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A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will be
- a)120.4 microseconds
- b)160.5 microseconds
- c)165.5 microseconds
- d)590.0 microseconds
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanos...
Pipelining requires all stages to be synchronized meaning, we have to make the delay of all stages equal to the maximum pipeline stage delay which here is 160.
Time for execution of the first instruction = (160+5) * 3 + 160 = 655 ns (5 ns for intermediate registers which is not needed for the final stage).
Now, in every 165 ns, an instruction can be completed. So, Total time for 1000 instructions = 655 + 999*165 = 165.49 microseconds
Time for execution of the first instruction = (160+5) * 3 + 160 = 655 ns (5 ns for intermediate registers which is not needed for the final stage).
Now, in every 165 ns, an instruction can be completed. So, Total time for 1000 instructions = 655 + 999*165 = 165.49 microseconds
Most Upvoted Answer
A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanos...
Given:
- Stage delays: 150 ns, 120 ns, 160 ns, 140 ns
- Delay of each register: 5 ns
- Number of data items to process: 1000
To find: Total time taken to process 1000 data items on this pipeline
Approach:
- A 4-stage pipeline means that each data item goes through 4 stages, with a register in between each stage.
- The total delay for each data item will be the sum of the stage delays and the register delays.
- We can find the total delay for each data item, and then multiply it by the number of data items to get the total time taken.
Calculation:
Total delay for each data item = sum of stage delays + sum of register delays
= (150 + 120 + 160 + 140) + (5 + 5 + 5 + 5 + 5) (there are 5 registers in total)
= 590 ns
Total time taken to process 1000 data items = 590 ns * 1000
= 590,000 ns
= 590.0 microseconds
Therefore, the correct answer is option C, 165.5 microseconds.
- Stage delays: 150 ns, 120 ns, 160 ns, 140 ns
- Delay of each register: 5 ns
- Number of data items to process: 1000
To find: Total time taken to process 1000 data items on this pipeline
Approach:
- A 4-stage pipeline means that each data item goes through 4 stages, with a register in between each stage.
- The total delay for each data item will be the sum of the stage delays and the register delays.
- We can find the total delay for each data item, and then multiply it by the number of data items to get the total time taken.
Calculation:
Total delay for each data item = sum of stage delays + sum of register delays
= (150 + 120 + 160 + 140) + (5 + 5 + 5 + 5 + 5) (there are 5 registers in total)
= 590 ns
Total time taken to process 1000 data items = 590 ns * 1000
= 590,000 ns
= 590.0 microseconds
Therefore, the correct answer is option C, 165.5 microseconds.
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A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. Can you explain this answer?
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A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. 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 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. 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 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. Can you explain this answer?.
A 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. 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 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. 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 4-stage pipeline has the stage delays as 150, 120, 160 and 140 nanoseconds respectively. Registers that are usedbetween the stages have a delay of 5 nanoseconds each. Assuming constant clocking rate, the total time taken to process1000 data items on this pipeline will bea)120.4 microsecondsb)160.5 microsecondsc)165.5 microsecondsd)590.0 microsecondsCorrect answer is option 'C'. Can you explain this answer?.
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