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Picture 011 - CHAPTER 4 Memory Management ———®...

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Unformatted text preview: CHAPTER 4 Memory Management ———® Answer: Eachentryeonlains 1 bittoindicateifthepageisvnlidand l'4bitatospecifythepage framennmben . 4.22 On a simple paging system with a page table containing 64 entries of 11 hits (including valid/invalid bit) each. and a page size of 512 bytes, hbv'vmany bits in the logical address specify the page number? - Answer: Sincethere are64 = 25pages, 6bi'ts ofthelogicaladdresaamreqtfiredtospecifythe page number. ' On a simple paging system with a page table containing. 64. ermine of 11 hits (including valid] invalid hit) each, and a page size of ‘5 12 bytes, how many hits in the logical address specify the offset within the page? ' Anpwer: Since a page is 512 = 29 bytes long, 9 bits are requiredtospecify theofl‘set within thepage. 4.24 On a simple paging system with a page table containing 64 entries cf 11 hits (including valid/invalid hit) each. and a page size of 5 12 bytes, how many hits are in a logical addmss? Answer: The 6 most significant bits of the logical address wmtld specify the page number, the 9 least significant bits would specify the page ofiset, making the total number of bits 15. On a simple paging system with a page table containing 64 entries of 11 hits (including valid/invalid hit) each, and a page size of 512 bytes, what isthe size of the logical address space? Answer: lo the previous problem, the logical address was deem-mined to have 15 bits. A ls-bit address creates an address space of 32,768 = 215. 4.26 On a simple paging system with a page table containing 64 entries of 11 hits (including valid/invalid hit) each, and a pagem'ze of 512 bytes, how many hits in the physical address specify the page flame number? ...
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