TCP wrap speed question - TCP-IP

This is a discussion on TCP wrap speed question - TCP-IP ; This is quoted from an old Stanford CS PhD document that I have "The key issue is being able to detect old delayed segments showing up. The 32 bits allow 4 gigabytes to be transmitted before wrap, which is 8 ...

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  1. TCP wrap speed question

    This is quoted from an old Stanford CS PhD document that I have

    "The key issue is being able to detect old delayed segments showing
    up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    a
    TCP segment is not expected to live in the net more than a few
    seconds so this should catch old delyated segment even just using
    half the range"

    What does the 8 represent in the 8 * 4G/10M equation?




  2. Re: TCP wrap speed question

    On Jun 13, 6:45*pm, grocery_stocker wrote:
    > This is quoted from an old Stanford CS PhD document that I have
    >
    > "The key issue is being able to detect old delayed segments showing
    > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > a
    > TCP segment is not expected to live in the net more than a few
    > seconds so this should catch old delyated segment even just using
    > half the range"
    >
    > What does the 8 represent in the 8 * 4G/10M equation?


    er delayed, not delyated. I typed the passage because there is no way
    for me to copy and section question.

  3. Re: TCP wrap speed question

    On Jun 13, 8:45*pm, grocery_stocker wrote:
    > This is quoted from an old Stanford CS PhD document that I have
    >
    > "The key issue is being able to detect old delayed segments showing
    > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > a
    > TCP segment is not expected to live in the net more than a few
    > seconds so this should catch old delyated segment even just using
    > half the range"
    >
    > What does the 8 represent in the 8 * 4G/10M equation?



    Sounds like homework.

    Think about it. What does the 10M represent, and what does the 4G
    represent?

  4. Re: TCP wrap speed question

    On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
    wrote:
    > On Jun 13, 8:45*pm, grocery_stocker wrote:
    >
    > > This is quoted from an old Stanford CS PhD document that I have

    >
    > > "The key issue is being able to detect old delayed segments showing
    > > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > > a
    > > TCP segment is not expected to live in the net more than a few
    > > seconds so this should catch old delyated segment even just using
    > > half the range"

    >
    > > What does the 8 represent in the 8 * 4G/10M equation?

    >
    > Sounds like homework.
    >
    > Think about it. *What does the 10M represent, and what does the 4G
    > represent?


    This isn't homework. But, 4G comes from 2^32 possible bits. I assume
    that 10Mbps is just the maximum speed of the medium that the data
    travels over.

  5. Re: TCP wrap speed question

    On Jun 13, 11:14*pm, grocery_stocker wrote:
    > On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
    >
    >
    >
    >
    >
    > wrote:
    > > On Jun 13, 8:45*pm, grocery_stocker wrote:

    >
    > > > This is quoted from an old Stanford CS PhD document that I have

    >
    > > > "The key issue is being able to detect old delayed segments showing
    > > > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > > > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > > > a
    > > > TCP segment is not expected to live in the net more than a few
    > > > seconds so this should catch old delyated segment even just using
    > > > half the range"

    >
    > > > What does the 8 represent in the 8 * 4G/10M equation?

    >
    > > Sounds like homework.

    >
    > > Think about it. *What does the 10M represent, and what does the 4G
    > > represent?

    >
    > This isn't homework. But, 4G comes from 2^32 possible bits. I assume
    > that 10Mbps is just the maximum speed of the medium that the data
    > travels over.- [Yes]


    2^32 possible bytes, you mean.

    So have to multiply by 8 for bits.

    -Le Chaud Lapin-

  6. Re: TCP wrap speed question

    On Jun 13, 11:14*pm, grocery_stocker wrote:
    > On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
    >
    >
    >
    >
    >
    > wrote:
    > > On Jun 13, 8:45*pm, grocery_stocker wrote:

    >
    > > > This is quoted from an old Stanford CS PhD document that I have

    >
    > > > "The key issue is being able to detect old delayed segments showing
    > > > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > > > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > > > a
    > > > TCP segment is not expected to live in the net more than a few
    > > > seconds so this should catch old delyated segment even just using
    > > > half the range"

    >
    > > > What does the 8 represent in the 8 * 4G/10M equation?

    >
    > > Sounds like homework.

    >
    > > Think about it. *What does the 10M represent, and what does the 4G
    > > represent?

    >
    > This isn't homework. But, 4G comes from 2^32 possible bits. I assume
    > that 10Mbps is just the maximum speed of the medium that the data
    > travels over.



    You're close - it's 4G *bytes*. Times eight to make it bits and
    divide it by a bit rate (10Mbps).

  7. Re: TCP wrap speed question

    On Jun 13, 9:56*pm, "robertwess...@yahoo.com"
    wrote:
    > On Jun 13, 11:14*pm, grocery_stocker wrote:
    >
    >
    >
    >
    >
    > > On Jun 13, 7:16*pm, "robertwess...@yahoo.com"

    >
    > > wrote:
    > > > On Jun 13, 8:45*pm, grocery_stocker wrote:

    >
    > > > > This is quoted from an old Stanford CS PhD document that I have

    >
    > > > > "The key issue is being able to detect old delayed segments showing
    > > > > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > > > > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > > > > a
    > > > > TCP segment is not expected to live in the net more than a few
    > > > > seconds so this should catch old delyated segment even just using
    > > > > half the range"

    >
    > > > > What does the 8 represent in the 8 * 4G/10M equation?

    >
    > > > Sounds like homework.

    >
    > > > Think about it. *What does the 10M represent, and what does the 4G
    > > > represent?

    >
    > > This isn't homework. But, 4G comes from 2^32 possible bits. I assume
    > > that 10Mbps is just the maximum speed of the medium that the data
    > > travels over.

    >
    > You're close - it's 4G *bytes*. *Times eight to make it bits and
    > divide it by a bit rate (10Mbps).- Hide quoted text -
    >
    > - Show quoted text -


    Gosh I feel dense. I should have read the problem a bit closer. Thanks.

  8. Re: TCP wrap speed question

    grocery_stocker wrote:
    > This is quoted from an old Stanford CS PhD document that I have
    >
    > "The key issue is being able to detect old delayed segments showing
    > up. The 32 bits allow 4 gigabytes to be transmitted before wrap,
    > which is 8 * 4G/10M = 3200 seconds or almost one hour at 10 Mbps. And
    > a
    > TCP segment is not expected to live in the net more than a few
    > seconds so this should catch old delyated segment even just using
    > half the range"
    >
    > What does the 8 represent in the 8 * 4G/10M equation?
    >
    >
    >


    The following is NOT intended to be an offensive remark, please don't
    take it as such:

    This reminds me of high school, when students used to try to remember
    formulas without paying attention to units of measure.

    If "8 * 4G/10M = 3200 seconds", was written as it should be: "8 *
    4GB/10Mb/sec. = 3200 sec.", the answer might be more apparent.

    If "K" is your constant (8), consider the following:

    K * (Bytes/bits/sec) = sec

    K * Bytes = sec * bits/sec

    K * Bytes = bits

    8 * Bytes = bits

    Bytes = bits/8

    Best Regards,
    News Reader

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