>What if the dish can actually hold 20 fruits in total, but only 2 apples
>and 8 oranges are present? What would be the occupancy?
Ok. In that case,
occ_percent::total =100% = totalocc_percent::apples +
occ_percent::oranges + remaining fruits.

Isn't that?

Back to the cache, if the items in the cache are only (writebacks,
switch_cpu::data, switch_cpu::inst,prefetcher, cpu::inst and
cpu::data) then the total occupancy of 97% means that there are 3%
*remaining items*.
So what is the remaining items then?

On 5/26/12, Nilay Vaish <[email protected]> wrote:
> On Sat, 26 May 2012, Mahmood Naderan wrote:
>
>>> Suppose the system is just about to start, that is number of cycles/ticks
>>> = 0. What >would be the value of all of these quantities?
>>
>> There is no problem with the starting point. At starting point there
>> is nothing in cache. The stats are calculated at the end of simulation
>> and we want to calculate the items in the cache.
>
> You are right, there is nothing in the cache when the simulation starts,
> which means that all the quantities will have a value of 0. How can that
> ever sum up to 1? In which case you are wrong in claiming that these
> values should add up to 1.
>
>>
>>
>>> Why do you expect these numbers should add up to 100%?
>>
>> Let me say this example:
>> There is a dish that contains 2 apples and 8 oranges. Then total
>> number of items are 10. Now the occupancy percentage for apples is
>> (2/10)*100=20% and the occupancy percentage for oranges is
>> (8/10)*100=80%
>> => occ_percent::apples + occ_percent::oranges = 100%
>
> What if the dish can actually hold 20 fruits in total, but only 2 apples
> and 8 oranges are present? What would be the occupancy?
>
> --
> Nilay
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>


-- 
// Naderan *Mahmood;
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