Practice Question - 1

Duration: 2 min

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  1. About the course: About the Course
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  7. Unit 1 | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation
  8. Unit 2 | COA: Basics of COA, Floating Point Rep, Cache Memory Organization, Input Output Organisation, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations, Programming the Basic Computer
  9. Unit 3 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  10. Unit 3 | Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
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  15. Unit 8 | Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
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  24. Full Length Mock Tests — General Paper:
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AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This lecture segment presents a practice problem on fully associative mapped cache memory organization. The instructor analyzes a scenario with 16 KB cache size, 256 bytes block size, and 128 KB main memory. The core task is determining the number of tag bits and total tag directory size. The instructor decomposes the physical address into Block Number (Tag) and Block Offset fields, calculating 8 bits for offset based on block size. For fully associative mapping, the entire Block Number becomes the Tag since there is no index field.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The instructor introduces a fully associative cache problem with on-screen text stating 'Q Consider a fully associative mapped cache of size 16 KB with block size 256 bytes.' The main memory is specified as 128 KB. He calculates the Block Offset bits as 8 (since log2(256) = 8). He writes '14' initially, likely referring to total cache lines or address bits, then corrects to '6' for Cache Line bits (log2(16KB/256B) = log2(64) = 6). The diagram shows Main Memory (Physical Address) split into Block Number and Block Offset, with Tag = Block Number.

  2. 2:00 – 2:07 02:00-02:07

    The instructor finalizes the address breakdown showing 17 total physical address bits. He explicitly writes 'Block Number 9' and 'Block Offset 8', confirming Tag = Block Number is 9 bits. He computes the tag directory size using the formula 2^6 x 9, where 2^6 represents the number of cache lines (64) and 9 is the tag bits per line. The screen displays 'Find out the: Number of bits in tag and Tag directory size?' as the final question to be answered.

The video demonstrates a standard method for solving fully associative cache problems. Key steps include calculating block offset from block size, determining the number of cache lines to find tag bits (since index is zero in fully associative), and computing directory size. The instructor emphasizes that Tag equals Block Number in this mapping scheme, distinguishing it from direct or set-associative mappings where an index field exists. The final calculation of 2^6 x 9 illustrates how directory size scales with the number of lines and tag width.

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