How to get number of ways of L2 cache based on the following performance graph?
I was reading this
article
And they have this graph they got programatically
They say in the article they can't use this graph to calculate the associativity of cache L2 because they need more strides.
What I don't get is why can't they simply divide 1048576 (or 524288,262144) for 32768 (16384,8192; respectively), achieving 32 ways - the same way they do to achieve the number of ways in L1 (through the following image) - where 1048576/262144 = 4 which is the number of ways of L1.
Also, not only I don't know why they can't do this, but they seem to be right, since the number of ways of L2 is not 32 but 4 (just like L1) (they have it in the article).
caching memory hardware processor associativity
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
add a comment |
I was reading this
article
And they have this graph they got programatically
They say in the article they can't use this graph to calculate the associativity of cache L2 because they need more strides.
What I don't get is why can't they simply divide 1048576 (or 524288,262144) for 32768 (16384,8192; respectively), achieving 32 ways - the same way they do to achieve the number of ways in L1 (through the following image) - where 1048576/262144 = 4 which is the number of ways of L1.
Also, not only I don't know why they can't do this, but they seem to be right, since the number of ways of L2 is not 32 but 4 (just like L1) (they have it in the article).
caching memory hardware processor associativity
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
add a comment |
I was reading this
article
And they have this graph they got programatically
They say in the article they can't use this graph to calculate the associativity of cache L2 because they need more strides.
What I don't get is why can't they simply divide 1048576 (or 524288,262144) for 32768 (16384,8192; respectively), achieving 32 ways - the same way they do to achieve the number of ways in L1 (through the following image) - where 1048576/262144 = 4 which is the number of ways of L1.
Also, not only I don't know why they can't do this, but they seem to be right, since the number of ways of L2 is not 32 but 4 (just like L1) (they have it in the article).
caching memory hardware processor associativity
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
I was reading this
article
And they have this graph they got programatically
They say in the article they can't use this graph to calculate the associativity of cache L2 because they need more strides.
What I don't get is why can't they simply divide 1048576 (or 524288,262144) for 32768 (16384,8192; respectively), achieving 32 ways - the same way they do to achieve the number of ways in L1 (through the following image) - where 1048576/262144 = 4 which is the number of ways of L1.
Also, not only I don't know why they can't do this, but they seem to be right, since the number of ways of L2 is not 32 but 4 (just like L1) (they have it in the article).
caching memory hardware processor associativity
caching memory hardware processor associativity
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
edited Nov 21 '18 at 0:15
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
asked Nov 20 '18 at 23:53
Tiago Oliveira
193
193
add a comment |
add a comment |
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