Question
In: Computer Science
Describe the number of bits required in each entry of a TLB that has the following...
Describe the number of bits required in each entry of a TLB
that has the following characteristics:
• Virtual addresses are 32 bits wide
• Physical addresses are 31 bits wide
• The page size is 2K bytes
• The TLB contains 16 entries of the page table
• The TLB is direct-mapped
Solutions
Expert Solution
virtual address =32 bit (given)
page size=2k bytes=
bytes
page offset=11 bit
total number of pages in process=virtual address space/page
size=
number of bits require to represent the page in process=log(total number of pages in process)=21bit
total number of frame in physical address=physical address
space/page size=
number of bits require to represent the frame in physical memory=log(number of frame in physical memory)
=log(
)=20
bit
As we know that TLB contains <page number, frame number>
total number of bit require to represent the TLB=21+20=41bit
number of entries in TLB =16(given)
size of TLB=16*41/8 bytes
size of TLB=82 bytes
in each entry of TLB contain <page number,frame number> so to represent it 41 bit is require .
venereology answered 1 year agoRelated Solutions
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