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3 votes
3 votes

A certain computer system has the physical address space as 216 bytes . And consists of 2byte page table entries . Assume that each page table entry contain (beside other information ) 1 bit for valid , 3 bit for protection and 1 bit for dirty . How many bits are available in page table entry for storing the aging information for the page ?

Assume that the page size is 512 bytes ?

a) 3  b) 4 c) 5 d) 6 

 

1 Answer

Best answer
2 votes
2 votes

Since physical address space = 2^16 

page size = frame size = 512 bytes 

So number of frames = 2^16/2^9= 2^7

Number of bits used to address frame = log2(2^7)= 7 bits 

since page table entry is 2 bytes = 16 bits 

so now page table entry bit = no of bit used for farme number + 1 valid + 3 bit for protection + 1 dirty bit + Required Aging information 

so 16 = 7+1+3+1+x

x=4 

is it right ?

They have given answer as 3 

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