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A computer system implements a $40\;\text{-bit}$ virtual address, page size of $8\;\text{kilobytes}$, and a $128\text{-entry}$ translation look-aside buffer $\text{(TLB)}$ organized into $32$ sets each having $4$ ways. Assume that the $\text{TLB}$ tag does not store any process id. The minimum length of the $\text{TLB}$ tag in bits is ______.
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Just think in terms of set-associative cache mapping used in cache organization.

Number of pages = (Total Virtual address space) / (Page size)

                                =(2^40) / (2^13) = (2)^27

No. of bits required to uniquely identify a page is 27

Assume page as a block for simplicity 

Like in the cache we take each main memory block and map it to a block in the cache

Number of blocks = 2^27,  Number of bits to uniquely identify a block =27

So here we are using 4 way set associative mapping and the total number of sets = 32

Number of bits required to uniquely identify a set =5 

So number of tag bits = 27-5 = 22 bits 

Here logic is the same just assume page bits as block bits and subtract set bits from them we can get tag bits.

The answer is 27-5 = 22 tag bits

Answer:

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