GATE Computer Science (CS) 2020 Solved Papers

Question : 61 of 65

Marks: +1, -0

A computer system with a word length of 32 bits has a 16 MB byte-addressable main memory and a 64 KB, 4-way set associative cache memory with a block size of 256 bytes. Consider the following four physical addresses represent hexadecimal notation.

{A} 1=0 × 42 {C} 8 {A} 4

{A} 2=0 × 546888

{A} 3=0 × 6 {A} 289 {C}

{A} 4=0 × 5 {E} 4880

Which one of the following is TRUE?

A1 and $A 3$ are mapped to the same cache set.
${A} 3$ and ${A} 4$ are mapped to the same cache set.
A 2 and $A 3$ are mapped to the same cache set.
A1 and $A 4$ are mapped to different cache sets.

Validate

Solution:

1 word = 4 bytes
Main memory = 16 MB
Bytes – Addressable
Cache size = 64 KB
4-way SAM Block size = 256 bytes

A_{1}=(42 C 8 A 4)_{i 6}

{A}_{2}=(546888)_{16}

{A}_{3}=(6 {A} 289 {C})_{16}

{A}_{4}=(5 {E} 4880)_{16}

Which two map to the same cache set?

TAG

SET

WORD

SAM:
Address length

=\;{\;\text " Main memory size "\;}/{\;\text " Byte "\;}=16 × 2^{20}

Bytes

=2^{24}

Bytes

=24

bits.
Word

\Rightarrow

Block size

=256

bytes

⇒ 8

bits for word
No. of sets

=\;{\;\text " Total Block "\;}/{\;\text " Blokes Per set "\;}=\;{2^{8}}/{2^{2}}=2^{6}

sets
No. of block in cache

=\;{\;\text " cache size "\;}/{\;\text " Blcok size "\;}=\;{64 {KB}}/{256 {B}}=\;{2^{16}}/{2^{8}}=2^{8}

So, SET field 6 bits
Remaining bits for

{TAG}=24-(8+6)=10

bits

Hence,

A_{1}

and

A_{4}

maps to same set

{A}_{2}

and

{A}_{3}

maps to same set

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