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