johnnyrocket Posted December 6, 2011 Share Posted December 6, 2011 The format of the Page Table Entries for a certain system that employs 32-bit virtual addresses and pages that are 8192 bytes in size is shown below: Bit23: Valid Bit Bit22: Modify Bit Bits 21-18: LRU Bits Bits 17-0: Frame Number - What is the maximum amount of physical memory the system could contain? My guess: We have 18 bits for the frame number. So 2^18 * 8192 bytes = 2 GB -What would be the total size in bytes of the page map table? -If an inverted page map table is used what would be its total size in bytes if the LRU field within each PTE is reduced to 3 bits? Quote Link to comment Share on other sites More sharing options...
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