Block Size = 32 byte => 2^5
Cache size = 64KB
Number of words = 2^16 / 2^5 => 2^11
Offset = 5
Index = 11
Tag = 32 - 5 - 11 => 16
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Question 17 12 points A direct mapped cache holds 64KB of useful data (not including tag...
12 points A direct mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block streis 32-byte and the address is 32-bit, find the number of bits needed for tag index and byte select fields of the address Number of bits for offset bits Number of bits for index bits Number of bits for tag Moving to another question will save this response. Question 17 of 18
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
Question 17 A direct-mapped cache holds 128KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag .. bits
Question 17 12 points Save Answer A direct-mapped cache holds 32KB of useful data (not including tag or control bits). Assuming that the block size is 16-byte and the address is 32-bit, find the number of bits needed for tag, index, and byte select fields of the address. Number of bits for offset bits Number of bits for index bits Number of bits for tag bits
For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram. Indicate the block size, number of blocks, and address field decomposition (block offset, index, and tag bit width) assuming a 32-bit memory address.
Design a 256KB (note the B) direct‐mapped data cache that uses a 32‐bit address and 8 words per block. Calculate the following: How many bits are used for the byte offset and why? How many bits are used for the set (index) field? How many bits are used for the tag? What’s the overhead for that cache?
Cache question computer architecture A cache holds 128 words where each word is 4 bytes. Assuming a 32-bit address, for each of the following organizations, complete the table. a.A direct-mapped cache with block size = 16words b.2-way set-associative cache with block size = 8words c.4-way set-associative cache with block size = 4words d.A fully associative cache with block size = 2words. Cache a Cache b Cache c Cache d total # bits for word & byte displacement # bits in...
For a direct-mapped cache with a 32-bit address and 32-bit words, the following address bits are used to access the cache. TAG INDEX OFFSET 31-15 14-8 7-0 a. What is the cache block size (in words)? [13 points] b. How many blocks does the cache have? [12 points]
The following figure shows the address that is going from a certain processor to a direct-mapped cache. The address is divided into fields. The index of the first bit and the last bit of each field is written below it. Calculate the size of the data that is stored in the cache, in Kibytes, and the total number of bits within the cache, in Kibits. TAG = 31 - 16 INDEX = 15 - 6 BLOCK OFFSET = 5 -...
1. A cache holds 64 words where each word is 4 bytes. Assume a 32 bit address. There are four different caches a. A direct-mapped cache with block size = 16 words b. 2-way set-associative cache with block size = 8 words c. 4-way set-associative cache with block size=4 words d. A fully associative cache with block size = 16 words. Complete the table for each cache. Cache a Cache be Cache Cache de 16 Number of bits needed for...