I am looking to calculate how many page-table entries can fit into a corresponding TLB space (Translation Look-aside Buffer ). I just want to make sure I am getting the calculation right, as I fear I could be wrong somewhere.

So, the data I have is as follows:

  • Page size $= 2^{12} bytes$

  • Physical address space $= 2^{24} bytes$

  • Logical address space $= 2^{32} bytes$
  • Translation Look-aside Buffer (TLB) size $=2^6 bytes$

How many page-table entries can fit into TLB space given that every entry carries the necessary information needed for the logical to physical address translation?

What I have done so far is finding the frame number doing the following:

$nframes = 2^{24}/2^{12} = 2^{12}$


$frame.number = 12 bits $

Side question: Do I have to divide the TLB size with the 12 bits as in order to find the size of a table you have to multiply the number of rows by the bytes that each row carries?

*The problem is that in this case (which I think is the correct answer) the TLB entries seem to derive from a decimal number which is $2^7/3$ so I take the closest integer approximation.


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