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in CO and Architecture by Active (1.1k points)
edited by | 105 views
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32 - (3+6+18) = 5 bits, 32 different instructions are possible..
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answer they have given to be 2^3=8. Even i got 2^5 = 32.
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is this a correct solution ... i just doubt.

+2
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.

2 Answers

+2 votes
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
by Loyal (6.7k points)
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got it thanks.
0 votes

8 instn is possible

by Boss (36.5k points)
0

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

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