Let's solve the problem step-by-step: --- ## Given: - Floating-point instructions: **60%** of total program time - Of floating-point time, **25%** is spent on square root calculations --- ## **Part a: Design Alternatives** ### **Option 1: Improve all floating-point instructions by 1.5x** #### **Step 1: Calculate total speedup using Amdahl's Law** - Fraction affected (\( f \)) = 0.6 (floating point) - Speedup (\( S \)) = 1.5 \[ \text{Speedup}_{\text{fp}} = \frac{1}{(1-f) + \frac{f}{S}} \] \[ = \frac{1}{0.4 + \frac{0.6}{1.5}} \] \[ = \frac{1}{0.4 + 0.4} \] \[ = \frac{1}{0.8} = 1.25 \] --- ### **Option 2: Speed up square root by 8x** #### **Step 2: Find total time spent on square root** - Fraction of time for square root: \(0.6 \times 0.25 = 0.15\) - Fraction of time for other floating-point: \(0.6 - 0.15 = 0.45\) #### **Step 3: Apply Amdahl's Law** - Fraction affected \( f = 0.15 \) - Speedup \( S = 8 \) \[ \text{Speedup}_{\text{sqrt}} = \frac{1}{(1-f) + \frac{f}{S}} \] \[ = \frac{1}{0.85 + \frac{0.15}{8}} \] \[ = \frac{1}{0.85 + 0.01875} \] \[ = \frac{1}{0.86875} \approx 1.15 \] --- #### **Final Comparison:** - **Speeding up all floating-point instructions:** **Speedup = 1.25** - **Speeding up square root:** **Speedup = 1.15** **Conclusion:** **Improving all floating-point instructions by 1.5× benefits the application more.** --- ## **Part b: Parallelization (Amdahl's Law)** ### **Step 1: Speedup with 16 CPUs, 90% parallelizable** Amdahl's Law: \[ \text{Speedup} = \frac{1}{(1 - p) + \frac{p}{N}} \] where \( p \) is the parallelizable fraction, \( N \) is the number of CPUs. \[ p = 0.9, \quad N = 16 \] \[ \text{Speedup} = \frac{1}{0.1 + \frac{0.9}{16}} \] \[ = \frac{1}{0.1 + 0.05625} \] \[ = \frac{1}{0.15625} \] \[ = 6.4 \] --- ### **Step 2: What fraction needs to be parallelized to get a speedup of 10?** Set Speedup = 10 and solve for \( p \): \[ 10 = \frac{1}{(1 - p) + \frac{p}{16}} \] \[ (1 - p) + \frac{p}{16} = 0.1 \] \[ 1 - p + \frac{p}{16} = 0.1 \] \[ 1 - p + 0.0625p = 0.1 \] \[ 1 - p + 0.0625p = 0.1 \] \[ 1 - 0.9375p = 0.1 \] \[ 0.9375p = 0.9 \] \[ p = \frac{0.9}{0.9375} \] \[ p \approx 0.96 \] --- ## **Final Answers** ### **(a)** - **Improving all floating-point instructions by 1.5× (Speedup = 1.25) is better** than speeding up square roots by 8× (Speedup = 1.15). ### **(b)** - **Speedup with 16 CPUs and 90% parallelizable code:** **6.4** - **Fraction of code to parallelize for speedup of 10:** **96%** --- **Let me know if you want the reasoning shown in a more detailed way or have additional questions!**

A: According to Amdahl's Law, achieving a speedup of 10 requires approximately 96% of the code to be perfectly parallelizable.