthermore, as

k0and thexk

x*

, it can be shown that j(k) j*

For our example (page 156) -k/gj(xk) =k/1

=

-2k/(1-1 3 4/3, as k0.


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Penalty and Barrier function methods have been referred to as SEQUENTIAL

UNCONSTRAINED MINIMIZATION TECHNIQUES (SUMT) and studied in

detail by Fiacco and McCormick. While they are attractive in the simplicity of the

principle on which they are based, they also possess several undesirable properties.

When the parameters are very large in value, the penalty functions tend to be

ill-behaved near the boundary of the constraint set where the optimum points usually

lie. Another problem is the choice of appropriate 1and values. The rate at which

change (i.e. the values) can seriously affect the computational effort to find a

solution. Also as increases, the Hessian of the unconstrained function becomes ill-

conditioned.

Some of these problems are addressed by the so-called "multiplier" methods.

Here there is no need forto go to infinity, and the unconstrained function is better

conditioned with no singularities. Furthermore, they also have faster rates of

convergence than SUMT.


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Ill-Conditioning of the Hessian Matrix

Consider the Hessian matrix of the Auxiliary function P(x)=x12+2x2

2+(1-x1-x2)

2:

H =

242

222. Suppose we want to find its eigenvalues by solving

|H-I| = (2+2-)*(4+2-) - 42

= 2

(6+4) + (8+12) = 0

This quadratic equation yields

= 14)23(2

Taking the ratio of the largest and the smallest eigenvalue yields

14)23(14)23(

2

2

. It should be clear that as , the limit of the preceding ratio

also goes to . This indicates that as the iterations proceed and we start to increase

the value of , the Hessian of the unconstrained function that we are minimizing

becomes increasingly ill-conditioned. This is a common situation and is especially

problematic if we are using a method for the unconstrained optimization that requires

the use of the Hessian.


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MULTIPLIER METHODS

Consider the problem: Minf(x), st gj(x)0,j=1,2,...,m.

In multiplier methods the auxiliary function is given by

P(x) =f(x) + Max0,

where j>0 and 0are parameters associated with thejth constraint. This is also

often referred to as theAUGMENTED LAGRANGIANfunction

Note that if=0 and j=, this reduces to the usual exterior penalty function.

The basic idea behind multiplier methods is as follows:

Letxibe the minimum of the auxiliary function Pi(x) at some iteration i. From the

optimality conditions we have

P 0,

Now, 0, , . Therefore P implies

,

Let us assume for a moment that are chosen at each iteration i so that they satisfy

the following:

(1)

It is easily verified that (1) is equivalent to requiring that

0, 0and 0

, 0


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Therefore as long as (2) is satisfied, we have for the pointxithat minimize Pi(x)

P (3)

If we let

and use the fact that j>0,then (2) reduces to

0, 0and 0, (A)and (3) reduces to

. (B)

It is readily seen that (A) and (B) are merely the KARUSH-KUHN-TUCKER

necessary conditions forxi to be a solution to the original problem below!!

Minf(x), st gj(x)0,j=1,2,...,m.

We may therefore conclude that

and

Herex*is the solution to the problem, and j

*is the optimum Lagrange multiplier

associated with constraintj.

From the previous discussion it should be clear that at each iteration jshould be

chosen in such a way that , 0, which results inxix*.Note thatxiisobtained here by minimizing the auxiliary function with respect tox.


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ALGORITHM: A general algorithmic approach for the multiplier method may now

be stated as follows:

STEP 0: Set i=0, choose vectorsxi, and .STEP 1: Set i=i+1.STEP 2: Starting atxi-1, Minimize Pi(x) to findxi.STEP 3: Check convergence criteria and go to Step 4 only if the criteria are not

satisfied.

STEP 4: Modify

based on satisfying (1). Also modify if necessary and go toStep 2.

(Usually is set to 0)

One example of a formula for changing is the following:

,


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PRIMAL METHODS

By a primal method we mean one that works directly on the original constrained

problem:

Minf(x)

st gj(x)0j=1,2,...,m.

Almost all methods are based on the following general strategy:

Letx

i

be a design at iteration i. A new designx

i+1

is found by the expressionx

i+1

=x

i

+ i, where iis a step size parameter and is a search direction (vector). Thedirection vectoris typically determined by an expression of the form:

where fand gj are gradient vectors of the objective and constraint functions andJis an active set" given byJ={j|gj(x

i)+ 0, >0}.

Unlike with transformation methods, it is evident that here the gradient vectors of

individual constraint functions need to be evaluated. This is a characteristic of all

primal methods.


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METHOD OF FEASIBLE DIRECTIONS

The first class of primal methods we look at are the feasible directions methods,

where eachxiis within the feasible region. An advantage is that if the process is

terminated before reaching the true optimum solution (as is usually necessary with

large-scale problems) the terminating point is feasible and hence acceptable. The

general strategy at iteration i is:

(1)Letxibe a feasible point.(2)

Choose a direction

so that

a)xi+ is feasible at least for some "sufficiently" small >0, andb)f(xi+)


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Further, if we also havef(x'+d)


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Geometric Interpretation

Minimizef(x), st gj(x) 0,j=1,2,,m

(i) (ii)In (ii)dany positive step in this direction is infeasible (however we will often

consider this a feasible direction)

FARKAS LEMMA: GivenARmn,bRm,xRn,yRm,the following statements

are equivalent to each other:

1. yTA 0 yTb 0

2. z such thatAz=b, z0

dTgj(x*)


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Application to NLP

Let b -f(x*) Ajgj(x*)y d(direction vector), zjjforjJx* {j |gj(x*)=0}

Farkas Lemma then implies that

(1) dTgj(x*) 0 jJx* -dTf(x*) 0, i.e.,dTf(x*) 0

(2) j0such that jJx*jgj(x*) = -f(x*)are equivalent! Note that (1) indicates that

directions satisfyingdTgj(x*) 0jJx* are feasible directions directions satisfying dTf(x*) 0 are ascent (non-improving) directions

On the other hand (2) indicates the K-K-T conditions. They are equivalent!

So at the optimum, there is no feasible direction that is improving

Similarly for the general NLP problem

g2(x*)

Cone generated by gradients

of tight constraints

K-K-T implies that the steepest descent direction is in

the cone generated by gradients of tight constraints

*

g3(x)=0

g2(x)=0

g3(x

*

)

-f(x*)

g1(x)=0

g1(x

*)

Feasible region


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Minimizef(x)

st gj(x) 0,j=1,2,...,m

hk(x) 0, k=1,2,...,p

the set of feasible directions is

Dx'= {d| gjT(x')d< 0 jJx', and hkT(x') d= 0 k}====================================================

In a feasible directions algorithm we have the following:

STEP 1 (direction finding)

To generate an improving feasible direction fromx', we need to finddsuch that

Fx'Dx', i.e.,

fT(x')d< 0 and gjT(x')d< 0 forjJx', hkT(x') d= 0 kWe therefore solve the subproblem

Minimize fT(x')dst gjT(x')d< 0 forjJx'

(0 forj gjlinear)

hkT (x')d= 0 for all k,along with some normalizing constraints such as

-1 di 1 fori=1,2,...,n or d2=dTd 1.

If the objective of this subproblem is negative, we have an improving feasible

directiondi.


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In practice, the actual subproblem we solve is slightly different since strict inequality

(


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NOTE: (1) In practice the setJx' is usually defined asJx'={ j |gj(xi)+0}, where the

tolerance iis referred to as the "constraint thickness," which may be

reduced as the algorithm proceeds.

(2) This procedure is usually attributed to ZOUTENDIJK

EXAMPLE

Minimize 2x12+x2

2- 2x1x2 - 4x1 - 6x2

st x1 + x2 8

-x1 + 2x2 10

-x1 0

-x2 0 (A QP)

For this, we have

1

1 , 1

2 , 1

0 , 0

1,

and 4 2 42 2 6.

Let us begin with 00,withf(x0)=0

ITERATION 1 J={3,4}. STEP 1 yields

Minz Minz

st fT(x1)dz d1(4x1-2x2-4) + d2(2x2-2x1 -6) zg3T(xi)dz -d1zg4T(xi)dz -d2z-1d1, d21 d1,d2 [-1,1]


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i.e., Minz

st -4d1- 6d2 z, -d1z, -d2z d1,d2 [-1,1]

yieldingz*

= -1 (


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Thus max = { | 8-8, -4+810, -40, -40} = 4

We therefore solve: Minimize 2(4-)2+ 4

2 2(4-)4 4(4-) 6(4), st 04.

=1=1

34

, withf(x2)= -26.

ITERATION 3 J={}. STEP 1 yields

Minz

st d1(4x1-2x2-4) + d2(2x2-2x1 -6) z

-1d1, d21

i.e., Minz

st -4d2 z, d1,d2 [-1,1]

yieldingz*

= -4 (


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ITERATION 4 J={1}. STEP 1 yields

Minz

st d1(4x1-2x2-4) + d2(2x2-2x1 -6) z

d1+ d2z

-1d1, d21

i.e., Minz

st -2d1- 10d2 z, d1d2z, d1,d2 [-1,1]

yieldingz* = 0, withd* = 00. STOP

Therefore the optimum solution is given by 35, withf(x*)= -29.

1 2 3 4 5 6 87

1

2

3

4

5

6

8

7

x0

x1

x2

x3

-x1+2 x2=10

x1+ x2=8

path followed by algorithm

feasible region


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GRADIENT PROJECTION METHOD (ROSEN)

Recall that the steepest descent direction is -f(x). However, for constrained

problems, moving along -f(x) may destroy feasibility. Rosen's method works by

projecting -f(x) on to the hyperplane tangent to the set of active constraints. By

doing so it tries to improve the objective while simultaneously maintaining

feasibility. It usesd=-Pf(x) as the search direction, whereP is a projection matrix.

Properties ofP (nn matrix)

1)PT =P andPTP = P (i.e.,Pis idempotent)2)P is positive semidefinite3)P is a projection matrix if, and only if,I-P is also a projection matrix.4) Let Q=I-P andp= Px1, q=Qx2 wherex1, x2Rn. Then

(a)pTq= q

Tp =0 (p and q are orthogonal)

(b) AnyxRn can be uniquely expressed asx=p+q, wherep=Px, q=(I-P)x

x2

x1

x

p

q

givenx, we can

findp andq


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Consider the simpler case where all constraints are linear. Assume that there are 2

linear constraints intersecting along a line as shown below.

Obviously, moving along -f would take us outside the feasible region. Say we

project -f on to the feasible region using the matrixP.

THEOREM: As long asP is a projection matrix andPf0,d=-Pfwill be animproving direction.

PROOF: fTd= fT(-Pf) = -fTPTPf= -Pf2 < 0. (QED)

For -Pf to also be feasible we must have g1T (-Pf) and g2T (-Pf) 0 (bydefinition).

-f

( -P)(-f)g2

g1

g1(x)=0 g2(x)=0

P(-f)


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Say we pickP such that g1T(-Pf) = g2T(-Pf) = 0, so that -Pf is a feasibledirection). Thus if we denoteMas the (2n) matrix where Row 1 is g1T andRow 2 is g2T, then we haveM(-Pf)= 0 (*)

To find the form of the matrixP, we make use of Property 4(b) to write the vector -fas -f= P(-f) + [I-P](-f) = -Pf - [I-P]f

Now, -g1 and -g2 are both orthogonal to -Pf, and so is [I-P](-f). Hence we must

have

[I-P](-f) =- [I-P]f= 1g1+ 2g2

-[I-P]f = MT, where = -M[I-P]f = MMT(MMT)-1M[I-P]f= (MMT)-1MMT= = { (MMT)-1Mf (MMT)-1 MPf}= { (MMT)-1Mf+(MMT)-1M(-Pf)}= (MMT)-1Mf (from (*) aboveM(-Pf)=0)

Hence we have

-f = -Pf - [I-P]f= -Pf +MT-Pf +MT(MMT)-1Mfi.e.,Pf = f -MT(MMT)-1Mf= [I-MT(MMT)-1M]fi.e., P = [I-M

T(MM

T)

-1]


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Question: What ifPf(x) = 0?Answer: Then 0 =Pf = [IMT(MMT)-1M]f= f +MTw, wherew = -(MM

T)

-1Mf. Ifw0,thenx satisfies the Karush-Kuhn-Tucker conditions and

we may stop; if not, a new projection matrix could be identified such thatd= -fis an improving feasible direction. In order to identify this matrix we merely pickany component ofw that is negative, and drop the corresponding row from the matrix

M. Then we use the usual formula to obtain .

====================

Now consider the general problem

Minimizef(x)

st gj(x) 0,j=1,2,...,m

hk(x) 0, k=1,2,...,p

The gradient projection algorithm can be generalized to nonlinear constraints by

projecting the gradient on to the tangent hyperplane atx (rather than on to the surface

of the feasible region itself).


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STEP 1: SEARCH DIRECTION

Letxibe a feasible point and letJbe the "active set", i.e.,J= {j | gj(x

i)=0}, or more

practically,J= {j | gj(xi)+0}.

Suppose eachjJis approximated using the Taylor series expansion by:

gj(x) = gj(xi) + (x-x

i)gj(xi).

Notice that this approximation is a linear function ofx with slope gj(xi). LetMmatrix whose rows are gjT(xi) forjJ(active constraints) and hkT(xi)

fork=1,2,...,p (equality constraints)

Let letP=[I-MT(MMT)-1M] (projection matrix). IfMdoes not exist, letP=I. Letdi = -Pf(xi).

a) Ifdi =0 andMdoes not exist STOPb) Ifdi =0 andMexists find

= -(MMT)

-1Mf(xi)

where u corresponds tog and v toh. Ifu>0 STOP, else delete the row ofM

corresponding to some uj


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STEP 2. STEP SIZE (Line Search)

SolveMinimize

where max= Supremum {| gj(xi+d

i) 0, hj(x

i+d

i) =0}.

Call the optimal solution *and findx

i+1= x

i+

*d

i. If all constraints are not linear

make a "correction move" to returnx into the feasible region. The need for this is

shown below. Generally, it could be done by solvingg(x)=0 starting atxi+1using the

Newton Raphson approach, or by moving orthogonally fromxi+1

etc.

g1, g2 are linear constraints g1 is a nonlinear constraint

NO CORRECTION REQD. REQUIRES CORRECTION

g1=0

g2=0 g1f

Pf

O.F. contours f

Pfg1=0

O.F. contours

correction move

xi+1


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LINEARIZATION METHODS

The general idea of linearization methods is to replace the solution of the NLP by the

solution of a series of linear programs which approximate the original NLP.

The simplest method is RECURSIVE LP:

Transform Minf(x), st gj(x), j=1,2,...,m,

into the LP

Min {f(x

i

) + (x-x

i

)

Tf(x

i

)}

st gj(xi) + (x-x

i)

Tgj(xi) 0, for allj.Thus we start with some initialx

iwhere the objective and constraints (usually only

the tight ones) are linearized, and then solve the LP to obtain a newxi+1

and

continue...

Note thatf(xi) is a constant andx-xi=dimplies the problem is equivalent to

Min dTf(xi)

st dTgj(xi) -gj(xi), for allj.

and this is a direction finding LP problem, andxi+1

=xi+dimplies that the step

size=1.0!

EXAMPLE: Minf(x)= 4x1 x22

-12,

st g1(x) =x12

+x22

- 25 0,

g2(x) = -x12

-x22

+10x1 + 10x2 - 34 0

If this is linearized at the pointxi=(2,4), then since


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42 4

8; 22

48;

2 102 10 62

it yields the following linear program (VERIFY)

Min(x) = 4x1 8x2 + 4,st 1(x) = 4x1 + 8x2 - 45 0,

2(x) = 6x1 + 2x2 - 14 0

The method however has limited application since it does not converge if the local

minimum occurs at a point that is not a vertex of the feasible region; in such cases it

oscillates between adjacent vertices. To avoid this one may limit the step size (for

instance, the Frank-Wolfe method minimizesfbetweenxiandx

i+dto getx

i+1).

f=-20

f=-10

f=0

=-30

=01=0

2=0

1=0

2=0

*


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RECURSIVE QUADRATIC PROGRAMMING (RQP)

RQP is similar in logic to RLP, except that it solves a sequence of quadratic

programming problems. RQP is a better approach because the second order terms

help capture curvature information.

Min dTL(xi) + dTBd

stdTgj (xi) -gj(xi) j

whereB is the Hessian (or an approximation of the Hessian) of the Lagrangian

function of the objective functionfat the pointxi. Notice that this a QP indwith

linear constraints. Once the above QP is solved to getdi, we obtainx

iby finding an

optimum step size alongxi+d

iand continue. The matrixB is never explicitly

computed --- it is usually updated by schemes such as the BFGS approach using

information from f(xi) and f(xi+1) only. (Recall the quasi-Newton methods forunconstrained optimization)


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KELLEY'S CUTTING PLANE METHOD

One of the better linearization methods is the cutting plane algorithm of J.E.Kelley

developed for convex programming problems.

Suppose we have a set of nonlinear constraints gj(x)0 forj=1,2,...,m that determine

the feasible region G. Suppose further that we can find a polyhedral set H that

entirely contains the region determined by these constraints, i.e., GH.

In general, a cutting plane algorithm would operate as follows:

1. Solve the problem with linear constraints.2. If the optimal solution to this problem is feasible in the original (nonlinear)

constraints then STOP; it is also optimal for the original problem.

3. Otherwise add an extra linear constraint (i.e., an extra line/hyperplane) so that thecurrent optimal point becomes infeasible for the new problem with the additional

constraint (we thus "cut out" part of the infeasibility).

Now return to Step 1.

1=0

2=0

3=0G

1=0

H

2=03=0

4=0

G:

j(x): nonlinearj(x): linear; hence

His polyhedral


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JUSTIFICATION

Step 1: It is in general easier to solve problems with linear constraints than ones with

nonlinear constraints.

Step 2: G is determined by gj(x)0,j=1,...,m, where each gjis convex, while

H is determined by hk0 k, where each hkis linear. G is wholly contained inside

the polyhedral set H.

Thus every point that satisfies gj(x)0 automatically satisfies hk(x)0, but not

vice-versa. Thus G places "extra" constraints on the design variables and therefore

the optimal solution with G can never be any better than the optimal solution with H.

Therefore, if for some region H, the optimal solutionx*

is also feasible in G, it

MUST be optimal forG also. (NOTE: In general, the optimal solution forH at some

iteration will not be feasible in G).

Step 3: Generating a CUT

In Step 2, let Maximum , whereJis the set of violated constraints,

J={j | gj(x)>0}, i.e., is the value of the most violated constraint atxi

. By the

Taylor series approximation aroundxi,

+

, for everyxRn (since is convex).


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If then

But since is violated atxi, 0. Therefore 0xRn.

is violated for allx,

the original problem is infeasible.

If , then consider the following linear constraint:

i.e., 0 ()

(Note thatxi, and are all known)

The current pointxiis obviously infeasible in this constraint --- since hk(x

i) 0

implies that 0, which contradicts the fact thatis violated atxi , (i.e, (xi) >0).

So, if we add this extra linear constraint, then the optimal solution would

change because the current optimal solution would no longer be feasible for the new

LP with the added constraint. Hence () defines a suitable cutting plane.

NOTE: If the objective for the original NLP was linear, then at each iteration we

merely solve a linear program!


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An Example (in R2) of how a cutting plane algorithm might proceed.

(--------- cutting plane).

Nonnegativity + 5 linear

constraints

G

3H

21

4=x*

5 6

G

3H

21

45

True optimum

x*

forG


constraints


constraints

G

3H

21

4


constraints

G

3H

21

h4

h5

h6

constrained non linear programming

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