Consider a two-level memory system (i.e., cache and main memory). Assume the cache access time is 3 clock cycles, and the hit rate is 95%. What should be the main memory access time (penalty) to ensure the memory access efficiency stays at 0.90
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Consider a two-level memory system (i.e., cache and main memory). Assume the cache access time is 3 clock cycles, and the hit rate is 95%. What should be the main memory access time (penalty) to ensure the memory access efficiency stays at 0.90?
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- Consider a memory system with a cache access time of 100ns and a memory access timeof 1200ns. If the effective access time is 10% greater than the cache access time, what is thehit ratio H?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?Suppose the cache access time is 1 ns, main memory access time is 100 ns and the cache hit rate is 98%. Assuming memory access is initiated with cash access, what is the effective memory access time (round up to 2 digits after the decimal point)
- You are asked to perform capacity planning for a two-level memory system. The first level, M1 is a cache with three capacity choices of 64Kbytes, 128Kbytes, and 256Kbytes. The second level, M2, is a main memory with a 4Mbyte capacity. Let c1 and c2 be the costs per byte and t1 and t2 be the access times for M1 and M2, respectively. Assume c1 = 20 * c2 and t2 = 10 * t1. The cache hit ratios for the 3 capacities are assumed to be 0.7, 0.9 and 0.95, respectively. What is the average memory access time, AMAT, if t1 = 20ns in the three cache designs? (Note that t1 is the time from M1 to CPU and t2 is that from M2 to CPU. This means that t2 includes t1).Consider the following situation: we have a byte-addressable computer with 2-way set associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. Based on the fact that each block has 8 bytes, figure out how big the offset field should be, and then show your work.Suppose a 1 MiB L1 cache has an access time of 1 clock cycle, the 16 MiB L2 has an access time of 4 clock cycles, and the 8 GiB DRAM has an average access time of 40 clock cycles. 80% of memory reads are satisfied by the L1 cache. Only 10% have to go to main memory. (a) What is the AMAT following a miss on L1? (b) What is the AMAT for memory overall?
- Suppose a byte-addressable computer using set associative cache has 2^24 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?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 cacheConsider a system with 2 level cache. Access times of Level 1 cache, Level 2 cache and main memory are 1 ns, 10 ns, and 500 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache?
- Consider a main memory with size 128 Mbytes with cache size 64 Kbytes and memory block is 4 bytes. Determine how to split the address (s-d, d, w) for set associative mapping. Assume each cache set is 2 lines of cache.If we have to design a 4 - way set - associative cache of 8 MB size that could work for a main memory of size 4 GB , determine the following : 1. The total number of cache locations 2. The size of tag Consider the cache block size as 4 bytes . Note:- Here he want the number of cache locations And, the size of tag show all the steps.Q. Consider a system with 4-way set associative cache of 256 KB. The cache line size is 8 words (32 bits per word). The smallest addressable unit is a byte, and memory addresses are 64 bits long. (a) How many bits are used for TAG and INDEX fields of cache mapping?