GATE CS 2014 Set 1 Q.19 (CS Q.9): Machine Instructions and Addressing Modes

A machine has a 32-bit architecture, with 1-word long instructions. It has 64 registers, each of which is 32 bits long. It needs to support 45 instructions, which have an immediate operand in addition to two register operands. Assuming that the immediate operand is an unsigned integer, the maximum value of the immediate operand is ____________.

Your answer

Official answer and solution

Official answer: 16383
From: Official GATE 2014 answer key

Solution

Determine the bit width of each instruction field and subtract from the 32-bit total instruction length to find the bits available for the immediate operand.

Instruction length = 1 word = 32 bits.
Number of distinct instructions = 45.
The number of bits required for the opcode is: $\lceil \log_2 45 \rceil = 6 \text{ bits} \quad (2^5 = 32 < 45 \le 64 = 2^6)$
Number of registers = 64.
The number of bits required to specify one register is: $\log_2 64 = 6 \text{ bits}$ There are 2 register operands, so the register fields take: $2 \times 6 = 12 \text{ bits}$
Remaining bits for the immediate operand: $\text{Immediate bits} = 32 - (\text{Opcode bits} + \text{Register bits})$ $\text{Immediate bits} = 32 - (6 + 12) = 32 - 18 = 14 \text{ bits}$
For an unsigned 14-bit integer, the maximum value is: $2^{14} - 1 = 16384 - 1 = 16383$

Final answer

16383

Common trap

Using 5 bits for opcode ( $2^5 = 32$ , which cannot support 45 instructions) or giving $2^{14}$ instead of $2^{14} - 1$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 19 of 65, printed as CS Q.9
Section: Computer Science and Information Technology

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