GATE Overflow Test Series | Operating Systems | Test 2 | Question: 21

Question

A CPU generates $64$-bit virtual addresses. The page size is $8$ KB. The processor has a translation look-aside buffer (TLB) which can hold a total of $256$ page table entries and is $8$-way set associative. The minimum size of the tag memory for TLB in bytes is _______

Answer 1

Number of sets in TLB $= 256/8 = 32.$

Number of PTEs $ =2^{64} B/8KB = 2^{51}  $

So, number of PTEs which can map to a TLB set $ = 2^{51}/32= 2^{46}.$

So, minimum size of TLB tag $ = 46$ bits.

We have $256$ entries in TLB each of $46$ bits. So, total tag memory size for TLB $ = 46 \times 256\; \text{bits} = 46 \times 32\; \text{bytes} = 1472\:\text{bytes.}$

Comments

Here for each entry 46 bits need to be approximated to 48 bits = 6 B ?

So tag memory must be equal to 256 * 6 = 1536 B ?

Please clarify.

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Whats the need for such approximation?