A multimedia streaming service uses TCP, UDP, SCTP, and QUIC for different components of its network communication during a live broadcast. The service architecture is designed to optimize the streaming of video data, control signals, user data, and real-time gaming commands.Transmission Details:UDP is used for real-time gaming commands, with 11 commands transmitted every second. Each UDP datagram has a payload of 500 bytes and a header of 8 bytes. TCP is employed for sending control signals with a transmission rate of 9 signals per second. Each TCP segment contains a payload of 1000 bytes and a header of 20 bytes. SCTP is chosen for transmitting user data, with 2 messages sent per second. Each message is divided into 4 chunks, with each chunk containing a payload of 500 bytes (2000 bytes total) and a 16-byte chunk header. QUIC is used for video data, sending 24 frames per second. Each QUIC packet holds a payload of 1500 bytes and a header of 32 bytes. Task:Calculate the total amount of data (in bytes) transmitted during a 7-second session across all protocols, accounting for both payload and header sizes for each type of data packet.

Question

BlackTom AI · Accepted Answer

Per-packet total = payload + header; multiply by the number of packets sent in 7 seconds for each protocol and sum the results.

1) Relevant knowledge points and concepts
- Total bytes transmitted = (payload bytes + header bytes) × number of packets.
- Number of packets = rate (packets/sec) × duration (seconds).
- For protocols that split a message into chunks (SCTP), sum chunk payloads and chunk headers per message before multiplying by message count.

2) Step-by-step reasoning and 3) calculations (showing intermediate steps)
- Session duration = 7 s.

UDP (gaming commands)
- Rate = 11 datagrams/s → total datagrams = 11 × 7 = 77
- Per datagram = 500 (payload) + 8 (header) = 508 bytes
- Total UDP = 77 × 508 = 77 × 500 + 77 × 8 = 38,500 + 616 = 39,116 bytes

TCP (control signals)
- Rate = 9 signals/s → total segments = 9 × 7 = 63
- Per segment = 1000 + 20 = 1,020 bytes
- Total TCP = 63 × 1,020 = 63,000 + 1,260 = 64,260 bytes

SCTP (user data)
- Rate = 2 messages/s → total messages = 2 × 7 = 14
- Each message = 4 chunks; per chunk = 500 (payload) + 16 (chunk header) = 516 bytes
- Per message = 4 × 516 = 2,064 bytes (equivalently 2,000 payload + 64 chunk headers)
- Total SCTP = 14 × 2,064 = 28,896 bytes

QUIC (video frames)
- Rate = 24 frames/s → total packets = 24 × 7 = 168
- Per packet = 1,500 + 32 = 1,532 bytes
- Total QUIC = 168 × 1,532 = 252,000 + 5,376 = 257,376 bytes

Sum all protocols:
- Total = 39,116 + 64,260 + 28,896 + 257,376
- Total = 103,376 + 28,896 + 257,376 = 132,272 + 257,376 = 389,648 bytes

4) Why the correct answer is correct
- Each term explicitly accounts for both payload and header bytes and multiplies by the exact count of packets/messages sent during 7 seconds. The arithmetic summation yields 389,648 bytes.

5) Other options
- Not applicable (numerical-answer problem with no alternative choices to compare).

6) Conclusion
- The total data transmitted in 7 seconds across all protocols is 389,648 bytes, which matches the provided answer.

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