A processor X1 operating at 2 GHz has a standard 5-stage RISC instruction pipeline having a base CPI (cycles per instruction) of one without any pipeline hazards. For a given program P that has 30% branch instructions, control hazards incur 2 cycles stall for every branch. A new version of the processor X2 operating at same clock frequency has an additional branch predictor unit (BPU) that completely eliminates stalls for correctly predicted branches. There is neither any savings nor any additional stalls for wrong predictions. There are no structural hazards and data hazards for X1 and X2. If the BPU has a prediction accuracy of 80%, the speed up (rounded off to two decimal places) obtained by X2 over X1 in executing P is ______.Correct answer is between '1.42,1.43'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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A processor X₁ operating at 2 GHz has a standard 5-stage RISC instruction pipeline having a base CPI (cycles per instruction) of one without any pipeline hazards. For a given program P that has 30% branch instructions, control hazards incur 2 cycles stall for every branch. A new version of the processor X₂ operating at same clock frequency has an additional branch predictor unit (BPU) that completely eliminates stalls for correctly predicted branches. There is neither any savings nor any additional stalls for wrong predictions. There are no structural hazards and data hazards for X₁ and X₂. If the BPU has a prediction accuracy of 80%, the speed up (rounded off to two decimal places) obtained by X₂ over X₁ in executing P is ______.

Correct answer is between '1.42,1.43'. Can you explain this answer?

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A processor X1 operating at 2 GHz has a standard 5-stage RISC instruct...

Concept:
The given data,
Cycle time (tp) = 2 GHz = 0.5 nanoseconds.
K = 5,
X1:
Without branch prediction.

CPI = 1 + Number of stalls instruction
CPI = 1 + (0.3) × 2
CPI = 1.6
Average serial instruction execution time = CPI * Cycle time
Average serial instruction execution time = 1.6* 0.5nsec
Average serial instruction execution time = 0.8 ns.
X2:
If BPU predicted the correct branch then it eliminate stalls but if BPU predicted the wrong branch then BPU does not add any additional stalls but the remaining stalls will be present.

CPI = 1+(0.3 ×0.2 ×2)
CPI =1.12
Average parallel instruction execution time = CPI × Cycle time
Average parallel instruction execution time = 1.12 × 0.5 ns
Average parallel instruction execution time = 0.56 ns.

Speedup:
A speedup is a number that measures the relative performance of two systems processing the same problem.
Speedup = Serial Execution Time / Parallel Execution Time.
Speed up= 0.8 / 0.56
Speed up=1.428
Hence the correct answer is 1.428.

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A processor X1 operating at 2 GHz has a standard 5-stage RISC instruct...

Given Information:
- Processor X1 operates at 2 GHz (clock frequency).
- X1 has a 5-stage RISC instruction pipeline with a base CPI of 1 (without any pipeline hazards).
- Program P has 30% branch instructions, and control hazards incur a 2-cycle stall for every branch.

Calculating Execution Time for X1:
The execution time for a program can be calculated using the following formula:
Execution Time = (Number of Instructions * CPI) / Clock Frequency

Since the base CPI for X1 is 1, we can calculate the number of instructions executed in Program P as follows:
Number of Instructions = Total Instructions * (1 + (Branch Instruction Percentage * Control Hazard Cycles))

Given that the total instructions in Program P are unknown, we can assume a value of 100 for simplicity:
Number of Instructions = 100 * (1 + (30% * 2)) = 100 * (1 + 0.6) = 100 * 1.6 = 160

Now, we can calculate the execution time for X1:
Execution Time for X1 = (160 * 1) / 2 GHz = 160 / 2 = 80 ns

Calculating Execution Time for X2:
The new processor X2 has an additional branch predictor unit (BPU) that eliminates stalls for correctly predicted branches with an accuracy of 80%.

To calculate the execution time for X2, we need to consider the number of stalls incurred due to branch instructions. Since the BPU eliminates stalls for correctly predicted branches, we only need to calculate the stalls for incorrectly predicted branches.

Stalls for Incorrectly Predicted Branches = Total Branch Instructions * (1 - Accuracy) * Control Hazard Cycles
Stalls for Incorrectly Predicted Branches = 30% * (1 - 80%) * 2 = 30% * 20% * 2 = 0.06 * 2 = 0.12

Now, we can calculate the execution time for X2:
Execution Time for X2 = (Number of Instructions * CPI + Stalls for Incorrectly Predicted Branches) / Clock Frequency
Execution Time for X2 = (160 * 1 + 0.12) / 2 GHz = (160 + 0.12) / 2 = 160.12 / 2 = 80.06 ns

Calculating Speedup:
The speedup is calculated by dividing the execution time of X1 by the execution time of X2:
Speedup = Execution Time for X1 / Execution Time for X2
Speedup = 80 ns / 80.06 ns ≈ 1.00

Explanation:
The speedup obtained by X2 over X1 in executing Program P is approximately 1.00, which means there is no significant improvement in performance. This is because the branch predictor unit (BPU) in X2 has an accuracy of 80%, which results in a small number of stalls for incorrectly predicted branches. As a result, the overall execution time for X2 remains almost the same as X1. The negligible speedup indicates that the BPU's prediction accuracy is not high enough to provide significant performance improvement in this scenario.

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A processor X1 operating at 2 GHz has a standard 5-stage RISC instruction pipeline having a base CPI (cycles per instruction) of one without any pipeline hazards. For a given program P that has 30% branch instructions, control hazards incur 2 cycles stall for every branch. A new version of the processor X2 operating at same clock frequency has an additional branch predictor unit (BPU) that completely eliminates stalls for correctly predicted branches. There is neither any savings nor any additional stalls for wrong predictions. There are no structural hazards and data hazards for X1 and X2. If the BPU has a prediction accuracy of 80%, the speed up (rounded off to two decimal places) obtained by X2 over X1 in executing P is ______.Correct answer is between '1.42,1.43'. Can you explain this answer?

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