Final answer:
In a 1024 byte, 16-way low-order interleaved memory system, the size of the memory address module offset field is 10 bits. So, the size of the memory address module offset field is 4 bits (option d).is the correct answer.
Step-by-step explanation:
To determine the size of the memory address module offset field in a 1024 byte, 16-way low-order interleaved memory system, you need to understand the structure of the memory system. Low-order interleaving means that consecutive bytes are spread across different memory modules. Since the memory is byte-addressable and there are 16 modules, the lower bits of the address are used to select one of the 16 modules. This requires 4 bits, because 24 = 16. The remaining bits are used to select the byte within the module. Since the memory is 1024 bytes in size, which is 210 bytes, you need 10 bits to uniquely address each byte within the module. Thus, the size of the offset field is 10 bits.
The size of the memory address module offset field can be calculated by finding the number of bits needed to represent the offset within each memory module and multiplying it by the number of modules in the low-order interleaved memory system.
In this case, we have a 1024 byte memory divided into 16 modules. To represent 1024 bytes, you would need 10 bits, which is the log base 2 of 1024. Since it is 16-way low-order interleaved, we need to divide the 10 bits by 16, resulting in a 4-bit offset field.
So, the size of the memory address module offset field is 4 bits (option d).