Question

Suppose we have a 4096 byte byte-addressable memory that is 64-way high-order interleaved, what is the bit-size of the memory address module offset field? question 17 options: a. 12

b. 6
c. 10
d. 8

Answer 1

a) 12

Step-by-step explanation:

In a 64-way high-order interleaved memory, the memory is divided into 64 equal-sized sections, and the memory address is divided into two parts: the index and the offset. The index selects the specific section of memory to be accessed, and the offset selects the specific byte within that section.

Since the memory is byte-addressable, the offset field needs to be able to address each of the 4096 bytes in the memory section.

4096 can be represented by 2^12, so we need 12 bits to address each byte.

Therefore, the bit-size of the memory address module offset field is 12.

So the correct answer is (a) 12.