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In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.
    Correct answer is '9.6'. Can you explain this answer?
    Verified Answer
    In TCP connection, When SYN segment is sent, the value of retransmissi...
    Concept:
    The measured round-trip time for a segment is the time required for the segment to reach the destination and be acknowledged, although the acknowledgment may include other segments. In TCP, there can be only one RTT measurement in progress at any time.
    Calculation:
    RTTM = 3.2
    RTTS = 3.2
    RTTD = RTTS/2 = 1.6
    RTO = RTTS + 4 × RTTD = 9.6
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    Most Upvoted Answer
    In TCP connection, When SYN segment is sent, the value of retransmissi...
    Time-out value for the next segment.

    There are different formulas and methods to calculate the retransmission timeout (RTO) in TCP, but one common approach is the Karn/Partridge algorithm, which uses the Estimated Round-Trip Time (ERTT) and the Deviation of RTT (DEVRTT) to adjust the RTO dynamically. The basic idea is to measure the time it takes for a segment to travel from sender to receiver and back, and use this information to estimate the expected time for future segments, while accounting for variations in network delay and congestion.

    In this scenario, we have some information about the time it takes for the SYN segment and the SYN ACK segment to travel between sender and receiver, but we don't have any previous measurements or statistics about the network conditions. Therefore, we can use a simplified version of the Karn/Partridge algorithm that assumes a fixed multiplier for the ERTT and a fixed factor for the DEVRTT.

    Let's call the RTO for the SYN segment RTO1, and the ERTT and DEVRTT for this segment as ERTT1 and DEVRTT1, respectively. We can calculate these values as follows:

    - ERTT1 = (1 - alpha) * ERTT0 + alpha * RTT1, where ERTT0 is the previous ERTT (or a default value if this is the first measurement), alpha is a smoothing factor (usually between 0.1 and 0.5), and RTT1 is the measured Round-Trip Time for the SYN segment, which is equal to the time it took for the SYN ACK segment to arrive minus the time it took for the SYN segment to travel to the receiver. In this case, we have RTT1 = 3.2 - 11 = -7.8 sec., which is negative because the SYN ACK arrived before the expected time. To avoid negative values, we can set a lower bound for the ERTT, such as 1 sec. Therefore, we get: ERTT1 = (1 - 0.5) * 1 + 0.5 * (-7.8) = -3.4 sec.

    - DEVRTT1 = (1 - beta) * DEVRTT0 + beta * abs(ERTT1 - ERTT0), where DEVRTT0 is the previous DEVRTT (or a default value if this is the first measurement), beta is another smoothing factor (usually between 0.1 and 0.5), and abs() is the absolute value function. The DEVRTT represents the degree of variation or fluctuation in the RTT values, and is used to adjust the RTO based on the estimated error in the ERTT. In this case, we can assume a default value for DEVRTT0, such as 3 sec. Therefore, we get: DEVRTT1 = (1 - 0.25) * 3 + 0.25 * abs(-3.4 - 1) = 2.1 sec.

    - RTO1 = ERTT1 + 4 * DEVRTT1, where 4 is a scaling factor that provides some safety margin for unexpected delays or losses. This formula calculates a new RTO value for the next segment, based on the estimated time it takes for a segment to travel from sender to receiver and back
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    In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.Correct answer is '9.6'. Can you explain this answer?
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    In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.Correct answer is '9.6'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.Correct answer is '9.6'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.Correct answer is '9.6'. Can you explain this answer?.
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