>The binary mask for last octet would be 10100000.
Actually, it's worse than that!!
10100000 = 128 + 32 = 160, not 148 :|
148 = 128 + 20 = 128 + 16 +4 = 1001 0100
I'll let you fill the rest in :)
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=================================================
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
John Neiberger
Sent: Thursday, March 15, 2001 12:41 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: Discontiguous networks
ouch! Please do not attempt this at home. Heck, please do not attempt
this anywhere! :-)
The binary mask for last octet would be 10100000. If you assume the use
of subnet zero, your host address go from 1 to 31, then skip to 64-94.
The next subnet in binary is 00100000 so the network address is
192.10.3.32 and host address are 33-63 and 95-126. This is painful,
it's making my head hurt....
Okay then, the next subnet in binary is 10000000, or .128 in decimal,
and your host addresses are 129-159 and 192-223. The final subnet is
10100000, or .148 and the host addresses are 149-180 and 224-254.
I think. That is really painful to think through and I was interrupted
multiple times while trying to write this. Please forgive me if these
numbers aren't correct.
And always remember: Friends don't let friends use discontiguous
subnet masks!
>>> "Arthur Simplina" <[EMAIL PROTECTED]> 3/15/01 10:04:17 AM >>>
I know that there was an earlier posting and a very good explanation on
this. So, kindly bear with me.
I am trying to figure out the subnets (and hosts) for this address:
192.10.3.0 with subnet mask 255.255.255.148.
I am asking this out of curiosity and to learn how to go about this.
Thanks for any help.
Arthur
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