A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses direct mapping with a block size of 4 words. i) How many bits are there in the tag, index, block & words fields of the address formats. ii) How many bits are there in each word of cache?iii) How many blocks can the cache accommodate?
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A: Here, I have to provide a solution to the above question.
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A: I have answered this question in step 2.
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A: Given question has asked to find the Tag field bits value. Information provided in question are as…
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Q: A digital computer has a memory unit of 64K X 16 and a cache memory of 1K words. The cache uses…
A: Given : A digital computer has a memory unit of 64K X 16 and a cachememory of 1K words. The cache…
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A: The answer is as follows.
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A: A) Total number of blocks in main memory = 2^22/2^4 = 2^18 B) Size of offset field =log(block size)…
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A: Since it is given that set index bits are 9, we have Total sets = 29 = 512
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Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…
A: Note - As per the guidelines, we are only allowed to answer 1 question with 3 sub-parts a time.…
Q: Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits.…
A: Given the cache's capacity of 128K bits. Cache capacity in bytes = 128K bits / 8 = 16KB Because this…
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Q: Assume we have a computer with 512 blocks of cache memory with a total capacity of 128K bits.…
A: Introduction: Assume we have a computer with 512 blocks of cache memory with a total capacity of…
Q: Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses…
A: i) Answer: Tag = 6 bitsIndex = 10 bitsBlock = 8 bitsWord = 2 bits Explanation: Given that:…
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Q: Suppose a computer using direct mapped cache has 232 words of main memory, and a cache of 1024…
A: Given: Suppose a computer using a direct-mapped cache has 232 words of main memory and a cache of…
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A: Explanation: Cache has 224 words main memory with 24 bits Each cache contains 16 words =24 words…
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A: According to the Question below the Solution:
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A: Note: In the BNED Guidance, only the first question can be answered at a time. Resend the question…
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A: Size of Cache block =8 byts No of bits for Block offset = log2(8)=3 bits No of cache lines are 32K…
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A: solution:
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Q: . suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory and…
A: Actually, cache is a fast access memory. Which located in between cpu and secondary memory.
Q: Consider a 64K L2 memory and a 4K L1 direct mapped cache with block sizes of 512 values. a. How…
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A: Given, size of L2 = 64K and size of L1 = 4K associativity = 4 - way and , block size = 512
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A: Ans.1: 16 bytesEach block has 16 bytes. Calculate the number of bits in the TAG, SET, and OFFSET…
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Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses…
A: Answer: ANSWER:-IN THE QUESTION IT IS GIVEN THAT DIGITAL COMPUTER HAS A MEMORY UNIT OF 64K X 16 AND…
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A: A) Block size = 32 words = 32*4 = 128B So block offset bits = 7 bits Total number of set inside…
Q: How many blocks of main memory are there? What is the format of a memory address as seen by the…
A: 1. Number of blocks of main memory: cache sizeblock size = 22424 = 220 2. There are 24-bit…
A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The
cache uses direct mapping with a block size of 4 words.
i) How many bits are there in the tag, index, block & words fields of the address formats.
ii) How many bits are there in each word of cache?iii) How many blocks can the cache accommodate?
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- A digital computer has a memory unit of 64K X 16 and a cache memory of 1K words. The cache uses direct mapping with a block size of four words.i) How many bits are there in the tag, index, block and word fields of the address format?ii) How many bits are there in each word of cache, and how are they divided into functions? Include a valid bit.iii) How many blocks can the cache accommodate?Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses direct mapping with a block size of 4 words. i) How many bits are there in the tag, index, block & words fields of the address formats. ii) How many bits are there in each word of cache? iii) How many blocks can the cache accommodate?A computer of 32 bits has a cache memory of 64 KB with a cache line size of 64 bytes. The cache access time is 20 ns, and the miss penalty is 120 ns. The cache is 2-way associative. a) What is the number of cache lines? b) What is the number of cache sets? c) What is the number of lines per set? d) Draw a scheme of this cache. e) Calculate the time to read a word in case of miss.
- Suppose a computer using direct-mapped cache has 2 bytes of byte-addressable main memory and a cache of 32 blocks, where each cache block contains 16 bytes.Q.) What is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, block, and offset fields?Q: A digital computer has a memory unit of 64k * 16 and a cache memory of 1k words. The cache uses direct mapping with a block size of 4 words. i) How many bits are there in the tag, index, block & words fields of the address formats.ii) How many bits are there in each word of cache? iii) How many blocks can the cache accommodate? Note: this question is related from computer architecture subject kindly solved this correctly and completly.Suppose a computer using fully associative cache has 216 bytes of byte-addressable main memory and a cache of 64 blocks, where each cache block contains 32 bytes.Q.) What is the format of a memory address as seen by the cache; that is, what are the sizes of the tag and offset fields?
- Suppose a computer using fully associative cache has 220220 words of main memory and a cache of 128 blocks, where each cache block contains 16 words. (a) How many blocks of main memory are there? (b) What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag and word fields? (c) To which cache block will the memory reference 01D872_{16}01D872_{16} map?Suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory and a cache of 128 blocks, where each block contains 64 bytes.Q.) What is the format of a memory address as seen by cache; that is, what are the sizes of the tag and offset fields?Suppose a computer using direct-mapped cache has 232 bytes of byte-addressable main memory and a cache size of 512 bytes, and each cache block contains 64 bytes.Q.) What is the format of a memory address as seen by cache; that is, what are the sizes of the tag, block, and offset fields?
- Suppose a computer using fully associative cache has 224 words of main memory and a cache of 512 blocks, where each cache block contains 16 words. How many blocks of main memory are there? What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? To which cache block will the memory reference 17042416 map?Suppose a computer using fully associative cache has 4G bytes of byte-addressable main memory and a cache of 512 blocks, where each cache block contains 128 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? c) To which cache block will the memory address 0x018072 map?Computer Science Consider a direct-mapped cache with 8 lines, each holding 16 bytes of data. The cache is byte-addressable and the main memory consists of 64 KB, which is also byte-addressable. Assume that a program reads 16KB of memory sequentially. Answer the following questions:a) How many bits are required for the tag, index, and offset fields of a cache address?b) What is the cache size in bytes?c) What is the block size in bytes?d) What is the total number of blocks in main memory?e) How many cache hits and misses will occur for the program, assuming that the cache is initially empty?f) What is the hit ratio?g) Give an example virtual address (in BINARY) that will be placed in cache line 5.