instruction-format

Question

Assume an instruction set that uses a fixed 31 bit instruction length. Operand specifies are 4 bits in length. If there are m-three operand instructions in total, then how many two instructions are possible at maximum?

• A.
• B.
• C.
• D.

Comments

is it D?

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yes

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Can you please share a simple solution?

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for 3 address instruction we need all 3 address and operand .

with 4 bit as operand we can have at most $2^{4}=16$ , $3$ address instruction but it is given only $m$ 3 address instruction is possible . 

Now we consumed $(2^{4}-m)$ instruction .

Two address instruction need 2 operand .

so,we can use extra 9 bits to say (add1’s 9 bit) to specify 2-address instruction .

so total we can have $(2^4-m)*2^{9}$ , $2$ address instruction .

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In the question, it’s given ”Operand specifies are 4 bits in length” then why we take 4 bits for opcode??

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I believe they mean opcode by that .

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Yes @Kabir5454 Otherwise Opcode comes out to be $19 bits$ which does not match the options

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here operand specifies means opcode

Answer 1

Ans:(2^19  -m)*2^4