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Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of page is 16 kB and the main memory size is 16 GB. The minimum size of page table with each page table entry need 2 protection need is ______ (in GB)

Do we take page table entry size as 3Bytes or 22 bits only?

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Step-1 Find the number of enteries in page table

No of enteries in page table = virtual address space / page size = 248/214 = 234

Step-2 Find number of frames in physical memory

No of Frames = 234/214=220

So we need 20 bits + 2 additional bits for protection = 22 bits

22 bits = 2.75 bytes

Therefore size of page table = 234 x 2.75 =44 GB Approx

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it should be 22 bits and Page table size = 2^34 * 22

                                                                =(16*22)/8 * 2^30

                                                                =44 GB

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