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consider the line y=-x+1 on xy plane. the shortest line from the origin to y=-x+1 intersects with it at (x*,y*)=(0.5,0.5). Confirm this result by formulating and solving and optimization problem. 

Here is a short wrapper which automates repeated calls to the DirectSearch 2 curve-fitting routine. It offers both time and repetition (solver restart) limits.

The global optimization package DirectSearch 2 (see Application Center link, and here) has some very nice features. One aspect which I really like is that it can do curve-fitting: to fit an expression using tabular data. By this, I mean that it can find optimal values of parameters present in an expression (formula) such that the residual error between that formula and the tabular data is minimized.

Maple itself has commands from the CurveFitting and Statistics packages for data regression, such as NonlinearFit, etc. But those use local optimization solvers, and quite often for the nonlinear case one may need a global optimizer in order to produce a good fit. The nonlinear problem may have local extrema which are not even close to being globally optimal or provide a close fit.

Maplesoft offers the (commercially available) GlobalOptimization package as an add-on to Maple, but its solvers are not hooked into those mentioned curve-fitting commands. One has to set up the proper residual-based objective function onself in order to use this for curve-fitting, and some of the bells and whistles may be harder to do.

So this is why I really like the fact that the DirectSearch 2 package has its own exported commands to do curve-fitting, integrated with its global solvers.

But as the DirectSearch package's author mentions, the fitting routine may sometimes exit too early. Repeat starts of the solver, for the very same parameter ranges, can produce varying results due to randomization steps performed by the solver. This post is branched off from another thread which involved such a problematic example.

Global optimization is often a dark art. Sometimes one may wish to simply have the engine work for 24 hours, and produce whatever best result it can. That's the basic enhancement this wrapper offers.

Here is the wrapper, and a few illustrative calls to it on the mentioned curve-fitting example that show informative  progress status messages, etc. I've tried to make the wrapper pretty generic. It could be reused for other similar purposes.

Other improvements are possible, but might make it less generic. A target option is possible, where attainment of the target would cause an immediate stop. The wrapper could be made into an appliable module, and the running best result could be stored in a module local so that any error (and ensuing halt) would not wipe out the best result from potentially hours and hours worth of conputation.

restart:
randomize():

repeater:=proc(  funccall::uneval
               , {maxtime::numeric:=60}
               , {maxiter::posint:=10}
               , {access::appliable:=proc(a) SFloat(a[1]); end proc}
               , {initial::anything:=[infinity]}
              )
          local best, current, elapsed, i, starttime;
            starttime:=time[real]();
            elapsed:=time[real]()-starttime;
            i:=1; best:=[infinity];
            while elapsed<maxtime and i<=maxiter do
              userinfo(2,repeater,`iteration `,i);
              try
                timelimit(maxtime-elapsed,assign('current',eval(funccall)));
              catch "time expired":
              end try;
              if is(access(current)<access(best)) then
                best:=current;
                userinfo(1,repeater,`new best `,access(best));
              end if;
              i:=i+1;
              elapsed:=time[real]()-starttime;
              userinfo(2,repeater,`elapsed time `,elapsed);
            end do;
            if best<>initial then
              return best;
            else
              error "time limit exceeded during first attempt";
            end if;
          end proc:


X := Vector([seq(.1*j, j = 0 .. 16), 1.65], datatype = float): 

Y := Vector([2.61, 2.62, 2.62, 2.62, 2.63, 2.63, 2.74, 2.98, 3.66,
             5.04, 7.52, 10.74, 12.62, 10.17, 5, 2.64, 11.5, 35.4],
            datatype = float):

F := a*cosh(b*x^c*sin(d*x^e));

                                    /   c    /   e\\
                         F := a cosh\b x  sin\d x //

infolevel[repeater]:=2: # or 1, or not at all (ie. 0)
interface(warnlevel=0): # disabling warnings. disable if you want.

repeater(DirectSearch:-DataFit(F
                      , [a=0..10, b=-10..10, c=0..100, d=0..7, e=0..4]
                      , X, Y, x
                      , strategy=globalsearch
                      , evaluationlimit=30000
                              ));
repeater: iteration  1
repeater: new best  9.81701944539358706
repeater: elapsed time  15.884
repeater: iteration  2
repeater: new best  2.30718902535293857
repeater: elapsed time  22.354
repeater: iteration  3
repeater: new best  0.627585701120743822e-4
repeater: elapsed time  30.777
repeater: iteration  4
repeater: elapsed time  47.959
repeater: iteration  5
repeater: new best  0.627585700905294148e-4
repeater: elapsed time  55.221
repeater: iteration  6
repeater: elapsed time  60.009
 [0.0000627585700905294, [a = 2.61748237902808, b = 1.71949329097179, 

   c = 2.30924401405164, d = 1.50333106110324, e = 1.84597267458055], 4333]


# without userinfo messages printed
infolevel[repeater]:=0:
repeater(DirectSearch:-DataFit(F
                      , [a=0..10, b=-10..10, c=0..100, d=0..7, e=0..4]
                      , X, Y, x
                      , strategy=globalsearch
                      , evaluationlimit=30000
                              ));

 [0.0000627585701341043, [a = 2.61748226209478, b = 1.71949332125427, 

   c = 2.30924369227236, d = 1.50333090706676, e = 1.84597294290477], 6050]


# illustrating early timeout
infolevel[repeater]:=2:
repeater(DirectSearch:-DataFit(F
                      , [a=0..10, b=-10..10, c=0..100, d=0..7, e=0..4]
                      , X, Y, x
                      , strategy=globalsearch
                      , evaluationlimit=30000
                              ),
         maxtime=2);

repeater: iteration  1
repeater: elapsed time  2.002
Error, (in repeater) time limit exceeded during first attempt

# illustrating iteration limit cutoff
infolevel[repeater]:=2:
repeater(DirectSearch:-DataFit(F
                      , [a=0..10, b=-10..10, c=0..100, d=0..7, e=0..4]
                      , X, Y, x
                      , strategy=globalsearch
                      , evaluationlimit=30000
                              ),
         maxiter=1);

repeater: iteration  1
repeater: new best  5.68594272127419575
repeater: elapsed time  7.084
 [5.68594272127420, [a = 3.51723075672918, b = -1.48456068506828, 

   c = 1.60544055207338, d = 6.99999999983179, e = 3.72070034285212], 2793]


# giving it a large total time limit, with reduced userinfo messages
infolevel[repeater]:=1:
Digits:=15:
repeater(DirectSearch:-DataFit(F
                      , [a=0..10, b=-10..10, c=0..100, d=0..7, e=0..4]
                      , X, Y, x
                      , strategy=globalsearch
                      , evaluationlimit=30000
                              ),
         maxtime=2000, maxiter=1000);

repeater: new best  3.10971990123465947
repeater: new best  0.627585701270853103e-4
repeater: new best  0.627585700896181428e-4
repeater: new best  0.627585700896051324e-4
repeater: new best  0.627585700895833535e-4
repeater: new best  0.627585700895607885e-4
 [0.0000627585700895608, [a = 2.61748239185387, b = -1.71949328487160, 

   c = 2.30924398692221, d = 1.50333104262348, e = 1.84597270535142], 6502]

Hi all,

I try to solve the following nonlinear programming problem:

NLPSolve((int((c[0]+c[1]*u+c[2]*u^2)^2, u = -a .. a))^(4/5)*(int(u^2*(c[0]+c[1]*u+c[2]*u^2), u = -a .. a))^(2/5), variables = [c[0], c[1], c[2]], [int(c[0]+c[1]*u+c[2]*u^2, u = -a .. a) = 1, int(c[0]*u+c[1]*u^2+c[2]*u^3, u = -a .. a) = 0])

but this error message appear:

I got an MS Excel file where the matrix (7054x60) is created and the last row of the matrix is used for optimization of its (the matrix) values. In Maple you can create the matrix as follows:

M := Matrix(7054, 60):
for i to 10 do M[1, i] := 1 end do:

Mat_proc := proc(x)
  local i, j;
  global M;
  for i from 2 to 7054 do

helo to all i want to know how to solve a lp model having 4 decision variables and 4 constraints means how to solve them to get  optimal solution . Plz guide me ,thanks  : )

Hi,

I would like to ask, what is the way to use the Direct Search (or other similar package) to solve not a static optimization problem for, say f(x_1,x_2), but the dynamic one, say for f(x_1(t),x_2(t),t)?

I think there must be the way of adapting the algorithm for that.

Any ideas welcome.

Best

I was solving one Math puzzle and found out a strange behaviour of the Optimization package commands in the Classic Worksheet.

Let us minimize the function 7*x+3*y under the constraint 6*x+2*y >= 49 for non-negative integers. When I type

Optimization:-Minimize(7*x+3*y, {6*x+2*y >= 49}, assume = nonnegint);

Let function f(x)=x(2011+√(2013-x2)).Find maximum and minimum value of function f(x)?

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 ...

HI TO ALL. I M IN THE THEISIS PHASE OF MY MS. I NEED CODES FOR SOLVING A JOB SHOP SCHEDULING PROBLEM USING ANT COLONY OPTIMIZATION (ACO). IF ANY ONE HAS SUCH AN ACO CODE DEVELOPED TO SOLVE ANY TYPE OF JOB SHOP SCHEDULIGN PROBLEM, PLZ PLZ SHARE THAT AT seoccsf7@hotmail.com

 

I WILL BE GREATFUL

 

REGARDS 

Hi,

how can i install the DirectSearch optimization package for mac version (MAPLE 15)?

Gil

I am confused by the below results.

Why does LSSolve (given an arbitrary expected return) produce a higher risk
adjusted return than QPSolve which explicity is given an objective to maximize
risk adjusted returns? ie minimize Transpose(W).Cov.W-Transpose(W).ev
This to me seems very strange?!

Also, how do you specify in the objective function for LSSolve to maximize
risk adjusted returns? Now we have simply provide LSSolve with some user specified

How to play the Maple worksheet Alkylation Process Model Using DirectSearch

(http://www.maplesoft.com/applications/view.aspx?SID=1675)

There is some equivalence between GlobalOptimization commands: GetLastSolution and GlovalSolve and DirectSearch commands?

Gracias

How to test for convexity

December 20 2011 by Carlos Mallen 5 Maple

Hi,

I'm using Maple 14 and have a function in several variables, which I want to minimize. As far as I know, to guarantee that NLPSolve (from the Optimization package) is really finding a global minimum, this function must be convex.

What is the best way to test for convexity in Maple?

Thanks in advance,

Carlos Mallen

raw_panel_cuttin.doc

 Given a large rectangular wooden panel, you want to cut specific quantities of smaller rectangles from it.

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