GATE CS 2014 Set 1 Q.37 (CS Q.27): Transport Layer: UDP, TCP Flow and Congestion Control, Sockets

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 _________.

Your answer

Official answer and solution

Accepted answers: 1100 to 1300
From: Official GATE 2014 answer key

Solution

Follow the TCP Tahoe/Reno congestion control cycle from timeout through slow start up to ssthresh, followed by linear congestion avoidance to 32 KB.

Given parameters:

Congestion window when timeout occurs: $\text{cwnd} = 32\text{ KB} = 16\text{ MSS}$ (since $\text{MSS} = 2\text{ KB}$ ).
RTT = $100\text{ msec}$ .

Upon timeout:

$\text{ssthresh} = \frac{\text{cwnd}}{2} = \frac{32\text{ KB}}{2} = 16\text{ KB} = 8\text{ MSS}$ .
$\text{cwnd}$ is reset to $1\text{ MSS} = 2\text{ KB}$ .

Phase 1: Slow Start (exponential growth, doubling cwnd every RTT):

Start ( $t = 0$ ): $\text{cwnd} = 1\text{ MSS} = 2\text{ KB}$ .
After 1 RTT ( $t = 100\text{ ms}$ ): $\text{cwnd} = 2\text{ MSS} = 4\text{ KB}$ .
After 2 RTT ( $t = 200\text{ ms}$ ): $\text{cwnd} = 4\text{ MSS} = 8\text{ KB}$ .
After 3 RTT ( $t = 300\text{ ms}$ ): $\text{cwnd} = 8\text{ MSS} = 16\text{ KB} = \text{ssthresh}$ .
Slow start ends at 3 RTTs.

Phase 2: Congestion Avoidance (linear growth, increasing by 1 MSS = 2 KB per RTT):

After 4 RTT ( $t = 400\text{ ms}$ ): $\text{cwnd} = 9\text{ MSS} = 18\text{ KB}$ .
After 5 RTT ( $t = 500\text{ ms}$ ): $\text{cwnd} = 10\text{ MSS} = 20\text{ KB}$ .
After 6 RTT ( $t = 600\text{ ms}$ ): $\text{cwnd} = 11\text{ MSS} = 22\text{ KB}$ .
After 7 RTT ( $t = 700\text{ ms}$ ): $\text{cwnd} = 12\text{ MSS} = 24\text{ KB}$ .
After 8 RTT ( $t = 800\text{ ms}$ ): $\text{cwnd} = 13\text{ MSS} = 26\text{ KB}$ .
After 9 RTT ( $t = 900\text{ ms}$ ): $\text{cwnd} = 14\text{ MSS} = 28\text{ KB}$ .
After 10 RTT ( $t = 1000\text{ ms}$ ): $\text{cwnd} = 15\text{ MSS} = 30\text{ KB}$ .
After 11 RTT ( $t = 1100\text{ ms}$ ): $\text{cwnd} = 16\text{ MSS} = 32\text{ KB}$ .

Total elapsed time = $11 \times 100\text{ msec} = 1100\text{ msec}$ .

Final answer

1100

Common trap

Continuing exponential doubling past ssthresh (16 KB), rather than switching to linear additive increase.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 37 of 65, printed as CS Q.27
Section: Computer Science and Information Technology

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