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Sometimes this method may be =
useful.

Let us consider the following polynomial

f(x)=3D 240000 * x - 25000 * x^2 + 3500 * x^3/3 - 25*x^4 + =
x^5/5

and the interval [x]=3D[-1000, 1]

First derivative of the function f has the following form

d(1)f(x)=3D 240000-50000*x+3500* x*x-100*x*x*x+x*x*x*x;

Interval extension of this function is equal to:

d(1)f([-1000, 1]) =3D [-1103310000,1103550240000]

We don’t know if this function is monotone.

Because of = that we can=20 calculate second derivative

Because of = that we can=20 calculate second derivative

d(2)f(x)=3D -50000 + 7000* x - 300* x*x + 4* x*x*x

Interval extension of this function is equal to:

d(2)f([-1000, 1]) =3D [-4307050000,4257000]

We don’t know if this function is monotone.

Because of = that we can=20 calculate third derivative

Because of = that we can=20 calculate third derivative

d(3)f(x)=3D 7000 - 600* x + 12*x*x

Interval extension of this function is equal to:

d(3)f([-1000,1])=3D[-5600,12607000]

We don’t know if this function is monotone.

Because of = that we can=20 calculate forth derivative

Because of = that we can=20 calculate forth derivative

d(4)f(x)=3D -600+24*x

Interval extension of this function is equal to:

d(4)f([-1000,1]) =3D [-24600,-576]

We can see that the sign of the forth derivative is =
constant.

Because of=20 that third derivative is monotone and

extreme value of this function = can be=20 calculated

using endpoints of given interval.

Because of=20 that third derivative is monotone and

extreme value of this function = can be=20 calculated

using endpoints of given interval.

d(3)f(-1000) =3D 12607000

d(3)f(1) =3D 6412

d(3)f(1) =3D 6412

We can see that the sign of the third derivative is =
constant.

Because of=20 that second derivative is monotone and

extreme value of this = function can be=20 calculated

using endpoints of given interval.

Because of=20 that second derivative is monotone and

extreme value of this = function can be=20 calculated

using endpoints of given interval.

d(2)f(-1000)=3D -4.3071e+009

d(2)f(1)=3D -43296

d(2)f(1)=3D -43296

We can see that the sign of the second derivative is =
constant.

Because=20 of that first derivative is monotone and

extreme value of this = function can=20 be calculated

using endpoints of given interval.

Because=20 of that first derivative is monotone and

extreme value of this = function can=20 be calculated

using endpoints of given interval.

d(1)f(-1000) =3D 1.1036e+012

d(1)f(-1) =3D = 193401

d(1)f(-1) =3D = 193401

We can see that the sign of the first derivative is =
constant.

Because of=20 that function f is monotone and

extreme value of this function can = be=20 calculated

using endpoints of given interval.

Because of=20 that function f is monotone and

extreme value of this function can = be=20 calculated

using endpoints of given interval.

f(-1000)=3D -2.2619e+014

f(1)=3D 2.1614e+005

f(1)=3D 2.1614e+005

Finally

f([-1000, 1])=3D[ -2.2619e+014, 2.1614e+005]

This is the exact range of the function f.

If this procedure fails,

we can divide the interval into = two parts=20 and repeat procedure again.

If the boxes are sufficiently small =

then we=20 can apply Taylor model.

we can divide the interval into = two parts=20 and repeat procedure again.

If the boxes are sufficiently small =

then we=20 can apply Taylor model.

This procedure can be also applied in multidimensional case.

Pownuk A., New inclusion functions in interval global optimization =
of=20
engineering structures.

EUROPEAN CONFERENCE ON COMPUTATIONAL=20 MECHANICS,

Cracow, 26 - 29 June 2001, pp.460-461

EUROPEAN CONFERENCE ON COMPUTATIONAL=20 MECHANICS,

Cracow, 26 - 29 June 2001, pp.460-461

Matlab code:

lowerX=3D-1000;

upperX=3D = 1;

x=3Dinterval(lowerX,upperX);

upperX=3D = 1;

x=3Dinterval(lowerX,upperX);

'First order derivative'

240000-50000*x+3500*=20 x*x-100*x*x*x+x*x*x*x

'Second order derivative'

-50000 + 7000* x - = 300*=20 x*x + 4* x*x*x

'Third order derivative'

7000 - 600* x + = 12*x*x

'Fourth=20 order derivative'

-600+24*x

240000-50000*x+3500*=20 x*x-100*x*x*x+x*x*x*x

'Second order derivative'

-50000 + 7000* x - = 300*=20 x*x + 4* x*x*x

'Third order derivative'

7000 - 600* x + = 12*x*x

'Fourth=20 order derivative'

-600+24*x

x=3DlowerX;

y4_lower=3D-600+24*x

x=3DupperX;

y4_upper=3D-60= 0+24*x

y4_lower=3D-600+24*x

x=3DupperX;

y4_upper=3D-60= 0+24*x

x=3DlowerX;

y3_lower=3D7000 - 600* x + = 12*x*x

x=3DupperX;

y3_upper=3D7000=20 - 600* x + 12*x*x

y3_lower=3D7000 - 600* x + = 12*x*x

x=3DupperX;

y3_upper=3D7000=20 - 600* x + 12*x*x

x=3DlowerX;

y2_lower=3D-50000 + 7000* x - 300* x*x + 4*=20 x*x*x

x=3DupperX;

y2_upper=3D-50000 + 7000* x - 300* x*x + 4* = x*x*x

y2_lower=3D-50000 + 7000* x - 300* x*x + 4*=20 x*x*x

x=3DupperX;

y2_upper=3D-50000 + 7000* x - 300* x*x + 4* = x*x*x

x=3DlowerX;

y1_lower=3D240000-50000 *x+3500* x*x-100*=20 x*x*x+x*x*x*x

x=3DupperX;

y1_upper=3D240000-50000 *x+3500* = x*x-100*=20 x*x*x+x*x*x*x

y1_lower=3D240000-50000 *x+3500* x*x-100*=20 x*x*x+x*x*x*x

x=3DupperX;

y1_upper=3D240000-50000 *x+3500* = x*x-100*=20 x*x*x+x*x*x*x

x=3DlowerX;

y0_lower=3D240000 * x - 25000 * x^2 + 3500 * x^3/3 - = 25*x^4 +=20 x^5/5

x=3DupperX;

y0_upper=3D240000 * x - 25000 * x^2 + 3500 * = x^3/3 - 25*x^4=20 + x^5/5

y0_lower=3D240000 * x - 25000 * x^2 + 3500 * x^3/3 - = 25*x^4 +=20 x^5/5

x=3DupperX;

y0_upper=3D240000 * x - 25000 * x^2 + 3500 * = x^3/3 - 25*x^4=20 + x^5/5

Regards,

Andrzej Pownuk

Ph.D., research associate (adjunct) at:

Chair of Theoretical = Mechanics

Faculty of Civil Engineering

Silesian University of=20 Technology

E-mail: pownuk [at] zeus [dot] polsl.gliwice.pl=

URL:=20 http://zeus.polsl.gliwice.p= l/~pownuk

I have worked on this problem since =
1995.

Now I=20 can solve system of 1000 equations with dependent interval = parameters.

(I=20 have solved system of 2300 equations

with two intervals = Young’s=20 modulus.)

Now I=20 can solve system of 1000 equations with dependent interval = parameters.

(I=20 have solved system of 2300 equations

with two intervals = Young’s=20 modulus.)

This algorithm can be downloaded =
directly from the=20
internet.

This algorithm was presented on the following = conferences:

1) Pownuk A., Modelling of structures with uncertain =
parameters.

XLI=20 Sympsion on Modeling in Mechanics,

18-22.02.2002, Wis=B3a,=20 Poland

(Theoretical background)

XLI=20 Sympsion on Modeling in Mechanics,

18-22.02.2002, Wis=B3a,=20 Poland

(Theoretical background)

2) Pownuk A., Numerical solutions of fuzzy partial differential = equation

and its application in computational = mechanics.

Assessment and=20 New Directions for Research.

FUZZY PARTIAL DIFFERENTIAL = EQUATIONS,

FUZZY=20 RELATIONAL EQUATIONS,

FUZZY DIFFERENCE EQUATIONS.

March 15-17,=20 2002

University of California-Berkeley ,

California 94720 -=20 USA

(Applications)

Regards,

&n=
bsp; =20
Andrzej Pownuk

...............................

Andrzej Pownuk

Chair of = Theoretical=20 Mechanics

Faculty of Civil Engineering

Silesian University of=20 Technology

E-mail: pownuk [at] zeus [dot] polsl.gliwice.pl=

URL:=20 http://zeus.polsl.gliwice.= pl/~pownuk/

...............................

Andrzej Pownuk

Chair of = Theoretical=20 Mechanics

Faculty of Civil Engineering

Silesian University of=20 Technology

E-mail: pownuk [at] zeus [dot] polsl.gliwice.pl=

URL:=20 http://zeus.polsl.gliwice.= pl/~pownuk/

...............................

>

> Dear colleagues,

>

> I am trying to = solve a 1D=20 heat conduction problem by FInite Elements

> Method ( = Ax=3Db). My=20 stiffness matrix (A) is 51 x 51 ( row, column). This

> example = accessories=20 my Phd thesis.

> I am using the thermal conductivity (k) as = interval=20 number, with a little

> variation in k I get a large = overestimation in the=20 results.

> I have already tried the Preconditioning Gauss = Elimination and=20 Gauss-Seidel

> Methods, the Combinatorial and the Powell=20 Optimization Method.

> Do anybody know how to treat and avoid this = so=20 large overestimation ?

> Papers, routines, books an examples are=20 welcome.

>

> Best Regards,

>

> Thank you = very=20 much.

>

> Sebastiao Pereira

> Petrobras - Brazilian = Oil=20 Company

> http://www.petrobras.com.br

&= gt;=20 phone: 55 21 38656436

> fax: 55 21 3865 6441

>=20

>

------=_NextPart_000_01CD_01C1D051.8ECE07C0--
From owner-reliable_computing [at] interval [dot] louisiana.edu Wed Mar 20 14:53:38 2002
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by interval.louisiana.edu (8.11.3/8.11.3/ull-interval-math-majordomo-1.2) with ESMTP id g2KKrWr01346
for > Dear colleagues,

>

> I am trying to = solve a 1D=20 heat conduction problem by FInite Elements

> Method ( = Ax=3Db). My=20 stiffness matrix (A) is 51 x 51 ( row, column). This

> example = accessories=20 my Phd thesis.

> I am using the thermal conductivity (k) as = interval=20 number, with a little

> variation in k I get a large = overestimation in the=20 results.

> I have already tried the Preconditioning Gauss = Elimination and=20 Gauss-Seidel

> Methods, the Combinatorial and the Powell=20 Optimization Method.

> Do anybody know how to treat and avoid this = so=20 large overestimation ?

> Papers, routines, books an examples are=20 welcome.

>

> Best Regards,

>

> Thank you = very=20 much.

>

> Sebastiao Pereira

> Petrobras - Brazilian = Oil=20 Company

> http://www.petrobras.com.br

&= gt;=20 phone: 55 21 38656436

> fax: 55 21 3865 6441

>=20

>

Dear Andrzej,

I am very interested in your work and would very =
much like=20
to read the publications you refer to below, however I am unable to open =
the=20
download from the URL you give below.

Perhaps there is another version as a pdf file =
that I=20
could open and read?

[Can anyone else in the reliable_computing group =
help with=20
this?]

best wishes,

Ramon Moore

------=_NextPart_000_0034_01C1D026.E9E024C0-- From owner-reliable_computing [at] interval [dot] louisiana.edu Wed Mar 20 15:48:58 2002 Received: (from daemon@localhost) by interval.louisiana.edu (8.11.3/8.11.3/ull-interval-math-majordomo-1.2) id g2KLmv301504 for reliable_computing-outgoing; Wed, 20 Mar 2002 15:48:57 -0600 (CST) Received: from clmboh1-smtp3.columbus.rr.com (clmboh1-smtp3.columbus.rr.com [65.24.0.112]) by interval.louisiana.edu (8.11.3/8.11.3/ull-interval-math-majordomo-1.2) with ESMTP id g2KLmqr01500 for----- Original Message -----From:=20 Andrzej PownukTo:reliable_comput= ing [at] interval [dot] louisiana.edu=20 ; sebastiaoc [at] cenpes [dot] petr= obras.com.br=20Cc:Andrzej PownukSent:Wednesday, March 20, 2002 = 2:55=20 PMSubject:Re: Interval Linear = Systems of=20 Equations - Overestimation - Help !!I have worked on this problem since = 1995.

Now=20 I can solve system of 1000 equations with dependent interval = parameters.

(I=20 have solved system of 2300 equations

with two intervals = Young’s=20 modulus.)This algorithm can be downloaded = directly from=20 the internet.

This algorithm was presented on the following = conferences:1) Pownuk A., Modelling of structures with uncertain = parameters.

XLI=20 Sympsion on Modeling in Mechanics,

18-22.02.2002, Wis=B3a,=20 Poland

(Theoretical background)

2) Pownuk A., Numerical solutions of fuzzy partial = differential=20 equation

and its application in computational = mechanics.

Assessment and=20 New Directions for Research.

FUZZY PARTIAL DIFFERENTIAL = EQUATIONS,

FUZZY=20 RELATIONAL EQUATIONS,

FUZZY DIFFERENCE EQUATIONS.

March 15-17,=20 2002

University of California-Berkeley ,

California 94720 -=20 USA

(Applications)Regards,=&n= bsp; =20 Andrzej Pownuk...............................

Andrzej Pownuk

Chair of = Theoretical=20 Mechanics

Faculty of Civil Engineering

Silesian University of=20 Technology

E-mail: pownuk [at] zeus [dot] polsl.gliwice.pl=

URL:=20 http://zeus.polsl.gliwice.= pl/~pownuk/

...............................>

> Dear colleagues,

>

> I am trying to = solve a 1D=20 heat conduction problem by FInite Elements

> Method ( = Ax=3Db). My=20 stiffness matrix (A) is 51 x 51 ( row, column). This

> example=20 accessories my Phd thesis.

> I am using the thermal conductivity = (k) as=20 interval number, with a little

> variation in k I get a large=20 overestimation in the results.

> I have already tried the=20 Preconditioning Gauss Elimination and Gauss-Seidel

> = Methods, the=20 Combinatorial and the Powell Optimization Method.

> Do anybody = know how=20 to treat and avoid this so large overestimation ?

> Papers, = routines,=20 books an examples are welcome.

>

> Best Regards,

> =

>=20 Thank you very much.

>

> Sebastiao Pereira

>=20 Petrobras - Brazilian Oil Company

> http://www.petrobras.com.br

&= gt;=20 phone: 55 21 38656436

> fax: 55 21 3865 6441

>=20

>

Dear People,

Sorry for the trouble. The problem was at my =
end. I have=20
the paper __Interval_FEM.pdf__

now.

Thanks for your help

Ramon Moore

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for Dear Colleagues,

This message is to inform you =
that my address=20
and

official address of Reliable =
Computing=20
journal changes tomorrow

from

to

First address will work properly =
during at=20
least one month and hopefully much more,

however please use the new =
address for all=20
kind of communications with me.

Best regards,

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