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Subnetting & CIDR

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  1. Subnetting & CIDR Tahir Azim Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  2. Announcements • Participate in NASCON, FAST-NU Islamabad • Assignment 1 deadline extended to Tuesday due to no BIT-7 classes on Monday • From last time: • Packet bursting: An approach to increasing the speed of 802.11g-based wireless networks by unwrapping short 802.11g packets and rebundling them into a larger packet to reduce the impact of mandatory gaps between packets (jwire.com) Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  3. Subnetting • Subnetting is a form of hierarchical routing. • Subnets are usually represented via an address plus a subnet mask or “netmask”. • e.g. nickm@elaine17.Stanford.EDU > ifconfig hme0 hme0: flags=863<UP,BROADCAST,NOTRAILERS,RUNNING,MULTICAST> mtu 1500 inet 171.64.15.82 netmask ffffff00 broadcast 171.64.15.255 • Netmask ffffff00: the first 24 bits are the subnet ID, and the last 8 bits are the host ID. • Can also be represented by a “prefix + length”, e.g. 171.64.15.0/24, or just 171.64.15/24. Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  4. Subnetting 16 2 14 CLASS “B” e.g. Company Host-ID 10 Net ID 16 16 2 14 2 14 e.g. Site 0000 1111 Host-ID Host-ID 10 Net ID 10 Net ID Subnet ID (20) Subnet Host ID (12) Subnet ID (20) Subnet Host ID (12) 16 16 2 14 2 14 e.g. Dept Host-ID 10 Net ID 1111011011 Host-ID 10 Net ID 000000 Subnet ID (26) Subnet Host ID (6) Subnet ID (22) Subnet Host ID (10) Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  5. Routing in the presence of subnets • The rest of the Internet is not aware of subnets within a network • Levels: site, subnet, host • Routing now involves delivery to the site, then the subnet and finally the host Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  6. Example of subnetting hpr1-rtr 171.64.74.0/24 To: cenic.net 171.64.1.131 171.64.1.132/30 171.64.1.133 171.64.1.178 171.64.74.1 171.64.1.161 Class B Address Gates-rtr bbr2-rtr 171.64.74.58 171.64.1.160/27 171.64.0.0/16 AS 32 EndHost 171.64.1.152 171.64.1.144/28 border2-rtr To: cogentco.com 171.64.1.148 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  7. 142.12/19 Classless Interdomain Routing (CIDR)Addressing • The IP address space is broken into line segments, or blocks • e.g. Block of 2 addresses, block of 128 addresses etc. • Each block is described by a prefix. • A prefix is of the form x/y where x indicates the prefix of all addresses in the block, and y indicates the length of the prefix. • e.g. The prefix 128.9/16 represents the block containing addresses in the range: 128.9.0.0 …128.9.255.255. 128.9.0.0 65/8 128.9/16 0 232-1 216 128.9.16.14 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  8. 128.9.19/24 128.9.25/24 128.9.16/20 128.9.176/20 Most specific route = “longest matching prefix” Classless Interdomain Routing (CIDR)Addressing 128.9/16 0 232-1 128.9.16.14 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  9. Classless Interdomain Routing (CIDR)Addressing Prefix aggregation: • If a service provider serves two organizations with prefixes, it can (sometimes) aggregate them to form a shorter prefix. Other routers can refer to this shorter prefix, and so reduce the size of their address table. • E.g. ISP serves 128.9.14.0/24 and 128.9.15.0/24, it can tell other routers to send it all packets belonging to the prefix 128.9.14.0/23. ISP Choice: • In principle, an organization can keep its prefix if it changes service providers. Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  10. Size of the Routing Table at the core of the Internet Source: http://www.cidr-report.org/ Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  11. Prefix Length Distribution Source: Geoff Huston, Jan 2006 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  12. Examples Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  13. Finding the first address • What is the first address in the block if one of the addresses is 167.199.170.82/27? • Solution: The prefix length is 27, which means that we must keep the first 27 bits as is and change the remaining bits (5) to 0s. The following shows the process: Address in binary: 10100111 11000111 10101010 01010010 Keep the left 27 bits: 10100111 11000111 10101010 01000000 Result in CIDR notation: 167.199.170.64/27 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  14. Finding the first address • What is the first address in the block if one of the addresses is 140.120.84.24/20? Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  15. Finding the last address in the block • To the first address, add the number of addresses, minus one • OR • Set all bits that are not part of the CIDR prefix to 1 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  16. Example • Find the number of addresses in the block if one of the addresses is 140.120.84.24/20. • Solution: The prefix length is 20. The number of addresses in the block is 232−20 or 212 or 4096. Note that this is a large block with 4096 addresses. Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)

  17. Example 2 • Find the last address in the block if one of the addresses is 140.120.84.24/20. • Solution • We found in the previous examples that the first address is 140.120.80.0/20 and the number of addresses is 4096. To find the last address, we need to add 4095 (4096 − 1) to the first address. • Or, set all bits that are not part of the CIDR prefix to 1 • 140.120.(0101 1111)2. (1111 1111)2 = 140.120.95.255 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)