Architecture

Homework1_Yue He

1.4

(a). Each component has the maximum load, so the power for each component should maximum.

Pserver = PIntel + PKingston + Phard drive

So 0.8 * Pserver = 66 + 2* 2.3 + 7.9

So the Pserver = 98.125 w

(b) For 7200 rpm disk drive, the power for read/seek is 7.9w, the power for idle is 4.0w.

P7200 rpm = Pread/seek + Pidle

So P7200 rpm = 0.4 * 7.9 + 0.6 * 4

So P7200 rpm = 5.56 w

(c) Given that the percentage of read time for 5400rpm disk is T5400 rpm, so the percentage of idle time should be (1 - T5400 rpm). The percentage of read time for 7200rpm disk is (0.75 * T5400 rpm), so the percentage of idle time should be (1 - 0.75 * T5400 rpm).

| Pread/seek | + | Pidel |

P5400rpm | 7.0 * T5400 rpm | + | (1 - T5400 rpm) * 2.9 |

P7200rpm | 7.9 *T5400 rpm *0.75 | + | (1 - 0.75 * T5400 rpm) * 4 |

Because of the power consumption for the two disks be equal.

So P5400rpm =P7200 rpm

7.0 * T5400 rpm + (1 - T5400 rpm) * 2.9 = 7.9 *T5400 rpm *0.75 + (1 - 0.75 * T5400 rpm) * 4

So 1.175 T5400 rpm = 1.1

So T5400 rpm = 0.93617

So T7200 idel = (1 - 0.75 * T5400 rpm) = 29.8%

1.7

(a). For dual-cores system, assuming the application is 80% parallelizable

补充:一定时间内有多少次，就是频率，因为双核，减少了60%，那么频率也要相应的减少，才能保持相同的performance。

（1-0.8）已经减少的，然后因为双核，那可以同时执行，所以实际是0.8/2

So 1 – 0.8 + 0.8/2 = 0.6

So we can decrease the 60% frequency to get the same performance.

(b). 1.5 Section: formula:

Powerdynamic ∝ ½ * Capacitive load * Voltage2 * Frequency switched.

Since the capacitance is unchanged and the voltage is decreased linearly with the frequency.

So PdualPsingle = (Voltage*0.6)2*(0.6*Frequency switched)Voltage2 * Frequency switched = 0.63 = 0.216

So the dual-cores system requires 21.6% dynamic power as compared to the single-core system.

(c). Assuming x is the percent of parallelization. Voltagedual = 0.75 * Voltagesingle

So 1 = 0.751-x+x/2 ; x = 50% ;

So the parallelization should be 50%

(d). The same as...