cache memory

Question

what will be size of main memory. when 4-way set associative mapping  of cache memory is done and cache size is 256 KB and Tag field has 7 bits( consider, memory is byte addresable )

Comments

8 MB ??

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@joshi_nitish,

can you tell how it will be 8MB?

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let y and x bits for set and offset field respectively

there will be 2^y sets and every set will have 4 lines and every line has a weight of 2^x bytes

therfore size of cache = 2^y*4*2^x = 256*2¹⁰

                                   => x+y = 16

total bits to represent MM address = 7(tag) + y(set) + x(offset) = 7 + 16 = 23bits

therefore MM = 2²³ bytes = 8MB

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@joshi_nitish , can you tell that what will be the size of tag directory and block size?

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@amrendra pal

no, we can not calculate size of tag directory or block size because,

for that we need to know individual value of x and y, here we only know x+y

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@joshi_nitish, can you tell that how many multiplexers are used here and what will be the size of that multiplexers

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it will require 

4 levels of MUXs with each level containing seven (2^y $\times$ 1) MUX

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@joshi_nitish,

what is the means of level here ?

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@joshi_nitish, what about the OR gate here?tell

Answer 1

Just Find Total number of bits in tag when direct mapped cache.

7bit tag given in 4way set associative so in direct mapped cache it is of 7-2 =5bits tag.

So main memory is 2⁵ times bigger than cache memory.

So main memory size = 32×256KB =8MB

If you get logic you can do directly without conversion.

Answer 2

tag	set	block size
7	x	y

cache size=no.of set*lines per set*size of set

256 KB=4*2^y*2^x

2¹⁶  = 2^y*2^x

x+y=16

The number of bits in physical address=7+x+y=23

The main memort size =2²³ = 8 MB