Re: HLASM Floating Point Field - Remainder?
- From: Hansedl <david.l.hansen@xxxxxxxxxxxxxxxxx>
- Date: Wed, 27 Jun 2007 08:27:03 -0700
On Jun 19, 6:30 pm, glen herrmannsfeldt <g...@xxxxxxxxxxxxxxxx> wrote:
Hansedl wrote:
(snip)
uses 32 bits of data from the registers. The field contains
x'43E88000'. This means the x'43' denotes the sign bit and the
characteristic or
exponent. The x'E88000' is the mantissa or fraction (binary '1110
1000 1000'). The exponent is binary '100 0011' (67 decimal). In
order to support
negative characteristics, the assembler increases the exponent by 64
before converting it to binary. The bias is 64 so 67 - 64 = 3 (the
exponent).
So the value is '111.010001000' because we move the radix point three
places to the right. This field contains the Elapsed time, in seconds
which is
The exponent is base 16, so you move three hex digits, or 12 bits,
such that the value us X'E88' or decimal 3720.
(snip)
Another number I have is x'423c0000'. So x'42' is binary '1000010'
or decimal 66. And 66 - 64 = 2. So x'3C' is binary '111100'
adjusted two
Again two hex digits to the right for X'42' or 66.
If this is homework, please reference the newsgroup.
-- glen
Glen,
Thank you for your response. The records are from our z/VM system
Performance Monitor. I was asked to cut the interval times down. I
started by looking at what was actually collected. The first field
contains elapsed time in seconds so x'E88' is 3,270 seconds or 62
minutes. Doesn't look like I can report by 15 minute intervals with
this data. The second field is the length of the sample period so
x'3C' is 60. This may be in seconds. The third field is from the
same data but shows x'528' or 1,320. Possibly 22 seconds. Now I'll
call IBM and ask then about the intervals which I have found the
duration of.
Thanks again, Dave H.
.
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