Testbook Test Series

Question

Answer 1

Instruction length = 1 word = 48 bits.

#instructions to be supported = 275  => opcode bits = 9bits.

#registers = 200 => #bits required to uniquely identify register = 8 bits.

Instruction format :-

opcode (9)

Immediate operand (x)

reg (8)

reg (8)

reg(8)

=> 9+x+8+8+8 = 48

=> x = 15 bits.

Assuming immediate operand’s value is in 2’s complement.

Min. value = $-2^{n-1} = -2^{14} = -16384$

Answer: -16384

Comments

For max, it would have been [2^(N – 1)] – 1 where N =  15?

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Yes @ankit3009

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Good explanation.

Answer 2

This can be solved in this way: