# Energy Consumption and Frequency Scaling Let's break down the problem step by step, clarify the confusion about the power-frequency relationship, and compute the energy at 3.2 GHz. --- ## 1. **Given Data** - **Original frequency (\(f_1\))**: 4 GHz - **Execution time (\(T_1\))**: 50 sec - **Dynamic power (\(P_{dyn,1}\))**: 70 W - **Leakage power (\(P_{leak}\))**: 30 W (assumed unchanged with frequency) --- ## 2. **Relationships** - **Dynamic Power:** \[ P_{dyn} = \alpha C V^2 f \] Where \(\alpha\) is activity factor, \(C\) is capacitance, \(V\) is supply voltage, and \(f\) is frequency. - **Total Power:** \[ P_{total} = P_{dyn} + P_{leak} \] - **Energy:** \[ E = P_{total} \times T \] - **Execution Time:** For the same instruction count, \[ T \propto \frac{1}{f} \] So at new frequency \(f_2\): \[ T_2 = T_1 \times \frac{f_1}{f_2} \] --- ## 3. **Why Dynamic Power Sometimes Scales with \(f^2\)** - If **voltage is also reduced** when frequency is reduced, and if \(V\) is roughly proportional to \(f\), then \[ P_{dyn} \propto V^2 f \propto f^2 f = f^3 \] or if \(V \propto f\), \(P_{dyn} \propto f^3\). - **However**, in many basic problems (like this one unless told otherwise), we assume **voltage is constant** as frequency changes. So, \[ P_{dyn} \propto f \] - If the solution you saw used \(f^2\), they probably assumed voltage also decreases with frequency (dynamic voltage and frequency scaling, DVFS). **Here, unless specified, we'll assume voltage is constant and use \(P_{dyn} \propto f\).** --- ## 4. **Calculating at 3.2 GHz (\(f_2\))** ### **Step 1: New Execution Time** \[ T_2 = T_1 \times \frac{f_1}{f_2} = 50 \times \frac{4}{3.2} = 50 \times 1.25 = 62.5\ \text{sec} \] ### **Step 2: New Dynamic Power** \[ P_{dyn,2} = P_{dyn,1} \times \frac{f_2}{f_1} = 70 \times \frac{3.2}{4} = 70 \times 0.8 = 56\ \text{W} \] ### **Step 3: Total Power at 3.2 GHz** \[ P_{total,2} = P_{dyn,2} + P_{leak} = 56 + 30 = 86\ \text{W} \] ### **Step 4: Energy Consumed** \[ E_2 = P_{total,2} \times T_2 = 86 \times 62.5 = 5375\ \text{J} \] --- ## 5. **Summary Table** | Frequency | Execution Time (s) | Dynamic Power (W) | Leakage Power (W) | Total Power (W) | Energy (J) | |-----------|--------------------|-------------------|-------------------|-----------------|------------| | 4.0 GHz | 50 | 70 | 30 | 100 | 5000 | | 3.2 GHz | 62.5 | 56 | 30 | 86 | 5375 | --- ## 6. **Key Takeaways** - **If voltage is constant:** \(P_{dyn} \propto f\) - **If voltage scales with frequency:** \(P_{dyn} \propto f^3\) (or \(f^2\) for some voltage scaling models) - **In this calculation**, we assumed **voltage constant**, so dynamic power scales linearly with frequency. --- ## 7. **Answer to Your Question** > **Why did they use \(f^2\)?** - They likely assumed voltage reduces with frequency (\(V \propto f\)), so \(P_{dyn} \propto V^2 f \propto f^3\). - If **voltage is NOT reduced**, use \(P_{dyn} \propto f\), which matches most straightforward homework problems. --- ## 8. **Final Answer (with calculations):** **At 3.2 GHz, the processor consumes 5375 Joules to execute the program.**