Re: hit percentage problem?
Jeremy Cowles
*special*.
This is why the page hit % is 100%, because the first loop guarantees that
the page will be loaded.
Without that, I'm not sure what would happen (as far as the page table hits
go).
"Jeremy Cowles" <jeremy...@berkeley.edu> wrote in message
news:g80grl$2p2e$1...@geode.berkeley.edu...
> "cs61c news" <cs61...@imail.eecs.berkeley.edu> wrote in message
> news:g80eqa$2oei$1...@geode.berkeley.edu...
>> I'm confused as to how to calculate the hit percentage for cache/TLB and
>> page table. (I'm looking at problem F2(c-e) in fall 06 final).
>> The solution only lists steps in fraction, doesn't tell me where the
>> numbers are from. Any help?
>
> c) Calculate hit % for cache
>
> Since this is a percentage, and both loops are doing the same thing in
> terms of memory access
> you can just look at one of them.
>
> Hit % for cache = hits / (total reads)
>
> reads = data / stride
> = 8K / 128 = 2^13 / 2^7 = 2^6
> = 64 reads
>
> The loop increments by 128 each iteration, which is in int's, so you
> multiply it by 4 to get bytes so it's compatible with the block size which
> is given in KiB:
>
> stride = 128*4
> = 2^7 * 2^2
> =2^9 Bytes
> = 512 Bytes
>
> Since the block size is 1 KiB, which is 1024 Bytes, and it was given that
> the data is block aligned, there will be one
> compulsory miss per block, and then one hit. Since there are 64
> iterations, each with 1 hit and 1 miss, you just count the hits, 32. So
>
> Hit % = 32 / 64 = 50%
>
>
> d) Calculate hit percentage for TLB
>
> Again, the loops are both processing in the same direction, so you can
> just look at one of them. So first find out how many reads there are per
> page:
>
> reads/page = page size / stride (128*4)
> = 2^12 / 2^9
> = 2^3
> = 8 reads / page
>
> So there are 8 reads per page, and one must be a compulsory miss, and 7
> will hit. So:
>
> TLB Hit % = 7/8
>
> e) Calculate the page hit % for the page table
>
> First, find the number of pages the data fits into:
>
> data = 8192 ints
> = 8KiB * 4
> = 32KiB
>
> The page size is 4 KiB, so
>
> #pages = 32 / 4
> = 8 pages of data
>
> And there is 1 MiB of physical address space, so
>
> # physical pages = 2^20 / 2^12
> = 2^8
> = 256 pages
>
> So the 8 pages fit in memory fine, every read will be a hit:
> 100%
>
>
> Sorry if I over explained this, I was half convincing myself at the same
> time :)
>
> Jeremy