Question 2 for Reference:

Answer to Question 2 for Reference:

You just need to answer question 3. Question 2 and its answer are there for reference since question 3 refers to it.

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3. (8 points) For the previous problem, assuming each page table entry is 32 bytes, what is the maximum number of pages needed for for the page tables? write answers as power of 2. ANSWERS Number of virtual pages = Number of pages for page tables = 2. (12 points) A memory system has 48 bits virtual address and 36 bits physical address. The page size is 4 KB. The cache is 128 KB 2-way set associative with 32 bytes per block and is physically addressed. The figure below shows the TLB translation and the addressing of the cache. Name each of the fields and indicate how many bits are needed, use the table: Virtual address field bits TUB name block offset tag Physical address physical page number page offset virtual page number index D E F to the cache Virtual address size = 48 bits page size = 4 KB = 212 Bytes page offset = log2(212) = 12 bits virtual page number = (48 - 12) = 36 bits physical page number = (36 - 12) = 24 bits Block size = 32 bytes Block offset = log232 = 5 bits Number of Blocks in cache = (217)/32 = 212 Number of sets in cache = (212)/2 = 211 index = log2(211) = 11 bits tag = (36 - (5 + 11)) = 20 bits Name = block offset field = F bits = 5 Name = tag field =D bits = 20 Name = Physical page Number field = 0 bits = 24 Name = page offset field = B bits = 12 Name = Virtual page Number field = A bits = 36 Name = Index field = E bits = 11