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A cache memory unit with capacity of $N$ words and block size of $B$ words is to be designed. If it is designed as a direct mapped cache, the length of the TAG field is 10 bits. If the cache unit is now designed as a 16-way set-associative cache, the length of the TAG field is ____________ bits.

retagged | 7.8k views

In set-associative 1 set = 16 lines. So the number of index bits will be $4$ less than the direct mapped case.

So, Tag bits increased to $14$ bits.

by Boss (12k points)
selected by
+10

In direct mapped cache. TAG field = 10 bits so 210 MM can be mapped to 1 cache line. Now we have 16 lines in 1 cache set so 24*210 = 214 MM blocks can be mapped to 1 cache set. So in a 16-way set-associative cache, the length of the TAG field is 14 bits.

+3
New concept i learned

Cache capacity =N words

Block size= B words,bits for block offset=log(B)

No of lines in Direct  Mapped cache =N/B,bits for no. of lines=log(N/B)

No of Sets in 16-way set associative cache=N/16B,,bits for no. of sets=log(N/16B)

In direct mapped length of tag =10 bits

 TAG-10 LINE.NO:-LOG(N/B) BLOCK OFFSET:-LOG(B)
 TAG-X SET.NO-LOG(N/16B) BLOCK OFFSET:-LOG(B)

As Physical address is same for both cache mappings,

10+LOG(N/B)+LOG(B)=X+LOG(N/16B)+LOG(B)

10+LOG(N/B)=X+LOG(N/16B)                              //AFTER CANCELLING LOG(B)

2^(10+LOG(N/B))=2^(X+LOG(N/16B))                 //EXPONENTIATION WITH 2 ,BOTH SIDES

(2^10) *(N/B)=(2^X) *(N/16B)

(2^10) =(2^X) *(1/2^4)

2^14=2^X

X=14                                                                 //APPLYING LOG TO BOTH SIDES

by Active (4.7k points)

hope it helps!!!

by Boss (41.4k points)
edited
+1
Very good explaination thanks.
0
Yes it does !

Explained as simple as it could be.

Thank you.
in Direct mapping length is 10 bits that means at each index of cache 2^10 lines can be mapped

after 16 way set associative....16*(2^10) lines can be mapped in each set so total 2^14 lines hence tag bit must be able to address these many lines so 14 bits required

by Active (3.4k points)

Since offset not given so we can neglect it because line offset in (direct cache) and set offset in (set - associative cache) is same.

In direct cache = Tag(10) + line (x, assume) + offset (neglect it)

In set associative = Tag(t assume) + Set(x - 4) + offset (neglect it) , since set = $\frac{Number -of- Lines }{16- way -set- associative}$ = 2x-4

Now , In direct cache = 16- way set associative

Tag(10) + line (x, assume) =  Tag(t assume) + Set(x - 4)

10 + x = Tag + (x - 4)

Tag = 10+4 = 14 bits

by Veteran (62.3k points)

So answer is tag bits = 14 bits.

by (203 points)
–1 vote

One approach will be we can find the number of blocks in main memory with the help of tag bits like given in the following picture

so total number of blocks in main memory is 131072 so we can find number of blocks in the cache as 131072/1024=128 blocks
now when mapping is set associative so total number of sets in cache is 128/16=8 sets so now total number of blocks in main memory is 131072/8=16384 which is 2^14 so 14 bits are needed in tag.

by (231 points)
–1 vote

No of TAG bits in Direct Map Cache : (No. of bits Required for Main Memory Address M)-(number of bits for index I+bits for offset O)

No of TAG bits in 16-way set associative : M -(set bits + offset bits)

number of bits for set : index bits - 4 // 16 way set associative

No of TAG bits in 16-way set associative : M -(index bits - 4 + offset bits)

: (M-I-O)+4=10+4=14

by Active (3.6k points)
–1 vote

Here we go

by (175 points)

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