Final answer:
To accommodate all the given instructions in a 32-bit instruction, we need to consider the bit requirements for each type of instruction, which sum up to 69 bits, exceeding the available 32 bits.
Step-by-step explanation:
The op-code needs to be designed in such a way that it can accommodate all the given instructions in a 32-bit instruction. Let's break it down for each type of instruction:
a. 14 instructions with one 22-bit address and 4-bit register number:
In this case, we need 14 op-codes (one for each instruction) with a 22-bit address field and a 4-bit register number field. This would require 22 + 4 = 26 bits from the 32-bit instruction.
b. 720 instructions with one 12-bit address and 5-bit register number:
For this set of instructions, we would need 720 op-codes with a 12-bit address field and a 5-bit register number field. This would require 12 + 5 = 17 bits from the 32-bit instruction.
c. 128 instructions with two 12-bit addresses and 2-bit register number:
For these instructions, we would need 128 op-codes with two 12-bit address fields and a 2-bit register number field. This would require 2 * 12 + 2 = 26 bits from the 32-bit instruction.
Adding the bit requirements for each type of instruction, we get 26 + 17 + 26 = 69 bits. This means that 32 bits are not sufficient to encode all the given instructions. Some trade-offs and optimizations would need to be made to fit the instructions within the available 32 bits.