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 10 months ago

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