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Typical conversion

1 byte = 8 bits  # The size of a char
2 - 4 bytes      # The size of integer
1 million int32 characters (8 bytes each) takes up 8MB.

1024 bytes = 1 kilobyte (K / Kb) 
1024 * 1024 = 1,048,576 bytes = 1024 kilobytes = 1 megabyte (M / MB)
1024 MB = 1 gigabyte (G / GB)
1024 GB = 1 terabyte (T / TB)
1024 TB = 1 petabyte (P / PB)

1 Kb = 1e3 byte
1 MB = 1e6 byte
1 GB = 1e9 byte
1 TB = 1e12 byte
1 PB = 1e15 byte

1 Kb to bytes = 1024 ^ 1
1 MB to bytes = 1024 ^ 2
1 GB to bytes = 1024 ^ 3
1 TB to bytes = 1024 ^ 4
1 PB to bytes = 1024 ^ 5

Unit

thousand = 1e3 / 1000
million = 1e6 / 1,000,000
billion = 1e9 / 1,000,000,000

Time

1 hr = 60 * 60 = 3600 seconds
1 day = 24 * 3600 = 86,400 seconds
1 month (30 days) = 720 hours = 2,592,000 seconds

Capacity Estimation and Constraints

Typical system design questions involves solving the Fermi Problem through Fermi Questions or Back-of-the-Envelope Calculations.

  • Traffic estimate
  • Storage estimates
  • Bandwidth estimates
  • Memory estimates

Example with url shortener. Assumption, read and write ratio is 100:1, and we have 500M new URL shortening per month (write). So:

Traffic Estimate

ratio read:write is 100:1
50B read:500M write 

Query per Second (QPS) for a month:
1 month (30 days) = 30 * 24 * 3600
Write QPS: 500e6/(30 * 24 * 3600) = 192.9012345679 ~200 URLs/s
Read QPS: Write QPS * 100 = ~19 K/s

Storage Estimate

Assume we are going to store the URLs for 5 years. Since we estimate 500M URLs per month, and we assume each of them will take 500 bytes:
500M * 12 months * 5 = 30B (30e9)
30 billion * 500 bytes = 30e9 * 500 = 15e12 bytes
Since 1 TB = ~ 1e12 bytes:
That leaves us with ~15 TB (15e12) total storage

Bandwidth Estimate

For every write requests, since we expect 200 new URLs per second, the transfer would be:
200 URLs/s * 500 bytes = 100,000 bytes
Total incoming data is: ~100 KB/s

For read requests, we expect ~19 K/s URL redirection, which makes it:
19e3 * 500 = 9,500,000 bytes
Total outgoing data is: ~9.5 MB/s

Memory Estimates

To cache hot URLs, we follow the 20-80 rule (pareto's principle), meaning 20% of URLs generate 80% of traffic.
Since we have 19K requests per second, in one day, we would have:

19e3 * 3600 * 24 = 1,641,600,000
~1.7 billion requests

To cache 20% of this, we would need at least 170 GB of memory:
0.2 * 1.7e9 * 500 bytes / 1e9 (1GB) = ~170 GB

To summarize:

Write: 200/s
Read: 19K/s
Incoming data: 100 KB/s
Outgoing data: 9MB/s
Storage for 5 years: 15TB
Memory for cache 170GB

References