A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.a. What is the number of sets in the cache?b. What is the size (in bits) of the tag field per cache block?c. What is the number and size of comparators required for tag matching?d. How many address bits are required to find the byte offset within a cache block?e. What is the total amount of extra memory (in bytes) required for the tag bits?Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.
a. What is the number of sets in the cache?
b. What is the size (in bits) of the tag field per cache block?
c. What is the number and size of comparators required for tag matching?
d. How many address bits are required to find the byte offset within a cache block?
e. What is the total amount of extra memory (in bytes) required for the tag bits?

Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer?

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A CPU has 32-bit memory address and a 256 KB cache memory. The cache i...

What is the number of sets in the cache?
Number of sets = Cache memory/(set associativity * cache block size)
= 256KB/(4*16 B)
= 4096
What is the size (in bits) of the tag field per cache block?
Memory address size = 32-bit
Number of bits required to identify a particular set = 12 (Number of sets = 4096)
Number of bits required to identify a paticular location in cache line = 4 (cache block size = 16)
size of tag field = 32 -12 -4 = 16-bit
What is the number and size of comparators required for tag matching?

We use 4-way set associate cache. So, we need 4 comparators each of size 16 bits

How many address bits are required to find the byte offset within a cache block?
Cache block size is 16 byte. so 4 bits are required to find the byte offset within a cache block.
What is the total amount of extra memory (in bytes) required for the tag bits?
size of tag = 16 bits
Number of sets = 4096
Set associativity = 4
Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 2¹⁸bytes

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A CPU has 32-bit memory address and a 256 KB cache memory. The cache i...

Given:

32-bit memory address

256 KB cache memory

4-way set associative cache

Cache block size of 16 bytes

a. Number of sets in the cache:

Number of cache blocks = Cache size / Cache block size = 256 KB / 16 bytes = 16,384 blocks

Number of sets = Number of blocks / Associativity = 16,384 / 4 = 4,096 sets

b. Size of tag field per cache block:

Size of cache block = 16 bytes = 2^4

Size of offset bits = log2(Size of cache block) = 4 bits

Size of index bits = log2(Number of sets) = log2(4,096) = 12 bits

Size of tag bits = Address bits - Offset bits - Index bits = 32 - 4 - 12 = 16 bits

c. Number and size of comparators required:

Number of comparators required = Number of cache blocks = 16,384

Size of each comparator = Size of tag bits = 16 bits

d. Address bits required for byte offset within a cache block:

Size of cache block = 16 bytes = 2^4

Address bits required for byte offset = log2(Size of cache block) = 4 bits

e. Total amount of extra memory required for tag bits:

Size of tag bits per cache block = 16 bits

Number of cache blocks = 16,384

Total size of tag bits = Size of tag bits per cache block * Number of cache blocks = 16 * 16,384 = 262,144 bits

Extra memory required to store the tag bits = Total size of tag bits / 8 = 32,768 bytes = 218 KB

Therefore, the correct answer is 'Extra memory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'.

Answered by Gate Gurus · View profile

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A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.a. What is the number of sets in the cache?b. What is the size (in bits) of the tag field per cache block?c. What is the number and size of comparators required for tag matching?d. How many address bits are required to find the byte offset within a cache block?e. What is the total amount of extra memory (in bytes) required for the tag bits?Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer?

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A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.a. What is the number of sets in the cache?b. What is the size (in bits) of the tag field per cache block?c. What is the number and size of comparators required for tag matching?d. How many address bits are required to find the byte offset within a cache block?e. What is the total amount of extra memory (in bytes) required for the tag bits?Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.a. What is the number of sets in the cache?b. What is the size (in bits) of the tag field per cache block?c. What is the number and size of comparators required for tag matching?d. How many address bits are required to find the byte offset within a cache block?e. What is the total amount of extra memory (in bytes) required for the tag bits?Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes.a. What is the number of sets in the cache?b. What is the size (in bits) of the tag field per cache block?c. What is the number and size of comparators required for tag matching?d. How many address bits are required to find the byte offset within a cache block?e. What is the total amount of extra memory (in bytes) required for the tag bits?Correct answer is 'Extramemory required to store the tag bits = 16 * 4096 * 4 bits = 218 bytes'. Can you explain this answer?.

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