Practice Question-1

Duration: 4 min

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Module outline

  1. About Course: About the Course
  2. Paper-2 | Programming & Data Structures: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  3. Paper-2 | Databases / DBMS & SQL: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, SQL, Relational Calculus, Transaction Management, Concurrency Control
  4. Paper-2 | Algorithms: Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos
  5. Paper-2 | Computer Networks & Security: 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 Access Technologies, Network Security
  6. Paper-2 | Digital Logic: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation, Floating Point Rep
  7. Paper-2 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management
  8. Paper-2 | Theory of Computation: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines
  9. Paper-2 | Computer Organization & Architecture: Basics of COA, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Programming the Basic Computer
  10. Paper-2 | 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. Paper-1 | General Awareness: Polity, Physics, Biology, Chemistry, Static General Knowledge, GK- Current Affairs, History, Geography, Indian Economy, Environment
  12. Paper-1 | Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Data Interpretation, Blood Relations, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Selection Decision Table (Decision Making), Clock, Calendar, Cubes & Dices, Directions (Direction Test), Analogy, Seating Arrangements, Analytical Reasoning (Counting Figures Reasoning)
  13. Paper-1 | Quantitative Aptitude: Ratio and Proportion (Ratios), Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Speed Time and Distance, Average, Age Problems, Percentage, Profit and Loss, Time and Work, Pipe and Cistern, Simple Interest and Compound Interest, Powers and Exponents (Surds and Indices), Mensuration and Geometry, Probability, Statistics, Arithmetic, Algebra
  14. Paper-1 | English: Vocabulary, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Construction, Cloze Test (Cloze Passage)
  15. Coal India Previous Year Paper:
  16. Engineering Mathematics: Set Theory, Relations, Functions, Graph Theory, Group Theory, Propositional and Predicate Logic, Linear Algebra, Calculus
  17. Coal India Mock Test:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

The video presents a step-by-step solution to a problem in computer architecture concerning a micro-programmed control unit. The instructor begins by analyzing the requirements: the system must support 256 instructions, 16 flag conditions, and generate 48 control signals. The first part of the solution calculates the number of bits required for each control word. This is done by determining the number of bits needed for the instruction address (log2(256) = 8 bits), the flag conditions (log2(16) = 4 bits), and the control signals (48 bits). The total is the sum of these: 8 + 4 + 48 = 60 bits. The second part calculates the total control memory size in bytes. The instructor multiplies the number of instructions (256) by the number of bits per control word (60) to get the total number of bits (15,360). This is then converted to bytes by dividing by 8, resulting in 1,920 bytes. The entire process is demonstrated on a whiteboard with clear, handwritten calculations.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video starts with the instructor presenting a problem about a micro-programmed control unit. The on-screen text states the system must support 256 instructions, 16 flag conditions, and generate 48 control signals. The instructor begins to break down the problem, writing '256 instruction' and '16 flag' on the whiteboard. He then calculates the number of bits required for the instruction address, writing 'log2(16)' and 'log2(16) = 4' to determine that 4 bits are needed for the flag conditions. He also writes '48 control signal' to account for the control signals. The instructor is seen writing 'log2(16)' and 'log2(16) = 4' on the whiteboard, indicating the calculation for the flag conditions.

  2. 2:00 – 3:56 02:00-03:56

    The instructor continues the calculation for the first part of the problem. He writes 'log2(256)' and calculates it as 8, determining that 8 bits are needed for the instruction address. He then adds the bits for the instruction address (8), flag conditions (4), and control signals (48), writing '8 + 4 + 48 = 60 bits' to find the total bits per control word. For the second part, he calculates the total control memory size. He multiplies the number of instructions (256) by the bits per control word (60), writing '256 x 60 = 15,360 bits'. He then converts this to bytes by dividing by 8, writing '15,360 / 8 = 1,920 bytes'. The final answer for the control memory size is clearly written as 1,920 bytes.

The video provides a clear, methodical walkthrough of calculating the control memory requirements for a micro-programmed control unit. It demonstrates the fundamental principle that the size of the control memory is determined by the number of instructions and the size of each control word. The control word size is the sum of the bits needed to address the instruction, the bits to represent the flag conditions, and the bits for the control signals. The solution is broken down into two logical steps: first, calculating the bits per control word, and second, calculating the total memory size by multiplying the number of instructions by the bits per word and converting to bytes. This approach effectively teaches the core concepts of control memory design in computer architecture.

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