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Test by Bikram | Computer Organization and Architecture | Test 2 | Question: 24

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A byte addressable computer can support maximum of $2^i$ KB memory and has $2^j$ instructions. An instruction involving $2$ operands and $1$ operator needs how many bits ?

  1. $3i$
  2. $2i + j$
  3. $2i + j + 20$
  4. $i + j$

1 Answer

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9 9 votes

   Memory size = 2 i KB = 2 (i+10)  byte 

so, (i+10) bits are required for 1 memory cell  .

here , it is  given that  "2j " instructions are  present , so there must be    "2j " opcode . so j bits for opcode .

intruction format  is:    !! opcode !!   address !!  address !!

 so, total bits needed = J   +  (i+10) +(i+10) = J +2i + 20 bits  

 option c is correct.

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