All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  Consider the propagation delay along the bus ... Start Learning for Free
SaveJoin the Discussion on the App
Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ ns
    Correct answer is '53'. Can you explain this answer?
    FREE This question is part of
    Download PDF Attempt Test
    		Download PDF Attempt this Test
    Verified Answer
    Consider the propagation delay along the bus and through the ALU is 35...
    Transfer time = propagation delay + copy time
    = 35 + 18 
    = 53 ns
    View all questions of this test
    Most Upvoted Answer
    Consider the propagation delay along the bus and through the ALU is 35...
    Given information:
    Propagation delay along the bus = 35 ns
    Propagation delay through the ALU = 120 ns
    Time taken for a register to copy data from the bus = 18 ns

    Calculating the total time required:
    To transfer data from one register to another, we need to consider the following steps:

    1. Copying data from the source register to the bus: This step takes 18 ns.

    2. Propagation delay along the bus: After the data is copied onto the bus, it takes 35 ns for the data to propagate along the bus to reach the ALU.

    3. Propagation delay through the ALU: Once the data reaches the ALU, it takes 120 ns for the ALU to process the data.

    4. Copying data from the bus to the destination register: After the data is processed by the ALU, it takes 18 ns for the data to be copied from the bus to the destination register.

    Calculating the total time:
    The total time required is the sum of the times taken for each step.

    Total time = Time to copy data from source register to the bus + Propagation delay along the bus + Propagation delay through the ALU + Time to copy data from the bus to the destination register

    Total time = 18 ns + 35 ns + 120 ns + 18 ns

    Total time = 191 ns

    Therefore, the total time required to transfer data from one register to another is 191 ns.
    < Previous Question Next Question >
    Explore Courses for Computer Science Engineering (CSE) exam

    Similar Computer Science Engineering (CSE) Doubts

    On a non-pipelined sequential processor, a program segment, which is a part of the interrupt service routine, is given to transfer 500 bytes from an I/O device to memory.Initialize the address registerInitialize the count to 500LOOP: Load a byte from deviceStore in memory at address given by address registerIncrement the address registerDecrement the countIf count != 0 go to LOOPAssume that each statement in this program is equivalent to a machine instruction which takes one clock cycle to execute if it is a non-load/store instruction. The load-store instructions take two clock cycles to execute.The designer of the system also has an alternate approach of using the DMA controller to implement the same transfer. The DMA controller requires 20 clock cycles for initialization and other overheads. Each DMA transfer cycle takes two clock cycles to transfer one byte of data from the device to the memory.Q. What is the approximate speedup when the DMA controller based design is used in place of the interrupt driven program based input-output?

    Top Courses for Computer Science Engineering (CSE)View all

    Top Courses for Computer Science Engineering (CSE)

    Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer?
    Question Description
    Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer?.
    Solutions for Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
    Here you can find the meaning of Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer?, a detailed solution for Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? has been provided alongside types of Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ nsCorrect answer is '53'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
    Explore Courses for Computer Science Engineering (CSE) exam

    Top Courses for Computer Science Engineering (CSE)

    Explore Courses

    Suggested Free Tests

    Signup for Free!
    Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
    10M+ students study on EduRev
    Get App
    © EduRev
    Education Revolution
    Signup to see your scores go up within 7 days!
    Access 1000+ FREE Docs, Videos and Tests
    Takes less than 10 seconds to signup