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1. Suppose a computer using direct mapped cache has 228 words of main memory and a cache of 512 blocks, where each cache block contains 8 words. a. How many blocks of main memory are there? b. What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block, and word fields? c. To which cache block will the memory reference 0081B2316 map? 2. Suppose a computer using fully associative cache has 226 words of main memory and a cache of 128 blocks, where each cache block contains 64 words. a. How many blocks of main memory are there? b. 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? c. To which cache block will the memory reference 02C47216 map? 3. Suppose a computer using set associative cache has 232 words of main memory and a cache of 64 blocks, and each cache block contains 16 words. 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 word fields? b. If this cache is 4-way set associative, what is the format of a memory address as seen by the cache? 4. Create a page translation table the meets the requirements of the virtual memory system shown below. Assume page (and frame) sizes of 10 with pages 0 through 3 in logical memory and frames 0 through 7 in physical memory. Logical Memory Physical Memory 0 9 A 0 19 10 39 B B 20 49 40 50 59 A 60 2 69 70 79 80 Page Frame 5. A computer system with 16K of memory, a Memory Management Unit with a page size of 2000, and the following page translation table (all numbers in hexadecimal): Logical Address Physical Start 0000 10A00 2000 1F200 4000 2CC00 6000 22400 8000 30300 A000 41600 C000 81E00 E000 3F500 (Note: All additions and subtractions should be done in hexadecimal) a. Indicate the physical memory location corresponding to logical address 2000. b. Indicate the logical address corresponding to physical memory location 41B55.

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5b)The closest mapping for 41B55 is 41600 (from physical location).
The difference is 41B55 – 41600 = 555, so we must add this value to the lower bound of A000. We obtain the logical address A000 + 555= A555...

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