Consider a two-tier memory system consisting of cache (SRAM) and main memory (DRAM). The cache access time is 1 nsec and the main memory access time is 50 nsecs. (1 nsec = 1 ́ 10-9 secs).
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Consider a two-tier memory system consisting of cache (SRAM) and main memory (DRAM).
The cache access time is 1 nsec and the main memory access time is 50 nsecs. (1 nsec = 1 ́
10-9 secs).
(a) What is the overall memory access time given a cache hit rate of 95%?
(b) What will the cache hit rate need to be if the overall memory access time in (a) is to be
halved?
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- A 32-bit computer has a memory of 256 KB and a cache line size of 64 bytes. The memory cache access time is 5ns. This cache is 4-way associative and use LRU as a replacement algorithm. a) What is the number of lines and sets of this memory cache? b) What is the block size transferred between the cache memory and the main memory? c) If the time to transfer a line to cache memory is 200 ns, what is the hit ratio needed to obtain an average access time of 20 ns?Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache?Suppose a byte-addressable computer using set associative cache has 224 bytes of main memory and a cache size of 64K bytes, and each cache block contains 32 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields? b) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache?
- If memory read cycle takes 100 ns and a cache read cycle takes 20 ns, then for four continuous references, the first one brings the main memory contents to cache and the next three from cache. Find the time taken for the Read cycle with and without Cache? What is the Percentage speedup obtained?Suppose a byte-addressable computer using set associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields? b) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache?Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, set, and offset fields?
- A CPU has 32-bit memory address and a 256 KB cache memory. The cache is organized as a 4-way set associative cache with cache block size of 16 bytes. a. What is the number of sets in the cache? b. What is the size (in bits) of the tag field per cache block? c. What is the number and size of comparators required for tag matching? d. How many address bits are required to find the byte offset within a cache block? e. What is the total amount of extra memory (in bytes) required for the tag bits?Suppose a byte-addressable computer using set associative cache has 8M byes of main memory and a cache of 128 blocks, where each cache block contains 64 bytes. a) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields? b) If this cache is 16-way set associative, what is the format of a memory address as seen by the cacheGiven that a 4-way set associative cache memory has 64 KB data and each block contains 32 bytes. The main memory capacity is 4 GB. a. Find the number of bits for the main memory address. ANSWER: b. How many blocks are there in a set? ANSWER: c. How many sets the cache has? ANSWER: d. The main memory address format is => | Tag: bits | blocks sets bits | Set: e. Which set will be mapped by the main memory address 458195h. ANSWER: decimal) bits bits | Word: (in
- In a two-level cache system, the access time of cache L₁ is 2 cycle and the access time of cache L2 is 7 cycle. The miss rate of L₁ is thrice the miss rate of L2. the miss penalty from the L2 cache to main memory is 20 clock cycles. The average memory access time of the system is 4 cycle. The hit rate of L2 is (correct up to 2 decimal places).Suppose a byte-addressable computer using set associative cache has 4Mbyes of main memory and a cache of 64 blocks, where each cache block contains 8 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, set, and offset fields? b) If this cache is 4-way set associative, what is the format of a memory address as seen by the cache? Show all work and explain how you got the answers please. ThanksA cache memory has a line size of eight 64-bit words and a capacity of 4K words. The main memory size that is cacheable is 1024 Mbits. Assuming 4-way set associative mapping and that the addressing is done at the byte level. What is the format of the main memory addresses (i.e s-d, d, and w)? For the hexadecimal main memory location 2BFACEDH, find the corresponding 4-way set-associative memory format