Analyzing Computer Instruction Format Bits

The question asks us to determine the number of bits allocated to different parts of a computer instruction: the operation code, the register code part, and the address part. We are given the total size of the memory, the size of each memory word, and the structure of the instruction code.

Understanding the Instruction Structure

We are told that a binary instruction code is stored in one word of memory. The memory unit has words of 32 bits each. Therefore, each instruction is 32 bits long.

The instruction is divided into four parts:

1. An indirect bit
2. An operation code (Opcode)
3. A register code part
4. An address part

The total number of bits for these four parts must sum up to the total instruction length, which is 32 bits.

Calculating Bits for Each Part

Indirect Bit

The question explicitly states there is an indirect bit. This typically occupies 1 bit.

Indirect bit bits = 1

Register Code Part

The register code part is used to specify one of 64 registers. To uniquely identify one out of 64 items, we need a certain number of bits. The number of bits required is given by $\text{log}_2(\text{number of items})$.

Number of registers = 64

Register code bits = $\text{log}_2(64)$

Since $2^6 = 64$, $\text{log}_2(64) = 6$.

Register code part bits = 6

Address Part

The address part is used to address memory locations. The memory unit has 256K words. First, let's convert 256K words into the total number of words.

$1\text{K} = 1024$ words.

$256\text{K} = 256 \times 1024 = 262144$ words.

To address one of the 262144 memory locations, we need a certain number of bits for the address part. Similar to the register code, the number of bits required is $\text{log}_2(\text{total number of memory words})$.

Total memory words = 262144

Address part bits = $\text{log}_2(262144)$

We know that $2^{10} = 1024$ and $2^8 = 256$.

So, $262144 = 256 \times 1024 = 2^8 \times 2^{10} = 2^{(8+10)} = 2^{18}$.

Therefore, $\text{log}_2(262144) = 18$.

Address part bits = 18

Operation Code (Opcode)

The operation code specifies the operation to be performed. The bits for the opcode are the remaining bits after allocating bits for the indirect bit, register code, and address part from the total instruction length.

Total instruction bits = 32

Bits used by other parts = Indirect bit bits + Register code part bits + Address part bits

Bits used by other parts = $1 + 6 + 18 = 25$ bits.

Operation code bits = Total instruction bits - Bits used by other parts

Operation code bits = $32 - 25 = 7$ bits.

Summary of Bit Allocation

Let's summarize the number of bits for each part of the computer instruction:

• Operation code: 7 bits
• Register code part: 6 bits
• Address part: 18 bits

The question asks for the number of bits in the order: operation code, the register code part, and the address part.

This order corresponds to 7, 6, 18.

Comparing with Options

Let's compare our calculated values (7, 6, 18) with the given options:

• Option 1: 7, 7, 18 (Incorrect - Register code is 6, not 7)
• Option 2: 18, 7, 7 (Incorrect - Order and values are wrong)
• Option 3: 7, 6, 18 (Correct - Matches our calculated values)
• Option 4: 6, 7, 18 (Incorrect - Opcode is 7, not 6; Register code is 6, not 7)

Our calculation confirms that the number of bits are 7 for the operation code, 6 for the register code part, and 18 for the address part.

Revision Table: Computer Instruction Bits

Additional Information: Computer Instruction Format

A computer instruction is a command that tells the computer's CPU what operation to perform. Instructions are typically stored in memory and fetched by the CPU during program execution. The format of an instruction defines how the bits of the instruction are organized into functional parts.

• Opcode (Operation Code): This is the most crucial part, specifying the type of operation to be performed (e.g., Add, Subtract, Load, Store). The number of bits in the opcode determines the maximum number of distinct operations a computer can perform ($2^{\text{number of opcode bits}}$).
• Address Part (Operand Address): This part specifies the memory address or the register number of the data (operands) that the operation will use. The number of bits depends on the size of the main memory or the number of registers.
• Mode bits: These bits specify the addressing mode, which determines how the operand's effective address is calculated (e.g., direct, indirect, immediate, register). In this question, an "indirect bit" is explicitly mentioned, indicating a specific addressing mode related to indirection.
• Register Code Part: This part identifies which CPU register is involved in the operation, either as a source of an operand or a destination for the result. The number of bits depends on the total number of general-purpose registers in the CPU.

Different computer architectures have varying instruction formats. The format significantly impacts the CPU design and the efficiency of program execution. Fixed-length instructions, like the 32-bit format described, are common in RISC (Reduced Instruction Set Computer) architectures, while variable-length instructions are found in CISC (Complex Instruction Set Computer) architectures.