MadeEasy Test Series: CO & Architecture - Instruction Format

Question

How will this question be solved?

Comments

is this a correct solution ... i just doubt.

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this is correct, because it's given that memory is byte addressable. hence, number of bits for accessing a memory address would be 20 instead of 18.

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got it thanks.

Answer 1

Total Memory = 2 ^ 18 * 32 bits =>2 ^(18+5)=2^23 bits => 2^23/2^3 B ( given byte addressable) =>2^20 B

Address length for Memory =20 bits

Mode = log 7= 3 bits

Registers= log 60 =6 bits

Opcode=32-(3+6+20)=> 32-29 =>3  

So no of instructions = 2^3 = 8 diff instructions are possible

Comments

got it thanks.

Answer 2

8 instn is possible

Comments

abhishekmehta4u Can you provide a source to learn this as I'm not able to solve these type of questions.