Cache memory - Find Cache size ( given K, line size, tag memory, physical address space )

Question

Consider a physically tagged, word addressable, 16-way set associative cache with the line size of 128 Bytes. What is the size of cache if tag memory size is 2Kbits. Further assume that physical address space is 24 bits and word size is 4 Bytes.

a. 1 KB

b. 2 KB

c. 3 KB

d. 4 KB

Comments

@utkarsh bhai please tell me that if it is word addressable then why we are not also dividingg $2^{24}$ with $2^2$ also i am not getting that tag memory size is 2k bits so why we are not converting it in bytes and then we should divide it with $2^2$ please help me

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please tell me that if it is word addressable then why we are not also dividing $2^{24}$ with $2^2$

${24}$ bits are needed for addressing physical memory and memory is word addressable so ask yourself first why do you need to divide anything?

${24}$ bits are addressing ${2^{24}}$ addresses which are word addressable. 

 

tag memory size is 2k bits so why we are not converting it in bytes

ok let's try converting in into bytes

Tag memory = $2^{11}$ bits = $2^8$ B

Tag memory size = Tag bits  $×$ Number of blocks

So now i must convert tag memory size to bits otherwise i can't move forward because tag bits are bits

and there is no point of converting tag bits into bytes also its unknown.

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Also these options are all wrong so don't be misguided 

i reported it in august

Answer 1

• 16 - way set Associative 
• Line size = 128B 
• Size of tag memory = 2Kbits
• Physical Address space = 24 bits
• Word size = 4B

Words in a line = $\large \frac{Line \ size}{ Word \ size} = \frac{128 B}{4 B} = 32 \ words$

Bits needed for indentifying a word from a line = 5 bits

Let number of bits for set = s

Then tag bits = 24 - 5 - s = 19 -s

 

Number of blocks = Number of sets $\times$ number of blocks per set = $2^s \times 16$

Tag memory size = Tag bits  $\times$ Number of blocks

$\Large 2^{11} = \left ( 19-s \right ) \times 2^s \times 16$

$\Large \frac{2^7}{2^s} = 19 - s$

$\Large 2^{7 -s} = 19 - s$

$\large s = 3$

Tag bits = 19 - 3 = 16

Number of blocks = $\frac{Tag \ Memory \ Size}{ Tag \ Size}$ = $\frac{2^{11}}{2^4} = 2^7$

Size of cache = Number of blocks $ \times $ Block size   $= 2^7 \times 2^7   = 2^{14} = 16 KB$

 

Comments

thanks :)

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@Mk Utkarsh

why have u not converted physical address space in words?

each word is of size 4B

so 2^22 words are present each of size 4 byte

so PAS=22 bits??

please clear the doubt

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himgta 

assume that physical address space is 24 bits 

this means $2^{24}$ words are addressed in physical memory.

Now you want how many bytes then you can convert it but i don't think its necessary in this question