BROCADE IP PRIMER

BI4XGBI4XGT X RX T X RX T X RX T X RXT X RX T X RX T X RX T X RX6 12 18 2430 36 42 48BigIron RX-8NETWORKS1 7 13 1925 31 37 43BI24CBI24CBI24CAC OK DC OK ALMEJECT SYS AC OK DC OK ALMEJECT SYS AC OK DC OK ALMEJECT SYS AC OK DC OK ALMBI24CBI24CBI24CBI24CBI24CBI4XGBI4XGT X RX T X RX T X RX T X RXT X RX T X RX T X RX T X RXEJECT SYSBI4XGBI4XG6 12 18 2430 36 42 48BigIron RX-8NETWORKS1 7 13 1925 31 37 43BI24CBI24CBI24CT X RX T X RX T X RX T X RXT X RX T X RX T X RX T X RXAC OK DC OK ALMEJECT SYS AC OK DC OK ALMEJECT SYS AC OK DC OK ALMEJECT SYS AC OK DC OK ALMBI24CBI24CBI24CBI24CBI24CBI4XGBI4XGT X RX T X RX T X RX T X RXT X RX T X RX T X RX T X RXEJECT SYSChapter 2: TCP/IPthe hosts. This gives us 254 usable hosts (remember to subtract the networkand broadcast address). We only need two hosts, so using the default subnetmask is pretty wasteful. What if we used 255.255.255.252 as the subnetmask? In binary, this would be 1111 1111 . 1111 1111 . 1111 1111 . 11111100. This means that 30 bits are used for the network portion and only twobits are left for the hosts. This gives us 22 - 2, or 4 - 2, which equals 2 hosts.Perfect! So, with the subnet mask of 255.255.255.252, the network addressis 192.168.100.0, one host is 192.168.100.1, the other host is192.168.100.2, and 192.168.100.3 is the broadcast address.RouterT1192.168.100.2mask 255.255.255.252Router4.0.0.1192.168.100.1mask 255.255.255.2525.0.0.14.0.0.05.0.0.0This is referred to as classless networking, as you're not conforming to classes.You can subdivide (or subnet) as much as you need. Remember that this is adivision. You can't make the subnet mask smaller than the default for theclass. For example, you can't define a Class C network to have a subnet maskof 255.255.0.0 (or 16 bits for the network portion). A Class C subnet musthave at least 24 bits for the network portion. Likewise, a Class B subnet musthave at least 16 bits for the network portion, and a Class A network must haveat least eight bits for the network portion.Let's try another example. Let's say we have a network that will require 400hosts. If we use a Class C, the largest number of hosts we can have per networkis 254. We could use a traditional Class B, but this would give us 65,534hosts, a lot more than we need. So, let's start with a Class B, say, 172.16.0.0with a subnet mask of 255.255.0.0. We've got to create a subnet to allow for400 hosts. We know that eight host bits gives us 254 hosts. How many doesnine host bits give us? This would be 29 - 2, or 512 - 2, which would equal 510hosts. This is still more than we need, but it's the smallest subnet we canmake that will still meet our requirement. If we have nine host bits, that leavesus 23 (32 - 9) network bits. With 23 network bits, this gives us a subnet maskof 1111 1111 . 1111 1111 . 1111 1110 . 0000 0000, or, in decimal,255.255.254.0. So, our final network address is 172.16.0.0 with a subnetmask of 255.255.254.0. The first usable address would be 172.16.0.1. Thelast usable address would be 172.16.1.254. The broadcast address would be172.16.1.255.46 Brocade IP Primer