Re: [Gems-users] cache size calculation

[Abdullah Kayi <apokayi@xxxxxxxxxxxxxx>]

Shoaib Altaf wrote:
> I want to have 1 MB private L2 cache with 16 banks for each processor in a 
> 16 core CMP.
> So which one is the right way to declare that,
>  either
>
> L2_CACHE_ASSOC: 4
> L2_CACHE_NUM_SETS_BITS :12
>  This will from a 1 MB cache but in the final report the modified value 
> comes out to be less then 1 MB , which I believe is due to
>  L2_CACHE_NUM_SETS_BITS = L2_CACHE_NUM_SETS_BITS- 
> log_int(g_NUM_L2_BANKS_PER_CHIP) (RubyConfig.C:137)
>  or
>
> L2_CACHE_ASSOC: 4
> L2_CACHE_NUM_SETS_BITS :16
>   
You need to use

L2_CACHE_NUM_SETS_BITS :16

as it will be reduced to 12 because of the exact calculation you pointed 
out (16 - log_2(16) = 12).
Regards,

Abdullah

> Regards
> Shoaib
>
>
> --------------------------------------------------
> From: "Abdullah Kayi" <apokayi@xxxxxxxxxxxxxx>
> Sent: Sunday, April 12, 2009 9:05 PM
> To: "Gems Users" <gems-users@xxxxxxxxxxx>
> Subject: Re: [Gems-users] cache size calculation
>
>   
> > Shoaib Altaf wrote:
> >     
> > > Hi,
> > >
> > > Are there any other parameters to be specified while designing the
> > > size of a cache other then the following
> > >
> > > LI_CACHE_ASSOC: 4
> > > L1_CACHE_NUM_SETS_BITS :8
> > >
> > > How these configurations make the size of the L1 cache to  be 64 KB in
> > > the statistics below.
> > >
> > >       
> > It should be something like this:
> >
> > 2^8 = 256 is the # of sets
> >
> > Cache Size = (# of sets) * (# of blocks / set) * (# of bytes / block)
> > Cache Size = 256 * 4 (cache associativity) * 64 (or cache line size) = 64 
> > KB
> > ~ AK
> >     
> > > num_cache_set: 256
> > > cache_size_Kbytes: 64
> > >
> > > Muhammad Shoaib
> > > ------------------------------------------------------------------------
> > >
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