Question

Suppose we have a 4096 byte byte-addressable memory that is 64-bank high-order interleaved, what is the size of the memory address module offset field?

Answer 1

The size of the memory address module offset field in a 64-bank high-order interleaved memory with a size of 4096 bytes is 6 bits.

Step-by-step explanation:

In a 64-bank high-order interleaved memory, the memory is divided into 64 banks and each bank has a block of memory. The address module offset field represents the number of bits needed to specify the byte within a block of memory. Since the memory is byte-addressable, each byte within a block needs a unique address.

If the memory has a size of 4096 bytes, and each bank is of equal size, then the size of each bank is 4096 / 64 = 64 bytes. Therefore, the number of bits needed to address each byte within a bank is log2(64) = 6 bits. Hence, the size of the memory address module offset field is 6 bits.