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A computer uses a memory unit with $256$ K words of $32$ bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of $64$ registers and an address part. How many bits are there in the operation code, the register code part and the address part?

  1. $7,6,18$
  2. $6,7,18$
  3. $7,7,18$
  4. $18,7,7$
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i am assuming it is word addressable, each word is 32 bits

Computer has 256 K words ===> 18 bits required to find a word in a Memory.

Given that there are 64 registers ==> 6 bits required to identify a register uniquely

Given that instruction code stored in 1 word of memory ===> there are 32 bits in a instruction.

Instruction has four parts, 1 bit for indirect, an operation code, register code part, and address part.

==> no.of bits reserved for operation code = 32-1-6-18 = 7
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$1$ indirect bit

$log_2 64 = 6$ bits for registers

$18$ bits for address ($2^{18} = 256K$)

Bits for operation code = $32-1-6-18 = 7$

$3. \ \ \  7, 6 , 18 $ is the answer
Answer:

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