The size of the physical address space of a processor is $2^{32}$ bytes. The capacity of a cache memory unit is $2^{23}$ bytes. The cache block size is 128 bytes. The cache memory unit can be built as a direct mapped cache or as a $K$-way set-associative cache, where $K = 2^L$ and $L \in \{1, 2, 3\}$. Let the length of the TAG field be $M$ bits for the direct mapped cache, and $N$ bits for the set-associative cache. Which one of the following options is true?

Given:

• Physical Address Space $= 2^{32}\text{ bytes} \implies$ Physical Address $(\text{PA}) = 32\text{ bits}$.

• Cache Capacity $= 2^{23}\text{ bytes}$.

• Cache Block Size $= 128\text{ bytes} = 2^7\text{ bytes} \implies$ Block offset field $= 7\text{ bits}$.

Total number of cache blocks / lines:

1. For Direct-Mapped Cache:

• Number of lines $= 2^{16} \implies$ Line Index field $= 16\text{ bits}$.

• TAG field length $M$ is given by:

2. For $K$-way Set-Associative Cache ($K = 2^L$):

• Number of sets $= \frac{\text{Total Blocks}}{K} = \frac{2^{16}}{2^L} = 2^{16 - L} \implies$ Set Index field $= (16 - L)\text{ bits}$.

• TAG field length $N$ is given by:

Substituting $M = 9$ into the equation for $N$:

Thus, (A) is the correct option.