Skip to main content

5 EXAMS APART FROM COLLEGE PLACEMENTS FOR B.TECH( COMPUTER SCIENCE)

To write these exams, you don't need support from the college if you have a courage and desire to achieve something big then pick any and go for better future after college.

1.               NET / JRF- This exam you can take for the admission in P.H.D or Research field. Qualifying this exam also makes a person eligible for applying lectureship in the universities.



2.               GATE-
            This exam you can take to get admission into M. Tech program in prestigious institutes such as IIT/IISC. The score is valid for two years and also helps in getting jobs in Public Service Union sectors.

            http://gate.iitk.ac.in/GATE2015/



3.               PSU-

            There are various PSU's like CIL, BEL, IOCL, SAIL e.t.c. which organize their exams separately or take based on the GATE score. One can simply crack them to complete the desire for working in most sought government engineering sector of the country.
           

           

4.               ELITMUS/AMCAT- Through these exams you can get a chance to sit in the off campus recruitment of the various companies. On getting higher percentile companies will call you further for their recruitment events.

            https://www.elitmus.com/jobs

            https://www.myamcat.com/

5.               CAT- Cracking this exam will open the door for the admission in management programs (MBA) after graduation. 




Comments

.

Popular posts from this blog

Basics of System Design

This article is first one from the series of articles dedicated to system design interviews. Here i am going to present the base scenario to consider before starting to solve system design problems.
Questions to ask?
1) what is the number of requests a website will recieve in a day/month/second? 2) what is the amount of memory a website will deal in a day/month/second? 3) what is the number of servers that can accomodate these requests?
To answer this , first we need to remember the below numbers:-
1 million = 10 lakh = 1000000 = 10^6 1 billion = 1000 million = 10^9
1 KB = 1024 B = 10^31 MB= 10^6 = 1024 KB 1 GB= 10^9 = 1024 MB1 TB = 10^12 = 1024 GB
Memory we need to see in BytesRequests we need to see in numbers
example :-
suppose a website recieves 100M requests every month then:-
requests per day = request per month /24 = 416700 requests requests per second = requests per day / (24*3600) = 4.8 requests per second
memory:-
if we take 20:80 principal where 20 percent is write and 80 percent is …

Calculating size of User and Cache storage

user storage:-
it can be solved using same scenario as mentioned in first article .

https://tech.nazarmubeenworks.com/2019/08/basics-of-system-design-chapter-0.html


Suppose there are 12 Million of users are adding every year meaning 1 Million per month.
So if we consider 5 year there will be around 60 million of users. Now in terms of data we can look to our DB table and get information about it. A user will be generally having name , id , address , some forien keys .
Lets assume 10 columns with each column on an average storing 4 byte of data.
10*4 = 40 bytes of data for one user.
60 * 10^6 * 40 = 2400 * 10^6 = 2.4 * 10^9 = 2.4 GB of data we need to store only user values.
Also while calculating storage for the user there is also one important point we need to remember is of ids. If we are going to have 60 Million of users which means 60*10^6 users so unique id’s will be
as we know below figures are almost equal
2^10=10^3 then 60*(10^3)^2 = 2^20 *2^6
which means we almost need 26 …