IP address assistance

[Ken Beames]

Please check my work:


Following the RFC 1897 format:


   | 3 |  5 bits  |  16 bits | 8 |   24 bits  | 8 | 16 bits|48 bits|

   +---+----------+----------+---+------------+---+--------+-------+

   |   |          |Autonomous|   |    IPv4    |   | Subnet | Intf. |

   |010|  11111   |  System  |RES|   Network  |RES|        |       |

   |   |          |  Number  |   |   Address  |   | Address|  ID   |

   +---+----------+----------+---+------------+---+--------+-------+


If my ISPs AS# is 1239 (Sprint)

my IPv4 network address is 199.0.193

Subnet zero

Interface 00-A0-D1-02-33-FA


Would be, in binary:


0101111100000100:1101011100000000:1100011100000000:1100000100000000:0000000000000000:0000000010100000:1101000100000010:0011001111111010 


Or,


5F04:D700:C700:C100:0:A0:D102:33FA


Right?


If I were subnetted at 255.255.255.192 and I was on the 3rd subnet (.128), would I represent that as 0080 in the 5th (double octet, word, what name do we call these, hexadectets?)


what is the common practice?


thanks.  -Ken.

-----------------------------------====================

Ken Beames				Phone: 408 542 0268

Network Systems Engineer   		Page:	800 INS 1INS

Information Technology Group	Cell:	415 602 3758

International Network Services	Fax:	408 542 0105

Sunnyvale, CA				Mail:	beames@ins.com

<color><param>0000,0000,ffff</param>Go Dragonslayers!!

</color>=============================--------------------------

"How come nostalgia isn't what it used to be?" 

	-Alfred.