Awesome q2a theme
0 votes
32 views
Consider the type addressable memory unit of a computer has 64 K words of 16 bit each. The computer has an instruction format with 4 fields namely opcode, mode field specify 9-addressing modes, register address field with 50 registers and a memory address field. If an instruction is 32 bits long then the number of different instructions are _____.
closed with the note: I found the solution of this question in comments.
in CO & Architecture by (7 points)
closed by | 32 views
0
32??
0
Nope, answer is 32. I'm not getting why did they took 17 bits for memory address field?
0
oh, yes, it will be 32

32-(4+6+17)=5

Calculation mistake occurred.

Hope you got it now.

right?
0
Yes, I got it. Thank you. :)

1 Answer

0 votes
I think 64  

no of bits required for opcode=32-(4+6+16)=6

(16  bits memory address field as there are 64k words)

so no of different instruction=2^6=64

is it 64?
by (41 points)
0
Nope, answer is 32. I'm not getting why did they took 17 bits for memory address field?
+1
word size=16 bit=2 Bytes

64k words=64k*2 Bytes=128KB

so,memory address field requires 17 bit
0
Thank you so much for the clairfication. :)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Welcome to GATE CSE Doubts, where you can ask questions and receive answers from other members of the community.
Top Users Jul 2020
  1. Shaik Masthan

    39 Points

  2. hiteshpujari

    9 Points

  3. Venkatesh Akhouri

    6 Points

  4. Meghana518

    6 Points

  5. bittujash

    6 Points

  6. Pawan_k

    6 Points

  7. rits78671

    6 Points

  8. Sumaiyas

    6 Points

  9. srestha

    6 Points

  10. RavGopal

    4 Points

7,545 questions
1,783 answers
10,867 comments
90,484 users