Printable View
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?
On Jun 13, 6:45*pm, grocery_stocker <cdal...@gmail.com> wrote:[color=blue]
> 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?[/color]
er delayed, not delyated. I typed the passage because there is no way
for me to copy and section question.
On Jun 13, 8:45*pm, grocery_stocker <cdal...@gmail.com> wrote:[color=blue]
> 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?[/color]
Sounds like homework.
Think about it. What does the 10M represent, and what does the 4G
represent?
On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
<robertwess...@yahoo.com> wrote:[color=blue]
> On Jun 13, 8:45*pm, grocery_stocker <cdal...@gmail.com> wrote:
>[color=green]
> > This is quoted from an old Stanford CS PhD document that I have[/color]
>[color=green]
> > "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"[/color]
>[color=green]
> > What does the 8 represent in the 8 * 4G/10M equation?[/color]
>
> Sounds like homework.
>
> Think about it. *What does the 10M represent, and what does the 4G
> represent?[/color]
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.
On Jun 13, 11:14*pm, grocery_stocker <cdal...@gmail.com> wrote:[color=blue]
> On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
>
>
>
>
>
> <robertwess...@yahoo.com> wrote:[color=green]
> > On Jun 13, 8:45*pm, grocery_stocker <cdal...@gmail.com> wrote:[/color]
>[color=green][color=darkred]
> > > This is quoted from an old Stanford CS PhD document that I have[/color][/color]
>[color=green][color=darkred]
> > > "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"[/color][/color]
>[color=green][color=darkred]
> > > What does the 8 represent in the 8 * 4G/10M equation?[/color][/color]
>[color=green]
> > Sounds like homework.[/color]
>[color=green]
> > Think about it. *What does the 10M represent, and what does the 4G
> > represent?[/color]
>
> 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][/color]
2^32 possible bytes, you mean. :)
So have to multiply by 8 for bits.
-Le Chaud Lapin-
On Jun 13, 11:14*pm, grocery_stocker <cdal...@gmail.com> wrote:[color=blue]
> On Jun 13, 7:16*pm, "robertwess...@yahoo.com"
>
>
>
>
>
> <robertwess...@yahoo.com> wrote:[color=green]
> > On Jun 13, 8:45*pm, grocery_stocker <cdal...@gmail.com> wrote:[/color]
>[color=green][color=darkred]
> > > This is quoted from an old Stanford CS PhD document that I have[/color][/color]
>[color=green][color=darkred]
> > > "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"[/color][/color]
>[color=green][color=darkred]
> > > What does the 8 represent in the 8 * 4G/10M equation?[/color][/color]
>[color=green]
> > Sounds like homework.[/color]
>[color=green]
> > Think about it. *What does the 10M represent, and what does the 4G
> > represent?[/color]
>
> 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.[/color]
You're close - it's 4G *bytes*. Times eight to make it bits and
divide it by a bit rate (10Mbps).
On Jun 13, 9:56*pm, "robertwess...@yahoo.com"
<robertwess...@yahoo.com> wrote:[color=blue]
> On Jun 13, 11:14*pm, grocery_stocker <cdal...@gmail.com> wrote:
>
>
>
>
>[color=green]
> > On Jun 13, 7:16*pm, "robertwess...@yahoo.com"[/color]
>[color=green]
> > <robertwess...@yahoo.com> wrote:[color=darkred]
> > > On Jun 13, 8:45*pm, grocery_stocker <cdal...@gmail.com> wrote:[/color][/color]
>[color=green][color=darkred]
> > > > This is quoted from an old Stanford CS PhD document that I have[/color][/color]
>[color=green][color=darkred]
> > > > "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"[/color][/color]
>[color=green][color=darkred]
> > > > What does the 8 represent in the 8 * 4G/10M equation?[/color][/color]
>[color=green][color=darkred]
> > > Sounds like homework.[/color][/color]
>[color=green][color=darkred]
> > > Think about it. *What does the 10M represent, and what does the 4G
> > > represent?[/color][/color]
>[color=green]
> > 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.[/color]
>
> 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 -[/color]
Gosh I feel dense. I should have read the problem a bit closer. Thanks.
grocery_stocker wrote:[color=blue]
> 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?
>
>
>[/color]
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