A different take, using a data step and a hash object:
data want;
set have;
by company_id;
if _n_ = 1
then do;
length month number 8;
declare hash sum ();
sum.definekey("month");
sum.definedata("number");
sum.definedone();
end;
if first.company_id then rc = sum.clear();
month = intnx('month',announcement,0,"b");
rc = sum.find();
number = sum(number,1);
if rc
then rc = sum.add();
else rc = sum.replace();
month = intnx('month',announcement,-12,"b");
do until (month ge intnx('month',announcement,0,"b"));
if sum.find() = 0
then Prior_12_Month_Ann = sum(Prior_12_Month_Ann,number);
month = intnx("month",month,1,"b");
end;
drop rc number month;
run;
Since the code counts per calendar month, not per "month counted backwards from the current date", it creates slightly different numbers.
OTOH, it will be faster than SQL with larger datasets, as it avoids the sub-select.
