The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average,

Answer:

1. The capacitive load for the Pentium 4 Prescott processor is 32 nF. For the Core i5 Ivy processor, it is 29.05 nF.

2. The static power makes up 10% of the total power dissipated for the Pentium 4 Prescott processor. The static to dynamic power ratio is 0.11.

For the Core i5 Ivy Bridge processor, the static power percentage is 42.86%. The ratio of static to dynamic power stands at 0.75.

3. The voltage reduction for the Pentium 4 Prescott processor equals a decrease of 5.9 %.

The Core i5 Ivy Bridge processor sees a 9.8 % reduction in voltage.

Explanation:

1. Recognizing dynamic power, P as approximately 1/2 CV²f, where C is the transistor's capacitive load, v denotes voltage, and f is frequency.

Thus, C is found using the formula C ≈ 2P/V²f.

For the Pentium 4 Prescott processor, with V₁ = 1.25 V, f₁ = 3.6 GHz, and P₁ = 90 W, we denote its capacitive load as C₁. Thus, we find C₁ ≈ 2P/V²f = 2 × 90 W/(1.25 V)² × 3.6 × 10⁹ Hz = 3.2 × 10⁻⁸ F = 32 × 10⁻⁹ F = 32 nF.

For the Core i5 Ivy Bridge processor, with V = 0.9 V, f = 3.4 GHz, and P = 40 W, we define C₂ as its load. Therefore, C₂ ≈ 2P/V²f = 2 × 40 W/(0.9 V)² × 3.4 × 10⁹ Hz = 2.905 × 10⁻⁸ F = 29.05 × 10⁻⁹ F = 29.05 nF.

2. The summation of total power is derived from static plus dynamic power.

For Pentium 4 Prescott, static power adds to 10 W and dynamic power is 90 W. Hence, the overall power, P = 10 W + 90 W = 100 W.

The fraction of this total attributed to static power is calculated as static power over total power multiplied by 100, thus static power/total power × 100 = 10/100 × 100 = 10%.

The ratio of static to dynamic power equals static power over dynamic power = 10/90 = 0.11.

For the Core i5 Ivy Bridge, static power figures at 30 W, and dynamic power at 40 W, meaning the total power becomes P = 30 W + 40 W = 70 W.

The portion of the total power that is static is computed as static power/total power × 100 = 30/70 × 100 = 42.86%.

That ratio of static to dynamic stands at static power/dynamic power = 30/40 = 0.75.

3. The total power comprises static and dynamic contributions and resulting leakage current arises from static power. We understand that P = IV, hence leakage current, I = P/V.

With an intended total power reduction of 10%, we have P₂ = (1 - 0.1)P₁ = 0.9P₁, where P₁ is the initial dissipated power before the 10% decrement and P₂ represents the new dissipated power.

Hence, new total dissipated power P₂ = new static power I₂V₂ + new dynamic power 1/2C₂V₂²f₂ = 0.9P₁.

For the Pentium 4 Prescott with P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁, given I₂ as leakage current which equals static power/voltage = 10 W/1.25 V = 8 A (since leakage remains constant), we determine

8 A × V₂ + 1/2 × 32 × 10⁻⁹ F × V₂² × 3.6 × 10⁹ Hz = 0.9 × 100.

This simplifies to 8V₂ + 57.6V₂² = 90, leading to the quadratic equation.

57.6V₂² + 8V₂ - 90 = 0, from which applying the quadratic formula yields

V₂ = .

Choosing the positive result, V₂ arrives at 1.18 V. The calculated reduction percentage is given by (new voltage - old voltage)/new voltage × 100% = (1.18 - 1.25)/1.18 × 100% = -0.07/1.18 × 100% = -5.9% with a 5.9% drop from 1.25V.

For the Core i5 Ivy Bridge processor, it follows that P₂ = I₂V₂ + 1/2C₂V₂²f₂ = 0.9P₁. With I₂ as leakage current equaling static power/voltage = 30 W/0.9 V = 33.33 A (again, leakage remains constant), we next evaluate

33.33 A × V₂ + 1/2 × 29.05 × 10⁻⁹ F × V₂² × 3.4 × 10⁹ Hz = 0.9 × 70.

This resolves to 33.33V₂ + 49.385V₂² = 63. Thus, it simplifies to the quadratic equation

49.385V₂² + 33.33V₂ - 63 = 0, whereby employing the quadratic formula lets us find

V₂ = .

Choosing the positive answer provides a new voltage of 0.82 V. The percentage reduction computes as (new voltage - old voltage)/new voltage × 100% = (0.82 - 0.9)/0.82 × 100% = -0.08/0.82 × 100% = -9.8% with a 9.8% decrease from 0.9V.