Simple Subnetting Question

Simple Subnetting Question

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Subject Author Date
Simple Subnetting Question jjbladester@gmail.com 10-30-2006
Posted by jjbladester@gmail.com on October 30, 2006, 10:48 pm
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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?

-Jeff


Posted by Joe Schmuckatelli on October 31, 2006, 12:58 pm
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On 30 Oct 2006 19:48:08 -0800, "jjbladester@gmail.com"

>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.

Back in the day, the first subnet was not usable, much like the first
address in a subnet is not usable (because it identifies the subnet).
Later versions of IOS do permit the use of subnet 0.

>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?

Given that subnet 0 is not available for the purposes of this
question, the first available subnet starts with .64, so its first
available address is .65; the second available subnet starts with
.128, so its first address is .129.



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