Test by Bikram | Computer Organization and Architecture | Test 1 | Question: 27

Question

The memory unit of a computer has $256$ K words of $32$ bits each. The computer has an instruction format with four fields: an operation code field, a mode field (to specify one of seven addressing modes), a register address field (to specify one of $60$ processor registers), and a memory address.

What are the number of bits, in each of the below given fields, when the instruction is $one$ memory word long?

1. Op-code
2. Mode
3. Register address
4. Memory address 

 

• A. $6,5,3,18$
• B. $5,6,3,18$
• C. $5,18,6,3$
• D. $5,3,6,18$

Answer 1

1.  Op-code => 32 - ( 3 + 6 + 18 ) = 5  bits
2.  Mode => we need to identify 1 of 7 addressing modes so 2³ = 8 hence 3 bits are required .
3. Register Address => 60 registers implies 2⁶ = 64 so 6 bits
4. Memory Address => 256 K = 2⁸ * 2¹⁰  = 2¹⁸  means 18 bits 

Hence option D , 5,3,6,18 .