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An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20% instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.
  • a)
    2.86
  • b)
    3.21
  • c)
    1.65
  • d)
    2.57
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An instruction pipeline has 4 stages Instruction Fetch(IF), Instructio...
Suppose each stage takes 1 s. 20 conditional – 20*3 s (Cycles per instruction for conditional instructions is 3 as branch is known at the third stage) 80 unconditional – 80 s (Cycles per instruction for unconditional is 1) So total time taken with pipeline = 20*3 + 80 = 140 s Time taken without pipeline = 4 * 100 (Cycle per instruction for all is 4) Speed up = 400 / 140 = 2.86
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An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20% instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.a)2.86b)3.21c)1.65d)2.57Correct answer is option 'A'. Can you explain this answer?
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An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20% instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.a)2.86b)3.21c)1.65d)2.57Correct answer is option 'A'. 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 An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20% instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.a)2.86b)3.21c)1.65d)2.57Correct answer is option 'A'. 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 An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20% instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.a)2.86b)3.21c)1.65d)2.57Correct answer is option 'A'. Can you explain this answer?.
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