computer organization

CO & Architecture Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 20 Comments

Materials for GATE preparation in Computer Organization and Architecture

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GATE Overflow
Aptitude Overflow

@Vaishnavi Gadhe Revision Strategy :: Let’s us we share here some General and Important Revision Strategy for GATE 2022 .. Make short notes :: Develop the habit of jotting down key concepts from every topic that you study which will make your revision easy... You can read the short notes when the concepts get blurry, having the key concepts ready will save you a lot of time. Having a book where you note down all your mistakes while solving practice problems will ensure that you do not repeat the same mistakes... Make yourself a compendium of all-important formulas so you… [...]

Answered: Made Easy Test Series

@LRU Optimal window size =(1+2a) where a =Tp/Tt ….. Full duplex link, so window size = 2* Tp * Band Width = 60 * 10-3 * 512*103 =30720 bytes… packet size = 64 bytes... window size in terms of packets = 30720 / 64 = 480…. optimal window size = 480+1= 481 ... 1. https://gateoverflow.in/144462/Cn-optimal-windows-size [...]

Answered: computer organization

@ramraj13 yes ... Non-Linear Pipeline : Non-Linear pipeline is a pipeline which is made of different pipelines that are present at different stages. The different pipelines are connected to perform multiple functions. It also has feedback and feed-forward connections. It is made such that it performs various function at different time intervals….. In Non-Linear pipeline the functions are dynamically assigned.... The transfer of control is nonlinear, A dynamic pipelining can start tasks from multiple reservation tables together to enable multiple numbers of initiations of multiple functions in the same pipeline.... 1. https://www.mediafire.com/folder/gp6z7khjzyl8d/gate_materials?fbclid=IwAR0ezzTYvJdobF2hXtr3xVNZpZOFtw96yHAywR2_j9BOKGe1mBNSgVUcsvw#vynf5o6pb4zxb and https://www.mediafire.com/folder/gp6z7khjzyl8d/gate_materials?fbclid=IwAR0ezzTYvJdobF2hXtr3xVNZpZOFtw96yHAywR2_j9BOKGe1mBNSgVUcsvw#mp3qyi2prh3fb 2. https://drive.google.com/file/d/0Byt7-j-JD0d0bmxlRkZGcjN2cjA/view?resourcekey=0-VfnVIJ2CEXaqv-0JK6w4CQ … [...]

Made Easy Test Series

“Consider L = {w011w | w ∈ (o+1)*}. Find minimum number of states that are required in DFA (L).” Found this question in Made Easy Test Series and I think the question is incorrect, as the language is not regular. The solution of this question given by Made Easy is “L is a language, which contains all the strings containing substring 011.” But for the given solution to be true L should be, L = {w011x | w,x ∈ (o+1)*}. Request you all to provide your comments. [...]

Answered: ISRO2007-11, GATE CSE 2001 | Question: 1.19

I have a different opinion on this old question : According to me, answer is B and D. Let me make my points : Throughput = #processes executed per unit time. If context switch time = 0; then no matter what algorithm you use, throughput is going to be same. Now, if CS time matters, throughput will be higher in non-preemptive algorithms.(as less #CS will happen here). So that rules out HRRN and RR as you may have more CS resulting in more time to complete all the n tasks, thus low throughput compared to non-preemptive algorithms. But, the rest 2… [...]

Answered: Made Easy Test Series

$S_{x}\ =\ \frac{t_{n}}{(1 \ + \ number \ of \ stalls \ per \ instruction)cycle time}$ $S_{x}\ =\ \frac{22}{1 \ + \ (0.2 \times 4)8 \ ns}$ $\Rightarrow \frac{22}{14.4} \ \Rightarrow 1.528$ $S_{y}\ =\ \frac{t_{n}}{(1 \ + \ number \ of \ stalls \ per \ instruction)cycle time}$ $S_{y}\ =\ \frac{22}{(1 \ + \ 1.6)5 \ ns}$ $\Rightarrow \frac{22}{13}\ \Rightarrow \ 1.692$ $\frac{S_{x}}{S_{y}} \ = \ \frac{1.528}{1.692} \ = \ 0.90$ [...]

CAT2021 Set 3 | Verbal Ability and Reading Comprehension Question: 1