Networking
Trivia
2 ** 32 = 4,294,967,296 is the number of Internet addresses available using the
current IPV4 standards
2 ** 128 = 3.40282367 × 10**38 is the number of Internet addresses that
will be available
using the IPV6
Windows 2000 & XP
Checking for network parameters at the command line:
(go to dos prompt via start/run cmd)
ipconfig/all
Release and renew IP number with DHCP
ipconfig/release
ipconfig/renew
This same trick is done by right clicking the My Network Places (usually on the
desktop or in the start menu) and
then
right
clicking the connection in need of repair and selecting repair connection.
Macintosh OS/9
Go to control panel and click TCP/IP and negotiate the menus as needed
Netmask Netmask (binary) CIDR Notes
_____________________________________________________________________________
255.255.255.255 11111111.11111111.11111111.11111111 /32 Host (single addr)
255.255.255.254 11111111.11111111.11111111.11111110 /31 Unuseable
255.255.255.252 11111111.11111111.11111111.11111100 /30 2 useable
255.255.255.248 11111111.11111111.11111111.11111000 /29 6 useable
255.255.255.240 11111111.11111111.11111111.11110000 /28 14 useable
255.255.255.224 11111111.11111111.11111111.11100000 /27 30 useable
255.255.255.192 11111111.11111111.11111111.11000000 /26 62 useable
255.255.255.128 11111111.11111111.11111111.10000000 /25 126 useable
255.255.255.0 11111111.11111111.11111111.00000000 /24 "Class C" 254 useable
255.255.254.0 11111111.11111111.11111110.00000000 /23 2 Class C's 255.255.252.0 11111111.11111111.11111100.00000000 /22 4 Class C's 255.255.248.0 11111111.11111111.11111000.00000000 /21 8 Class C's 255.255.240.0 11111111.11111111.11110000.00000000 /20 16 Class C's 255.255.224.0 11111111.11111111.11100000.00000000 /19 32 Class C's 255.255.192.0 11111111.11111111.11000000.00000000 /18 64 Class C's 255.255.128.0 11111111.11111111.10000000.00000000 /17 128 Class C's 255.255.0.0 11111111.11111111.00000000.00000000 /16 "Class B" 255.254.0.0 11111111.11111110.00000000.00000000 /15 2 Class B's 255.252.0.0 11111111.11111100.00000000.00000000 /14 4 Class B's 255.248.0.0 11111111.11111000.00000000.00000000 /13 8 Class B's 255.240.0.0 11111111.11110000.00000000.00000000 /12 16 Class B's 255.224.0.0 11111111.11100000.00000000.00000000 /11 32 Class B's 255.192.0.0 11111111.11000000.00000000.00000000 /10 64 Class B's 255.128.0.0 11111111.10000000.00000000.00000000 /9 128 Class B's 255.0.0.0 11111111.00000000.00000000.00000000 /8 "Class A" 254.0.0.0 11111110.00000000.00000000.00000000 /7 252.0.0.0 11111100.00000000.00000000.00000000 /6 248.0.0.0 11111000.00000000.00000000.00000000 /5 240.0.0.0 11110000.00000000.00000000.00000000 /4 224.0.0.0 11100000.00000000.00000000.00000000 /3 192.0.0.0 11000000.00000000.00000000.00000000 /2 128.0.0.0 10000000.00000000.00000000.00000000 /1 0.0.0.0 00000000.00000000.00000000.00000000 /0 IP space
Net Host Total
Net Addr Addr Addr Number
Class Range NetMask Bits Bits of hosts
----------------------------------------------------------
A 0-127 255.0.0.0 8 24 16777216 (i.e. 114.0.0.0)
B 128-191 255.255.0.0 16 16 65536 (i.e. 150.0.0.0)
C 192-254 255.255.255.0 24 8 256 (i.e. 199.0.0.0)
D 224-239 (multicast)
E 240-255 (reserved)
F 208-215 255.255.255.240 28 4 16
G 216/8 ARIN - North America
G 217/8 RIPE NCC - Europe
G 218-219/8 APNIC
H 220-221 255.255.255.248 29 3 8 (reserved)
K 222-223 255.255.255.254 31 1 2 (reserved)
Souce of tables: http://www.oav.net/mirrors/cidr.html
(ref: RFC1375 & http://www.iana.org/assignments/ipv4-address-space )
( http://www.iana.org/numbers.htm )
When refering to a subnet mask in the /x notation you are saying that the number
of hosts that can have addresses on a given network and can be calculated using
this formula:
hosts = 2**(32-x)-2
where x is the netmask.
Example using a netmask of 24 gives us:
2**(32-42)-2=254 hosts on a subnet