Hi all,
In my problem, I have two non linear constrains ; one of them in is arl1 <= 4 and the other one arl0 >= 500 and this is my problem because when I wrote these constraints using proc iml, the SAS considers them as equality constraints and I get an arl0 value=500, so how can I write this constraint in SAS so that it is considered as a greater than constraint? please note that I set optn[10]=2;optn[11]=0;
This is part of my program :
start
arl(x);
bb0=j(2,1,0.);
m=100;
meuw=0;
L=x[2]*SQRT(x[1]/(2-x[1]));
t=2*m+1;
delta=2*L/t;
z=-L+.5*delta;
do
i=2 to t;
k=-L+(i-.5)*delta;
z=z//k;
end;
do
i=1 to t;
zz=0;
do
j=1 to t;
kone=((z[j,]+delta/2)-(1-x[1])*z[i,])/x[1];
ktwo=((z[j,]-delta/2)-(1-x[1])*z[i,])/x[1];
pij=probnorm(kone)-probnorm(ktwo);
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);
arl=inv(I(t)-r)*j(t,1,1);
arl0=(v*arl);
bb0[1]=arl0-500;
m=100;
shift=2;
L=x[2]*SQRT(x[1]/((2-x[1])));
t=2*m+1;
delta=2*L/t;
z=-L+.5*delta;
do
i=2 to t;
k=-L+(i-.5)*delta;
z=z//k;
end;
do
i=1 to t;
zz=0;
do
j=1 to t;
kone=((z[j,]+delta/2)-(1-x[1])*z[i,])/x[1];
ktwo=((z[j,]-delta/2)-(1-x[1])*z[i,])/x[1];
pij=probnorm(kone-(shift*sqrt(n)))-probnorm(ktwo-(shift*sqrt(n)));
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);
arl=inv((I(t)-r))*j(t,1,1);
arl1=(v*arl);
bb0[2]=4-arl1;
return (bb0);
finish
arl;
con
= { 0. 0.,
1. .};
x={0.3 2.5};
optn=j(1,11,.);optn[1]=0;optn[2]=2;optn[10]=2;optn[11]=0;
CALL nlpnms(rc,xres,"Cost",x,optn,con)nlc="arl";
quit;
Thanks,
Aya
