How to get number of ways of L2 cache based on the following performance graph?












0















I was reading this
article



And they have this graph they got programatically



enter image description here



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.



enter image description here



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).










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    0















    I was reading this
    article



    And they have this graph they got programatically



    enter image description here



    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.



    enter image description here



    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).










    share|improve this question



























      0












      0








      0








      I was reading this
      article



      And they have this graph they got programatically



      enter image description here



      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.



      enter image description here



      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).










      share|improve this question
















      I was reading this
      article



      And they have this graph they got programatically



      enter image description here



      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.



      enter image description here



      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






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      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 21 '18 at 0:15







      Tiago Oliveira

















      asked Nov 20 '18 at 23:53









      Tiago OliveiraTiago Oliveira

      193




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