Reputation: 1537
Given a set of variable v(1) - v(k)
, a function f
is defined as f(v1,v2,...vk)
.
The target is to have a set of v(i) that maximize f
given v(1)+v(2)+....+v(k)=n
. All elements are restricted to non-negative integers.
Note: I don't have SAS/IML
or SAS/OR
.
If k
is known, say 2
, then I can do sth like this.
data out;
set in;
maxf = 0;
n1 = 0;
n2 = 0;
do i = 0 to n;
do j = 0 to n;
if i + j ne n then continue;
_max = f(i,j);
if _max > maxf then do;
maxf = max(maxf,_max);
n1 = i;
n2 = j;
end;
end;
end;
drop i j;
run;
However, this solution has several issues.
Using loops seems to be very inefficient.
It doesn't know how may nested loops needed when k is unknown.
It's exactly the "Allocate n balls into k bins" problem where k
is determined by # of columns in data in
with specific prefix and n
is determined by macro variable.
Function f
is known, e.g f(i,j) = 2*i+3*j
;
Is this possible to be done in data step?
Upvotes: 0
Views: 529
Reputation: 12465
As said in the comments, general non-linear integer programs are hard to solve. The method below will solve for continuous parameters. You will have to take the output and find the nearest integer values that maximize your function. However, the loop will now be much smaller and quicker to run.
First let's make a function. This function has an extra parameter and is linear in that parameter. Wrap your function inside something like this.
proc fcmp outlib=work.fns.fns;
function f(x1,x2,a);
out = -10*(x1-5)*(x1-5) + -2*(x2-2)*(x2-2) + 2*(x1-5) + 3*(x2-2);
return(out+a);
endsub;
run;quit;
options cmplib=work.fns;
We need to add the a
parameter so that we can have a value that SAS can pass besides the actual parameters. SAS will think it's solving the likelihood of A
, based on x1
and x2
.
Generate a Data Set with an A
value.
data temp;
a = 1;
run;
Now use PROC NLMIXED
to maximize the likelihood of A
.
ods output ParameterEstimates=Parameters;
ods select ParameterEstimates;
proc nlmixed data=temp;
parms x1=1 x2=1;
bounds x1>0, x2>0;
y = f(x1,x2,a);
model a ~ general(y);
run;
ods select default;
I get output of x1=5.1
and x2=2.75
. You can then search "around" that to see where the maximum comes out.
Here's my attempt at a Data Step to search around the value:
%macro call_fn(fn,n,parr);
%local i;
&fn(&parr[1]
%do i=2 %to &n;
, &parr[&i]
%end;
,0)
%mend;
%let n=2;
%let c=%sysevalf(2**&n);
data max;
set Parameters end=last;
array parms[&n] _temporary_;
array start[&n] _temporary_;
array pmax[&n];
max = -9.99e256;
parms[_n_] = estimate;
if last then do;
do i=1 to &n;
start[i] = floor(parms[i]);
end;
do i=1 to &c;
x = put(i,$binary2.);
do j=1 to &n;
parms[j] = input(substr(x,j,1),best.) + start[j];
end;
/*You need a macro to write this dynamically*/
val = %call_fn(f,&n,parms);
*put i= max= val=;
if val > max then do;
do j=1 to &n;
pmax[j] = parms[j];
end;
max = val;
end;
end;
output;
end;
Upvotes: 1