SAP BW Pre-calculation Server Not Working as Expected

Question

Hi All,
It has already been 2 months since we have installed our pre-calculation servers. We have been bumping to issue with our server showing as "At Capacity" after a few runs of workbooks. Then we need to restart the service in order to resume the pre-calculation. Something really dumb is that any workbooks that failed during that time will not get pre-calculated again.
This is really bugging me as every morning when I know, there is some broadcasting, then I need to start monitoring the queue, to see if the symptoms of only a few workbooks (usually 10 workbooks) being able to calculate before I need to restart the service.
Please help as I am really in a dire situation.
David Yee

Answer 1

dk00111 wrote:
Ah, I forgot to cast the correctAnswer as a double. Thanks, it works now! (disregard the username change)I sense you're still misunderstanding, so I decided to follow up.
The issue isn't that you weren't casting your expression to a double before assigning to correctAnswer. That cast is implicit as an integer is automatically promoted to a double, since it is a widening conversion (see here).
What you had before was like this:
double d = 1 / 5;
System.out.println(d); //0.0In this case, we have an integer divided by an integer, and so the result is (naturally!) an integer. Since 1 / 5 is truncated to 0, that's the result of the expression.
"Casting to a double" would be this:
d = (double)(1 / 5);
System.out.println(d); //0.0But that's no different! The cast is pointless, as it happens anyway when you assign the result of the expression to d. 1 / 5 is still occurring with integer operands.
Instead, based on the fact that it worked, I'm guessing what you did was this:
d = (double)1 / 5;
System.out.println(d); //0.2 That's not casting the result of (1 / 5), it's casting 1 to a double, and then dividing by 5. It's equivalent to this:
d = ( (double) 1 ) / 5;Since now one of the operands is a double, the result of the expression is a double, so you get the expected 0.2.
You could get the same result by doing this in your code:
double correctAnswer = 1.0 / x;As the literal 1.0 is obviously not an integer, but a double, so x is promoted to a double and the division occurs on two doubles instead.