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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.
 
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- 5
 - 28
 - 14
 - 7
 
 
Correct Option: C
In Direct Mapped :
| No. of line in direct mapped cache = | ![]()  | ![]()  | bits | |
| B | 
| For No. of line = log | ![]()  | ![]()  | ||
| B | 
In Set Associative :

| TAG1 + log | ![]()  | ![]()  | + log(B) = TAG2 + log | ![]()  | ![]()  | + log(B) | ||
| B | 16 B | 
By cancelling the same terms
As given that
TAG1 = 10 bits for Direct Mapping.
TAG2 let say X bits.
| Then , 10 + log | ![]()  | ![]()  | = X + log | ![]()  | ![]()  | |||
| B | 16 B | 
By taking exponentation both side w.r.t 2.

| (2^10)* | ![]()  | ![]()  | = (2^X)* | ![]()  | ![]()  | |||
| B | 16 B | 
| (2^10) = (2^X) | ![]()  | ![]()  | ||
| 16 | 
| (2^10) = (2^X) | ![]()  | ![]()  | ||
| 2^4 | 
(2^14) = (2^X)
| X = 14 | 
So, the required length of TAG field in Set Associative mapped is = 14.



