The figure below shows a virtual memory system with a virtual address space of 6
ID: 3840314 • Letter: T
Question
The figure below shows a virtual memory system with a virtual address space of 64K, a physical address space of 32K and a page size of 4K. Addresses are in base 10 form. Showing workings, state the:
i. number of entries in the page table.
ii. number of bits required to specify a location within a page.
iii. number of bits required to specify a location in physical memory.
iv. physical address of virtual address 43320.
3. Derive the complete page table for the virtual memory system of the figure below. The page table entries must be in binary. Assume that the page entry is zero if the page frame is not mapped to a physical page.
Virtual Address Space 0- 4K. 4K-8K 8K 12K 12K 16K 16K 20K 20K 24K 24K -28K 28K 32K 32K 36K 36K -40K 40K 44K 44K 48K 48K 52K 52K 56K 56K- 60K 60K 64K Virtual Page Physical Memory Addresses 0- 4K 4K-8K 8K-12K 12K 16K 16K 20K 20K 24K 24K -28K 28K 32K Page FrameExplanation / Answer
Virtual Address Space(V.A.S.) = 64K
Physical Address Space(P.A.S.) = 32K
Page Size = 4K
i) Number of entries in the page table
Number of entries in the page table = Number of Pages in V.A.S.
And, Number of pages in L.A.S.= (V.A.S. Size)/Page Size
= 64 K/4 K
= 16
ii) number of bits required to specify a location within a page
= Number of bits required to represent the page size
= 12 (Since Page Size= 4K =22+10=212)
iii) number of bits required to specify a location in physical memory
Number of bits required to specify a location in physical memory = Number of bits required to represent the frames of P.A.S.
And, Number of Frames in P.A.S. = P.A.S. Size / Frame Size
= 32 K / 4 K (Since Frame Size = Page Size)
= 8
Number of bits required to specify a location in physical memory = Number of bits required to represent the frames of P.A.S.
=> Number of bits required to specify a location in physical memory = 3 (Since 8=23)
iv) physical address of virtual address 43320
40K-44K Page from V.A.S. ,i.e., Page number 10 is in Frame number 5 (20K - 24K)
and the page offset or frame offset , d= 3320
=> physical address of virtual address 43320 = 23320
3) The complete page table for the virtual memory system of the figure is given below :
Page Number Frame Number 0 000 1 011 2 010 3 001 4 000 5 000 6 000 7 100 8 110 9 000 10 101 11 111 12 000 13 000 14 000 15 000Related Questions
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