PUERTA | GATE-CS-2014-(Conjunto-1) | Pregunta 65 – Part 3

Deje que el tamaño de la ventana de congestión de una conexión TCP sea de 32 KB cuando se agote el tiempo de espera. El tiempo de ida y vuelta de la conexión es de 100 milisegundos y el tamaño máximo de segmento utilizado es de 2 KB. El tiempo que tarda (en mseg) la conexión TCP en volver a la ventana de congestión de 32 KB es _________.
(A) 1100 a 1300
(B) 800 a 1000
(C) 1400 a 1600
(D) 1500 a 1700

Respuesta: (A)
Explicación:

Current size of congestion window in terms of number of segments
                            = (Size in Bytes)/(Maximum Segment Size)
                            = 32KB / 2KB 
                            = 16 MSS
          
When timeout occurs, in TCP's Slow Start algorithm, threshold is 
reduced to half which is 16KB or 8MSS. Also, slow start phase begins 
where congestion window is increased twice. 
So from 1MSS to 8 MSS window size will grow exponentially. 
Congestion window becomes 2MSS after one RTT and becomes 4MSS after
2 RTTs and 8MSS after 3 RTTs.  At 8MSS, threshold is reached and
congestion avoidance phase begins.  In congestion avoidance phase,
window is increased linearly. So to cover from 8MSS to 16MSS, it needs
8 RTTs

Together, 11RTTs are needed (3 in slow start phase and 8 in congestion
avoidance phase).

Cuestionario de esta pregunta