Consider a system with 1 MB physical memory and a word length of 1 byte. The system uses a direct mapped cache, with block numbers starting from 0. The word with physical address 0xA2C28 is mapped to the cache block number $176_{10}$. The maximum possible size of the cache (in KB) for this configuration is ________. (answer in integer) Note: 1K=$2^{10}$ and 1M=$2^{20}$

Physical Address Representation:

• Physical memory = $1\text{ MB} = 2^{20}\text{ bytes} \implies 20\text{-bit}$ physical address.

• Given physical address: $\text{0xA2C28}_{16}$.

• In 20-bit binary:

  $$\underbrace{1010}_{\text{A}} \; \underbrace{0010}_{\text{2}} \; \underbrace{1100}_{\text{C}} \; \underbrace{0010}_{\text{2}} \; \underbrace{1000}_{\text{8}}$$

  Writing bit-by-bit from $b_{19}$ to $b_0$:

  $$b_{19..0} = 1010\,0010\,1100\,0010\,1000$$

Direct Mapped Cache Addressing:

• Physical Address format: $[\text{Tag} \mid \text{Line Index} \mid \text{Block Offset}]$

• Let the block offset have $w$ bits and line index have $c$ bits.

• Given cache block number is $176_{10} = 10110000_2$.

Locating the Index Bits:

• Looking at the bit string of the physical address:

  • Bits $b_{13..6} = 10110000_2 = 176_{10}$.

• Thus, the block offset must end at bit 5, giving $w = 6$ bits (Block size $= 2^6 = 64\text{ bytes}$).

• To maximize the cache size, the number of index bits $c$ must be maximized.

• Notice that bits $b_{14}, b_{15}, b_{16}$ are all $0$, so including them in the index still keeps the value as $00010110000_2 = 176_{10}$.

• However, bit $b_{17} = 1$; including $b_{17}$ would add $2^{11}$ to the index value, changing it from 176.

• Therefore, the maximum number of index bits is $c = 11$ (spanning bits $b_{16..6}$).

Maximum Cache Size:

• Maximum number of blocks $= 2^{11} = 2048$.

• Maximum cache capacity $= 2^{11} \times 2^6\text{ bytes} = 2^{17}\text{ bytes} = 128\text{ KB}$.