Optimization of thermal processes2007/2008 Optimization of thermal processes Maciej Marek...

Optimization of thermal processes 2007/2008

Optimization of thermal processes

Maciej Marek

Czestochowa University of TechnologyInstitute of Thermal Machinery

Lecture 5


Overview of the lecture

• Optimization with inequality constraints− Discussion of Kuhn-Tucker conditions

• Convex programming

• Example of problem with inequality constraints

• A design problem: optimal parameters for drying of sugar


Optimization with inequality constraints

Minimize ( )f X

subject to( ) 0 , 1, 2,...,jg j m X

1x

2x

1( ) 0g X

2 ( ) 0g X

3( ) 0g X

Feasible region

Constraint surfaces

Free stationary point

Bound sationary point

Active constraint


Kuhn-Tucker conditions for minimization problems(necesary conditions)

1

0 , 1, 2,...,m

jj

ji i i

gL fi n

x x x

0 , 1, 2,...,j jg j m

0 , 1, 2,...,jg j m

0 , 1, 2,...,j j m

1

( , ) ( ) ( )m

m mj

L f g

X λ X X Lagrange function

We can use constraintsthemselves instead ofthe slack variables

For minimization

These conditions are also sufficient forconvex programming problems.

Non-convex (concave)region


Convex programming problem

• Optimization problem is called convex programming

problem if− the objective function is convex− and the constraint functions are convex

( )f X( )jg X

x

( )f x

1x 2x

For any x1 and x2 the lineis above the graph of thefunction

Convex region

• A given function f is convex if the Hessian matrix is positive

semidefinite 2 ( )( )

i j

fH

x x

XX



The matrix H is positive semidefinite when:

• each of the principle minors H1, H2, ..., Hn is non-negative

2 2 2

1 1 1 2 1

2 2 2

2 1 2 2 2

2 2 2

1 2

n

n

n n n n

f f f

x x x x x x

f f f

x x x x x x

f f f

x x x x x x

H

...2

11 1

fH

x x

2 2

1 1 1 22 2 2

2 1 2 2

f f

x x x xH

f f

x x x x

2 2 2

1 1 1 2 1

2 2 2

2 1 2 2 2

2 2 2

1 2

n

n n

n n n n

f f f

x x x x x x

f f f

H x x x x x x

f f f

x x x x x x

0jH for 1, 2,...,j n



Notes:

• If the constraint functions are linear then the feasible

region is convex (every component of the Hessian is

equal to zero)

• Linear programming problems (linear objective function

and linear constraints) are convex programming

problems

• It is very often difficult to ascertain whether the

objective and constraint functions involved in a practical

problem are convexFor convex functions:

relative extreme point = global extreme point


Kuhn-Tucker conditions (methodology)

0 , 1, 2,...,j jg j m

j-th constraint functionj-th Lagrange multiplier

Example: suppose we have two constraints

1 1 2 1 1 1 2 1 2( , ) 0, ( , ) 0g x x x g x x x x

• For each of the equations assume the constraint is

active

or inactive

• Try all possibilities, solve the equations and make sure

the constraints are satisfied in each case

( 0)jg ( 0, 0)j jg

active - inactive

case

1

2

3

4

1g 2g

-

-

- -

Then there are four cases.


Example of optimization with inequality constraintsFind the point in the area D described by inequalities

which is the nearest to the point A=[3,5].

1 2 1 20, 1 0, 2x x x x

First, transform the inequalities into the standard form:

1 2 1 20, 1 0, 2 0x x x x

1x

2x

3

5(3,5)A

2 1 0x

1 0x

1 2 2 0x x

Feasible region D


Example of optimization with inequality constraintsWhat is the objective function?

2 21 2 1 2( , ) ( 3) ( 5)f x x x x The distance from point A to 1 2[ , ]x x

This form is inconvenient. Let’s see if there is other objective function withthe same extreme point.

2 2

1 2 1 2 1 2

1min ( , ) min ( , ) min ( , )

2f x x f x x f x x

For non-negative function

So, finally we choose the following objective function:

2 21 2 1 2

1 1( , ) ( 3) ( 5)

2 2f x x x x

Is it convex function?


Example of optimization with inequality constraints

2 2

21 1 2

2 2

22 1 2

f f

x x xH

f f

x x x

The Hess matrix 2 21 2 1 2

1 1( , ) ( 3) ( 5)

2 2f x x x x

1 21 2

3, 5f f

x xx x

2

121 1

3 1f

xx x

2

222 2

5 1f

xx x

2 2

11 2 2 1 2

3 0f f

xx x x x x

1 0

0 1H

So, f is a convex function.

And the constraints are linear.Thus, the problem is convex.

1 1 1 0H

2

1 01 0

0 1H



2 21 2

1 1( , ) ( 3) ( 5)

2 2L x x X λ

1 1 2 2 3 1 2( 1) ( 2)x x x x Lagrange function

Constraint functions

Kuhn-Tucker conditions (in this case also sufficient):

1 1 31

3 0,L

xx

2 2 32

5 0L

xx

1 1 2 2 3 1 20, ( 1) 0, ( 2) 0x x x x

1 2 1 20, 1 0, 2 0x x x x Constraints

1 2 30, 0, 0 Minimum

Let’s make some order:



1 1 33 0x

2 2 35 0x

1 1 0x

2 2( 1) 0x

3 1 2( 2) 0x x

1 0x

2 1 0x

1 2 2 0x x

1 2 30, 0, 0

A1

A2

B2

B3

B4

C1

C2

C3

D1

C1 C1

Active constraint

Inactive constraint

Let’s employ the followingnotation:



C1 1 0x

2 1x (C2)

2 2x (C3)

2 1x Thus C2 is inactive

2 0

C2

2 32, 0x

2 5 0x (A2)

2 5x This violates (C2)

We reject C1 C2 C3

C3

C32 2x

32 5 0 (A2)

1 33 0 (A1)

1 4 which violates (D1)

We reject C1 C2 C3



C1 1 10, 0x

1 33 0x (A1)

3 13 0x 1 0x as

3 0 C3 has to be active

C31 2 2 0x x

2 21 0, 0x

1 33 0x (A1)

2 35 0x (A2)

1 2 2 0x x with

(A1) and (A2) give2 0x

Which violates (C2)

We reject C1 C2 C3

C2

C2 2 1 0x

We get:

1 2 1 2 31, 1, 0, 2, 4x x And all conditions are satisfied.

We accept C1 C2 C3

Solution: point [-1,-1]


Optimal parameters for drying of sugar

Sugar that leaves the dryer has temperature T [oC] and humidity W [kg/kg]. The sugar is then stored in a silo, where some amount of the sugar may lump, if the conditions (T,W) are not appropriate.

The total cost of: drying, storage and loss of sugar because of lumping (caking) is described by the function [PLN/kg]:

3 3 21 0.0005 0.0005 15 15

( , ) 10 2 2 1510 0.0005 0.0005 15 15

W W T Tf T W

The function takes large values for:•too large humidity (considerable lumping)•too large temperature (high cost of drying)

This is the objective function we want to minimize.

What are the constraints?



The thermodynamic parameters of sugar (T,W) cannot be above the melting curve. It was experimentally verified that the following inequalityshould be fulfilled:

( 15)( 0.0005) 0.075T W

T

W

Lumped sugar

Loose sugar

Melting curve

The lowest humidity that can be achieved in a dryer and the lowest temperature of sugar are, respectively:

* *0.0005, 15W T

Find the optimum parameters of the sugar.



Let’s introduce the following variables:

1 2

0.0005 15,

0.0005 15

W Tx x

Our optimization problem can be stated as:

3 2 31 2 1 1 2 2

1( , ) 10 2 2 15

10f x x x x x x Minimize

subject to

1 1 2 1 2( , ) 10 0g x x x x

2 1 2 1( , ) 0g x x x

2 1 2 2( , ) 0g x x x

Now, we can apply Kuhn-Tucker conditions.



3 2 31 2 1 2 3 1 1 2 2 1 1 2 2 1 3 2

1( , , , , ) 10 2 2 15 ( 10) ( ) ( )

10L x x x x x x x x x x

21 1 2 2

1

0.3 1 0L

x xx

22 2 1 1 3

2

0.4 0.6 0L

x x xx

1 1 2 2 1 3 2( 10) 0, ( ) 0, ( ) 0x x x x

1 2 1 210 0, 0, 0x x x x

1 2 30, 0, 0

C1 C2 C3



C1 1 2 110, 0x x

1 2, 0x x 2 30, 0

C2 C3

1 2 10x x

22 2 1 10.4 0.6 0x x x

21 1 20.3 1 0x x

1 22.6, 2.802x x +

1 0

So, we reject the case 1 0

C1 1 0

21 20.3 1 0x

22 2 30.4 0.6 0x x

C3 C3

2 0x

3 0

1 0x

2 1 0 2 0

We reject the case.

1 1.825x We take the positive value.

C2C2

1

2

1.825

0

x

x



For values:

all conditions are satisfied. Thus, the solution is:

1

2

1.825

0

x

x

0.0005 1.825 0.005 0.00141 ( 0.14%)optW o15 0 15 optT C

The minimum cost is:

( , ) 0.283 PLN/kgopt optf W T


Thank you for your attention

Optimization of thermal processes2007/2008 Optimization of thermal processes Maciej Marek...

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