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###### Question 8427 – Computer-Networks
May 3, 2024

Assume that the bandwidth for a TCP connection is 1048560 bits/sec. Let α be the value of RTT in milliseconds (rounded off to the nearest integer) after which the TCP window scale option is needed. Let β be the maximum possible window size the window scale option. Then the values of α and β are

Correct Answer: C

Question 53 Explanation:
TCP header sequence number field consist 16 bits. The maximum number of sequence numbers possible = 216 = 65,535.
The wrap around time for given link = 1048560 * α.
The TCP window scale option is an option to increase the receive window size. TCP allows scaling of windows when wrap around time > 65,535.
==> 1048560 * α > 65,535*8 bits
==> α = 0.5 sec = 500 mss
Scaling is done by specifying a one byte shift count in the header options field. The true receiver window size is left shifted by the value in shift count. A maximum value of 14 may be used for the shift count value. Therefore maximum window size with scaling option is 65535 × 214.
63 milliseconds, 65535×214
63 milliseconds, 65535×216
500 milliseconds, 65535×214
500 milliseconds, 65535×216
##### SONAL Nagariya
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