Page size=2KB=2 Power 11 so bits for page offset=11 bits for logical address=26bits bits for page offset=11 bits so bits for number of pages=26-11=15bits
Q2.A system has 64 bit logical address and 43 bit physical address. if the pages are 8KB in size the number of bits required for logical address and physical address will be.
Bits for Logical address=64 bit Bits for physical address=43bit bits for pages=13bit(because page size is 8KB) number of bits for logical address=64-13=51 number of bits for physical address=43-13=30
Q3.Consider a logical address space of 8 pages with page size 1024bytes. The physical memory contains 32 frames.(Find bits for LA,bits for PA,Page table size).
number of bits for Page offset=10bits(because page size is 1024bytes) bits for number of pages=3(because 8 pages) so bits for logical address=10+3=12bits number of bits of frame=5bits(32 frame) number of bits for page offset=10(page size is 1024 bytes) so bits for physical address=10+5=15bits. Pages table size=no of pages*1 entry size=8*5=40bits.
Q4.A system supports 4K pages of size 256bytes each in demand paging system. Main memory contain 1K frames.Number of bits required for logical address and physical address are?
number of bits for pages=12bits.(number of pages 4k) number of bits for frames=10(1k frames) number of bits for page offset=8bits(page size is 256bytes) so number of bits for Logical address=12+8=20bits so number of bits for physical address=10+8=18bits.
Q5.What is the size of the physical address space and logical address space in a paging system which has a page table containing 64 entries of 11bits(including a valid bit) each and page size of 512bytes.
number of bits for page offset=9bits(512bytes) number of bits for pages=6(64 entries) entrysize=11bits f+1=11 f=10bits. number of bits for Logical address=9+6=15bits number of bits for physical address=10+9=19bits
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