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equal.) We can solve directly for k without knowing the other terms: So, no matter what the clock rate, instruction count, or CPI, you would need to increase the clock
rate to 171% to decrease time by 30% if you pay a 20% CPI penalty. The answers become: Answer for followup question:
Different processors may have different instructions. If a processor has more powerful
instructions, it is likely to have a slower clock rate. (Some processors get around this by
internally converting complex instructions into simpler ones.) On the other hand, a processor
with more powerful instructions may be able to finish a task in a small number of instructions,
where a simpler processor may require more instructions for the same task. So, when comparing
processors, a processor with a slower cycle time would finish a task more quickly if it has more
powerful instructions and requires fewer instructions for the same task.
2. Exercise 1.11:
Each part of this problem relies on the formulas for the cost of an integrated circuit (from lecture
and page 46) and simple formulas for computing area. Note that the book formulas approximate
the actual formulas  we are only interested in this approximation for this problem.
1.11.1: First, compute the area of each wafer: Next, compute area of each die: Next, calculate the yield using the formula from page 46: ( ⁄) ( ⁄) 1.11.2: To find the cost per die, use the formula from page 46: 1.11.3: This is a repeat of part #1, except that the original data is scaled slightly. This
approximates what would happen as feature size gets smaller: Compute area of each die assuming there are 10% more dies: Next, calculate the yield using the formula from page 46, except assume defects have increased
by 15%: ( ⁄) ( ⁄) So, even though the defect rate went up significantly, the decrease in feature size compensated for
it (in this example).
Answer for followup question:
A very large die is very likely to include a defect. Consider the smaller defect rate from the
problem: 0.018 defects/cm2 on averag...
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 Fall '11
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