Jeremy Cowles

2008-08-14 05:46:50 UTC

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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 cachepage 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?

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