A CPU uses a 16-bit instruction format. If 4 bits are used for the opcode and the remaining bits specify a single memory address, what is the maximum addressable memory space for this instruction format?
Step-by-step Solution:
Total size of the instruction = 16 bits. Bits used for the opcode = 4 bits. The remaining bits are used to specify the memory address: \[ 16 \text{ bits} - 4 \text{ bits} = 12 \text{ bits} \] The maximum number of unique memory addresses that can be represented with 12 bits is calculated as \( 2^{12} \). \[ 2^{12} = 4096 \] Therefore, the maximum addressable memory space is 4,096 locations.