Subnettig question VLSM

Unanswered Question
May 1st, 2008

Hi,

In Odom ICND1 2nd Edition, Figure 15.5 the subsequent text says that there are 2 errors on the diagram. But I can't get my head around what I see as a third. Please help : -

There's a PC with address 10.10.23.11 /21 in a subnet 10.10.24.0 /21.

To my mind the /21 mask leaves 3 host bits in the 3rd octet or 2**3 = 8 hosts (multiplied by 256 in the 4th octet, minus 2). Therefore the 3rd octet would have values for each subnet of (0 - 7, 8 - 15, 16 - 23, 24 to 31,...)

So, in the 10.10.24.0 /21 subnet, the 1st host address is 10.10.24.1 and the last is 10.10.31.254. The PC is in subnet 10.10.16.0, with 1st address 10.10.16.1 and last 10.10.23.254.

Where am I going wrong?

Thanks,

Dave

I have this problem too.
0 votes
  • 1
  • 2
  • 3
  • 4
  • 5
Overall Rating: 0 (0 ratings)
Loading.
Jon Marshall Thu, 05/01/2008 - 03:18

Dave

Your are not going wrong

/21 = 255.255.248.0

Quick way to work out how subnets go up -

256 - 248 = 8 so your networks go up in 8 ie

10.10.0.0/21

10.10.8.0/21

10.10.16.0/21

10.10.24.0/21

10.10.32.0/21

You are right.

Jon

d.brett Thu, 05/01/2008 - 05:53

Jon,

Thanks for putting that straight. Your method looks like a good one for the actual exam.

Cheers,

Dave

Jon Marshall Thu, 05/01/2008 - 06:17

Dave

No problem, glad to have helped.

Good luck with your exam when you take it - you certainly seem to have subnetting covered :-)

Jon

Actions

This Discussion