ClassLess Interdomain Routing - CIDR With Example
Duration: 4 min
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Inside: a video lesson and guided study material.
Module outline
- Quantitative Ability: Speed Mathematics and Basic Mathematics, Percentage, Ratio and Proportion (Ratios), Average, Profit and Loss, Simple Interest and Compound Interest, Time and Work, Pipe and Cistern, Speed Time and Distance, Number System, HCF LCM, Data Interpretation, Age Problems, Mixture and Alligation, Partnership, Algebra, Sequence and Series (Series), Mensuration and Geometry, Arithmetic
- Reasoning Ability: Puzzle Solving (Puzzles), Seating Arrangements, Coding Decoding, Blood Relations, Directions (Direction Test), Syllogisms, Data Sufficiency, Ranking, Order and Sequence, Series (Number and Letter Series) (Numerical Relations and Reasoning), Logical Sequence Of Words, Calendar, Clock, Cubes & Dices, Analytical Reasoning (Counting Figures Reasoning), Data Arrangement, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning)
- Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Cloze Test (Cloze Passage), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Vocabulary, Sentence Completion (Fill in the blanks), Active Passive Voice (Voice), Direct and Indirect Speech, Subject Verb Agreement (Verb Noun Agreement), Tenses, Preposition, Articles, Sentence Correction (Error Correction)
- Programming Basics (MCQs + Interview): C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir
- OOPs (MCQs + Interview): Object Oriented Programming (OOPS)
- DBMS (MCQs + Interview): Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Transaction Management, Decomposition Properties & 4NF, Concurrency Control, Database Recovery, File Organization & Indexing, Relational Algebra, Database Security & Authorization
- SQL (MCQs + Interview): SQL
- Operating Systems (MCQs + Interview): Introduction to OS, Process Management, CPU Scheduling, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Process Synchronization, Disc Scheduling, File Management, Security
- Computer Networks (MCQs + Interview): Introduction to CN, Data Communication, 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, Application Layer, Hardware basics, Network Security, Mobile Technology, Cloud Computing and IoT, Cyber and Info Security
- Coding: Coding Questions by Yash Sir
- HR Interview: MR & HR Interview Preparation
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
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.
2:00 – 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.