Practice Question

Duration: 3 min

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Inside: a video lesson and guided study material.

Module outline

  1. 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)
  2. 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
  3. 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)
  4. Section B | C Programming: C Programming by Yash Sir, Coding Questions by Yash Sir
  5. Section B | Data Structures: Algorithm Analysis, Greedy Algorithms, Time Complexity Analysis, Sorting Algorithms, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs
  6. Section B | OOP (C++): Object Oriented Programming (OOPS), Programming in C++
  7. 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
  8. 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
  9. Section B | AI & Big Data Basics: AI & ML by Arun Sir, Big Data Systems
  10. 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
  11. 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
  12. MOCK TEST:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

The video features an instructor solving a computer architecture problem regarding instruction set architecture (ISA) and opcode expansion. The problem involves a 16-bit instruction format with a 128-word memory size, supporting both 1-address and 2-address instructions. Given two 2-address instructions, the goal is finding supported 1-address instructions. The instructor derives x * 2^7 + y * 2^14 = 2^16, where x and y are instruction counts. He substitutes y=2 and solves for x, finding the answer to be 256. He breaks down bit allocation for addresses and opcodes.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The instructor begins by reading the problem statement displayed on the screen, highlighting critical constraints: 16 bit instruction size, 128 words memory, and the existence of two, 2-address instruction. He establishes the total instruction space as 2^16. He explains that since memory has 128 words, each address field requires 7 bits (2^7 = 128). He formulates the equation x * 2^7 + y * 2^14 = 2^16, where x represents the number of 1-address instructions and y represents the number of 2-address instructions. He substitutes y = 2 into the equation, leading to x * 2^7 + 2 * 2^14 = 2^16. He simplifies 2 * 2^14 to 2^15. The equation becomes x * 2^7 = 2^16 - 2^15. He factors out 2^15 to get x * 2^7 = 2^15(2 - 1), which simplifies to x * 2^7 = 2^15. Solving for x, he finds x = 2^8, which equals 256. He circles the result x = 256 on the board.

  2. 2:00 – 2:39 02:00-02:39

    In the final segment, the instructor double-checks his work. He writes 256 / 2^7 at the top left, dividing the number of 1-address instructions by the address space per instruction. This yields 2, representing the number of available opcode prefixes. This matches the unused 2-address opcodes (4 total - 2 used). He confirms the steps, and the video ends with x = 256 visible.

The lecture demonstrates opcode expansion in instruction set design. By allocating bits for addresses based on memory size (7 bits for 128 words), remaining bits are used for opcodes. The instructor calculates capacity for different instruction types by subtracting used space from total available space. The verification step confirms the derived instructions align with available opcode prefixes.

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