A four-way set-associative cache has lines of 32 bytes and a total size of 4 kB. The 32-MB main memory is byte addressable. Show the format of main memory addresses.
Number of block at the cache is = 4KB /32 BYTES= 4 * 1024 / 32 = 128 BLOCK
SET LINE = 128 /4 = 32 SET OF 4 LINE EACH
32 = 2^5 = 5 BIT FOR SET
MAIN MEMEORY = 32MB = 2^25 = 25 BIT
NUMBER OF BLOCK AT MEMORY = 32 MB / 32 BYTES= 2^20 = 20 BITS
TAG FILED= 20- 5=15
BLOCK SIZE = 32 BYTE= 5 BIT FOR WORD
SO MEMEORY FORMAT
TAG -> 15 BITS
SET -> 5
WORD(W)->5
A four-way set-associative cache has lines of 32 bytes and a total size of 4 kB....
A two-way set-associative cache has lines of 4 bytes and a total size of 4 kB. The 32-MB main memory is byte addressable. Show the format of main memory addresses.
Q2. Consider a four-way set associative cache with a data size of 64 KB. The CPU generates a 32-bit byte addressable memory address. Each memory word contains 4 bytes. The block size is 16 bytes. Show the logical partitioning of the memory address into byte offset, cache index, and tag components.
3. (12 points) Consider a cache has lines of 16 bytes and a total size of 16 kB. The main memory is 16MB and a word takes 4 bytes. For the hexadecimal main memory addresses FFF666, show the following information in hexadecimal format a. Tag and word values for associative cache b. Tag, set and word values for a two-way set-associative cache 3. (12 points) Consider a cache has lines of 16 bytes and a total size of 16 kB....
Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of cache, and a block size of 16 bytes. For each configuration below, determine the memory address format, indicating the number of bits needed for each appropriate field (i.e. tag, block, set, offset). Show any relevant calculations. Direct cache mapping and memory is byte-addressable a) Direct cache mapping and memory is word-addressable with a word size of 16 bits b) c) 2-way set associative cache...
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A 256kiB (2^18 bytes) cache has a block size of 32 bytes and is 32-way set-associative. How many bits of a 32-bit address will be in the Tag, Index, and Bock Offset?
Suppose a computer using a fully associative cache has 232 bytes of byte-addressable main memory and a cache of 1024 blocks, were each cache block contains 32 bytes. Consider a memory address as seen by the cache. How many bits are in the tag field?
Consider a 2-way set associative cache consisting of 8 blocks total of byte-addressable memory with 4 bytes per block. Assume that the cache is initially empty. Given the following address sequence, fill in the table below. Time Access Tag Set Offset 3 10010001 11001001 10110110 10101011 10110010
32 bytes of memory. 16 bytes of 2-way set-associative cache, where blocks can go anywhere within the set. Block is 2 bytes, set in cache is two blocks. Populate memory starting with upper-case letters, then 0-5. Hint- with full associativity in the set: each block has its own set of Tag bits in the cache. Memory is not organized by sets, though blocks get assigned to sets, and load in the cache per set. 1) Break down the addressing: Tag...