This should be helpful for your review of IP numbers and subnetting by using subnet masks. The concept of the subnet mask on the Internet involves knowing a little bit about binary. An IP number consists of 4 octets, for convenience, we'll denote it as w.x.y.z and realize that each letter refers to one byte, or 8 bits, hence the term octet. The three classes of addresses on the Internet are class A, class B and class C addresses. Class A addresses have the 1st bit off for the 1st octet. 0xxxxxxx Class B addresses have the 1st bit on and the 2nd bit off. 10xxxxxx Class C addresses have the 1st two bits on, the 3rd off. 110xxxxx An IP address is usually shown in dotted decimal notation. The IP address for COBRA is 134.161.1.11 and for VIPER is 134.161.1.16 and for chaos is 134.161.241.2 and for nova.cs.uni.edu is 134.161.70.20. The IP address for iscssun.uni.edu is 134.161.14.20. The subnet mask here is 255.255.252.0 UNI Network ID is 134.161.0.0 If the subnet mask is 255.255.252.0 the in binary, each octet expands and the subnect mask looks like this: 255.255.252.0 is 11111111.11111111.11111100.00000000 ******** ******** ------ The underlined bits specify a subnetwork address here at UNI. The * bits specify UNI itself (on the Internet), but UNI decided to take 6 of the 16 bits it has available as a class B IP address network, and use those 6 bits to specify subnetworks within UNI. The remaining 10 bits can be used to specify the host id of a particular machine. With 6 bits we can have a total of 64 minus 2 or 62 different subnets. With 10 bits remaining for the host id, each subnet at UNI can have up to 1024 minus 2 or 1022 machines. So for my machines, here is the output of ipconfig: Windows NT IP Configuration IP Address. . . . . . . . . : 134.161.242.243 Subnet Mask . . . . . . . . : 255.255.252.0 Default Gateway . . . . . . : 134.161.240.1 The UNI network is 134.161.0.0 since it is a class B network, and just uses the 1st two octets, w and x of the w.x.y.z scheme. 134 in binary is 10000110 so the IP address is 10000110.161.242.243 and as you can see, the 1st two bits of w are -- 10xxxxxx which means that we have a class B address. Of the w.x.y.z, the w and x octets represent UNI, i.e. 134.161 is uni.edu for the Internet. Class B addresses leave octets y and z, a total of 16 bits, open for the host id. By subnetting and using 255.255.252.0 as the subnet mask, UNI says: Let us take 6 of our 16 bits and use them to specify which UNI subnetwork. Recall 252.0 = 11111100.00000000 Thus we have 2 to the 6th power minus 2 or 64 - 2 or 62 possible subnets we can define within UNI. Each of those subnets can specify up to 2 to the 10th minus 2 or 1,024 - 2 or 1,022 different host ids. yyyyyy yy.zzzzzzzz Seeing the w.x.y.z bits broken down 111111 00.00000000 by the subnet mask into host and subnet ids. 123456 12 34567890 subnet host id 1 w.x wwwwwwww.xxxxxxxx 184.161 10111000.10100001 is UNI network id.... 242.243 looks like this in binary: 11110010 = 242 and 11110011 = 243 so the y.z part of my machines IP address looks like this: 111100 10.11110011 and the subnet mask is: 111111 00.00000000 so the subnetwork address is 111100 and the host id is 1011110011 What is 111100 in base 10? 32 + 16 + 8 + 4 = 60, so I am on subnetwork 60 at UNI. What is 1011110011 in base 10. 1 0 1 1 1 1 0 0 1 1 512 256 128 64 32 16 8 4 2 1 512 + 128+64+32 + 16 + 2 + 1 = 755, so my machine or host id number is 755 on subnet 60 at UNI which has Internet network address 184.161 ------------------------------------------------------------------------------ ******** Following here is a binary review ********* ------------------------------------------------------------------------------ Today is January 1101 but most would write it as January 13th, or Jan 13 2 10 The subscript 2 in the 1101 indicates the base, which is base 2 or binary. 2 The subscript 10 after and below the 13 in Jan 13 10 indicates the base, base 10 or decimal. 1101 in base 2 is 1 8s plus 1 4s plus 0 2s plus 1 1s or 8 + 4 + 0 + 1 = 13 3 2 1 0 1101 is 1 * 2 + 1 * 2 + 0 * 2 + 1 * 2 8 + 4 + 0 + 1 = 13 in base 10 8543 in base 10 is: 3 2 1 0 8 * 10 + 5 * 10 + 4 * 10 + 3 * 10 8 1000s + 5 100s + 4 10s + 3 1s = 8,543 in base 10 You don't want to see what 8,543 looks like in binary. It would have 14 binary digits! Here is what 8,191 looks like: 1111111111111 8,191 is the largest possible 2 13 digit binary number. Base 10 Base 2 Base 8 Base 16 1 1 1 1 2 10 2 2 Base 16 = Hexadecimal 3 11 3 3 Base 8 = Octal 4 100 4 4 Base 2 = Binary 5 101 5 5 Base 10 = Decimal 6 110 6 6 7 111 7 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F 16 10000 20 10 C:\>ping www.uni.edu Pinging goofy.uni.edu [134.161.1.36] with 32 bytes of data: Reply from 134.161.1.36: bytes=32 time=10ms TTL=253 Reply from 134.161.1.36: bytes=32 time<10ms TTL=254 Reply from 134.161.1.36: bytes=32 time<10ms TTL=254 Reply from 134.161.1.36: bytes=32 time<10ms TTL=254 C:\>ping www.cns.uni.edu Pinging spider.cns.uni.edu [134.161.241.20] with 32 bytes of data: Reply from 134.161.241.20: bytes=32 time<10ms TTL=255 Reply from 134.161.241.20: bytes=32 time<10ms TTL=255 Reply from 134.161.241.20: bytes=32 time<10ms TTL=255 Reply from 134.161.241.20: bytes=32 time<10ms TTL=255