Made easy test

Question

How to solve?

Answer 1

Address field = 9 bits.

For two address instructions, 18 bits will be for address. Remaining (32-18) = 14 bits will be for opcode.

Using 14 bit opcode, 2¹⁴ opcodes are possible. But there are only 400 two address instructions.

So, remaining (2¹⁴-400) opcodes can be used for one address instructions.

Possible one address instructions = (2¹⁴-400) * 2⁹ [using opcode expand technique]

Comments

what exactly this 

opcode expand technique

 is??

i googled bt can't find proper meaning and use of that. 

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See,

First of all the assumption here is, suppose there are two instructions ADD A,B and ADD C,D.

These two are considered the same instructions. 

Likewise, if there are 14 bits for opcodes, then 2¹⁴ two address instructions are possible.

But only 400 two address instructions are allowed.

So, (2¹⁴-400) opcodes are not used by two address instructions. These can be given to one address instructions. Now for an instruction to be one address, we dedicate 9 bits for address part. Rest 9 bits is free. So we use these 9 unused bits for opcodes (this is opcode expand technique).

So, no of opcodes for one address instructions = (2¹⁴-400) * 2⁹

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Nicely explained.. thank u.. :)