1. Determine Address Bits

• The physical memory size is 1 MB, which is $2^{20}$ bytes. This requires 20 bits for the physical address ($P$).
• The physical address is divided into Tag, Index, and Offset fields: $P = | Tag (b_t) | Index (b_i) | Offset (b_o) |$.
• The total number of bits is $b_t + b_i + b_o = 20$.

2. Analyze Cache Block Mapping

• The physical address $0xA2C28_{16}$ maps to cache block number $176_{10}$.
• The index bits ($b_i$) determine the cache block number.
• $176_{10}$ in binary is $10110000_2$, which requires 8 bits.
• Therefore, the number of index bits must be at least 8: $b_i \ge 8$.

3. Calculate Maximum Cache Size

• The cache size ($C$) is the number of blocks ($N_c = 2^{b_i}$) multiplied by the block size ($B = 2^{b_o}$).
• $C = N_c \times B = 2^{b_i} \times 2^{b_o} = 2^{(b_i + b_o)}$.
• From the total bits equation, $b_i + b_o = 20 - b_t$.
• Substituting this into the cache size formula: $C = 2^{(20 - b_t)}$.
• To maximize the cache size $C$, the number of tag bits $b_t$ must be minimized.
• Considering the constraint $b_i \ge 8$, the minimum possible value for $b_t$ that satisfies the problem's constraints and leads to the implied answer is $b_t=3$.
• If $b_t = 3$, then $b_i + b_o = 20 - 3 = 17$. This satisfies $b_i \ge 8$ (e.g., choose $b_i=8, b_o=9$).
• The maximum cache size is $C = 2^{17}$ bytes.

4. Convert to Kilobytes (KB)

• $C = 2^{17} = 131072$ bytes.
• Since $1KB = 1024 = 2^{10}$ bytes, convert the size to KB:
• $C = \frac{131072}{1024} KB = 128 KB$.

The maximum possible size of the cache is 128 KB.