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

Question

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 ?

• A. $3i$
• B. $2i + j$
• C. $2i + j + 20$
• D. $i + j$

Comments

@Bikram sir,

sir, please explain the answer.

Answer 1

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

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

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

Comments

@bharti

it is not given that 2^j distinct instruction.so opcode may repeat.am i right?