3- Minimization Review_DA.pdf

ECE 573 2001 - 2013 George Gross, University of Illinois at Urbana-Champaign; All Rights Reserved 1

ECE 573 Power System Operations and Control

3. Review of Minimization Problem Solution

Techniques

George Gross

Department of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign


UNCONSTRAINED MINIMIZATION

Consider the simple minimization problem:

and is continuously differentiable

A necessary condition for a minimum at is

is determined by finding the root of by

solving the set of n equations in n unknowns

nmin f x x

* Tf x 0

x*

(UMP )

: nf

f

x*


NOTATION

Consider a continuously differentiable function

; for any , we

write

is always a row vector and so is a

column vector and is called the gradient of f

nf :

n, , ,

1 2

T nx x x x

1 2

, , ,n

f f ff x

x x x

T

f x f x


NOTATION

For the mapping ,

is an matrix with each being a

row vector in

n mg :

1

2

x

x

x

x m

g x

g g xg

x

g x

m n x ig x

n


NOTATION

The Hessian is the second derivative of f

2

2

fH x f x

xx

2

n

n

n n n n nx x

2 2 2

1 1 1 2 11

2 2 2

2 2 1 2 2

2 2 2

1 2

f f ff x

x x x x x xx

f f ff x

x x x x x x xH x

f f ff x

x x x x x


NOTATION

We note that, by definition, is always

a symmetric matrix since

for a twice continuously differentiable function

f :

2 2

, 1, 2, ,i j j i

f fi j n

x x x x

H x

n


THE GRADIENT DIRECTION

The Taylor series expansion for obtains

for small , we neglect the h.o.t. so that

Suppose we set ; then,

xf x x f x f x higher order terms in x

xf x x f x f x

T

xx f x

2

xf x x f x f x

for > 0

f


STEEPEST DESCENT

Thus is a direction of descent and is

called the steepest descent direction; is called the

step size

There is a large collection of optimization

techniques for general nonlinear functions; the

simplest is the steepest descent scheme

T

xf x


STEEPEST DESCENT ALGORITHM

Step 0: determine an initial point x (0) ; set ,

define the convergence tolerances 1 > 0, 2 > 0

Step 1: compute

Step 2: if , stop; else evaluate

Step 3: set

Step 4: if ,stop and is the

solution; else, set and go to Step 1

T

x

vf x

1x vf x 1

T

x

v v vx x f x

21v vf x f x 1vx

1v v

v 0


STEEPEST DESCENT ITERATIONS

f x c 1

f x c c 2 1

3 2f x c c

x

3

x

4 x

2

x

1

0x

x2

contours

. of constant

value of

direction of the

negative of the

gradient

f xx1

21 43>c cc c

34f x c < c


SLOW CONVERGENCE OF THE STEEPEST DESCENT METHOD

0x


We use Newtons method to find the root of

Note that in this case the Jacobian of is

the Hessian of

NEWTONS METHOD FOR MINIMIZATION

T

f x 0

f x

f x

2

2

f xH x

x


Consider the problem

with continuously differentiable

We convert ( ECMP ) into the form of ( UMP ) by

defining a multiplier and the Lagrangian

EQUALITY - CONSTRAINED MINIMIZATION

( ECMP )

. .

min f x

s t

g x 0

n mg :

, Tx f x g x L

m

penalty for

g x 0violating


The necessary conditions for an optimum at

The presence of the m constraints leads

to the augmentation of the dimension of the

decision problem from n to n + m

A scheme for (UMP ) solution may be deployed to

solve for the (n + m) - dimensional optimal

decision variables

EQUALITY - CONSTRAINED MINIMIZATION

TT

x x

T T

x f x g x 0

x g x 0

* * * * *

* * *

,

,

L

L

* *,x

0xg

* *,x


Consider the minimization problem

with

One way to solve (ICMP ) is by transforming

(ICMP ) into the form of (UMP )

INEQUALITY - CONSTRAINED MINIMIZATION

(ICMP )

. .

min f x

s t

x 0h

nh :



We introduce for each inequality

the penalty function

and add the term to the objective

function with

2

i

i

i i

0 if h x 0p x

h x if h x 0

ih x

i ip x

i 0


The penalty coefficient is chosen large enough

so as to force to satisfy each constraint

Thus we convert the problem to the (UMP ) form

We can use appropriate (UMP ) solution schemes

for determining


1

i ii

min f x p x

x

*x

i


GENERAL MINIMIZATION PROBLEM

We consider the constrained minimization

problem

with and continuously

differentiable functions

min f x,u

s.t.

g x,u 0

h x,u 0

(CMP)

, m nf :

, m n mg :

n mu x and

and


The inequality is treated by

appending the penalty functions

, to the objective to construct

Thus we focus on the resulting (ECMP)


, m nh :

i 1,2, ,

1

, , ,i i

i

f x u f x u p x u

s.t.

g x ,u 0

, ,i ip x u

min f x ,u



We may view the constraint as the

functional means by which is defined

We construct the Lagrangian

, , , ,Tx u f x u g x u L

penalty for violating

,g x u 0

g x,u 0

x x u


NECESSARY CONDITIONS OF OPTIMALITY

For minimizing the unconstrained L , the

necessary conditions for optimality are

We consider a point at which

so that the total derivative

,

T T

T T

T T

x x x

u u u

f g 0

f g 0

g x u 0

L

L

L

x,u g x , u 0

d g g gx0

du x u u


We introduce the assumption that is

nonsingular in the region of interest; then

expresses the sensitivity of with respect to

which is obtained from the implicit functional

relationship


x

g

1g gx

u x u

x u

0u,xg


From the necessary conditions

it follows that

We use this information for constructing the

reduced gradient as a descent direction


T T

x x xf g 0 L

T

x xf g

1


THE REDUCED GRADIENT

We call the reduced gradient the total derivative

Geometrically is the projection of the total

derivative of f on the u-subspace of the space

We adapt the steepest descent to solve (CMP )

-

1 2

1

, , ,

T

n

T

TT T

u x u u

df df df df

du du du du

g gf f f g

x u

df

d u

u x


EXAMPLE: MINIMIZE ACTIVE POWER LOSSES

y13 = 4 j 10 y23 = 4 j 5

P3 + j Q3 = 2.0 - j 1.0 p.u.

P2 = 1.7 p.u.

P,V bus

reference/swing

bus

P,Q bus


VARIABLES

Control variables:

State variables:

u =V1

V2


OBJECTIVE FUNCTION

We notice that P2, P3 are fixed, so any changes in

the losses will be reflected in changes in P1, so

2Reboth lines

f I R


EQUALITY CONSTRAINTS

2 2

3 3

3 3

,

( , ) ,

,

P V P

g x u P V P

Q V Q

where

The AC power flow equations for bus 2 and 3


REDUCED GRADIENT METHOD

, , , ,TL x u f x u g x u

2 2

1 2

3 3

1 2

3 3

1 2

P P

V V

g P P

u V V

Q Q

V V

2 2 2

2 3 3

3 3 3

2 3 3

3 3 3

2 3 3

P P P

V

g P P P

x V

Q Q Q

V



1

T

T

u x

g gdff f

du x u



We pick

We solve the AC power flow and determine ,

we select a step size and iterate until

we find the solution

( )

( 1) ( )

u

dfu u

du

x(0)

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