The Pentium 4 Prescott processor, released in 2004, has a clock rate of 3.6 GHz
ID: 3770755 • Letter: T
Question
The Pentium 4 Prescott processor, released in 2004, has a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average, it consumed 10 W of static power and 90 W of dynamic power.
The Core i5 Ivy Bridge, released in 2012, had a clock rate of 3.4 GHz and voltage of 0.9 V. Assume that, on average, it consumed 30 W of static power and 40 W of dynamic power.
a) For each processor find the average capacitive loads. [4 points]
b) Find the percentage of the total dissipated power comprised by the static power and the ratio of
static power to dynamic power for each technology. [4 points]
c) If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to
maintain the same leakage current? Note: power is defined as the product of voltage and current. [6 points]
Explanation / Answer
1.CL = DynamicPower / (0.5 X V^2 X ClockRate)
P4: CL = 90 / (0.5 * 1.25^2 * 3.6E9) = 3.2*10^-8
I5: CL = 40 / (0.5 * 0.9^2 * 3.4E9) = 2.90486565*10^-8
2.P4
the total dissipated power = 10 + 90 = 100W
static power / the total dissipated power = (10 / 90) * 100 = 11%
the ratio of static power to dynamic power = 10 / 90 = 0.11
i5
the total dissipated power = 30 + 40 = 70W
static power / the total dissipated power = (30 / 70) * 100 = 43%
the ratio of static power to dynamic power = 30 / 40 = 0.75
3.
the voltage of both P4 and i5 should be reduced by 10%.
the total dissipated power = Voltage x leakage current(I)
P4
100 = 1.25 x I
I = 80A
the total dissipated power * 0.9 = 90W
90 = 1.25 x 80, = 0.9
i5
70 = 0.9 x I
I = 78 (rounded up)
the total dissipated power * 0.9 = 63W
63 = 0.9 x 78, = 0.9 (rounded up)
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