Assume a multi-level cache. The L1 cache has a miss-rate of 5%. There is miss penalty of 30 cycles for reaching e L2 cache. The L2 cache has a miss-rate of 3%. There is a 200 cycle miss penalty for reaching main memory. hat is the average miss penalty for this cache?

Keep in mind the difference between a fully associative cache and a directly mapped cache.

Cache Performance Please answer the following questions regarding average memory access time (AMAT). a. If a specific cache design has 2% miss rate for a set of benchmarks with a hit latency of 20 cycles and a miss penalty of 200 cycles. What is the AMAT for this cache? b. If another cache design has 5% miss rate for the same set of benchmarks with a hit latency of 13 cycles and a miss penalty of 200 cycles. What is the AMAT for this cache? c. Which cache design is better?

In a split L1 cache organization the I-cache size is 16-KB and the D-cache is 32-KB; the access time for both caches is 1 cycle. The miss rates are given in the table below. There is a 256KB second level cache (L2) that is a unified cache and has a miss rate given in the table. Please assume that miss penalty is 20 cycles to get data from L2 cache. There is a 4MB third level cache (L3) that has a latency of 42 cycles; its miss rate is 1.28%. 124 cycles are required to get data from main memory. A program executes 8.6×10° instructions; 32.8% of the executed instructions are data references (i.e. loads or stores). AMAT = Hit Time + Miss Rate x Miss Penalty Cache miss rates a) Please determine the AMAT for this memory system. [7 points] Cache Size L1 I-Cache L1 D-Cache L2 Unified Cache 8 KB 0.82% 12.21% 8.63% 16 KB 0.38% 11.36% 6.75% L1: I D 32 KB 0.14% 10.67% 5.18% 64 KB 0.06% 10.15% 4.89% L2: U_cache 128 KB 0.03% 9.81% 3.66% 256 KB 0.002% 9.06% 2.76% L3: U_cache M: Main

Asap please

Consider a 64K L2 memory and a 4K L1 2-way associative cache with block sizes of 512. a. How many blocks are in each set in L1? b. How many offset bits are in the L2 address? c. How many index bits are in the L2 address? d. How many tag bits are in the L2 address?

Explain the differences between a directly mapped cache and a fully associative cache.

Can I get some help.plz

Consider a cache with 32KİB data, 16-word blocks, and 24-bit addresses, answer the following questions: a) For the direct-mapped configuration, determine the number of index bits and tag bits in the 24-bit address. b) For 4-way set associative configuration, determine the number of index bits and tag bits in the 24-bit address. c) For fully associative contiguration, determine the number of index bits and tag bits in the 24-bit address. d) For an 8-way set associative configuration, identify the set number in the cache to which the following 24-bit memory address maps: Ox001Z00 (Hexadecimal notation) where Z is the least significant digit in your student ID (written as a decimal number)

Consider a 32-bit physical memory space and a 32 KiB 2-way associative cache with LRU replacement. You are told the cache uses 5 bits for the offset field. Write in the number of bits in the tag and index fields. Tag length in bits = Index length in bits =