Power Question from me...

[CompArcStudent]

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% of maximum 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 start but requires 20% of the maximum power while in this “barely alive” state.
a) How much power savings would be achieved 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?


a) is the answer 90% ? the question says 60% servers consume 90% power. So if 60% servers turned off, does that mean 90% power saved?

b) is the answer 70%? barely alive state requires 20% power, originally it consumes 90% power, so power saved = 90% - 70% = 20% ?

Any suggestion is appreciated. Thanks

 

[Audioguru]

The amount of money saved by reducing wasted power is almost nothing compared with their income from advertising.
They probably will not risk hibernating their unused servers during off peak hours.

 

[chuckey]

It is not possible to calculate unless we can find out what "most of the time is". You could work out the break even "most of the time" value and hence come to some sort of answer.
Frank

 

[Venkadesh_M]

A, The 60% of the server takes about 90% of maximum power. but if you turn off the remaining 40% of server then you will get 40% of 90% saving that is about 36% total savings on maximum power. in the same time the 60% will operate at 100% power so it is more (60% * 10%) 6% so the answer is 30% saving.

B, with the remaining taking 20% power when turned off would save the power about 40% * 70% of maximum power saving
so it is 28% - 6% is about 22% saving.

as of my opinion.