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Consider execution of 100 instructions on a 5 stage pipeline. Let P be the probability of an instruction being a branch. What must be the value of P such that speed up is atleast 4?
(Assume each stage takes 1 cycle to perform it’s task and branch is predicted on fourth stage of the pipeline)
- a)0.05
- b)0.08
- c)0.06
- d)0.04
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider execution of 100 instructions on a 5 stage pipeline. Let P b...
12P ≤ 1
= 0.08
Most Upvoted Answer
Consider execution of 100 instructions on a 5 stage pipeline. Let P b...
To calculate the value of P, we need to consider the speedup achieved in a pipeline. The speedup is given by the formula:
Speedup = (Pipeline Execution Time without Branches) / (Pipeline Execution Time with Branches)
Let's break down the calculation step by step:
1. Pipeline Execution Time without Branches:
- We have 100 instructions to execute.
- Each instruction takes 5 cycles to complete in the 5-stage pipeline.
- So, the total number of cycles required to execute 100 instructions without branches = 100 * 5 = 500 cycles.
2. Pipeline Execution Time with Branches:
- We are given that the branch is predicted on the fourth stage of the pipeline.
- This means that if the branch prediction is correct, the pipeline does not need to be flushed and restarted.
- Only when the branch prediction is incorrect, the pipeline needs to be flushed and restarted, resulting in a delay of 3 cycles.
- So, in the worst case, for every branch instruction, we have an additional delay of 3 cycles.
- Therefore, the total number of cycles required to execute 100 instructions with branches = 100 * 5 + 3 * (P * 100), where P is the probability of an instruction being a branch.
3. Speedup:
- We are given that the speedup is at least 4.
- So, Speedup >= 4.
- Using the formula for speedup, we have:
Speedup = (Pipeline Execution Time without Branches) / (Pipeline Execution Time with Branches)
4 <= 500="" (500="" +="" 3="" *="" (p="" *="">
4. Solving for P:
- Rearranging the inequality, we get:
4 <= 500="" (500="" +="">
2000 + 1200P <=>
1200P <=>
P <= -1500="">
P <=>
- Since P represents the probability, it cannot be negative. Therefore, we need to find the smallest positive value of P that satisfies the inequality.
- The closest positive option is 0.08, which satisfies the inequality.
- Therefore, the correct answer is option 'B' (P = 0.08).
Hence, the value of P that must be chosen to achieve a speedup of at least 4 is 0.08.
Speedup = (Pipeline Execution Time without Branches) / (Pipeline Execution Time with Branches)
Let's break down the calculation step by step:
1. Pipeline Execution Time without Branches:
- We have 100 instructions to execute.
- Each instruction takes 5 cycles to complete in the 5-stage pipeline.
- So, the total number of cycles required to execute 100 instructions without branches = 100 * 5 = 500 cycles.
2. Pipeline Execution Time with Branches:
- We are given that the branch is predicted on the fourth stage of the pipeline.
- This means that if the branch prediction is correct, the pipeline does not need to be flushed and restarted.
- Only when the branch prediction is incorrect, the pipeline needs to be flushed and restarted, resulting in a delay of 3 cycles.
- So, in the worst case, for every branch instruction, we have an additional delay of 3 cycles.
- Therefore, the total number of cycles required to execute 100 instructions with branches = 100 * 5 + 3 * (P * 100), where P is the probability of an instruction being a branch.
3. Speedup:
- We are given that the speedup is at least 4.
- So, Speedup >= 4.
- Using the formula for speedup, we have:
Speedup = (Pipeline Execution Time without Branches) / (Pipeline Execution Time with Branches)
4 <= 500="" (500="" +="" 3="" *="" (p="" *="">
4. Solving for P:
- Rearranging the inequality, we get:
4 <= 500="" (500="" +="">
2000 + 1200P <=>
1200P <=>
P <= -1500="">
P <=>
- Since P represents the probability, it cannot be negative. Therefore, we need to find the smallest positive value of P that satisfies the inequality.
- The closest positive option is 0.08, which satisfies the inequality.
- Therefore, the correct answer is option 'B' (P = 0.08).
Hence, the value of P that must be chosen to achieve a speedup of at least 4 is 0.08.
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Consider execution of 100 instructions on a 5 stage pipeline. Let P be the probability of an instruction being a branch. What must be the value of P such that speed up is atleast 4?(Assume each stage takes 1 cycle to perform it’s task and branch is predicted on fourth stage of the pipeline)a)0.05b)0.08c)0.06d)0.04Correct answer is option 'B'. Can you explain this answer?
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