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Subnetting & CIDR. Tahir Azim. Announcements. Participate in NASCON, FAST-NU Islamabad Assignment 1 deadline extended to Tuesday due to no BIT-7 classes on Monday From last time:
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Subnetting & CIDR Tahir Azim Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)
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)
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)
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)
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)
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)
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)
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)
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)
Size of the Routing Table at the core of the Internet Source: http://www.cidr-report.org/ Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)
Prefix Length Distribution Source: Geoff Huston, Jan 2006 Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)
Examples Courtesy Nick McKeown (Stanford), Umar Kalim (NIIT)
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)
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)
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)
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)
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)