Simple 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:

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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

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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

Reply to
jjbladester
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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.

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.

Reply to
Joe Schmuckatelli

Joe,

Much thanks. It's hard to make it through the self-study but little hits al> > >My question is... why is 192.168.1.65 the lowest available IP for the

first subnet was not usable, much like the first

Reply to
jjbladester

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