Question

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 17610. The
maximum possible size of the cache (in KB) for this configuration is ___________.
(answer in integer)
Note: 1K=210 and 1M=220

Hint:

Write 0xA2C28 in binary and choose the largest block-offset plus index split for which the line-index bits equal decimal 176.

Answer 1

Correct Answer: 128

Step 1: Convert the given physical address to binary form. The address is \(0xA2C28\), and physical memory is 1 MB = \(2^{20}\) bytes, so the physical address has 20 bits.

Step 2: In a direct mapped cache, the cache block number is obtained from the line-index bits of the physical address after removing the block-offset bits. The given cache block number is \(176_{10}\). Since \(176 = 128 + 32 + 16 = (10110000)_2\), the visible line-index suffix must be \(10110000\).

Step 3: The address \(0xA2C28\) in binary is \(1010\,0010\,1100\,0010\,1000\). To get decimal 176 as the cache block number, take a block offset of 6 bits. Then the index field begins after the lower 6 bits, and the next 11 index bits end with \(00010110000_2 = 176\).

Step 4: Therefore, the largest possible cache uses 6 offset bits and 11 index bits. The cache size is \[2^6 imes 2^{11} = 2^{17} ext{ bytes} = 128 ext{ KB}.\]

\[\boxed{128\text{ KB}}\]