Link to home
Start Free TrialLog in
Avatar of MarkSal
MarkSal

asked on

Subnetting Question

I have this question in my CCNA book that I understand somewhat but not
all the way:

Question:
Create a minimal configuration enabling IP on each interface on a 2600
series router (two serial, one Ethernet).  The Network Information
Center (NIC) assigns you network 192.168.1.0.  Your boss says that you
need, at most, 60 hosts per LAN subnet.  You also have point-to-point
links attached to the serial interfaces.  When choosing the IP address
values and subnet numbers, you decide to start with the lowest
numerical values.  Assume that point-to-point serial links will be
attached to this router and that EIGRP is the routing protocol.

Answer:
192.168.1.0 is a Class C network number, so it will have 24 network
bits and 8 host bits (by default). This examples calls for at most 60
hosts per LAN subnet.  This means that 2^6 - 2 = 62... so there must be
6 host bits.  That means that we need 24 network bits, 2 subnet bits,
and 6 host bits.  The mask that works for this need is 255.255.255.192,
because it has exactly 6 host bits.  For point-to-point links, we use
mask 255.255.255.252, because it only allows 3 different IPs.

This is the router config:

--------------------------------------------------------------------------------

interface ethernet 0/0
ip address 192.168.1.65 255.255.255.192

interface serial 0/0
ip address 192.168.1.129 255.255.255.252

interface serial 0/1
ip address 192.168.1.133 255.255.255.252

router eigrp 1
network 192.168.1.0

--------------------------------------------------------------------------------

My question is... why is 192.168.1.65 the lowest available IP for the
LAN, given network 192.168.1.0 & subnet 255.255.255.192.  Also, why
does the answer specify 192.168.1.129/133 with network 192.168.1.0 &
subnet 255.255.255.252 as the lowest values for the IP addresses of the
serial interfaces?
Avatar of gplana
gplana
Flag of Spain image

192.168.1.65 is the lowest available IP because 65 is 01 000001, so subnet 01, host 000001
Remember 0 value for net and for host is reserved (0 and the maximum -all 1s- are the reserved for net address and broadcast respectively).

Hope it helps.
ASKER CERTIFIED SOLUTION
Avatar of kdearing
kdearing
Flag of United States of America image

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
It is valid but a bit strange.
The first 64 addresses are the same as the next set they have picked.
Maybe they kept the first set for network things.
Maybe it is just unusual to make you think more :)
Avatar of MarkSal
MarkSal

ASKER

kdearing,

Yes thats exactly as I thought....ok not going crazy and thanks that really helped confirm my answer compaired to the answer the the book.

Mark
Avatar of MarkSal

ASKER

Awesome assistance
>My question is... why is 192.168.1.65 the lowest available IP for the
LAN, given network 192.168.1.0 & subnet 255.255.255.192

Because you cannot use 192.168.1.0/26 (Technically you can with IP subnet zero, but not in the CCNA, at least when I took it.)

http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml

>Also, why
does the answer specify 192.168.1.129/133 with network 192.168.1.0 &
subnet 255.255.255.252 as the lowest values for the IP addresses of the
serial interfaces?

192.168.0.64 + network incremental (2^6=64)
.64 +64 = 128
They just used the next available network 192.168.1.128/26 and just Subnetted that network into /30s to limit the exposure to wastefulness of IP space.

192.168.1.128/26
-192.168.1.128/30
192.168.1.129 - 192.168.1.130 (broadcast 192.168.1.131)

Next network: 192.168.1.132/30
192.168.1.133 - 192.168.1.134 (broadcast 192.168.1.135)

Billy