# Checking calculation

Can someone confirm if the below calculation looks good?
Assumptions:
DAU = 300 M
Requests per day on average = 20
Users post 2 tweets per day on average.
10% of tweets contain media.
Data is stored for 5 years.
Twitter server can handle approximately 6k requests per second (RPS)
Around 50% of tweets contain images and around 5% contain videos.
Assume that a single user views 50 tweets in a day

Estimating the Number of Servers (web):
Total Requests per day = 300 M * 50% * 20 = 3B
Total Requests per sec (t) = 3B / 86400 = 35000 (approx)
Peak hours Requests per sec = 2 * t = 70000
Number of servers required in a single data center = Number of requests per sec / RPS = 70000 / 6000 = 12
Accounting for redundancy (3x Redundancy) = 12 * 3 = 36
Accounting for multiple geographic distribution, number of servers required = 36 * 4 = 144

Query per second (QPS) estimate:
Daily active users (DAU) = 300 million * 50% = 150 million
Tweets QPS = 150 million * 2 tweets / 24 hour / 3600 seconds = ~3500
Peek QPS = 2 * QPS = ~7000

Storage estimate:
Total Requests per day = 300 M * 50% * 2 (tweets per day) = 300 M tweets per day
Storage required per tweet = 150 B
Storage required per image = 250 KB
Storage required per video = 5 MB
Storage for tweets = 300 M * 150B = 45 GB
Storage for images = 300 M * 50% * 250KB = 37.5 TB
Storage for videos = 300 M * 5% * 5MB = 75 TB
Total storage per day = 112.5 TB
5-year media storage = 112.5 TB * 365 * 5 = 205 PB

Bandwidth estimate:
Incoming traffic bandwidth = (112.5TB * 8) / 86400 = 10.5 Gbps
Outgoing traffic:
DAU = 150 million
Daily tweets viewed = 50 per user
Tweets viewed / second = (150M * 50) / 86400 = 87K
Bandwidth required for tweets = 87000 * 150B * 8 = 0.1Gbps
Bandwidth required for images = 87000 * 0.5 * 250KB = 87Gbps
Bandwidth required for videos = 87000 * 0.05 * 5MB = 174Gbps
Total Outgoing bandwidth = (0.1 + 87 + 174) Gbps = 261Gbps
Total bandwidth requirements = 261 + 10.5 = 271 Gbps

Hi Ritesh.