Course Solutions Uncategorized (Solved) : Let Virtual Addresses 16 Bit Wide Physical Addresses 15 Bit Wide Page Size 212byte 4 Kb Pa Q36572691 . . . .

(Solved) : Let Virtual Addresses 16 Bit Wide Physical Addresses 15 Bit Wide Page Size 212byte 4 Kb Pa Q36572691 . . . .

 

Let the virtual addresses be 16 bit wide, and the physicaladdresses 15 bit wide. The page size is 212byte = 4 kB, and thepage table is a simple single-level page table with 8 bit wideentries.
(a) In the virtual address, how many bits correspond to the pagenumber and the position inside the page (offset)?

(b) What is the total size of the page table?
(c) How many pages fit into the physical memory?
(d) Assume the current state of the page table is as follows. Whereis page 3, 6, and 11 located?
(e) Modify the content of the page table according to thefollowing.
• Page 6

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