RE: Correlating *SAVF "records" to FTP job I/O count

Dan,

The way I usually estimate the time required to send a file is as follows:

1.  Multiply the object size by 8 to get the number is bits being sent.  In
your case:
                59,785,216 * 8 = 478,281,728

2.  Calculate the number of seconds to send by dividing the number of bits
by the connection speed (in bits/second):
                478,281,728 / 50,000 = 9,566 seconds

3.  Calculate the number of hours by dividing the total number of seconds by
the number of seconds in an hour (3600):
                9,566 / 3600 = 2.66 hours   (2 hours 40 minutes)

The variable in this is the connection speed.  I actually determined your
speed at 50,000 bits/second by working backwards and then plugged that into
the example above.  Were you sending this file over a 56KB modem?  There are
a lot of variables in transmitting files as you know, such as noise on the
lines, retries, failures, conflicting traffic, etc.  But for most normal
transmissions, this has always worked fairly well for me.

When using a save file, be sure to compress the data when saving the
objects.  This will help send it as fast as possible.  This might explain
the difference you see in the object size versus the number of records in
your save file.

Bill

William K. Reger
Senior Project Manager
Levitz Furniture Corporation
Phone:  (561) 994-5114
E-mail:  breger@levitz.com <mailto:breger@levitz.com> 

                -----Original Message-----
                From:   Bale, Dan [mailto:DBale@lear.com]
                Sent:   Tuesday, May 02, 2000 11:47 AM
                To:     'MIDRANGE-L@midrange.com'
                Subject:        RE: Correlating *SAVF "records" to FTP job
I/O count

                That bugger took 3 hours and 40 minutes to transmit.  It was
59.7MB.  From
                the log file:

                    61202064 bytes transferred in 12555.306 seconds.
Transfer rate 4.875
                KB/sec.

                The save file had 115,913 records.  115,913 * 528 =
61,202,064.  The DSPOBJD
                size was 59,785,216 (????).

                Using 1480 bytes per frame, would you calculate the number
of puts as:
                   1)  61,202,064 / 1480 = 41,352.7         *or*
                   2)  1480 / 528 = 2 whole records per frame; 115,913 / 2 =
57,957
                Based on the "guesstimate" that the number of puts was
around 17,000 about
                two hours into the job, I'm not sure either of these
calculations work.

                I think I'm going to set up a test whereby I submit a batch
job to do an FTP
                and another batch job to do a DSPJOB OPTION(*OPNF) in a loop
that runs every
                15 seconds and run some stats on the collected data to see
if there's a
                pattern I can use.

                Other suggestions are greatly appreciated!

                - Dan Bale
                
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