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- A cache is set up with a block size of 32 words. There are 64 blocks in cache and set up to be 4-way set associative. You have byte address 0x8923. Show the word address, block address, tag, and index Show each access being filled in with a note of hit or miss. You are given word address and the access are: 0xff, 0x08, 0x22, 0x00, 0x39, 0xF3, 0x07, 0xc0.A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag isA cache is set up with a block size of 128 words. There are 32 blocks in cache and set up to be 4-way set associative. You have word address 0x3f42. Show the word address, block address, tag, and set.
- 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 isA cache memory system with capacity of N words and block size of B words is to be designed. If it is designed as a direct mapped cache, the length of the TAG field is 14 bits. If it is designed as a 4-way set associative cache, the length of the TAG field will be ………… bits.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. What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields?
- A computer has a 256 KB, K-way set associative write-back data cache with block size of 32 B. The address sent to the cache controller by the processor is of 32 bits. In addition to the address tag, each cache tag directory contains 2 valid bits and 1 modified bit. If 16 bits are used to address tag. What is the minimum value of K?Cache memory systems are designed such that the computer first checks the L1 cache for the desired memory. If the data is there, it accesses it and is done. It only checks the L2 cache if the data is not found in the L1 cache. Likewise, the computer checks the L3 cache if the desired data is not found in the L2 cache, and finally it on only checks main memory if the data is not found in the L3 cache. Imagine a computer system with the following cache access times: L1 cache: 3 processor cycles L2 cache: 10 processor cycles L3 cache: 25 processor cycles Main memory: 100 processor cycles So, in the best case, the desired data would be immediately found in the L1 cache, which requires only 3 cycles to check. Conversely, in the worst case, the desired data would only be in main memory, which would require 138 cycles to access (3 cycles to check L1 cache + 10 cycles to check L2 + 25 cycles to check L3 + 100 cycles to access main memory). What is important, however, is the average…A message slot is a cache line that contains status flags and the message itself. The state can be empty or ready, and communication occurs through a single cache line. Both the sender and the receiver write the message to the cache line. The receiver updates the state from empty to ready, then polls the cache line in a loop waiting for the sender to send a message. The sender waits in a tight loop until the receiver acknowledges the cache line, changes the state from ready to empty, and acknowledges receipt. Explain that the system uses a snooping cache coherence protocol and describe the cache-coherence transaction taking place on the coherence bus when a message transfer occurs.
- Suppose a computer using fully associative cache has 224 words of main memory and a cache of 128 blocks, where each cache block contains 64 words. How many blocks of main memory are there? 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? To which cache block will the memory reference 01D87216 map?The effective access time in a virtual memory system depends on the TLB hit rate but does not depend on whether the page table contains a valid translation for the page. O True O FalseA processor uses 2-level page tables for virtal to physical address translation. Page tables for hoth levels are stored in the main memory, Virtual and physical addreses are both 12 bits wide. The memory is byte adressable. For virtual to physical address translation, the 10 most significant bits for the virmal address are used as index into the first level poge table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 bytes wide. Further, the processor has a translation look-aside a buffer (TLB), with a hit a rate of 96% The TLB caches recently used virtual page numbers and the corresponding physical page numbers The processor also has a physically addressed cache with a hit rate of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and TLB access time is also 10 ns. 26. Assuming that no…