Given that the main memory access time is 1200 ns and cache access time is 100 ns. The average memory access time is not to exceed 120 ns. What is the hit ratio for the cache must be at least?
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- 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?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 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)
- Given 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: (inSuppose that in 1000 memory references there are 50 misses in the first-level cache, 20 misses in the second-level cache, and 5 misses in the third-level cache. Assume the miss penalty from the L3 cache to memory is 100 clock cycle, the hit time of the L3 cache is 10 clocks, the hit time of the L2 cache is 4 clocks, the hit time of L1 is 1 clock cycle. What is the average memory access time?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?
- In a Direct Mapped Cache Memory Physical Address format the Cache line offset field size and word offset field size are same (with word size of one Byte). The number of tag bits in the Physical Address format is equal to the number of blocks in Cache Memory. If the Tag field Size is Mega words. 16 bits, the size of the physical Memory isThe memory access time is I nanosecond for read operation with a hit in cache. * nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 anoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (in nanoseconds) in executing the sequence of instructions isSuppose 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?
- CO. A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes. Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the Tag and the Index ticlds respectively in the addresses generated by the processor?The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (in nanoseconds) in executing the sequence of instructions will be ?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?