Question

Assume a 32-bit microprocessor having 32-bit instructions collected of two fields: the first 8 bit contains the opcode and the remainder the immediate operand or an operand address. a. Determine the maximum directly addressable memory capacity (in bytes)?

Answer 1

The 32-bit microprocessor can directly address 2^24 memory locations, amounting to 16,777,216 bytes, which is commonly referred to as 16MB (megabytes).

Step-by-step explanation:

The student is asking about the maximum directly addressable memory capacity of a 32-bit microprocessor with 32-bit instructions that are divided into two fields: an 8-bit opcode and the remainder for an immediate operand or an operand address. In this architecture, the maximum directly addressable memory is determined by the size of the field used for the address. As the address field is 24 bits (32 bits total minus the 8-bit opcode), the microprocessor can directly address 224 memory locations.

Since each memory location is 1 byte, the total directly addressable memory is 224 bytes. Converting this to a more understandable unit, 224 bytes is equal to 16,777,216 bytes, which is often referred to by the multiplicative prefix as 16 megabytes (MB). The basic unit of computer memory is the byte, and the unit for one million bytes is called a megabyte.