MadeEasy Test Series: CO & Architecture - Instruction Format

Question

Answer 1

1 instruction size = 1 word = 4 bytes = 32 bits

170 operations, so it need 8 bits for opcode

37 registers need 6 bits for register operand

So remaining bits will be for memory operand i.e., 32 bits - (8 + 6) bits = 18 bits.

With 18 bit memory address, memory address possible is 2^18.

As word size is 4 bytes, memory size = 2¹⁸ * 4 bytes = 2²⁰ bytes = 1 MB.

Comments

 prathams

that's what i did , is it correct ?

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Sorry, i have corrected the answer. Please verify.