Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  On a TCP connection, current congestion windo... Start Learning for Free
On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)
    Correct answer is '4096'. Can you explain this answer?
    Verified Answer
    On a TCP connection, current congestion window size is 4 KB. The windo...
    As Receiver window size is 6KB and network congestion window size is 4KB so sender has to send only 4KB window.
    Sender window size = min(Window_congestion,Window_adverstised) = min(4KB,6KB) = 4KB
    Go through this guide.
    View all questions of this test
    Most Upvoted Answer
    On a TCP connection, current congestion window size is 4 KB. The windo...
    Explanation:

    The congestion window size determines the amount of data that can be sent by the sender without receiving acknowledgments from the receiver. It is used to control the flow of data and avoid network congestion.

    Given:
    - Congestion window size (CWND) = 4 KB (4096 bytes)
    - Receiver window size (RWND) = 6 KB (6144 bytes)
    - Last byte sent (LastByteSent) = 10240
    - Last byte acknowledged (LastByteAcked) = 8192

    Calculating the current window size at the sender:

    To calculate the current window size at the sender, we need to find the number of bytes that have been acknowledged by the receiver. This can be calculated by subtracting the last byte acknowledged from the last byte sent.

    BytesAcked = LastByteSent - LastByteAcked
    = 10240 - 8192
    = 2048 bytes

    Calculating the new window size:

    The new window size is determined by the minimum of the congestion window size and the receiver window size.

    NewWindow = min(CWND, RWND)
    = min(4096, 6144)
    = 4096 bytes

    Explanation:
    The current window size at the sender is 4096 bytes. This means that the sender can send up to 4096 bytes of data without receiving acknowledgments from the receiver. Once the sender receives acknowledgments for all the sent data within the current window, the window size can be increased to allow more data to be sent.

    It is important to note that the receiver's advertised window size (RWND) does not directly affect the congestion window size (CWND). The congestion window size is determined by the sender's congestion control algorithm and may be adjusted based on network conditions and congestion signals.

    In this scenario, the sender has a congestion window size of 4096 bytes, which is the minimum of its own window size and the receiver's advertised window size. This helps in preventing congestion and optimizing the network performance.
    Explore Courses for Computer Science Engineering (CSE) exam

    Top Courses for Computer Science Engineering (CSE)

    On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer?
    Question Description
    On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. 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 On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. 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 On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer?.
    Solutions for On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
    Here you can find the meaning of On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer?, a detailed solution for On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer? has been provided alongside types of On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice On a TCP connection, current congestion window size is 4 KB. The window size advertised by the receiver is 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is___(in bytes)Correct answer is '4096'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
    Explore Courses for Computer Science Engineering (CSE) exam

    Top Courses for Computer Science Engineering (CSE)

    Explore Courses
    Signup for Free!
    Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
    10M+ students study on EduRev