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From: Christopher A. Lee on 10 Jan 2010 12:33 On Sun, 10 Jan 2010 07:40:32 -0800 (PST), species8350 <not_here.5.species8350(a)xoxy.net> wrote: >On Jan 10, 1:44�pm, Bob <b...(a)invalid.invalid> wrote: >> On 10/01/2010 13:19, species8350 wrote: >> >> >> >> > Note my router has not yet arrived, hence all my questions may see >> > very basic. But I will learn in time and with experience. >> >> > Thanks >> >> The router seems to be taking an awfully long time to arrive. You had a >> thread going in Aug 2009 about a router and wireless card. >> <http://groups.google.com/group/alt.comp.hardware/browse_thread/thread...> > >Regarding bandwidth and channel. > >As an example, if two people on different networks are uising the same >channel, and say that both have a signal of 20%, does this mean that >for this channel, %40 of the bandwith is being used. Hence, %60 is >left. Speed will be reduced by %40. Worse than that. Queueing theory raises its ugly head. The more bandwidth used, the greater the likelihood of having to wait to send a message. This was standard fare for designers of on-line systems in the days of slower computers and 4800 baud modems. Thirty-odd years ago I made my living applying it to computer systems (comm lines, transactions queued up for a task, even disk I/O). Basically, the greater the channel utilisation, the more likely you are to wait to get on to it. Average wait is governed by a factor of p/(1-p) where p is the utilisation. For low utilisation p is small so the result is very small but it's a curve that gets a lot bigger faster than p does. If there is nothing on the channel then the p/(1-p) is zero so there is no wait. 50% utilisation gives p of 0.5 so the average wait is 0.5/(1-0.5) = 1, ie the same as the transmission time of the message. So including the wait the total time is twice the original message time. 75% utilisation gives p as 0.75 so the average wait is 0.75/(1-0.75) = 3, ie three times the message transmission time. So including the wait the total time is four times the original message time. This is an extreme example because everything is a lot faster these days but it shows the relationship isn't linear. And there is a heck of a lot of bandwidth wasted with poorly designed web sites, animated ads etc. Bottom line: if somebody else is on the same channel, use a different one. >Thanks
From: species8350 on 10 Jan 2010 16:16 On Jan 10, 5:33 pm, Christopher A. Lee <ca...(a)optonline.net> wrote: > On Sun, 10 Jan 2010 07:40:32 -0800 (PST), species8350 > > > > > > <not_here.5.species8...(a)xoxy.net> wrote: > >On Jan 10, 1:44 pm, Bob <b...(a)invalid.invalid> wrote: > >> On 10/01/2010 13:19, species8350 wrote: > > >> > Note my router has not yet arrived, hence all my questions may see > >> > very basic. But I will learn in time and with experience. > > >> > Thanks > > >> The router seems to be taking an awfully long time to arrive. You had a > >> thread going in Aug 2009 about a router and wireless card. > >> <http://groups.google.com/group/alt.comp.hardware/browse_thread/thread....> > > >Regarding bandwidth and channel. > > >As an example, if two people on different networks are uising the same > >channel, and say that both have a signal of 20%, does this mean that > >for this channel, %40 of the bandwith is being used. Hence, %60 is > >left. Speed will be reduced by %40. > > Worse than that. Queueing theory raises its ugly head. The more > bandwidth used, the greater the likelihood of having to wait to send a > message. > > This was standard fare for designers of on-line systems in the days of > slower computers and 4800 baud modems. Thirty-odd years ago I made my > living applying it to computer systems (comm lines, transactions > queued up for a task, even disk I/O). > > Basically, the greater the channel utilisation, the more likely you > are to wait to get on to it. > > Average wait is governed by a factor of p/(1-p) where p is the > utilisation. For low utilisation p is small so the result is very > small but it's a curve that gets a lot bigger faster than p does. > > If there is nothing on the channel then the p/(1-p) is zero so there > is no wait. > > 50% utilisation gives p of 0.5 so the average wait is > 0.5/(1-0.5) = 1, ie the same as the transmission time of the message. > > So including the wait the total time is twice the original message > time. > > 75% utilisation gives p as 0.75 so the average wait is > 0.75/(1-0.75) = 3, ie three times the message transmission time. > > So including the wait the total time is four times the original > message time. > > This is an extreme example because everything is a lot faster these > days but it shows the relationship isn't linear. > > And there is a heck of a lot of bandwidth wasted with poorly designed > web sites, animated ads etc. > > Bottom line: if somebody else is on the same channel, use a different > one. > > > > >Thanks- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - OK, point made A related point occurs to me. If someone has a router that has both a wired connection and wifi connections, but the person concerned does not use wifi, they simply use the wired connection, then presumably the wifi ports will still transmit. Is this correct? So if I se a wif connection somewhere, I might be seeing simply a transmitting port. Have I got this right? I assume that all channels broadcast and receive at the same power? Therefore I can make free choices. Is a computer that is connected to a router by a wire known as an Ethernet connenction? Thanks Ps. As a matter of interest, I assume that there is no way that I can know where these other transmitters are. In the same way, others cannot know of my location.
From: Christopher A. Lee on 10 Jan 2010 16:42 On Sun, 10 Jan 2010 13:16:46 -0800 (PST), species8350 <not_here.5.species8350(a)xoxy.net> wrote: >On Jan 10, 5:33�pm, Christopher A. Lee <ca...(a)optonline.net> wrote: >> On Sun, 10 Jan 2010 07:40:32 -0800 (PST), species8350 >> >> >> >> >> >> <not_here.5.species8...(a)xoxy.net> wrote: >> >On Jan 10, 1:44�pm, Bob <b...(a)invalid.invalid> wrote: >> >> On 10/01/2010 13:19, species8350 wrote: >> >> >> > Note my router has not yet arrived, hence all my questions may see >> >> > very basic. But I will learn in time and with experience. >> >> >> > Thanks >> >> >> The router seems to be taking an awfully long time to arrive. You had a >> >> thread going in Aug 2009 about a router and wireless card. >> >> <http://groups.google.com/group/alt.comp.hardware/browse_thread/thread...> >> >> >Regarding bandwidth and channel. >> >> >As an example, if two people on different networks are uising the same >> >channel, and say that both have a signal of 20%, does this mean that >> >for this channel, %40 of the bandwith is being used. Hence, %60 is >> >left. Speed will be reduced by %40. >> >> Worse than that. Queueing theory raises its ugly head. The more >> bandwidth used, the greater the likelihood of having to wait to send a >> message. >> >> This was standard fare for designers of on-line systems in the days of >> slower computers and 4800 baud modems. Thirty-odd years ago I made my >> living applying it to computer systems (comm lines, transactions >> queued up for a task, even disk I/O). >> >> Basically, the greater the channel utilisation, the more likely you >> are to wait to get on to it. >> >> Average wait is governed by a factor of p/(1-p) where p is the >> utilisation. For low utilisation p is small so the result is very >> small but it's a curve that gets a lot bigger faster than p does. >> >> If there is nothing on the channel then the p/(1-p) is zero so there >> is no wait. >> >> 50% utilisation gives p of 0.5 so the average wait is >> 0.5/(1-0.5) = 1, ie the same as the transmission time of the message. >> >> So including the wait the total time is twice the original message >> time. >> >> 75% utilisation gives p as 0.75 so the average wait is >> 0.75/(1-0.75) = 3, ie three times the message transmission time. >> >> So including the wait the total time is four times the original >> message time. >> >> This is an extreme example because everything is a lot faster these >> days but it shows the relationship isn't linear. >> >> And there is a heck of a lot of bandwidth wasted with poorly designed >> web sites, animated ads etc. >> >> Bottom line: if somebody else is on the same channel, use a different >> one. >> >> >> >> >Thanks- Hide quoted text - >> >> - Show quoted text -- Hide quoted text - >> >> - Show quoted text - > >OK, point made > >A related point occurs to me. > >If someone has a router that has both a wired connection and wifi >connections, but the person concerned does not use wifi, they simply >use the wired connection, then presumably the wifi ports will still >transmit. Is this correct? So if I se a wif connection somewhere, I >might be seeing simply a transmitting port. Have I got this right? Yes. But you should disable the wireless port from the device manager. >I assume that all channels broadcast and receive at the same power? >Therefore I can make free choices. For the same adapter and receiver. But remember the overlapping bell curves which can cause interference. >Is a computer that is connected to a router by a wire known as an >Ethernet connenction? Colloquially but it's better to say Wired Ethernet to avoid ambiguity. Ethernet is a multilayered protocol with at the bottom a physical transmission layer. The cable oe the wireless stream are the physical layer. >Thanks > >Ps. As a matter of interest, I assume that there is no way that I can >know where these other transmitters are. In the same way, others >cannot know of my location. It depends what you want. When I visited friends in Silicon Valley a while back their city had municipally provided wired internet. There was an on-line map showing their locations. You can by instruments which detect transmissions and use them. Incidentally it surprised me how many of their neighbours had no encryption on their routers so anybody could leech off them if they wanted to. But does it really matter if they know your location? Use a neutral name for the SSID that is distinctive like "Battersea Dog's Home", not "Fred's Network" or "Home Network". Use the strongest encryption your router will support. One of the problems with public wifi is that it is not encrypted. The latest home routers from Netgear and other companies allow two different levels of use - your own and guest with different passkeys so guests can use the router to get to the broadband connection but not access your private stuff eg your networked backup disk.
From: species8350 on 10 Jan 2010 20:16 On Jan 10, 9:42 pm, Christopher A. Lee <ca...(a)optonline.net> wrote: > On Sun, 10 Jan 2010 13:16:46 -0800 (PST), species8350 > > > > > > <not_here.5.species8...(a)xoxy.net> wrote: > >On Jan 10, 5:33 pm, Christopher A. Lee <ca...(a)optonline.net> wrote: > >> On Sun, 10 Jan 2010 07:40:32 -0800 (PST), species8350 > > >> <not_here.5.species8...(a)xoxy.net> wrote: > >> >On Jan 10, 1:44 pm, Bob <b...(a)invalid.invalid> wrote: > >> >> On 10/01/2010 13:19, species8350 wrote: > > >> >> > Note my router has not yet arrived, hence all my questions may see > >> >> > very basic. But I will learn in time and with experience. > > >> >> > Thanks > > >> >> The router seems to be taking an awfully long time to arrive. You had a > >> >> thread going in Aug 2009 about a router and wireless card. > >> >> <http://groups.google.com/group/alt.comp.hardware/browse_thread/thread...> > > >> >Regarding bandwidth and channel. > > >> >As an example, if two people on different networks are uising the same > >> >channel, and say that both have a signal of 20%, does this mean that > >> >for this channel, %40 of the bandwith is being used. Hence, %60 is > >> >left. Speed will be reduced by %40. > > >> Worse than that. Queueing theory raises its ugly head. The more > >> bandwidth used, the greater the likelihood of having to wait to send a > >> message. > > >> This was standard fare for designers of on-line systems in the days of > >> slower computers and 4800 baud modems. Thirty-odd years ago I made my > >> living applying it to computer systems (comm lines, transactions > >> queued up for a task, even disk I/O). > > >> Basically, the greater the channel utilisation, the more likely you > >> are to wait to get on to it. > > >> Average wait is governed by a factor of p/(1-p) where p is the > >> utilisation. For low utilisation p is small so the result is very > >> small but it's a curve that gets a lot bigger faster than p does. > > >> If there is nothing on the channel then the p/(1-p) is zero so there > >> is no wait. > > >> 50% utilisation gives p of 0.5 so the average wait is > >> 0.5/(1-0.5) = 1, ie the same as the transmission time of the message.. > > >> So including the wait the total time is twice the original message > >> time. > > >> 75% utilisation gives p as 0.75 so the average wait is > >> 0.75/(1-0.75) = 3, ie three times the message transmission time. > > >> So including the wait the total time is four times the original > >> message time. > > >> This is an extreme example because everything is a lot faster these > >> days but it shows the relationship isn't linear. > > >> And there is a heck of a lot of bandwidth wasted with poorly designed > >> web sites, animated ads etc. > > >> Bottom line: if somebody else is on the same channel, use a different > >> one. > > >> >Thanks- Hide quoted text - > > >> - Show quoted text -- Hide quoted text - > > >> - Show quoted text - > > >OK, point made > > >A related point occurs to me. > > >If someone has a router that has both a wired connection and wifi > >connections, but the person concerned does not use wifi, they simply > >use the wired connection, then presumably the wifi ports will still > >transmit. Is this correct? So if I se a wif connection somewhere, I > >might be seeing simply a transmitting port. Have I got this right? > > Yes. But you should disable the wireless port from the device manager. > > >I assume that all channels broadcast and receive at the same power? > >Therefore I can make free choices. > > For the same adapter and receiver. But remember the overlapping bell > curves which can cause interference. > > >Is a computer that is connected to a router by a wire known as an > >Ethernet connenction? > > Colloquially but it's better to say Wired Ethernet to avoid ambiguity. > > Ethernet is a multilayered protocol with at the bottom a physical > transmission layer. The cable oe the wireless stream are the physical > layer. > > >Thanks > > >Ps. As a matter of interest, I assume that there is no way that I can > >know where these other transmitters are. In the same way, others > >cannot know of my location. > > It depends what you want. When I visited friends in Silicon Valley a > while back their city had municipally provided wired internet. There > was an on-line map showing their locations. You can by instruments > which detect transmissions and use them. > > Incidentally it surprised me how many of their neighbours had no > encryption on their routers so anybody could leech off them if they > wanted to. > > But does it really matter if they know your location? Use a neutral > name for the SSID that is distinctive like "Battersea Dog's Home", not > "Fred's Network" or "Home Network". Use the strongest encryption your > router will support. > > One of the problems with public wifi is that it is not encrypted. > > The latest home routers from Netgear and other companies allow two > different levels of use - your own and guest with different passkeys > so guests can use the router to get to the broadband connection but > not access your private stuff eg your networked backup disk.- Hide quoted text - > > - Show quoted text - Thank you. When the wireless utility manager indicates signal, does it really mean signal, or does it mean a mixture of signal and noise. For example, if the latter, then %80 might mean %80 noise. Regarding knowing a person's location. It doesn't matter as far as I am concerned. But I can understand why some people might regard this as a privacy issue. You mentiont hat on a visit you were surprised by the number of unencrypted connections. I have a few domestic connections around me, all weak signals, and all encrypted. I understand that the router that is 'on the way' is a Thomson 585. Don't know anything about these. Hopefully radiates plenty of signal. Best wishes S
From: Christopher A. Lee on 10 Jan 2010 21:46
On Sun, 10 Jan 2010 17:16:48 -0800 (PST), species8350 <not_here.5.species8350(a)xoxy.net> wrote: >When the wireless utility manager indicates signal, does it really >mean signal, or does it mean a mixture of signal and noise. For >example, if the latter, then %80 might mean %80 noise. It's the actual radio signal level in decibels. |