A processor has 40 distinct instructions and 24general purpose registers. A 32 bits instructionword has an opcode, two register operands andan immediate operand. The number of bitsavailable for the immediate operand is

To solve this problem, we need to understand how the 32 bits of the instruction word are divided.

1. First, we need to calculate the number of bits needed for the opcode. Since there are 40 distinct instructions, we need enough bits to represent 40 different values. The smallest number of bits that can do this is 6, because 2^6 = 64, which is more than 40. So, we need 6 bits for the opcode.

2. Next, we need to calculate the number of bits needed for the two register operands. Since there are 24 general purpose registers, we need enough bits to represent 24 different values. The smallest number of bits that can do this is 5, because 2^5 = 32, which is more than 24. So, we need 5 bits for each register operand, which totals to 10 bits for the two register operands.

3. Now, we subtract the number of bits used for the opcode and the register operands from the total number of bits in the instruction word. This will give us the number of bits available for the immediate operand.

So, the calculation is: 32 bits (total) - 6 bits (opcode) - 10 bits (register operands) = 16 bits.

Therefore, the number of bits available for the immediate operand is 16.

To solve this problem, we need to understand how the 32 bits of the instruction word are divided.

1. First, we need to calculate the number of bits needed for the opcode. Since there are 40 distinct instructions, we need enough bits to represent 40 different values. The smallest number of bits that can do this is 6 because 2^6 = 64, which is more than 40. So, we need 6 bits for the opcode.

2. Next, we need to calculate the number of bits needed for the two register operands. Since there are 24 general purpose registers, we need enough bits to represent 24 different values. The smallest number of bits that can do this is 5 because 2^5 = 32, which is more than 24. So, we need 5 bits for each register operand, which totals to