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A processor provides an instruction which transfers 64 bytes of data from one register to another register. Instruction fetch (IF) and Instruction decode (ID) takes 20 clock cycle. Then it takes 30 clock cycles to transfer each byte. The processor is clocked at a rate of 12 GHz. What is the delay in acknowledging an interrupt if the instruction is non-interruptible? (Compute value rounding to two decimal places.)
    Correct answer is between '161.66,161.67'. Can you explain this answer?
    Verified Answer
    A processor provides an instruction which transfers 64 bytes of data f...
    Length of clock cycle = 1/12
    Length of instruction cycle = (20 + (30 x 64)) x 1/12 = 161.67 ns
    Worst case delay = length of instruction cycle = 161.67 ns
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    Most Upvoted Answer
    A processor provides an instruction which transfers 64 bytes of data f...
    Given Information:
    - The processor transfers 64 bytes of data from one register to another register.
    - Instruction fetch (IF) and Instruction decode (ID) take 20 clock cycles.
    - It takes 30 clock cycles to transfer each byte.
    - The processor is clocked at a rate of 12 GHz.

    Calculating the Delay:
    To calculate the delay in acknowledging an interrupt, we need to consider the time taken for the instruction to complete and the time taken for the interrupt to be acknowledged.

    Time taken for the instruction to complete:
    - The instruction transfers 64 bytes of data.
    - It takes 30 clock cycles to transfer each byte.
    - So the total time taken to transfer 64 bytes is 64 * 30 clock cycles.

    Time taken for IF and ID:
    - IF and ID take 20 clock cycles.
    - Since the processor is clocked at a rate of 12 GHz, the time taken for IF and ID is 20 / 12 GHz seconds.

    Total time taken:
    - The total time taken is the sum of the time taken for the instruction to complete and the time taken for IF and ID.
    - Total time taken = (64 * 30) clock cycles + (20 / 12 GHz) seconds.

    Converting to seconds:
    - The clock cycle time is the inverse of the clock rate.
    - Clock cycle time = 1 / 12 GHz seconds.

    Calculating the delay:
    - Substitute the values in the equation: Total time taken = (64 * 30) clock cycles + (20 / 12 GHz) seconds.
    - Convert the clock cycles to seconds using the clock cycle time.
    - Delay = (64 * 30) * (1 / 12 GHz) + (20 / 12 GHz) seconds.

    Rounding the answer:
    - Round the answer to two decimal places.

    Final Answer:
    The delay in acknowledging an interrupt is between 161.66 and 161.67 seconds.
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    A processor provides an instruction which transfers 64 bytes of data from one register to another register. Instruction fetch (IF) and Instruction decode (ID) takes 20 clock cycle. Then it takes 30 clock cycles to transfer each byte. The processor is clocked at a rate of 12 GHz. What is the delay in acknowledging an interrupt if the instruction is non-interruptible? (Compute value rounding to two decimal places.)Correct answer is between '161.66,161.67'. Can you explain this answer?
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    A processor provides an instruction which transfers 64 bytes of data from one register to another register. Instruction fetch (IF) and Instruction decode (ID) takes 20 clock cycle. Then it takes 30 clock cycles to transfer each byte. The processor is clocked at a rate of 12 GHz. What is the delay in acknowledging an interrupt if the instruction is non-interruptible? (Compute value rounding to two decimal places.)Correct answer is between '161.66,161.67'. 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 A processor provides an instruction which transfers 64 bytes of data from one register to another register. Instruction fetch (IF) and Instruction decode (ID) takes 20 clock cycle. Then it takes 30 clock cycles to transfer each byte. The processor is clocked at a rate of 12 GHz. What is the delay in acknowledging an interrupt if the instruction is non-interruptible? (Compute value rounding to two decimal places.)Correct answer is between '161.66,161.67'. 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 A processor provides an instruction which transfers 64 bytes of data from one register to another register. Instruction fetch (IF) and Instruction decode (ID) takes 20 clock cycle. Then it takes 30 clock cycles to transfer each byte. The processor is clocked at a rate of 12 GHz. What is the delay in acknowledging an interrupt if the instruction is non-interruptible? (Compute value rounding to two decimal places.)Correct answer is between '161.66,161.67'. Can you explain this answer?.
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