A) Dynamic power = capacitance x voltage2 x frequency
for version 1
let capacitance be C
Dynamic power0 = C x (1.5) ^ 2 x 2.4 = 5.4 C
FOr version 2
let capacitance be C1
Dynamic power1 = C1 x (1.2) ^ 2 x 3 = 4.32C1
We know that Dynamic power1 is 15 % less than Dynamic power0
so Dynamic power1 = 0.85 x Dynamic power0
4.32C1 = 5.4 C
C1 = 5.4 / 4.32 C = 1.25
B)
Let Capacitance of version 1 will be C1
let capacitance of version2 C2
we know C2 = 0.7 x C1
let Dynamic power of version 1 = Dp1
Dynamic power of version 2 = Dp2
Dp2 = 0.65xDp1
C2 x V22 x frequency = 0.65 x (C1 x V12 x frequency1)
0.7x C1 x V22 x 3 = 0.65 x C1 x 1.5 ^ 2 x 2.4
V2 = 1.29 volt
C)
According to amdahl's law
speedup achieved =
fraction enhanced = 30 %
speedup enhanced = 2
speed up achieved = = 1.33
D)
According to amdahl's law
speedup achieved =
fraction enhanced = 30 %
speedup required = 1.5
speedup enhanced = ?
speed up enhanced = 4.5
so we need to improve the alu 4.5 times to get 1.5x system wide speedup
Suppose we have developed new versions of a processor with the following characteristics: Version Voltage 1.5 V 1.2V Clock rate 2.4 GHz 3GHZ a) How much has the capacitive load varied between version...
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