GATE CS 2024 Set 2 — Question 61

NAT+2 / -0MediumInstruction FormatsInstructions & Addressing ModesComputer Organization & Architecture

Computer Organization & Architecture → Instructions & Addressing Modes → Instruction Formats

Last updated 20 August 2026

Question

A processor uses a 32-bit instruction format and supports byte-addressable memory access. The ISA of the processor has 150 distinct instructions. The instructions are equally divided into two types, namely R-type and I-type, whose formats are shown below.

R-type Instruction Format:

OPCODE	UNUSED	DST Register	SRC Register1	SRC Register 2

I-type Instruction Format:

OPCODE	DST Register	SRC Register	# Immediate value/address

In the OPCODE, 1 bit is used to distinguish between I-type and R-type instructions and the remaining bits indicate the operation. The processor has 50 architectural registers, and all register fields in the instructions are of equal size.

Let

X

be the number of bits used to encode the UNUSED field,

Y

be the number of bits used to encode the OPCODE field, and

Z

be the number of bits used to encode the immediate value/address field. The value of

X + 2Y + Z

is ________

Correct answer

34 to 34

Solution

1.Determine the number of bits for the OPCODE field ( $Y$ ):

Total instructions = 150.
Instructions are equally divided: 75 R-type and 75 I-type.
1 bit is used to distinguish between types (R-type vs I-type).
To represent 75 distinct operations within each type, we need $\lceil \log_2(75) \rceil = 7$ bits.
Therefore, total bits for the OPCODE field, $Y = 1 + 7 = 8$ bits.

2. Determine the number of bits for register fields:

The processor has 50 architectural registers.
Bits required to uniquely identify each register = $\lceil \log_2(50) \rceil = 6$ bits.
All register fields (DST, SRC1, SRC2, SRC) are of equal size, so each is 6 bits.

3. Determine the number of bits for the UNUSED field ( $X$ ) from the R-type format:

Total instruction length = 32 bits.
R-type format: $\text{OPCODE} + \text{UNUSED} + \text{DST Register} + \text{SRC Register1} + \text{SRC Register2} = 32$
$8 + X + 6 + 6 + 6 = 32$
$26 + X = 32 \implies X = 6$ bits.

4. Determine the number of bits for the immediate field ( $Z$ ) from the I-type format:

I-type format: $\text{OPCODE} + \text{DST Register} + \text{SRC Register} + \text{Immediate} = 32$
$8 + 6 + 6 + Z = 32$
$20 + Z = 32 \implies Z = 12$ bits.

5. Calculate the final value:

$X + 2Y + Z = 6 + 2(8) + 12 = 6 + 16 + 12 = 34$ .

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