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Which change would give the largest saving in dynamic power for a fixed amount of computation?

  • A.reducing the supply voltage and lengthening the time allowed
  • B.raising the clock frequency to finish sooner
  • C.moving the whole of the computation into an interrupt handler instead
  • D.increasing the size of the instruction cache

Answer: A

Energy per operation falls as the square of the supply, so doing the same work more slowly at a lower voltage wins where the deadline permits. Raising the clock reduces the time but raises the power in proportion, leaving the energy much as it was.

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