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zhenek [66]
2 years ago
11

Я люблю есть гнезда петух

Computers and Technology
1 answer:
Vsevolod [243]2 years ago
6 0

Explanation:

xwvrmmymyjhrh ery6jfdegb

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Answer:  

For 32 bits Instruction Format:

OPCODE   DR               SR1                   SR2      Unused bits

a) Minimum number of bits required to represent the OPCODE = 3 bits

There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.

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b) Minimum number of bits For Destination Register(DR) = 4 bits

There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value.  4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.  

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c) Maximum number of UNUSED bits in Instruction encoding = 17 bits

Total number of bits used = bits used for registers + bits used for OPCODE  

     = 12 + 3 = 15  

Total  number of bits for instruction format = 32  

Maximum  No. of Unused bits = 32 – 15 = 17 bits  

OPCODE                DR              SR1             SR2              Unused bits

  3 bits              4 bits          4 bits           4 bits                17 bits

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