The physical address of memory on a machine is 32 bits. The machine has a direct...
The physical address of memory on a machine is 32 bits. The machine has a direct mapped cache of size 512 KB with a block size of 8 bytes. What is the size of the tag field in bits? Can you please explain this step by step and what the importance of direct mapped is?
Homework Answers
It has only one importance that it is easy to implement .
it has many disadvantages over other cache mapping techniques.
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The physical address of memory on a machine is 32 bits. The
machine has a direct...
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,...
Text: Explain how a 32-bit byte memory address should be divided into Tag/Index/Offset fields for each...
Text:
Explain how a 32-bit byte memory address should be divided into
Tag/Index/Offset fields for each of the cache configurations below.
Note: 1KB = 210 bytes. You must explain how many bits to assign to
each field and the ordering of the three fields. You get at most
50% of the credit if you give the length of each field without an
explanation.
1) A fully associative cache with cache block size = 2 words and
cache size = 512KB....
A primary memory system consists of 128 address bits. Each block within the cache is 256...
A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size....
Question 6 For the following figure shows a hypothetical memory hierarchy going from a virtual address to L2 cache acce...
Question 6 For the following figure shows a hypothetical memory hierarchy going from a virtual address to L2 cache access. The page size is 8KB, the TLB is direct mapped with 128 entries. The L1 cache is a direct mapped 8 KB, and the L2 cache is 2MB and direct mapped. Both use 64 byte blocks. The virtual address is 64 bits and the physical address is 41 bits. For each block in the figure below, fill in the number...
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]
This problem concerns a physical memory cache. Recall that m is the number of physical address...
This problem concerns a physical memory cache. Recall that m is the number of physical address bits, C is the cache size (number of bytes), B is the block size in bytes, E is the associativity, S is the number of cache sets, t is the number of tag bits, s is the number of set index bits, and b is the number of block offset bits. Suppose we have a cache with the following characteristics m = 32 C...
4. Assume that the machine has a 128kB cache with a 32-bit address. Show the address...
4. Assume that the machine has a 128kB cache with a 32-bit address. Show the address decomposition of a direct-mapped cache (assume 16 bytes per block). Tag Index Offset
question 2 and 3 2. Determine how many sets of cache blocks will be there for...
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...
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.
2. A computer uses a memory with addresses of 8 bits. (What's the size of the...
2. A computer uses a memory with addresses of 8 bits. (What's the size of the MM?) This computer has a 16-byte cache with 4 bytes per block. (How many blocks in the cache?) The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. a. What's the format of a memory address as seen by the cache ? Tag ? bits Block ? bits Offset ? bits b. The...
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,...
Text:
Explain how a 32-bit byte memory address should be divided into
Tag/Index/Offset fields for each of the cache configurations below.
Note: 1KB = 210 bytes. You must explain how many bits to assign to
each field and the ordering of the three fields. You get at most
50% of the credit if you give the length of each field without an
explanation.
1) A fully associative cache with cache block size = 2 words and
cache size = 512KB....
A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size....
Question 6 For the following figure shows a hypothetical memory hierarchy going from a virtual address to L2 cache access. The page size is 8KB, the TLB is direct mapped with 128 entries. The L1 cache is a direct mapped 8 KB, and the L2 cache is 2MB and direct mapped. Both use 64 byte blocks. The virtual address is 64 bits and the physical address is 41 bits. For each block in the figure below, fill in the number...
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...
2. A computer uses a memory with addresses of 8 bits. (What's the size of the MM?) This computer has a 16-byte cache with 4 bytes per block. (How many blocks in the cache?) The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. a. What's the format of a memory address as seen by the cache ? Tag ? bits Block ? bits Offset ? bits b. The...
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