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Hi,

On page 32 (PDF)

 

Two different results were obtained using the Global optimization.

Log likelihood does not differ much. BUT the estimates vary a lot, such as mu[p].

tmp.mw

tmp.pdf

 

When I tried to use one of the answer from a particular run, I get the HFLOATING error, see picture.

So how reliable is this? Could there be a better way to optimize this ?

 

Thanks!

 

As an additional note, if I have Matlab R2014a, could I use Matlab to optimize the target function? DO I need to purchase a seperate addon?

 

Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

Download work1.mwswork1.mws

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I need to maximize two multivariate objective functions (f(x1,y1,z1,t1) and g(x2,y2,z2,t2)) with inequality and nonnegativity constraints (x1, x2>0 and y1, z1, t1, y2, z2, t2 >=0). I am looking for parametric not numerical solutions.


What is the best way to find the solution to such a problem using maple?

Birthday_LP.mw

I am attempting to solve the following using C.Loves Logicproblem package. But i do not know how to tell it either/or constaints. being a circular arrangement might also throw a spanner in the works.

Alans Birthday Logic Problem

It is Alan’s birthday and he is having a party. Seven other people will attend: Amy, Brad, Beth, Charles, Debbie, Emily and Frances.

Everyone will sit around the circular dining table. The seating arrangement must meet the following conditions:

• Amy and Alan sit together

• Brad and Beth sit together

• Charles sits next to either Debbie or Emily

• Frances sits next to Debbie

• Amy and Alan do not sit next to either Brad or Beth

• Brad does not sit next to Charles or Frances

• Debbie and Emily do not sit next to each other

• Alan does not sit next to either Debbie or Emily

• Amy does not sit next to Charles

Arrange the guests around the table to meet all of the conditions listed above.

restart:

Vars:= [Name,PN]:

Name:= [Alan,Amy, Brad, Beth, Charles, Debbie, Emily,Frances]:

PN:=[$1..8]:

Cons:= [NextTo(Amy,Alan,PN),NextTo(Brad,Beth,PN),NextTo(Charles,Debbie,PN),NextTo(Charl es,Emily,PN),NextTo(Frances,Debbie,PN),Rel(NotNextTo,Amy,Brad,PN),Rel(NotNextTo, Amy,Beth,PN),Rel(NotNextTo,Alan,Brad,PN),Rel(NotNextTo,Alan,Beth,PN),Rel(NotNext To,Brad,Charles,PN),Rel(NotNextTo,Brad,Frances,PN),

Rel(NotNextTo,Debbie,Emily,PN),Rel(NotNextTo,Alan,Debbie,PN),Rel(NotNextTo,Alan,Emily,PN),

Rel(NotNextTo,Amy,Charles,PN)]:

read "LogicProblem.mpl";

Birthday:= LogicProblem(Vars):

with(Birthday):

 

 

Following previous question at

http://www.mapleprimes.com/questions/149581-Improve-Algorithm-Dsolve

and also

http://www.mapleprimes.com/questions/149243-BVP-With-Constraining-Integrals

I wrote the following code

***********************

restart:

gama1:=0:


phi0:=0.00789:


rhocu:=2/(1-zet^2)*int((1-eta)*rho(eta)*c(eta)*u(eta),eta=0..1-zet):

eq1:=diff(u(eta),eta,eta)+1/(mu(eta)/mu1[w])+((1/(eta-1)+1/mu(eta)*(mu_phi*diff(phi(eta),eta)))*diff(u(eta),eta)):
eq2:=diff(T(eta),eta,eta)+1/(k(eta)/k1[w])*(2/(1-zet^2)*rho(eta)*c(eta)*u(eta)/(p2*10000)+( (a[k1]+2*b[k1]*phi(eta))/(1+a[k1]*phi1[w]+b[k1]*phi1[w]^2)*diff(phi(eta),eta)-k(eta)/k1[w]/(1-eta)*diff(T(eta),eta) )):
eq3:=diff(phi(eta),eta)-phi(eta)/(N[bt]*(1-gama1*T(eta))^2)*diff(T(eta),eta):
mu:=unapply(mu1[bf]*(1+a[mu1]*phi(eta)+b[mu1]*phi(eta)^2),eta):
k:=unapply(k1[bf]*(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2),eta):
rhop:=3880:
rhobf:=998.2:
cp:=773:
cbf:=4182:
rho:=unapply(  phi(eta)*rhop+(1-phi(eta))*rhobf ,eta):
c:=unapply(  (phi(eta)*rhop*cp+(1-phi(eta))*rhobf*cbf )/rho(eta) ,eta):
mu_phi:=mu1[bf]*(a[mu1]+2*b[mu1]*phi(eta)):

a[mu1]:=39.11:
b[mu1]:=533.9:
mu1[bf]:=9.93/10000:
a[k1]:=7.47:
b[k1]:=0:
k1[bf]:=0.597:
zet:=0.5:
#phi(0):=1:
#u(0):=0:
phi1[w]:=phi0:
N[bt]:=0.2:
mu1[w]:=mu(0):
k1[w]:=k(0):

eq1:=subs(phi(0)=phi0,u(0)=0,eq1):
eq2:=subs(phi(0)=phi0,u(0)=0,eq2):
eq3:=subs(phi(0)=phi0,u(0)=0,eq3):

p:=proc(pp2) global res,F0,F1,F2:
if not type([pp2],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({eq1=0,subs(p2=pp2,eq2)=0,eq3=0,u(0)=0,u(1-zet)=0,phi(0)=phi0,T(0)=0,D(T)(0)=1}, numeric,output=listprocedure):
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
evalf(2/(1-zet^2)*Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*( F1(eta)*rhop*cp+(1-F1(eta))*rhobf*cbf )/(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))-pp2*10000:
end proc:


s1:=Student:-NumericalAnalysis:-Secant(p(pp2),pp2=[6,7],tolerance=1e-6);

                   HFloat(6.600456858832996)

p2:=%:



ruu:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))):
phb:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))) :
TTb:=evalf(2/(1-zet^2)*(Int((1-eta)*F2(eta),eta=0..1-zet))):
rhouu:=evalf(2/(1-zet^2)*(Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))):
with(plots):
res(parameters=[R0,R1]):
odeplot(res,[[eta,u(eta)/ruu],[eta,phi(eta)/phb],[eta,T(eta)/TTb]],0..zet);

 

*************************************

as you can see at the second line of the code, the value of phi0:=0.00789. however, I want to modify the code in a way that phi0 is calculated with the following addition constraint

evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet)))-0.02=0

I would be most grateful if you could help me in this problem.

Thanks for your attention in advance

Amir

In MapleSim, it is possible to constrain a rigid body in a multibody system to an arbitrary constraint? 

Similarly to how a mass at the end of a simple pendulum can only move along a circle defined by the length of the pendulum, I was hoping to constrain one to a shape like an ellipse or sine wave.  I could then give it initial velocities and see how it would move along the shape over time.

I'd like to be able to define this constraint with an equation. ...

Hi all,

First of all, I would like to mention that I'm new to Maple, so please excuse my lack of knowledge.

I have a rather complicated expression, let us call it expr, and I have a constraint equation, let us call it constr = 0. I would like to simplify expr knowing that constr = 0.

Stupid example:

expr: 2*x+4*y+3

constr: x+2*y = 0

In that case:

expr = 2*(x+2*y)+3 = 2*0+3 = 3

How can I do that?

Thanks a lot.

Hello, I'm creating a rolling thin disk (rolling without slip) in the MapleSim, the model like the picture below (Fig. 1).

 

       ...

Hi people!

Why there is not a "Modelica.Mechanics.MultiBody.Joints.RollingWheel" (different from 1-D mechanics rolling wheel) in MapleSim 5? It is previewed in the Modelica language and SystemModeler has this implementation.

Is there a way to import the model o this joint from somewhere? If not, how could I model this non-holonimic constraint in MapleSim? 

Thanks! 

I have a non linear Sharpe ratio with 3 portfolio weights w1,w2 and w2. I want to (globally) maximize the sharpe ratio by choosing w1,w2 and w2 subject to the constraints that each of the variables is in the range of 0 to 1, and that their summation is equal to 1. I also want the maximization to start at an initial point of [w1=0.35,w2=0.6,w3=0.05].

The function is:

SR:= (0.012w1+0.007w2+0.0384w3-0.009)/(stdev)

where stdev is the standard deviation of the portfolio ...

I'm having trouble maximizing this funciton:

 

S=  (w1*E1+w2*E2+w3*E3)/(CoVar[1,2]), subject to the constraint w1+w2+w3=1

 

I need it to choose w1, w2, and w3 to maximize S

Does anybody know how to configure the model to obtain the position constraint equations of a slider-crank mechanism as:

l1·cos(theta)+l2·sin(beta)-s = 0
l1·sin(theta)-l2·cos(beta) = 0

instead as obtained in the help-example on:
http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=MapleSim/Multibody/Kinematic_Exports

Thanks in advance

Ok, when I run the below code which maximize the risk adjuested portfolio returns
(long and short positions) in QP matrix form on empirical data I get very strange
allocations ie we go 100% or 100% short in almost all stocks except for a few
where the allocations are more appropriate like 0.2 etc.


# Maximize Risk Adjuested Return Matrix Form
# Minimize W'.Cov.W−W'.EV
# R=Return Matrix

EV := Vector([seq(ExpectedValue(Column(R, i)), i = 1 .. N)], datatype = float[8]):

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