Question
A system uses a paging memory management scheme with a page size of 4 KB. The logical address space of a
process is 64 KB. (i) How many pages are in the logical address?
(ii) If the physical memory size is 128 KB, how many frames are in the physical memory?
For the logical address 20500, calculate the page number and the offset.
Answer :
Word Count : 453
In a paging memory management scheme, memory is divided into fixed-size blocks called pages for the logical address space and frames for the physical memory. The page size given is 4 KB, which is equivalent to 4096 bytes. The logical address space of the process is 64 KB, which equals 65536 bytes. To calculate the number of pages in the logical address space, we divide the total logical address space by the page size. That is, 65536 ÷ 4096, which equals 16. Therefore, the logical address space is divided into _________ ___ ___ _______ ____.
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In a paging memory management scheme, memory is divided into fixed-size blocks called pages for the logical address space and frames for the physical memory. The page size given is 4 KB, which is equivalent to 4096 bytes. The logical address space of the process is 64 KB, which equals 65536 bytes. To calculate the number of pages in the logical address space, we divide the total logical address space by the page size. That is, 65536 ÷ 4096, which equals 16. Therefore, the logical address space is divided into _________ ___ ___ _______ ____.
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