Please answer all parts correctly and show your work
Starting from power on, the following byte addressed cache reference are recorded. Address 0 16 132 232 160 1024 30 140 3100180 2180 d. How many blocks are replaced e. What is the hit ratio f. List final state of the cache, with each valid entry represented as a record of
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3- for a direct mapped cache design with a 32 bit address, the following bits of address are used...
For a direct-mapped cache with a 32-bit address and 32-bit words, the following address bits are...
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]
For a direct-mapped cache design with a 32-bit address, the following bits of the address are...
For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. Tag Index Offset 31-10 9-4 3-0 How many entries does the cache have?
Design a 256KB (note the B) direct‐mapped data cache that uses a 32‐bit address and 8...
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?
Question 3: Consider a 32-bit physical address memory system with block size 16 bytes and a...
Question 3: Consider a 32-bit physical address memory system with block size 16 bytes and a 32 blocks direct mapped cache. The cache is initially empty. The following decimal memory addresses are referenced 1020, 1006, 1022, 5106, 994, and 2019 Map the addresses to cache blocks and indicate whether hit or miss. Note: You must use the hexadecimal approach in solving this question. You must also show the computations of dividing the memory address into tag bits, cache index bits,...
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that...
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)....
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
a) Suppose we have a 64 KB, direct-mapped cache with 8-word blocks. Determine how many bits...
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,...
For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram....
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.
*Need answer for 5.3.6, information is in 5.3* 5.3 For a direct-mapped cache design with a...
*Need answer for 5.3.6, information is in 5.3*
5.3 For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. Tag Index Offset 31-10 9-5 4-0
The following figure shows the address that is going from a certain processor to a direct-mapped...
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 -...
Question 3: Consider a 32-bit physical address memory system with block size 16 bytes and a 32 blocks direct mapped cache. The cache is initially empty. The following decimal memory addresses are referenced 1020, 1006, 1022, 5106, 994, and 2019 Map the addresses to cache blocks and indicate whether hit or miss. Note: You must use the hexadecimal approach in solving this question. You must also show the computations of dividing the memory address into tag bits, cache index bits,...
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
*Need answer for 5.3.6, information is in 5.3*
5.3 For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. Tag Index Offset 31-10 9-5 4-0
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 -...
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