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qid 10610 · computer science

Question: A TCP entity sends 6 segments across the Internet. The measured round-trip times (RTTM) for the 6 segments are 68ms, 42ms, 65ms, 80ms, 38ms, and 75ms, respectively. Assume that the smooth averaged RTT (RTTs) and Deviation (RTTD) was respectively 70ms and 10ms just before the first of these six samples. According to the Jacobson's algorithm, the retransmission timeout (RTO) is given by one RTTs plus 4 times the value of RTTD. Determine the value of RTO (in ms) after the six segments using the Jacobson's algorithm if the exponential smoothing parameters (a and B) are 0.15 and 0.2 for calculating RTTs and RTTD respectively.

  1. 140.00
  2. 97.56
  3. 114.28
  4. 138.32
  5. 150.34
  6. 130.45
  7. 120.48
  8. 110.22
  9. 105.62
  10. 125.78

Our answer: C. 114.28 Source quote machine-checked (exact quote)

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How it was answered

Multi-step solver (maze), replayed by code

card: maze · card sha256 9830aeb64def1ea7…

Current source

GeeksforGeeks, "Algorithm for Dynamic Time out timer Calculation"

https://www.geeksforgeeks.org/algorithm-for-dynamic-time-out-timer-calculation/

“Jacobson's algorithm - Calculates TOT value more intuitively than basic algorithm. We assume initial round trip time i.e. PRTT.”

Source quote machine-checked (exact quote)

retrieved 2026-09-18T01:33:50.745Z

page text sha256 7f98610e52bbae1d… · content sha256 0bc0748d32e2d090…

addendum maze_qid10610_b4_maze_input_ADDENDUM_source_rs20260918T013350Z · sha256 c087c095c322a71b… · replaces the version below, addendum sha256 0de4d4ec8d0ed8bc…

Earlier version (superseded)

No public source has been found for this card yet (3 places checked internally).

addendum maze_qid10610_b4_maze_input_ADDENDUM_source · sha256 0de4d4ec8d0ed8bc…

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