Institute of Eng. and Comp. Mechanics Prof. Dr.-Ing. Prof. E.h. P. Eberhard
Optimization of Mechanical Systems WT 10/11 A1.1
Motivational Example 1 In medieval times catapults where used in the siege of castles. These catapults where capable of throwing heavy stones over a distance of several hundred meters. The catapult consists of an undercarriage, a boom with counter mass and the stone. The stone is connected to the boom by a rope. Optimize!
stone s boom with counter weight

undercarriage

Optimization of Mechanical Systems WT 10/11 A1.2
Motivational Example 2 For different TVs the table below summarizes the screen diagonal and price.
Diagonal [cm] 132 Price [Euro] 2156
A B C D E F G H I J K L M
LCD TV Sony KDL-26U3000 Sharp LC-26 D44 E-BK Sharp LC-32 D44 EBK Philips 32PFL9603D/10 Samsung LE-37R86BD Panasonic TXLZD70F Sony KDL-40V3000 Samsung LE40F86BD Panasonic TXLZD70F Panasonic TH-42PZ70E Pioneer PDP-4280XA Sharp LC-52 X 20 E Samsung LE 52 A 759 R1
a.) Order the TVs with respect of the screen diameters. Find the TV with the largest screen diagonal.
b.) Order the TVs with respect of the price. Find the TV with the minimal price.
c.) Which TV is a compromise in respect of both criteria?
Optimization of Mechanical Systems WT 10/11 A2
Example: Optimization of the rear suspension of a car

technical system

mechanical model M m c d cw s : car body mass : wheel and axel mass : stiffness of suspension : damping of suspension : stiffness of tire : street profile

mathematical model

& & M && = c( x xw ) d ( x xw ) x & & m &&w = c( x xw ) + d ( x xw ) cw ( xw s (t )) x
Optimization of Mechanical Systems WT 10/11 A3

Notation

a R , b R m , C R m n
matrix notation 1) derivative with respect to a scalar index notation
2) derivative with respect to a vector
3) derivative of a scalar product

x Rm bi = const

4) derivative of a quadratic form

A = AT R n n

Aij = const
Optimization of Mechanical Systems WT 10/11 A4.1
Automatic Differentiation
Graph for f ( x1 , x 2 ) = x1 x 2 e 2 x 2

Forward mode:

x1 =

x2 =

, x3 =

= x3 x2 x2 = ,

_ _ _ _ _ _

x x3 x j j j J3 = { 2 }

, x4 =

j J4 = {

} _ _ _

= _ _ _ _ _ _

, x5 =

j J5 = {

= _ _ _ _ _ _ ,

= _ _ _ _ _ _ x6 =

, x6 =

j J6 = {

= _ _ _ _ _ _
Optimization of Mechanical Systems WT 10/11 A4.2

Reverse mode:

_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ d x6 , = d x6 _ _ _ _ _ _ _ _

f = x6 =

j I 5 =
= _ _ _ _ _ _ _ _ = _ _ _ _ _ _ _ _ = _ _ _ _ _ _ _ _ = _ _ _ _ _ _ _ _
= _ _ _ _ _ _ _ _ = _ _ _ _ _ _ _ _ = _ _ _ _ _ _ _ _ + _ _ _ _ _ _ _ _

}_ _ _ _ _ _

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ x1 =

= _ _ _ _ _ _ _ _

df dx 1 df = dx df dx 2
x1 _ _ _ _ _ _ _ _ _ _ = = . x2 _ _ _ _ _ _ _ _ _ _
Optimization of Mechanical Systems WT 10/11 A4.3
Optimization of Mechanical Systems WT 10/11 A5.1
Automatic differentiation
For optimization purposes the sensitivities of the function f (x) = x1 sin ( x 2 ) 5 x 3 , with

x = [ x1 , x 2 , x 3 ] , must be calculated.
a) Complete the graph for the function f (x) and write down the derivatives along the arcs.

x 4 = x 3 x1 x2

Optimization of Mechanical Systems WT 10/11 A5.2
Firstly, use the forward mode to compute the gradient of the function. b) Give the index sets J i for your introduced intermediate and output quantities.
c) Compute the gradients for all input and intermediate quantities.
d) Give the gradient of the function f (x) in terms of the input quantities x = [ x1 , x 2 , x 3 ].
Optimization of Mechanical Systems WT 10/11 A5.3
Use also the reverse mode to compute the gradient of the function. e) Give the index sets I i for the input quantities and the introduced intermediate quantities.
f) Compute the scalar gradients x j for the output quantity and intermediate quantities.
g) Give the gradient of the function f (x) in terms of the input quantities x = [ x1 , x 2 , x 3 ].
h) Check your result by direct differentiation of the function f (x).
Optimization of Mechanical Systems WT 10/11 A6
Minima of the Rosenbrock-function
A common testing example for optimization algorithms is the Rosenbrock-function
2 f ( p) = 100 (p 2 p1 ) 2 + (1 p1 ) 2.
For this function many optimization algorithms show convergence problems. This is due to the slight curvature of the valley and the very slowly decreasing bottom of the valley.
a) Complete the gradient and the Hessian matrix for an arbitrary point p.
_ f ( p) = _ _ 2 f ( p) = _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ___________ ___________ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
b) Give the necessary condition of first order for minima. Compute from this a potential * minimizer p.
c) Does p satisfy the necessary condition of second order?
Optimization of Mechanical Systems WT 10/11 A7.1
Minimization of a Quadratic Function
Minimize the quadratic function
2 f ( p) = 10 p1 + p 2. 2

Analytical Solution

Using the gradient and the Hessian matrix,
_ _ _ _ f ( p) = , _ _ _ _
_ _ _ _ _ _ f ( p) = , _ _ _ _ _ _
we get the solution from the necessary conditions for a local minimizer, i.e.

__________ ,

Here, the solution p * =

____________________.

is also a global minimizer of the function.

Newton Method

Starting from the initial point p
= [ 0.1, 1 ] the search direction is

s ( 0) = f

2 ( 0 ) 1

f ( 0)

_ _ _ _ _ _ _ _ _ _ _ _ _ = = . _ _ _ _ _ _ _ _ _ _ _ _ _
Along the corresponding line

( 0) (0)

p ( ) = p
_ _ _ _ _ _ _ _ = , _ _ _ _ _ _ _ _

the criterion function is

f ( ) = f ( p ( ) ) =

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
The minimum along this line follows from

___________

( 0) =
Optimization of Mechanical Systems WT 10/11 A7.2
which yields the improved solution

p (1 ) = p ( 0 ) =

and so we got the minimizer in the first step.
Conjugate Gradient Method
= [ 0.1, 1 ] we get the following iteration steps:

_ _ _ _ _ _

exact line search
_ _ _ s = f ( 0) = _ _ _ _ _ _ _ _ _ _ _ p ( ) = _ _ _ _ _ _ _ _

f ( ) = f =

________________ =0

= f (1) +

f (1) f ( 0)

p () =

(1) =

convergence in two steps

Optimization of Mechanical Systems WT 10/11 A8.1

Quasi Newton Method

f () _ _ _ _ _ _
_ _ _ _ _ _ H ( 0) : = E = _ _ _ _ _ _

s ( 0) = H ( 0) f ( 0)

_ _ _ = _ _ _
_ _ _ _ _ _ _ _ p ( ) = _ _ _ _ _ _ _ _

f ( ) =

f = _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

(0) =

p = p (1) p ( 0)

_ _ _ = _ _ _ _ _ _

= f (1) f ( 0)

_ _ _ = _ _ _ _ _ _

Optimization of Mechanical Systems WT 10/11 A8.2
_ _ _ _ _ _ _ _ (1) H = _ _ _ _ _ _ _ _ _ _ _ _

f (1)

_ _ _ _ _ _ _ _ _ _ _ _ p ( ) = _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Optimization of Mechanical Systems WT 10/11 A9.1
Constrained Minimization of a Quadratic Function

2 f ( p ) = p1 + p 2 2

subject to the constraint

p1 p 2 4

or written in the standard form

h1 (p) =

_________________

Graphical Solution:

For this simple problem, the solution can be found graphically in the design parameter space. The dimension of the design parameter space is.

_ _ _ _ _ _ _

Plot the contour lines of the function f and the constraint h 1 and mark the optimal solution p.
Optimization of Mechanical Systems WT 10/11 A9.2

Analytical Solution:

The inequality constraint can be handled as an equality constraint by introducing the offset (slack) variable u 1

________________

The Lagrange function for an optimization problem with constraints generally reads as

L(p, , u ) = f (p)

It follows

i (h i + u i2 )

L(p, 1 , u1 ) =
_____________________________
The necessary condition of 1 order for a minimum is L(p, 1 , u1 ) = 0 , which leads to
the following four equations
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ! _ _ _ L = = p _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
! L = = __________________ ___ 1 ! L = = __________________ ___ u1
From the last equation we can distinguish two cases, either

_ _ _ _ _

Institute of Eng. and Comp. Mechanics Prof. Dr.-Ing. Prof. E.h. P. Eberhard Case 1:
Optimization of Mechanical Systems WT 10/11 A9.3
If 1 = 0 , the constraint is not considered and therefore inactive. This is only true, if the constraint is fulfilled. It follows from L(p, 1 , u1 ) = 0

, p2 =

This last equation cannot be satisfied and subsequently the constraint is not fulfilled at the minimum of f at p =

and case 1 is not true.

Case 2:
If u 1 = 0 , the distance from the limit of the constraint is 0 and at p we have an active constraint. The condition L(p, 1 , u1 ) = 0 leaves us with three equations
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

and it follows

Optimization of Mechanical Systems WT 10/11 A10.1
Constrained Minimization of a Nonlinear Function
3 f ( p) = ( p1 + 10 p 2 ) p1
subject to the equality constraint

2 g (p ) = p1 + p = 0 2

Solution by the LagrangeNewtonMethod:
The Lagrange function for this optimization problem can be written as

L(p, ) =

. _____________________________
The Karush-Kuhn-Tucker (KKT) equations represent necessary conditions for an optimum. They can be written as
L = ___________________ =
These nonlinear equations can be solved by the Newton-Raphson method. This yields the update equation of the Lagrange-Newton method
2 _ _ _ _ _ _ _ _ _ g (p) p _ _ _ _ _ _ _ _ _
g (p) f (p) p ( +1) p p _ _ _ _ _ _ _ _ _ _ _ _ _ = _ _ _ _. ( +1) g (p) _ _ _ _ _ _ _ _ p( ) _ _ _ _ _ _ _ _ _ p( ) , ( )

= A(p, )

= b(p)
Optimization of Mechanical Systems WT 10/11 A10.2
With the objective and constraint function given above, the matrix A and the right hand side b follows as
30p 4 + 1 _ _ _ _ _ _ _ _ _ _ _ _ _ A(p, ) = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ b(p) = _ _

Starting with p

_________ _________

_________

_ _ _ _ _ _ _ _ _ , 2 p2 _ _ _ _ _ _ _ _ _ 0 _ _ _ _ _ _ _ _ _
_____________ 3. 20 p1 _____________ _____________
= [ ] and ( 0) = 0 the update equation reads as
94 _ _ _ _ _ _ _ _ _ (1) p _ _ _ _. = _ _ _ _ _ _ _ _ _ (1) _ _ _ _ 8 _ _ _ _ _ _ _ _ _

_ _ _

30 ____ 60 ____ _____ 200 _____
Optimization of Mechanical Systems WT 10/11 A10.3
From these three scalar equations, the updated optimization variables and the new Lagrange multiplier are given by

+ p = [ _ _ _ _ _

_ _ _ _ _ ], (1) = _ _ _ _ _
The next iteration yields
18022 _ _ _ _ _ 1500 _ _ _ _ _ _ _ _ _ _

which results in

_____ 2.6 _____
16700 _ _ _ _ _____ ( 2) p _ _ _ _ = 2240 , _ _ _ _ _ _ _ _ _ ( 2) _ _ _ _ 17.69 _ _ _ _ _ _ _ _ _
= [ 3.73482 0.63777 ] and ( 2) = 319.442.
The following table summarizes the first 10 iterations of the Lagrange-Newton method: Iteration
1,0000E+00 5,0000E+00 3,7348E+00 3,1496E+00 2,8840E+00 2,5959E+00 2,5655E+00 2,5667E+00 2,5667E+00 2,5667E+00 2,5667E+00
0,0000E+00 -1,3000E+00 6,3777E-01 -1,3357E-01 -2,8855E+00 -1,8490E+00 -1,5787E+00 -1,5533E+00 -1,5531E+00 -1,5531E+00 -1,5531E+00
0,0000E+00 -2,6000E+01 -3,1944E+02 2,5548E+01 -1,1348E+02 -3,4899E+01 -1,0998E+01 -8,8899E+00 -8,8713E+00 -8,8713E+00 -8,8713E+00
-9,9000E+01 1,2044E+04 3,0458E+03 5,7946E+02 -2,6471E+02 -2,5859E+02 -2,5534E+02 -2,5478E+02 -2,5477E+02 -2,5477E+02 -2,5477E+02
-8,0000E+00 1,7690E+01 5,3556E+00 9,3751E-01 7,6436E+00 1,1574E+00 7,3970E-02 6,4691E-04 4,6864E-08 3,5527E-15 1,0011E-15

Draw the points p

and the final point p
in the contour plot on the next page.
Another solution process was started with the initial point p

= [ ] and ( 0) = 0. The

following table summarizes the corresponding optimization history:
Optimization of Mechanical Systems WT 10/11 A10.4

Iteration

-1,0000E+00 -5,0000E+00 -3,7693E+00 -3,2153E+00 -2,6628E+00 -2,3117E+00 -2,4700E+00 -2,4825E+00 -2,4838E+00 -2,4838E+00 -2,4838E+00
0,0000E+00 1,3000E+00 -7,7022E-01 2,8425E-01 4,0384E+00 2,4872E+00 1,8314E+00 1,6904E+00 1,6826E+00 1,6825E+00 1,6825E+00

0,0000E+00 7,4000E+01 -1,9290E+02 8,7719E+01 -3,0435E+02 -8,3576E+01 -8,2745E+00 8,1362E+00 8,9300E+00 8,9323E+00 8,9323E+00
1,0100E+02 1,3044E+04 4,1289E+03 1,2455E+03 7,2870E+02 3,8788E+02 2,5753E+02 2,5082E+02 2,5064E+02 2,5064E+02 2,5064E+02
-8,0000E+00 1,7690E+01 5,8005E+00 1,4188E+00 1,4399E+01 2,5296E+00 4,5508E-01 2,0039E-02 6,3143E-05 4,5733E-10 -3,5527E-15
of this run in the contour plot, too.
Contour plot of the constrained problem:

g(p) = 0

As we can see, the solution depends on the starting point p
Visit the webpage http://www.itm.uni-stuttgart.de/research/alpso and try to solve the problem with the stochastic particle swarm optimizer which is available there. What can you observe? What are the differences with respect to the Lagrange-Newton method?
Optimization of Mechanical Systems WT 10/11 A11.1
Application: Optimization of a Double Pendulum
An actively controlled double pendulum is described by the generalized coordinates

y = [ ].

Without the torque actuator T , the dynamic behavior is described by the nonlinear equations of motion
& & M ( t , y ) && + k ( t , y , y ) = q ( t , y , y ) y
and the initial conditions are

y ( 0) = y 0 ,

& y (0) = 0.
Linearization of the equations of motion about the equilibrium position results in

y (t) = 0 + (t) ,

& & y (t) = 0 + (t) ,
&& ( t ) = 0 + && ( t ) y
and the following system of linear equations
& M ( t ) && ( t ) + P ( t ) ( t ) + Q ( t ) ( t ) = h ( t ) ,

2mL2 M = 2 mL

0 mL0 2mgL , P = . ,Q = mgL mL
The generalized force vector q has to be considered with the control torque T ( t ) of the torque actuator

T( t ) h (t) = 0

and the PDcontrol law is

& T ( t ) = k 1 k 2

Optimization of Mechanical Systems WT 10/11 A11.2
By an optimization process the control parameters k1 and k 2 are to be optimized in such a way, that the vibrations of the pendulum, evaluated by the integral of y y within the first 10 s, are optimally damped. In order to ensure the stability of the system, both of the control parameters must not be negative. First we formulate the problem in standard form: design variables kinetics

= _____

_______________

initial conditions

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 0 &0 = 0, = = 0. = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

final condition

1 = 1 = 2 = 3 =

______

criteria functions
___________ ___________ ___________ _____

_____ _____

optimization criterion

h1 = h2 =

inequality constraints
The constrained optimization task then reads as find and

, where

___________________________
Optimization of Mechanical Systems WT 10/11 A11.3
The described optimization task is solved by a SQPmethod. The parameters of the double pendulum are
0.2 (0) m = 1 kg , L = 0.3 m , y 0 = , and the initial design variable vector is p = 0.2

5.0 5.0 .

The optimization algorithm needs 10 gradient evaluations and 16 function evaluations in order to find the optimizer

0.0 * p = . 1.63

The optimization history shows the convergence after 10 iterations:
The optimal dynamic behavior shows a significantly improved damping in comparison to the passive and the initial design:
Optimization of Mechanical Systems WT 10/11 A12.1
Multicriteria Optimization of a Truss
For the presented truss the cross sections A1 and A2 of the two bars are to be designed. Simultaneously, the overall mass and the vertical displacement of the point P shall be minimized, while an external load F is applied at P. This is a typical problem of multicriteria optimization, since the overall mass corresponds to the costs of material and the vertical displacement is a measure for the function of the framework. Using the density the overall mass is

___________________

For the computation of the vertical displacement the mass of the truss can be neglected in comparison to the external load. Using finite element modeling we get for the displacement y of the joint
A E 2 K y = q where K = l A 2
which yields the vertical displacement
u 0 , y= , q= F v A1 + A 2 A2
If A is a given minimal cross section of the bars, we get dimensionless design variables by normalization:

, p 2 :=

and the dimensionless optimization criteria

M v = , f 2 := = (Fl) (AE ) _ _ _ _ _ _ lA _ _ _ _ _ _ _ _ _ _
Optimization of Mechanical Systems WT 10/11 A12.2
If the possible cross section has a maximum of A, then we get the vector optimization problem
_ _ _ _ _ _ _ _ _ opt pP _ _ _ _ _ _ _ _ _

where P = p 2

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Sketch the problem in the design space and in the criteria space for =4.
Optimization of Mechanical Systems WT 10/11 A12.3
From these sketches we get the EdgeworthPareto optimal solutions

, ____________

For the numerical determination of the EPoptimal solutions we generally formulate several scalar optimization problems. Formulate the following scalar optimization problems and determine the corresponding solution graphically.
1) Weighted Criteria Method
(Weighting factors: w1 = w 2 = )

__________

2) Distance Method
(Minimization of the Euklidean distance to the ideal solution
_________________________
Optimization of Mechanical Systems WT 10/11 A12.4

3) Hierarchical Method

If the minimal vertical displacement is more important than the minimal overall mass of the truss, then the priorities are l1 = and l 2 =. In step 1, the minimization of the vertical displacement only leads to

f 2 = min f 2 (p) = 0

In step 2, we try to additionally minimize the overall mass. For this, we have to allow some increase of the vertical displacement, e.g. 1 = 100 %. The second optimization step then reads
f1 = min f1 (p) where P1 = pP1 ___________________
Optimization of Mechanical Systems WT 10/11 A13.1

Numerical Tools

Optimization with MATLAB
MATLAB basics The program MATLAB is a commercially available program. It uses a numeric representation of vectors and matrices and allows a simple programming of matrix operations and provides many different methods to analyze systems. Many so-called Toolboxes are available to extend the functionality of MATLAB. The commands can be entered directly at the MATLAB prompt and executed immediately. Alternatively the commands can be summarized in a text file <name>.m. This socalled m-file is simply loaded and executed in MATLAB by typing <name> at the MATLAB prompt. helpdesk a = [ 3 -] b = [ 3 -]' A = [ 4; 23; 77 ] A A(2,3) A(:,3) A(1,:) zeros(n,m) ones(n,m) diag([3]) c = A * b c = A \ b inv(A) [ EV, ew ] = eig(A) plot(t,x,g),hold on; starts online help defines a row vector defines a column vector defines matrix transpose of matrix A matrix element A23 3rd column of matrix A 1st row of matrix A defines a zero matrix defines a matrix of ones defines a diagonal matrix matrix-vector multiplication solves linear system inverse of a matrix matrix EV of eigenvectors and vector ew of eigenvalues of matrix A plot vector x over vector t

For the numerical integration MATLAB provides two explicit single-step methods (ode23, ode45), one implicit single-step method (ode23s), an Adams-Bashfort-Moulton predictor corrector method (ode113) and an implicit numerical differentiation formula (ode15s). All methods have implemented a step size control. [t,X] = ode23('<name>',tspan,x0) simulation of a system of ordinary differential equations, given in the form dx/dt = f(t,x). Outputs are the vector t of time steps and
Optimization of Mechanical Systems WT 10/11 A13.2
the matrix X of state trajectories. The number of columns in X corresponds to the number of states and the number of rows to the number of time steps. Input is the <name> of a function specifying dx/dt, a (2x1) vector tspan = [t0, tend] contains the initial and the final time and the ini& & tial state vector is x0 = [ x1 ( t 0 ) x 2 ( t 0 ) L x1 ( t 0 ) x 2 ( t 0 ) K ]. The function <name> must be written in a separate file <name.m> with the corresponding file name and must follow the syntax: function dx = <name>(t, x). dx =. Example: Linearized double pendulum function dx = doublependulum(t,x) M1 = 1.0; IZZ = 0.0; GE = 9.81; L = 0.3; M = [ 2.0*M1*L^2+IZZ M1*L^2; M1*L^2 M1*L^2+IZZ ]; Q = [ 2*GE*M1*L 0; 0 GE*M1*L ]; A = [ zeros(2) diag([ 1 1]); -inv(M)*Q zeros(2)]; dx = A*x; MATLAB Optimization Toolbox The MATALAB Optimization Toolbox provides deterministic optimization algorithms for linear and non-nonlinear problems. The toolbox includes different solution methods for minimization problems, equations of nonlinear systems (fsolve, fzero) and least squares problems (curve fitting). Some important commands (described for medium scale optimization problems): fminsearch fminunc Finds minimum of unconstrained multivariable function using the derivative-free Simplex Search Method. Finds minimum of unconstrained multivariable function. The algorithm uses a Quasi-Newton method with a cubic line search procedure. This Quasi-Newton method uses the Broyden-Fletcher-Goldfarb-Shanno formula for updating the approximation of the Hessian matrix or the DavidsonFletcher-Powell formula for approximating the inverse Hessian matrix.
Institute of Eng. and Comp. Mechanics Prof. Dr.-Ing. Prof. E.h. P. Eberhard fmincon
Optimization of Mechanical Systems WT 10/11 A13.3
Finds minimum of constrained nonlinear multivariable function. The algorithm uses a SQP method. During each iteration step an estimate of the Hessian of the Lagrangian is updated and a line search is performed.
Example: Multicriteria Optimization of a truss structure using weighted criteria (A11) Main m-file truss.m for optimization: clear all; global w1 w2 % bounds of design variables alpha = 3; A = [1 0; 0 1; -; 0 -1]; b = [alpha alpha -1 -1]'; % start values x0 = [1 1]; % options for optimization algorithm options = optimset('LargeScale','off'); % weighting factors w1 = 0.3; w2 = 1-w1; % optimization [x,fval] = fmincon(@criteria,x0,A,b,[],[],[],[],[],options) M-file criteria.m for function evaluation: function f = criteria(x) global w1 w2 % calculation of optimization criteria % criteria f1 and f2 f1 = x(1)+sqrt(2)*x(2); f2 = 1/x(1); % scalar optimization criteria f = w1*f1+w2*f2; Example: Computation of the Pareto-Front of a truss structure (A11) The file truss.m is extended in order to compute the entire Pareto-Front. Thereby the weighted criteria method with variable weights is used. clear all; global w1 w2 alpha = 3; % bounds of design variables A = [1 0; 0 1; -; 0 -1]; b = [alpha alpha -1 -1]'; % start values x0 = [1 1]; % options for optimization algorithm

Optimization of Mechanical Systems WT 10/11 A13.4
options = optimset('LargeScale','off'); for i=1:51; % weighting factors w1 = (i-1)/50; w2 = 1-w1; % optimization [x(:,i),fval] = fmincon(@criteria,x0,A,b,. [],[],[],[],[],options); % use previous results as next starting values x0 = x(:,i); % criteria f1 and f2 f1(i) = x(1,i)+sqrt(2)*x(2,i); f2(i) = 1/x(1,i); end figure(1) subplot(1,2,1) plot([1 alpha],[1 1],'b'), hold on; plot([1 1],[1 alpha],'b'); plot([1 alpha],[alpha alpha],'b'); plot([alpha alpha],[1 alpha],'b'); plot(x(1,:),x(2,:),'r+'); axis([0.8 alpha+0.2 0.8 alpha+0.2]) xlabel('p_1') ylabel('p_2') subplot(1,2,2) plot(f1,f2,'r+'), hold on; axis([0 1.2]) xlabel('f_1') ylabel('f_2')
Figure 1: Pareto-Front of the truss structure. What problem do you see?
MATLAB Genetic Algorithm and Direct Search Toolbox The MATLAB Genetic Algorithm and Direct Search Toolbox is a collection of stochastic optimization algorithms. This toolbox includes routines for solving optimization problems using Direct Search, Genetic Algorithm and Simulated Annealing.
Optimization of Mechanical Systems WT 10/11 A13.5

Optimization with EXCEL

For optimizing nonlinear problems EXCEL uses the Generalized Reduced Gradient (GRG2) algorithm developed by Leon Lasdon and Allan Waren. Other solvers can be installed, see e.g. www.solver.com/exceluse.htm.