The measured average round trip delay on a connection is 50 msec. If the link has a bandwidth of 1Gb/sec, what is the minimum number of Bytes that should be available in the window so that the transfer of the data does not become stop and wait?
Try an answer before revealing the guidance below.
Key Concepts
- Bandwidth-delay product
- RTT
- Window size
- Bits to bytes
Answer Approach
- Convert 50 ms to seconds.
- Multiply rate by RTT to get bits in flight.
- Divide by eight to express the window in bytes.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
The bandwidth-delay product is 1,000,000,000 bits/s × 0.050 s = 50,000,000 bits. Dividing by 8 gives 6,250,000 bytes, or about 6.25 MB of window space to keep the link busy.
Bandwidth-delay product = bandwidth × RTTWindow ≈ bandwidth × RTT / 8 bytesShortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next