GATE CS 2014 Set 1 — Question 37
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Computer Networks → Transport Layer → TCP Congestion Control
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Question
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.
Correct answer
1100 to 1300
Solution
When a timeout occurs in TCP:
In this phase, increases by 1 MSS (2 KB) per RTT.
Target is 32 KB. Current is 16 KB.
Number of increments needed = increments.
This takes 8 RTTs.Total Time:
Total RTTs = 3 (Slow Start) + 8 (Congestion Avoidance) = 11 RTTs.
Given RTT = 100 msec.
Total Time = msec.(Note: Depending on implementation details regarding when the window is checked/incremented, ranges like 1100-1500 are sometimes accepted, but 1100 is the precise theoretical minimum.)
1.The threshold () is set to half the current congestion window size.
2.The congestion window () is reset to 1 Maximum Segment Size (MSS).
3.The system enters Slow Start phase until reaches .
4.Then it enters Congestion Avoidance phase.
Step-by-step progression:- Start: KB
- Round 1 (Slow Start): doubles KB (Time: 1 RTT)
- Round 2 (Slow Start): doubles KB (Time: 2 RTTs)
- Round 3 (Slow Start): doubles KB (Time: 3 RTTs). Now , switch to Congestion Avoidance.
In this phase, increases by 1 MSS (2 KB) per RTT.
Target is 32 KB. Current is 16 KB.
Number of increments needed = increments.
This takes 8 RTTs.Total Time:
Total RTTs = 3 (Slow Start) + 8 (Congestion Avoidance) = 11 RTTs.
Given RTT = 100 msec.
Total Time = msec.(Note: Depending on implementation details regarding when the window is checked/incremented, ranges like 1100-1500 are sometimes accepted, but 1100 is the precise theoretical minimum.)
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