Computer architecture power savings question

[SpiffyEh]

Homework Statement

Server farms such as Google and Yahoo! provide enough compute capacity for the highest request rate of the day. Imagine that most of the time these servers operate at only 60% capacity. Assume further that the power does not scale linearly with the load; that is, when the servers are operating at 60% capacity, they consume 90% max power. The servers could be turned off, but they would take too long to restart in response to more load. A new system has been proposed that allows for a quick restart but requires 20% of the max power while in this "barely alive" state.
a) How much power savings would be achieced by turning off 60% of the servers?
b) How much power savings would be achieved by placing 60% of the servers in the "barely alive" state?
c) How much power would be saved by reducing the voltage by 20% and frequency by 40%?
d) How much power savings would be achieced by placing 30% of the servers in the "barely alive" state and 30% off?

Homework Equations

Power dynamic = 1/2 * capacitive load * voltage^2 * frequency switched

The Attempt at a Solution

For part a) the servers that are on are running at 100% their power so I'm guessing 60% of the power would be saved?

I don't understand how to go about trying to solve the rest of them

 

[KataKoniK]

.

Hello,

You are correct in your assumption for part a). The servers that are turned off would save 60% of their power consumption, resulting in a 60% power savings.

For part b), the servers that are placed in the "barely alive" state would consume 20% of their max power. So, the power savings would be 80% (100% - 20%) for those servers.

For part c), the power dynamic equation can be used to calculate the power savings. Since the voltage is reduced by 20% and the frequency is reduced by 40%, the power savings would be:

Power savings = 1/2 * capacitive load * (0.8*V)^2 * (0.6*f)^2

= 0.288 * capacitive load * V^2 * f^2

This means that the power savings would be about 28.8% of the original power consumption.

For part d), the power savings would be a combination of the calculations from parts a) and b). So, the power savings for placing 30% of the servers in the "barely alive" state would be 24% (30% of 80%) and the power savings for turning off 30% of the servers would be 18% (30% of 60%). Together, this would result in a 42% power savings.