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On a simple paging system with 2^24 bytes of physical memory ,256 pages of logical address space, and a page size of 2^10 bytes .

i)How many bytes are in page frame?

ii)how many bytes in the physical address specify the page frame?

iii)how many entries in the page table?

iv)how many bits in logical address?

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Given

physical memory: $2^{24}$ bytes

logical address space: 256 pages = $2^8$

page size: $2^{10}$ bytes

1. How many bytes are in the page frame? $2^{10}$ bytes
2. How many bytes in the physical address specify the page frame? We have $2^{24}$ bytes of physical memory, and a frame is $2^{10}$ big, so $2^{14}$ bytes in the physical address specify the page frame.
3. How many entries are in the page table? Since there are $2^{8}$ pages, there are 2^8 entries in the page table.
4. How many bits in logical address?

total logical address space = number of pages * page size = $2^{8}$ * $2^{10}$ = $2^{18}$ bytes

for $2^{18}$ bytes address space, 18-bit address is required.