viewuser/1652/Lee5941

The Eduladder is a community of students, teachers, and programmers just interested to make you pass any exams. So we solve previous year question papers for you.
See Our team
Wondering how we keep quality?
Got unsolved questions?

Ask Questions

Advance-Computer-Architecture-10CS74-unit-6-->View question


Q.65) Instruction execution in a processor is divided into 5 stages. Instruction Fetch(IF), Instruction Decode (ID), Operand Fetch(OF), Execute(EX), and Write Back(WB), These stages take 5,4,20, 10 and 3 nanoseconds (ns) respectively. A pipelined implementation of the processor requires buffering between each pair of consecutive stages with a delay of 2 ns . -gate computer science 2017

Two pipelined implementations of the processor are contemplated: 
(i) a na├»ve pipeline implementation (NP) with 5 stages and 
(ii) an efficient pipeline (EP) where the OF stage id divided into stages OF1 and OF2 with execution times of 12 ns and 8 ns respectively. 

The speedup (correct to two decimals places) achieved by EP over NP in executing 20 independent instructions with no hazards is ________________. 

A) 1.50-1.51
B) 1.51-1.52
C) 1.52-1.53
D) 1.53-1.54


By:satyashiromani

Taged users:
|aksingh1818|LupiN|Johnny|milan-ransingh|Jessika-K|milanyoyoyogmailcom|Purnima|satyashiromani|DEEPAK202|vaishnavi-Deshpande|Umang|Aparna-Dasgupta|iamknown|ThreeRed|Msshikhil|harshshah822|Lee5941|dillu550|deepuckraj|123|Manisha12|pankaj|prajwalamv|gagandeep|Raaghav-singhal|metaphor|Govt|Aaditi|madachod|tichaona-garaidenga

Likes:
|satyashiromani|aksingh1818|deepuckraj

Dislikes:
Be first to dislike this question

Talk about thisDelete|Like|Dislike|

You may also like our videos



Answers

A) 1.50-1.51

Explanation :-
 
Given, total number of instructions (n) = 20 
For naive pipeline (NP):
Number of stages(k) = 5
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 22, 12, 5 } = 22 nsec
Execution time (Enp) = ( k + n - 1 )*Tp = ( 5 + 20 - 1 )*22 = 528 nsec
For efficient pipeline (EP):
number of stages(k) = 6 ( delay with 20 nsec stage is divided into 12 nsec and 8 nsec )
Clock time (Tp) = max { (stage delay+buffer delay) } = { 7, 6, 14, 10, 14, 5 } = 14 nsec
Execution time (Eep) =  ( k + n - 1 )*Tp = ( 6 + 20 - 1 )*14 = 350 nsec
Therefore, Speedup = (Enp) / (Eep) = 528 / 350 = 1.508 

This explanation is contributed by Satya Shiromani

deepuckraj

Likes:
Be first to like this answer

Dislikes:
Be first to dislike this answer
Talk about this|Once you have earned teacher badge you can edit this questionDelete|Like|Dislike|
------------------------------------


Lets together make the web is a better place

We made eduladder by keeping the ideology of building a supermarket of all the educational material available under one roof. We are doing it with the help of individual contributors like you, interns and employees. So the resources you are looking for can be easily available and accessible also with the freedom of remix reuse and reshare our content under the terms of creative commons license with attribution required close.

You can also contribute to our vision of "Helping student to pass any exams" with these.
Answer a question: You can answer the questions not yet answered in eduladder.
Career: Work or do your internship with us.
Create a video: You can teach anything and everything each video should be less than five minutes should cover the idea less than five min.
Donate: Ad revenue alone is not able to take care of our server cost consider donating at least a dollar Click here to donate.

If you want the help of doing any of these never hesitate to drop an email to me arunwebber[at]gmail[dot]com, Or you can ping us on WhatsApp ph: 8147433408

Can you help us to add better answer here? Please see this



Not the answer you're looking for? Browse other questions from this Question paper or ask your own question.

Join eduladder!