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TCP Congestion Control Dynamics (Slow Start) (Variation 8)

Asked by Ananya_Sharma | Textbook Reference: Kurose & Ross CN

A TCP connection is using Slow Start. The initial Congestion Window (cwnd) is 1 MSS (Maximum Segment Size). The Threshold (ssthresh) is set to 8 MSS. Assuming no segment losses or timeouts occur, what will be the size of cwnd (in MSS) after the successful completion of 4 Round Trip Times (RTT)?

Community Explanations (3)

[Variation 8 Explanation] Let's trace slow start progression: - Start RTT 1: cwnd = 1 - Start RTT 2: cwnd = 2 - Start RTT 3: cwnd = 4 - Start RTT 4: cwnd = 8 (ssthresh reached! slow start ends, enters congestion avoidance) - Start RTT 5 (After 4 RTTs completed): cwnd increases linearly by 1 MSS: $8 + 1 = 9$ MSS. So the answer is **9**.

Answered by NileshNama | Agreed by 32 peers | ✓ Selected Solution

[Variation 8 Explanation] ### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!

Answered by NileshNama | Agreed by 8 peers

[Variation 8 Explanation] ### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!