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in CO and Architecture by Loyal (7.6k points) | 189 views

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GIven instruction size  =  1 memory word  =  32 bits

  No of bits needed to refer to a memory word = log(No of memory addresses/words) 

                                                                   = log2 (256K)

                                                                   = 18 bits

 No of bits needed to refer to processor registers  =  ceil ( log2 ( No of registers ))

                                                                        =   6 bits

 No of bits needed for addressing modes              =   ceil( log2 7 )

                                                                         =   3 bits

Hence no of bits left for opcode                          =     32 - 18 - 6 - 3

                                                                        =     5 bits...

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