ClassLess Interdomain Routing - CIDR With Example

Duration: 4 min

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  1. About Course: About the Course
  2. Paper 1 | General Studies: State Wise GK
  3. Paper 1 | Information and Communication Technology: Role of Internet and Web Services, Security in Mobile and Wireless Computing
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  15. Paper 2 | Computer Network And Information Security: Introduction to CN, Data Communication, Application Layer, Hardware basics, Cyber Threats and Malware Attacks, Network Security, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP
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  18. Paper 2 | Full Length Mock Tests:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This lecture segment teaches how to determine the number of addresses, first address, and last address for a CIDR block using an address mask. The instructor works through the example 167.199.170.82/27, showing its binary form 10100111 11000111 10101010 01010010. The mask is formed by setting the 27 leftmost bits to 1 and the remaining 5 bits to 0. The number of addresses is calculated as N = NOT(mask) + 1, which equals 2^5 = 32. The first address is found by ANDing the given address with the mask, yielding 167.199.170.64. The last address is found by ORing the given address with NOT(mask), yielding 167.199.170.95. A summary slide titled “Address Mask” defines the mask as a 32-bit number with n leftmost bits set to 1 and the rest 0, and lists the three bitwise operations NOT, AND, and OR used to compute block properties.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The slide presents the question “Find the Number of addresses, first and last address of the CIDR block to which” with “Address: 167.199.170.82/27 (10100111 11000111 10101010 01010010) belongs to.” The instructor circles the address line and underlines binary octets, isolating the final group “01010010.” He then writes “2^5 = 32” on the whiteboard, indicating that with a /27 prefix there are 5 host bits and therefore 32 addresses in the block.

  2. 2:00 – 4:15 02:00-04:15

    A second slide states “The mask by setting 27 leftmost bits to 1 and 5 rightmost bits to 0 in dotted-decimal notation is:” and “Number of addresses in the block: N = NOT (mask) + 1 =”, with “NOT” circled in red. The instructor sketches tally marks and five circles representing the zero bits. Later slides show “First address: First = (address) AND (mask) =” and “Last address: Last = (address) OR (NOT mask) =”, with binary rows for the AND and OR operations. The computed first address 167.199.170.64 and last address 167.199.170.95 appear on screen. A final “Address Mask” slide defines the mask as a 32-bit number with n leftmost bits set to 1 and lists N = NOT(mask) + 1, first address = (any address in block) AND mask, and last address = (any address in block) OR NOT(mask).

The lesson centers on CIDR address-mask arithmetic. The key rule is that a /n prefix creates a mask with n leading 1-bits and (32-n) trailing 0-bits. From this mask, three bitwise formulas give block properties: N = NOT(mask) + 1 for the number of addresses, first address = address AND mask, and last address = address OR NOT(mask). In the worked example 167.199.170.82/27, the five trailing zero bits give 2^5 = 32 addresses. ANDing with the mask clears host bits to produce the first address 167.199.170.64, while ORing with NOT(mask) sets host bits to 1 to produce the last address 167.199.170.95. The progression moves from identifying the prefix length, to constructing the mask, to applying NOT/AND/OR operations, and finally to a summary slide that generalizes the method.

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