Assume a 48 bit virtual address space (typical for a 64 bit Xeon achitecture). A

Question

Assume a 48 bit virtual address space (typical for a 64 bit Xeon achitecture). Assume 16KB pages, and real memory of 32GB How many entries in the page table? How many bits required to find a page in real memory? Do we need a dirty bit? A valid bit? How many bits to find a byte in the page? Assume 1 (8 byte) word for each page table entry, how many bytes does the page table need? If the page table is considered too big, we can use hierarchical or split page tables to allow parts of the page table to sit on disk or only be allocated when needed. These strategies add complexity and delay page lookup. Do we need to do this here?

Explanation / Answer

64-bit xeon architecture means each word is of size 64 bits

Virtual address space= 48 bits

page size= 16KB

main memory=32 GB

<-------- 48 bits ------------------->

#entries in the page table:

total virtual address space= 2^48*2^6 bits

page size = 2^4*2^10*8 bits

= virtual address space/page size = (2^48*2^6)/(2^4*2^10*8)=2^(54-17)=2^37

#bits required to find a page in main memory:

main memory size =2^30*2^5*2^3 bits

frame size = page size = 2^4*2^10*8 bits

No of frames in main memory =2^38/2^17 =2^21

so bits requiewd to identify the page is log(#frames)= 21

do we need a dirty bit or valid bits??

A dirty bit or modified bit is a bit that is associated with a block of computer memory and indicates whether or not the corresponding block of memory has been modified. The dirty bit is set when the processor writes to (modifies) this memory.

#bits to find byte in page??

page size is 16KB= 2^14 bytes

so 14 bits are required to find a byte in a page

Assume 1 word (8 bytes)for each page table entries , how many bytes does the page table needs??

# entries in page table is 2^37

each entry is 1 word means 8 byte

so total size of a page is 2^37*8 bytes=2^40bytes

Part2:

if the page size is too high then the there will be more than one page needed to store the page table.

So it can either be present in the main memory or it can be bought from the disk.so there will be an overhead

involved in this strategy in the page lookup.

The alternative to that method is to use hierarchial strategy for storing the page table.

Here in this case the page table size is bigger then the page size so we should use hierarchial page tables .

page no offset

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