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Consider an instance of TCPโ€™s $\text{Additive Increase Multiplicative Decrease (AIMD)}$ algorithm where the window size at the start of the slow start phase is $\text{1 MSS}$ and the threshold at the start of the first transmission is $\text{8 MSS}$. Assume that a timeout occurs during the sixth transmission and tenth transmission then what would be the window size at the $15^{th} \hspace{0.1cm} transmission $?
 

  1. $\text{10 MSS}$
  2. $\text{6 MSS}$
  3. $\text{5 MSS}$
  4. $\text{7 MSS}$
asked in Computer Networks by Boss (17.2k points) | 83 views

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Window sizes (in MSS) at the beginning of each time interval are as follows:

$t_1 \to 1$
$t_2 \to 2$
$t_3 \to 4$
$t_4 \to 8$ (threshold reached, additive increase)
$t_5 \to 9$
$t_6 \to 10$ (timeout, new window size = 1, threshold = 10/2 = 5)
$t_7 \to 1$
$t_8 \to 2$
$t_9 \to 4$
$t_{10} \to 5$ (timeout, new window size 1, threshold = 5/2 = 2)
$t_{11} \to 1$
$t_{12} \to 2$ (threshold reached, additive increase)
$t_{13} \to 3$
$t_{14} \to 4$
$t_{15} \to 5$
answered by Veteran (395k points)
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shouldn't the window size for $t_{10}$ be 5MSS as the threshold is 5MSS??
0
yes. Corrected ๐Ÿ‘
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