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Processor Thermal Power Dissipation (Computer & Electrical Engineering) Solution

In modern computer microchip architecture, the local dynamic thermal power dissipation PP (in milliwatts, mW\text{mW}) of a processing core transistor sub-grid depends on its clock frequency ff (in gigahertz, GHz\text{GHz}) and logic supply voltage VV (in volts, V\text{V}), governed by:

P(f,V)=12fV2P(f, V) = 12fV^2

To handle a demanding computational workload spike, an embedded Dynamic Voltage and Frequency Scaling (DVFS) microcontroller algorithm adjusts both operational parameters concurrently over run-time tt (in milliseconds, ms\text{ms}):

  • f(t)=2+ln⁡(t+1)f(t) = 2 + \ln(t + 1)
  • V(t)=1.0+0.05t2V(t) = 1.0 + 0.05t^2
  1. Compute the operating frequency f(1)f(1) and voltage V(1)V(1) at t=1 mst = 1\text{ ms}.
  2. State the multivariable chain rule formula for the total time derivative dPdt\frac{dP}{dt}.
  3. Calculate the instantaneous rate of change of power dissipation dPdt\frac{dP}{dt} at t=1 mst = 1\text{ ms}.
  4. Explain how modern thermal management firmware uses the partial sensitivities ∂P∂f\frac{\partial P}{\partial f} and ∂P∂V\frac{\partial P}{\partial V} to prevent processor thermal throttling.

Step-by-Step Solution:

  1. Evaluate Operating State at t=1 mst = 1\text{ ms}:

    • Frequency: f(1)=2+ln⁡(1+1)=2+ln⁡(2)≈2+0.69315=2.69315 GHzf(1) = 2 + \ln(1 + 1) = 2 + \ln(2) \approx 2 + 0.69315 = 2.69315\text{ GHz}
    • Voltage: V(1)=1.0+0.05(1)2=1.05 VV(1) = 1.0 + 0.05(1)^2 = 1.05\text{ V}
    • Baseline Power: P=12(2.69315)(1.05)2≈12(2.69315)(1.1025)≈35.63 mWP = 12(2.69315)(1.05)^2 \approx 12(2.69315)(1.1025) \approx 35.63\text{ mW}
  2. State Governing Multivariable Chain Rule Formula: Since PP depends on intermediate parameters ff and VV, which are parameterized by run-time tt:

    dPdt=∂P∂fdfdt+∂P∂VdVdt\frac{dP}{dt} = \frac{\partial P}{\partial f}\frac{df}{dt} + \frac{\partial P}{\partial V}\frac{dV}{dt}
    (Defined in Multivariable Chain Rule.)

  3. Compute Partial and Parametric Derivatives:

    • Partial Derivatives:
      ∂P∂f=12V2\frac{\partial P}{\partial f} = 12V^2
      ∂P∂V=24fV\frac{\partial P}{\partial V} = 24fV
    • Time Derivatives:
      dfdt=ddt[2+ln⁡(t+1)]=1t+1\frac{df}{dt} = \frac{d}{dt}[2 + \ln(t + 1)] = \frac{1}{t + 1}
      dVdt=ddt[1.0+0.05t2]=0.10t\frac{dV}{dt} = \frac{d}{dt}[1.0 + 0.05t^2] = 0.10t
  4. Evaluate Individual Derivatives at t=1 mst = 1\text{ ms}:

    • ∂P∂f=12(1.05)2=12(1.1025)=13.23 mW/GHz\frac{\partial P}{\partial f} = 12(1.05)^2 = 12(1.1025) = 13.23\text{ mW/GHz}
    • ∂P∂V=24(2.69315)(1.05)≈67.867 mW/V\frac{\partial P}{\partial V} = 24(2.69315)(1.05) \approx 67.867\text{ mW/V}
    • dfdt=11+1=0.50 GHz/ms\frac{df}{dt} = \frac{1}{1 + 1} = 0.50\text{ GHz/ms}
    • dVdt=0.10(1)=0.10 V/ms\frac{dV}{dt} = 0.10(1) = 0.10\text{ V/ms}
  5. Assemble Total Rate of Change:

    dPdt=(13.23)(0.50)+(67.867)(0.10)\frac{dP}{dt} = (13.23)(0.50) + (67.867)(0.10)
    dPdt=6.615+6.787=13.40 mW/ms\frac{dP}{dt} = 6.615 + 6.787 = 13.40\text{ mW/ms}

  6. Firmware Sensitivity Analysis: Notice that ∂P∂V=67.87 mW/V\frac{\partial P}{\partial V} = 67.87\text{ mW/V} is more than 5 times larger than ∂P∂f=13.23 mW/GHz\frac{\partial P}{\partial f} = 13.23\text{ mW/GHz}. Because dynamic power scales quadratically with voltage (V2V^2) but only linearly with frequency (ff), DVFS algorithms prioritize reducing voltage first during thermal emergencies to maximize temperature drops while minimizing clock frequency penalties.