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Unformatted text preview: ers 2 1.9a How much energy do you save if you execute at the current speed and turn off the system when the computation is complete. The equation for dynamic energy is: E = * CL * V2 (page 23 in text). However, this question is not changing C or V so energy has not changed, or has it? The equation represents the energy used in a single transition from one logic state to another. So this equation is applied for every clock edge. If the processor is allowed to run continuously, the total energy will be ET1 = * CL * V2 (* # clock edges). If the processor is turned off (or the clock stopped) when the computation is completed, which is way through the allowed time, the number of clock edges is cut in half, therefore, ET2 = * CL * V2 (* # clock edges) / 2. Thus ET2 = ET1. So the energy saved is 50%. 1.9b How much energy do you save if you set the voltage and frequency to be half as much? Enew = * CL * 0.5 V2 Eold = * CL * V2 Enew / Eold = .52 or 0.25 New energy is 0.25 * Eold. Therefore the energy saved is 75%, where Eold is equivalent to the ET2 of part a which is the energy to execute the given task. CS...
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 Spring '12
 kavi
 Computer Architecture

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