Imagine a TCP session over wireless where the congestion window is fixed at 5 segments (congestion control is turned off and no fast retransmits). Segments may get lost but are not reordered. The receiver has an infinite buffer and it sends an acknowledgment as soon as it receives a segment, i. e., acknowledgments are not deferred. Similarly, the sender transmits a segment as soon as it is allowed to. Each segment carries 1000 bytes and the time to transmit a segment is 2 ms. Assume that transmission of ACK takes negligible time. Note that the retransmission timer for a segment is started after the last bit of the segment is sent. Assume Go-Back-5, and accumulative ACK is used.

Suppose two data segments with byte sequence numbers 3000 and 15000 are lost once during the transmission. How many segments get retransmitted under each of the following conditions?

A. Round trip time = 100 ms, Timeout = 101 ms

B. Round trip time = 100 ms, Timeout = 152 ms

Question

Escobar

Computers & Technology

14 August, 03:02

Imagine a TCP session over wireless where the congestion window is fixed at 5 segments (congestion control is turned off and no fast retransmits). Segments may get lost but are not reordered. The receiver has an infinite buffer and it sends an acknowledgment as soon as it receives a segment, i. e., acknowledgments are not deferred. Similarly, the sender transmits a segment as soon as it is allowed to. Each segment carries 1000 bytes and the time to transmit a segment is 2 ms. Assume that transmission of ACK takes negligible time. Note that the retransmission timer for a segment is started after the last bit of the segment is sent. Assume Go-Back-5, and accumulative ACK is used.

Suppose two data segments with byte sequence numbers 3000 and 15000 are lost once during the transmission. How many segments get retransmitted under each of the following conditions?

A. Round trip time = 100 ms, Timeout = 101 ms

B. Round trip time = 100 ms, Timeout = 152 ms

+2

Answers (1)

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Nathen · Answer 1

a) 179 segments

b) 28 segments

Explanation:

Given data:

A) Round trip time = 100 ms, Timeout = 101 ms

number of segments for round trip = 3000/100 = 30 segments

number of segments for timeout = 15000/100 = 150 segments

total number of segments = 150 + 30 = 180

segments that get re-transmitted under A

Total segments - timeout = 180 - 101 = 179 segments

B) Round trip = 100 ms Timeout = 152 ms

number of segments for round trip = 3000/100 = 30 segments

number of segments for timeout = 15000/100 = 150 segments

total segments = 150 + 30 = 180

segments that get re-transmitted under B

total segments - timeout = 180 - 152 = 28 segments