Consider a cross-country file transfer from host A to host B. The RTT between th

Question

Consider a cross-country file transfer from host A to host B. The RTT between the hosts is 140 ms and there is 90 Mbps of available bandwidth on the path from A to B. Assume that packets are 1500 bytes inclusive of headers and data. Assume A has an infinite supply of bits to send to B and that there is no other traffic on the path.
- How large must the congestion window be (in bytes) for the path utilization to be at least 50%?

- Assume the TCP connection begins in TCP slow start with an initial congestion window of 1 MSS (maximum segment size). Assume the TCP does not exit slow start and that there is no loss. For approximately how long (in seconds) must the TCP flow be sending data before the path utilization reaches 50%?

Explanation / Answer

Sender Link Utilisation= windowsize * Transmission Time*100/(Total Time)

  Total time=Transmission Time + RTT

  Transmission Time= Data Size/Bandwith

Transmission time = 1500B/90MbPS * 8 = 0.13ms(here 90Mbps b is considered as bit to convert into bytes it is divided by 8)

total time = 140+0.0167=140.13ms

50 = windowsize *0.13*100 /140.13

the window size must be atleast 539 that will be 539*1500 bytes = 808500bytes

now starting from 1 MSS, it will go till 539 MSS(525 MSS is a window size at stability or when utilization become 50%), since it is never leaving slow start phase, it will keep growing exponentially

1,2,4,8,16,32,64,128,256,512.1024..to reach 539 MSS it will take 10 RTT=1.4s, and to reach 1024 MSS, it will take 11RTT= 1.54s, therefore to reach 525 MSS it will take b/w (1.4s,1.54s)

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