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-----Original Message-----
From: Dean [mailto:dean.deano AT gmail DOT com]=20
=09
=09
Paul,
=09
You certainly can create multiple st-units in one library. You
*have to* if you have more than one drive type in there.
=09
But your second point won't work. You don't "bind" a drive to a
st-unit. You just define the *maximum* number of drives that st-unit can
use. So if you have a st-unit with max-drives =3D 15, but 20 identical
drives in the library, the st-unit could use *any* of those drives, up
to 15 at a maximum.=20
=20
Yup.....I'm aware of that.....actually, I wasn't trying to
express "binding" in my second, point, my second point was this below:=20
=09
If your total st-unit drives is greater than the physical drives
available, the 20th job will just sit in the queue until one of the
other 19 active jobs completes, regardless of which st-unit it is going
to. If you have 15 drives in use in the PROD st-unit, and 4 in the DEV
st-unit, if the top job in the queue is a DEV job, once a drive in the
PROD st-unit is freed up, it will be "grabbed" for the DEV st-unit.
=09
So giving all your PROD policies a high job priority will help
achieve what you want, but it won't always work.=20
Yes, priority is just a tie breaker when several jobs are
queued, and one drive becomes available.=20
=09
=09
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<DIV class=3DOutlookMessageHeader lang=3Den-us dir=3Dltr =
align=3Dleft><FONT=20
face=3DTahoma size=3D2>-----Original Message-----<BR><B>From:</B> Dean =
[mailto:dean.deano AT gmail DOT com] <BR><BR></FONT></DIV>
<DIV>Paul,<BR><BR>You certainly can create multiple st-units in one =
library.=20
You *have to* if you have more than one drive type in =
there.<BR><BR>But your=20
second point won't work. You don't "bind" a drive to a st-unit. You =
just=20
define the *maximum* number of drives that st-unit can use. So if you =
have a=20
st-unit with max-drives =3D 15, but 20 identical drives in the =
library, the=20
st-unit could use *any* of those drives, up to 15 at a maximum.<SPAN=20
class=3D015265511-30062005><FONT face=3DArial color=3D#800000=20
size=3D2> </FONT></SPAN></DIV>
<DIV><SPAN class=3D015265511-30062005></SPAN> </DIV>
<DIV><SPAN class=3D015265511-30062005><FONT face=3DArial =
color=3D#800000=20
size=3D2>Yup.....I'm aware of that.....actually, I wasn't trying to =
express=20
"binding" in my second, point, my second point was this=20
below:</FONT> </SPAN><BR><BR>If your total st-unit drives is =
greater than=20
the physical drives available, the 20th job will just sit in the queue =
until=20
one of the other 19 active jobs completes, regardless of which st-unit =
it is=20
going to. If you have 15 drives in use in the PROD st-unit, and 4 in =
the DEV=20
st-unit, if the top job in the queue is a DEV job, once a drive in the =
PROD=20
st-unit is freed up, it will be "grabbed" for the DEV =
st-unit.<BR><BR>So=20
giving all your PROD policies a high job priority will help achieve =
what you=20
want, but it won't always work.<SPAN class=3D015265511-30062005><FONT =
face=3DArial=20
color=3D#800000 size=3D2> </FONT></SPAN></DIV>
<DIV><SPAN class=3D015265511-30062005><FONT face=3DArial =
color=3D#800000 size=3D2>Yes,=20
priority is just a tie breaker when several jobs are queued, =
and one drive becomes=20
available.</FONT> </SPAN><BR><BR></DIV></BLOCKQUOTE></BODY></HTML>
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