UGC NET CSE | December 2018 | Part 2 | Question: 14

Question

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?

• A. $7,6,18$
• B. $6,7,18$
• C. $7,7,18$
• D. $18,7,7$

Comments

64 registers => 6 bits => option 3.

---

@Shaik Masthan sir,

if it would have been a NAT question, then how can we make it clear that the system is word addressable & not byte-addressable, with the given language, as you’ve also mentioned that you’ve assumed memory to be word addressable as by byte-addressable none of the options are matching

---

in GATE exam, they will clearly specify. If not marks will allot to all possible cases.

Answer 1

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

Comments

can you tell how indirect is 1 bit please? looking everywhere for that closure.

---

mentioned in question itself.

The instruction has four parts: an indirect bit,

Answer 2

$1$ 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