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Question

Please answer both questions

Comments

For both option a

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The answer is given option b for both

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answer is b for both you have to consider the mode bit (which is 1bit)

Answer 1

The answer for Q.15 should be (b) and for Q.16 it is (b)

Answer 2

15. For register 7 bits

for memory 28 bits

remaining bits =40-(28+7)= 5

out of 5 bits one bit is used for mode and remaining 4 bits are used for opcode so 2^4 operations possible

so ans is b(16)

 

 

 

 

16.... out of 16 instructions n  instructions are 2 address instruction

free opcode=16-n

one address instructions which uses memory operand is (16-n)*2^7

so option b