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Consider the type addressable memory unit of a computer has 64 K words of 16 bit each. The computer has an instruction format with 4 fields namely opcode, mode field specify 9-addressing modes, register address field with 50 registers and a memory address field. If an instruction is 32 bits long then the number of different instructions are _____.
closed with the note: I found the solution of this question in comments.

closed | 32 views
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32??
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Nope, answer is 32. I'm not getting why did they took 17 bits for memory address field?
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oh, yes, it will be 32

32-(4+6+17)=5

Calculation mistake occurred.

Hope you got it now.

right?
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Yes, I got it. Thank you. :)

I think 64

no of bits required for opcode=32-(4+6+16)=6

(16  bits memory address field as there are 64k words)

so no of different instruction=2^6=64

is it 64?
by (41 points)
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Nope, answer is 32. I'm not getting why did they took 17 bits for memory address field?
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word size=16 bit=2 Bytes

64k words=64k*2 Bytes=128KB

so,memory address field requires 17 bit
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Thank you so much for the clairfication. :)