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.

Let the main memory physical address have $A$ bits.

1. Direct-Mapped Cache:

   • Cache capacity $= N$ words

   • Block size $= B$ words

   • Number of cache lines/blocks $= \frac{N}{B}$

   • Block offset bits $= \log_2(B)$

   • Line index bits $= \log_2\left(\frac{N}{B}\right) = \log_2(N) - \log_2(B)$

   • $\text{Tag bits}_{\text{direct}} = A - (\text{Line index bits} + \text{Block offset bits}) = A - \log_2(N)$

   • Given $\text{Tag bits}_{\text{direct}} = 10$, so:

     $$A - \log_2(N) = 10$$

2. 16-way Set-Associative Cache:

   • Set size $= 16$ blocks/set

   • Number of sets $= \frac{\text{Total blocks}}{16} = \frac{N/B}{16} = \frac{N}{16B}$

   • Set index bits $= \log_2\left(\frac{N}{16B}\right) = \log_2(N) - \log_2(B) - 4$

   • Block offset bits $= \log_2(B)$

   • $\text{Tag bits}_{\text{set-assoc}} = A - (\text{Set index bits} + \text{Block offset bits})$

     $$\text{Tag bits}_{\text{set-assoc}} = A - (\log_2(N) - 4) = (A - \log_2(N)) + 4$$

Substituting $A - \log_2(N) = 10$:

Thus, the length of the TAG field is 14 bits.