GATE CSE 2015 Set 3 | Question: 14

Question

Consider a machine with a byte addressable main memory of $2^{20}$ bytes, block size of $16$ bytes and a direct mapped cache having $2^{12}$ cache lines. Let the addresses of two consecutive bytes in main memory be $\textsf{(E201F)}_{16}$ and $\textsf{(E2020)}_{16}$. What are the tag and cache line addresses ( in hex) for main memory address $\textsf{(E201F)}_{16}$?

• A. $\textsf{E, 201}$
• B. $\textsf{F, 201}$
• C. $\textsf{E, E20}$
• D. $\textsf{2, 01F}$

Comments

repetitive concept

GATE 2020 also

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this is a beautiful question

Answer 1

Block size of $16$ bytes means we need $4$ offset bits. (The lowest $4$ digits of memory address are offset bits)

Number of sets in cache (cache lines) $= 2^{12}$ so the next lower $12$ bits are used for set indexing. 

The top $4\;\text{bits}\; ($out of $20)$ are tag bits.

So, the answer is A. 

Comments

If you convert the given hexadecimal numbers into binary you will come to know that these are two adjacent bytes in main memeory .

E201F- 1110 0010 0000 0001 1111

E2020- 1110 0010 0000 0010 0000

I think it is just given as a representation and doesnt have much significance in solving this question .

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@ Abhishek

for this question for block or for cache line u have to consider 12 bits while u consider only 4 bit

As MM size given 2^20 B so , to represent physical address we need 20 bits.

Here block size given 16 B so to represent block offset we need 4 bit which will be 4 LSB of PA

Here cache line given 2^12 so we need 12 bits to represent line number

Since 4 bit for block offset and 12 bit for line number so 4 MSB will represent tag (as tag + line + offset =20)

So Tag = 4 MSB =E

cache line =12 bit after tag= 201

 to solve this question we are not getting any information from (E201F) and (E2020) and we not required too, to solve this question

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@Upasana singh @Swati Rauniyar 

If some information is given in question its not that we have to use it to get the answer. Many a times such information are given just to confuse the aspirants.

Answer 2

Let the addresses of two consecutive bytes in main memory be $(E201F)_{16}$ and $(E2020)_{16}$

$E20$ is common to both hex numbers, so ignore it.

$1F=0001 \ 1111=31 \ (decimal)$

$20=0010 \ 0000=32 \ (decimal)$

 

So, difference between two consecutive memory locations = $1 \ B$ Hence, the memory is byte addressable.

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First $4$ bits would be tag bits, and next $12$ bits would be index bits.

Tag	Line	Offset
4	12	4

So, for $(E201F)_{16}$

Tag = $E$; Line = $201$

Answer 3

Block Size = 16 bytes
Block Offset = 4 


No. of sets or cache lines = 2

¹² Number of index bits = 12 Size of main memory = 2²⁰ Number of tag bits = 20 - 12 - 4 = 4 Let us consider the hex address E201F Tag lines = First 4 bits = E (in hex) Cache lines = Next 12 bits = 201 (In Hex)

Answer 4

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