Concepts of Computer Science

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

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 = 10^3 1 MB= 10^6 = 1024 KB 1 GB= 10^9 = 1024 MB 1 TB = 10^12 = 1024 GB Memory we need to see in Bytes Requests 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*360