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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.



