GATE CSE 2023 | Question: 54

Question

An $8$-way set associative cache of size $64 \mathrm{~KB} \;(1 \mathrm{~KB}=1024\; \text{bytes})$ is used in a system with $32$-bit address. The address is sub-divided into $\text{TAG, INDEX},$ and $\text{BLOCK OFFSET.}$

The number of bits in the $\text{TAG}$ is ___________.

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Detailed Video Explanation:

8-way Set Associative Cache, Tag Bits

Answer 1

System is 32-bit address. 

Cache Size : 64 KB = $2^{6} * 2^{10} = 2^{16}$. So, cache bits = 16.

So Tag Bits = 32 – 16 = 16 bit.

 

NOTE : If cache is k-way set associative then we have to transfer $log _{2}k$ bits from cache line bits to Tag bits.

 

As, Cache is 8-way set associative. So, we have to transfer 3 bits to tag side.

So, final Tag bits = 16 + 3 = 19.

Number of bits in Tag Field = 19.

Comments

Why are we not counting Bytes?

Cache size = 64KB = 2^6 * 2^10 * 2^3 = 2^19 bits.

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@SurajPatil2909 Here I am taking those number of bits that I need to specify the address.

By default, we consider memory as Byte Addressable memory(1B of data stored in each address). 

Here size of cache is 64 KB, which can be written as 64K * 1B. Here 64K specifies the number

of locations. That's why 16-bits.

Answer 2

Given that for set associative mapping technique:

• Cache memory size $(CM’s)=64 KB=2^{16} $ bytes
• Main memory size $(MM’s)=32$ bits
• $P=8$way SAM
• Tag bit size (y) =?

since block size is not given we assume it is $2^x$ byte.

Number of cache block $(N)= \frac{CM’s}{B’s}\implies \frac{2^16}{2^x}=2^{16-x}$

Number of sets $(S) = \frac{N}{P}= \frac{2^{16-x}}{2^3}=2^{13-x}$

as we know that for set associative mapping technique:

$\text{TAG+SLO+B’S=MM’s}$ 

$\implies y+13-x+x=32$bits

$\implies y=32-13$ bits

$\implies y=19$ bits

so the correct value of the tag size is $19$ bits.

 

Answer 3

Answer: 19 bits

Let size of block be $2^x$

Let tag bits be $t$

$\text{Number of sets} = {\large \frac{64*2^{10}}{2^x *2^3}} = 2^{13-x}$

$t$

$13-x$

$x$

Given address = $32$ bits

Tag bits = $32 – (13-x+x) = 32-13 = 19$ bits

Answer 4

Given that for set associative mapping technique:

• Cache memory size $(CM’s)=64 KB=2^{16} $ bytes
• Main memory size $(MM’s)=32$ bits
• $P=8$way SAM
• Tag bit size (y) =?

since block size is not given we assume it is $2^x$ byte.

Number of cache block $(N)= \frac{CM’s}{B’s}\implies \frac{2^16}{2^x}=2^{16-x}$

Number of sets $(S) = \frac{N}{P}= \frac{2^{16-x}}{2^3}=2^{13-x}$

as we know that for set associative mapping technique:

$\text{TAG+SLO+B’S=MM’s}$ 

$\implies y+13-x+x=32$bits

$\implies y=32-13$ bits

$\implies y=19$ bits

so the correct value of the tag size is $19$ bits.

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https://gateoverflow.in/398912/gate-cse-2023-memory-based-question-29