Hello... Here is my SAS program. I want to point out that it is not only the termination criteria values that differ from the SAS documentation, but their names also ( ABSXCONV versus ABSXTOL) prociml;   start ARL(hr); shift=5;    m=100;   t=2*m+1;   delta=2*hr[1]/t;    z=-hr[1]+.5*delta; do i=2 to t;   k=-hr[1]+(i-.5)*delta;   z=z//k;   end;         do i=1 to t;          zz=0;                 do j=1 to t;                  tau1=z[j, ]+(delta/2)-z[i,];                tau2=z[j, ]-(delta/2)-z[i,];   if tau1<-(hr[2]*hr[3])then fun1=tau1-(1-hr[2])*hr[3];      if abs(tau1)<=(hr[2]*hr[3])then fun1=tau1/hr[2];     if tau1>(hr[2]*hr[3])then fun1=tau1+(1-hr[2])*hr[3];    if tau2<-(hr[2]*hr[3])then fun2=tau2-(1-hr[2])*hr[3];    if abs(tau2)<=(hr[2]*hr[3])then fun2=tau2/hr[2];    if tau2>(hr[2]*hr[3])then fun2=tau2+(1-hr[2])*hr[3];                 kone=z[i,]+fun1;                 ktwo=z[i,]+fun2;              pij=probnorm(kone-shift)-probnorm(ktwo-shift);              zz=zz//pij;               end;          if i=1 then r=zz;          else        r=r||zz;          end;     v=j(t,1,0);          v[((t+1)/2),]=1;       v=t(v);   r=r[2:t+1,];   r=t(r);   avrl=inv((I(t)-r))*j(t,1,1);   bb=(v*avrl);   return(bb);   finish;    start ARL0(hr);   m1=100;   shift=0;   bb0=j(1,1,0.);   t1=2*m1+1;   delta1=2*hr[1]/t1;   z1=-hr[1]+.5*delta1;   do i1=2 to t1;   k1=-hr[1]+(i1-.5)*delta1;   z1=z1//k1;   end;   do i1=1 to t1;         zz1=0;                 do j1=1 to t1;                  tau11=z1[j1, ]+(delta1/2)-z1[i1,];                 tau21=z1[j1, ]-(delta1/2)-z1[i1,];   if tau11<-(hr[2]*hr[3])then fun11=tau11-(1-hr[2])*hr[3];   if abs(tau11)<=(hr[2]*hr[3])then fun11=tau11/hr[2];    if tau11>(hr[2]*hr[3])then fun11=tau11+(1-hr[2])*hr[3]; if tau21<-(hr[2]*hr[3])then fun21=tau21-(1-hr[2])*hr[3];    if abs(tau21)<=(hr[2]*hr[3])then fun21=tau21/hr[2];    if tau21>(hr[2]*hr[3])then fun21=tau21+(1-hr[2])*hr[3];            kone1=z1[i1,]+fun11;            two1=z1[i1,]+fun21;            pij1=probnorm(kone1-shift)-probnorm(ktwo1-shift);                 zz1=zz1//pij1;                 end;         if i1=1 then r1=zz1;          else          r1=r1||zz1;         end;   v1=j(t1,1,0);         v1[((t1+1)/2),]=1;        v1=t(v1);   r1=r1[2:t1+1,];   r1=t(r1);   avrl1=inv((I(t1)-r1))*j(t1,1,1);   bb0[1]=1000-(v1*avrl1);   return(bb0);   finish;   con = {   0    0    0,                     .    1    .    };   hr={2.015  0.354  1.989};    optn= j(1,11,.);optn[2]=5;optn[10]=1;optn[11]=0;   CALL NLPNMS(rc,hrres,"ARL",hr) blc=con opt=optn nlc="ARL0";   quit; ... View more

The uniqueness of a solution depends on the function being optimized, not the particular choice of an optimization routine. In general, if you start near a local optimum, you will converge to that optimum (but the beautiful fractals for Newton's iteration show that this is not always the case!)  If your function has several local maxima and you want to find the one that is the global maximum, you can try using initial conditions on a uniform grid in parameter space, as detailed in this article: How to find an initial guess for an optimization - The DO Loop ... View more

Oops, I missed the fact that you were still at 9.0.  RANDGEN and RANDSEED were introduced in 9.1. I don't believe the old random number generation functions allow the arbitrary re-initialization of the random number stream. In order to get the same sampling, you will have to create a table of random variates at the beginning of your program, and re-use that stream at each call. If you need more samples than can be stored in memory, you might have to store them in a data set and read them in as called for. ... View more

I don't know much about IML but you can always move your matrix to a data set then use a standard PROC EXPORT. Writing data from a matrix to a SAS data set - The DO Loop ... View more

Documentation and an example of using a linear constraint matrix is available at SAS/IML(R) 12.3 User's Guide However, you need to debug your function before you can optimize it.  It looks to me like the 'n' in sqrt(n) is not defined. Also, you are dividing by x[1], so you need to introduce the constraint that x1>0. ... View more

You have an infinite loop. There is no reason to expect the DO UNTIL loop to ever terminate. In fact, if you let it run for 10,000 iterations, you get a plot that looks like the one I've attached. For these iterations |z| is usually in [-0.5, 0.5]. A good programming practice is to put a maximum number of iterations on the DO/UNTIL loop, like this:    do rl = 1 to 10000 until(zz>h|zz<-h);    ...   end; That way, you never have an infinite loop. ... View more

Hello Rick,I tried the following bounds for the inner integral [-10,10] . I got errors concerning matrix inverse and convergence. The log file is attached ... View more

I run my model once again and it took 10 minutes to get the output i attached before, my questions are: 1-does the output indicate problems in the model?does the output indicate that my constraint is violated or that the optimal parameter estimates won't result in the displayed optimal objective value? 2-why the optimal parameter estimates don't satisfy the constraints and don't result in the optimal objective value? 3-why every time i run my model i get different answers? 4-Does this part of the output indicate that the constraint is satisfied? if so, why when i substitute with the optimal values in my constraint, i find that the constraint is not satisfied?                        Values of Nonlinear Constraints                                                                                                                                              Constraint          Residual                                                                                                                                                        [            1 ]           0 Active  NLEC Waiting for your reply, Emara ... View more

Here is the paper "Optimization designs of the combined Shewhart-CUSUM control charts" Waiting for your reply, Emara ... View more

Hello.......... 1- FUN3  represents the constraint of my model. FUN3 has the same formula as FUN "the objective function" but at s=0. Thus my problem is to  find the optimal "M" value that minimizes the integration of FUN subject to the constraint that FUN3=10. Thus my optimal M should give me the minimum integration value of FUN but at the same time FUN3 should be equal to 10 at the optimal M. That is I am restricting the value of FUN to be equal to 10 at s=0 **I should note that i have a mistake in my first post as I should have replaced   "X2=normal(-1)+0.5;"    by     "X2=normal(-1);"    in FUN3 4)When i run your code  , I didn't get a 10 but I got different values of Y for different M values. What i need is to get a 10 at the optimal M since FUN3 represents my constraint. 5) No, I am not trying to do so. Thanks alot, A.Emara ... View more

Emara, There's a quite similar posting about this on comp.soft-sys.sas which I believe is from you as well. I've responded in the newsgroup but will respond here also in the interest of getting the information to as many interested parties as possible. Thanks for posting this question.  The best way to do this currently is to use PROC NLP along with PROC FCMP (a Base SAS procedure) to define the objective function.  I consulted with one of our developers on this question and he sent some sample code which uses this approach for a problem that's pretty similar to yours.  It's included at the end of this email. PROC OPTMODEL in SAS/OR is the successor to PROC NLP (it handles not only nonlinear but also linear, mixed integer, and quadratic optimization). However OPTMODEL is not yet FCMP-enabled and so it's best, given your requirements, for you to stick with PROC NLP for now. If you have any questions please feel free to contact me. Thanks, Ed Hughes SAS/OR Product Manager 919-531-6916 Ed.Hughes@sas.com Sample code follows: data barddata; input y @@; datalines; 0.14 0.18 0.22 0.25 0.29 0.32 0.35 0.39 0.37 0.58 0.73 0.96 1.34 2.10 4.39 ; proc fcmp outlib=work.myfuncs.test; function bard(x1, x2, x3);   array y[15] /nosymbols ;   rc = read_array('barddata', y);   fx = 0;   do k = 1 to 15;    vk = 16 - k;    wk = min(k,vk);    fxk = y - (x1 + k/(vk*x2 + wk*x3));    fx = fx + fxk**2;   end;   return (0.5*fx); endsub; run; options cmplib = work.myfuncs; proc nlp; parms x1 x2 x3 = 1.0;   bounds  0 <= x1 <= 10,           0 <= x2 <= 10,           0 <= x3 <= 10; min bardfn; bardfn = bard(x1,x2,x3); run; ... View more