Practice Question - 2
Duration: 2 min
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Inside: a video lesson and guided study material.
Module outline
- Section A | English: Vocabulary, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Preposition, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Articles, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Verbal Analogy (Word Based Analogy)
- Section A | Quant Aptitude: Ratio and Proportion (Ratios), Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Arithmetic, Algebra, Powers and Exponents (Surds and Indices), Average, Age Problems, Percentage, Profit and Loss, Partnership, Time and Work, Simple Interest and Compound Interest, Mensuration and Geometry, Probability, Pipe and Cistern
- Section A | Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Calendar, Clock, Blood Relations, Directions (Direction Test), Seating Arrangements, Ranking, Order and Sequence, Puzzle Solving (Puzzles)
- Section B | C Programming: C Programming by Yash Sir, Coding Questions by Yash Sir
- Section B | Data Structures: Algorithm Analysis, Greedy Algorithms, Time Complexity Analysis, Sorting Algorithms, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs
- Section B | OOP (C++): Object Oriented Programming (OOPS), Programming in C++
- Section B | OS: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management, Windows OS
- Section B | Networking: 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
- Section B | AI & Big Data Basics: AI & ML by Arun Sir, Big Data Systems
- Section C | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation, Analog-Digital Conversion
- Section A, C | Comp. Fundamentals, COA: Basics of COA, Floating Point Rep, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations, Multiprocessors, Programming the Basic Computer
- MOCK TEST:
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This educational video segment presents a solved problem from the GATE 2017 examination concerning computer architecture, specifically focusing on direct-mapped cache memory calculations. The instructor analyzes a byte-addressable machine with 2^32 bytes of main memory and a block size of 32 bytes. The core task is to determine the bit size of the tag field given a cache with 512 lines. The visual content displays structured diagrams partitioning the physical address into Tag, Cache Line (Index), and Block Offset components. The solution involves calculating the offset bits from the block size, deriving the index bits from the number of cache lines, and subtracting these from the total address width to find the tag size.
Chapters
0:00 – 1:51 00:00-01:51
The instructor introduces a GATE 2017 problem on direct-mapped cache memory, displaying the question text: 'Consider a machine with byte addressable memory of 2^32 bytes divided into blocks of size 32 bytes.' The visual diagrams illustrate the physical address structure, explicitly labeling 'Main Memory (Physical Address) 32' bits. The instructor breaks down the address fields, showing that a block size of 32 bytes requires 5 bits for the Block Offset (since 2^5 = 32). The diagram further indicates that with 512 cache lines, the Cache Line index requires 9 bits (since 2^9 = 512). The final calculation partitions the 32-bit address into Block Number (27 bits) and Block Offset (5 bits), leading to the conclusion that the Tag field size is 18 bits, derived by subtracting the index and offset bits from the total address width.
The video provides a concise walkthrough of cache address mapping, emphasizing the relationship between memory parameters and bit allocation. Key evidence includes the on-screen text specifying '2^32 bytes' memory size and '512 cache lines,' alongside diagrams showing the 32-bit address split. The instructor demonstrates that Block Offset is determined by log2(block size) and Cache Line Index by log2(number of lines). The Tag field is the remainder. This problem reinforces the standard formula: Total Address Bits = Tag + Index + Offset. The solution confirms that for a 32-bit address, with 5 offset bits and 9 index bits, the Tag must be 18 bits. This aligns with option (C) in the multiple-choice question presented.