How many total bits (total cache size) are required for a cache with 32KB of data, block size: 2 words, and 32-bit address?
How many total bits (total cache size) are required for a cache with 32KB of data,...
Please calculate the total number of bits needed for the cache with listed below for a direct-mapped cache. Size of Cache Data: 32KB Size of Cache Block: 2 Words
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]
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
Computer architecture How many SRAM bits are needed to implement an 8KB two-way set associative cache with 64B block size? Assume that each line (entry) has a single valid bit and no dirty bits. There is one bit per set for true LRU. Assume that the address size of the machine is 32-bits and that the machine allows for byte addressing.
a) Suppose we have a 64 KB, direct-mapped cache with 8-word blocks. Determine how many bits are required for the tag, index, and offset fields for a 32-bit memory address. b) If instead, we use a 64 KB, 4-way set-associative cache with 8-word blocks, how many bits will be required for the tag, index, and offset fields for a 32-bit address? c) What type of cache is shown in problem 2? How many bits are required for this cache’s tag,...
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
If you were to build the cache memory from problem 1 (provided below), what is the total size (in bytes) required for this cache memory? The total size includes data, tag, and valid bit chips. Hint: your solution might not be a power of 2. Total cache memory is 128k bytes Cache block size is 4 words (1 word = 4 bytes) CPU address window = 32 bits Cache memory chip size = 4k x 8-bits
question 2 and 3 2. Determine how many sets of cache blocks will be there for the following Cache memory size (in bytes) Direct Mapped Blocks Size (in bits) 32 64 218 2-way Set Associative Block Size (in bits) 32 64 A 2A6 [0.5 * 16 = 8] 4-way Set Associative Block Size (in bits) 32 64 SK 64K 256K 3. The physical memory address generated by a CPU is converted into cache memory addressing scheme using the following mapping...
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
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?