Practice Question

Duration: 3 min

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Inside: a video lesson and guided study material.

Module outline

  1. HPSC Interview Preparation: Interview Preparation
  2. Python Language: Introduction to Python, Tokens, Variables, Data Types, I/O, Operators & Expressions, Control Flow & Looping, Collections & Strings, Functions & Modules, File Handling & Data Analysis, DSA using Python
  3. Data Structure & Algorithms: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing, Algorithm Analysis, Sorting Algorithms, Minimum Spanning Trees, Time Complexity Analysis, Graph Theory, Shortest Path Algos
  4. Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security
  5. Computer System & Organisation: Basics of COA, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Programming the Basic Computer
  6. Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, SQL, File Organization & Indexing, Transaction Management, Concurrency Control
  7. Operating Systems: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling
  8. Digital Logic & Circuits: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation
  9. Theory Of Computation: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Pushdown Automata & CFG, Turing Machines, Moore & Mealy Machines
  10. Compiler Design: Intro to Compilers, Lexical Analysis, Grammar & CFG, Syntax Analysis: Top-Down, Syntax Analysis: Bottom-Up, Semantic Analysis & SDT, Intermediate Code Gen, Code Optimization, Run Time Environment
  11. Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management, Data Warehousing & Mining
  12. Introduction to Emerging Trends: Emerging Tech Trends
  13. Digital Society & Cyber Law: Society, Law & Ethics
  14. HPSC PGT Previous Year Paper:
  15. HPSC Assistant Professor Previous Year Paper:
  16. Live Sessions: HPSC Live Sessions, EMRS TGT/PGT Live Class
  17. Full Mock Test:
  18. HPSC PGT - MOCK Test Series:
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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