UGC NET CSE | December 2013 | Part 2 | Question: 47

Question

The virtual address generated by a CPU is 32 bits. The Translation Look-aside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4-cache lines in the set). The page size is 4 KB. The minimum  size of TLB tag is

• A. 12 bits
• B. 15 bits
• C. 16 bits
• D. 20 bits

Answer 1

page size of 4KB ,so offset bits are 12 bits.

 Remaining bits of virtual address 32 - 12 = 20 bits will be used for indexing.

number of sets =$\frac{64}{4}$ = 16 (4-way set) => 4 bits.

So tag bits = 20 - 4 = 16 bits.

Hence,Option(C)16.

Comments

@vijay ,

64 entries are there . And 4 way set associative means 4 entries in one set .

Hence, total sets 64/4= 16 sets, which can be represented by 4 bits .

Page size is 4 KB , so 12 bits to represent a page .

hence, 32 - (4 + 12) = 16 tag bits .

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@Kapil Sir ..  I am not talking about TLB tag .. I am talking abt cache tag ..

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but, here we need TLB tag...