Hey guys, was jus wondering if any1 had a rough estimate (I know it aint easy) but a rough estimate of how much bandwith/traffic is used by an 800 user hub in a month?
Any help is appreciated!
QuoteOriginally posted by QuikThinker
Hey guys, was jus wondering if any1 had a rough estimate (I know it aint easy) but a rough estimate of how much bandwith/traffic is used by an 800 user hub in a month?
Any help is appreciated!
roughly 107GB I think
DAMN!!! 107GB/month??? 8o
QuoteOriginally posted by QuikThinker
DAMN!!! 107GB/month??? 8o
well, i have a hub with 80-90 users, which takes 4kb/s.
800/80 = 10
4kb/s * 10 = 40kb/s
3600sec * 24 * 31 = 2678400 seconds
2678400 * 40kb = 107136000 KB = roughly 102 GB (the number I said before was a bit off, but almost right :P)
Thats not right.
Increase of bandwidth is more quadratic than linear.
So here, 800 / 80 "=" 100.
So its rather 1 TB a month, not 100 GB. :-)
QuoteOriginally posted by dove
Thats not right.
Increase of bandwidth is more quadratic than linear.
So here, 800 / 80 "=" 100.
So its rather 1 TB a month, not 100 GB. :-)
but then it would take about 370kb/s and that is allot more then when I hosted a hub with 1100-1200 users.
Well:
Suppose you have n users in a hub: w.l.o.g. assume that every user sends an equal amount of X bytes for broadcasting to the hub in every time interval.
So for n user the hub has to send around nrbytes=n*(n-1)*X bytes.
Now suppose that the n increases by the factor M. That means almost nrbytes2= (Mn)*(Mn-1)*X.
lim (n-> inf.) of (nrbytes2/nrbytes1) is M*M.
So, if you increase the amount of users by the factor M,then you'll have to expect M*M more datato be sent. Well, not exactly. But its for sure more close to M*M than to M.
Thats all i can tell you. :-)
QuoteOriginally posted by dove
Well:
Suppose you have n users in a hub: w.l.o.g. assume that every user sends an equal amount of X bytes for broadcasting to the hub in every time interval.
So for n user the hub has to send around nrbytes=n*(n-1)*X bytes.
Now suppose that the n increases by the factor M. That means almost nrbytes2= (Mn)*(Mn-M)*X.
lim (n-> inf.) of (nrbytes2/nrbytes1) is M*M.
So, if you increase the amount of users by the factor M,then you'll have to expect M*M more datato be sent. Well, not exactly. But its for sure more close to M*M than to M.
Thats all i can tell you. :-)
ok , a bit more advance math then what I can :P
I simply read in the logs from my firewall that my hub takes 1-4kb/s.
So I took the max speed and multiplied it with 10.
Anyway I was wrong, but you weren't is exactly correct either ;)
yes, i am not able to give a correct answer either. And i think no one alive can either - without spending 2 days of investigation into the protocol and doing lots of statistics ;)
btw: 1 or 4 kb/s is a big difference if you are multiplying with such a big 9 digit number ;)
I think that is also the problem of the reasoning to 370kb/s for 1100 user. Suppose your current hub actually uses only 1 kb/s. Then you could run the 1000-user-hub with around 100 kb/s (assuming quadratic correltaion again). But somehow i think that is way below reallity ;)
http://dcdev.net/yhub/forum/viewtopic.php?p=11879&sid=99639ab8449076ec7462a56adcf6dd52
Maximum upload speed of 1.66 Mbit's
can handle maximum 824 users
-------------------------------------------------
Upload per second: 1.66 Mbits 212.50 KB
Upload per minute: 48.32 Mbits 6.04 MB
Upload per hour: 2.83 Gbits 362.38 MB
Upload per day: 67.95 Gbits 8.49 GB
Upload per month: 1.99 Tbits 254.80 GB
-------------------------------------------------
This is just the uploaded data
-------------------------------------------------calculated with the formula made by verliba.
and made into a script by me.
Mb = 1024
Upload_speed = 1700-- in Kbit's
N=sqrt(Upload_speed)*20
function con(num)
local t = floor(num)
if strlen(t) >= 10 then
num = format("%0.2f", (num/1024/1024/1024)) .." T"
elseif strlen(t) >= 7 then
num = format("%0.2f", (num/1024/1024)) .." G"
elseif strlen(t) >= 4 then
num = format("%0.2f", (num/1024)) .." M"
else
num = format("%0.2f", num).." K"
end
return num
end
print("")
print("Maximum upload speed of "..con(Upload_speed).."bit's")
print("can handle maximum "..floor(N).." users")
print("-------------------------------------------------")
print("Upload per second: \t"..con(Upload_speed).."bits \t"..con(Upload_speed / 8).."B")
Umin = N *60
print("Upload per minute: \t"..con(Umin).."bits \t"..con(Umin / 8).."B")
Umin = Umin *60
print("Upload per hour: \t"..con(Umin).."bits \t"..con(Umin / 8).."B")
Umin = Umin *24
print("Upload per day: \t"..con(Umin).."bits \t"..con(Umin / 8).."B")
Umin = Umin *30
print("Upload per month: \t"..con(Umin).."bits \t"..con(Umin / 8).."B")
print("-------------------------------------------------")
print("This is just the uploaded data")
print("-------------------------------------------------")btw a search would have done as i posted this before.
plop
Damn, thanx guys. I honestly ad no idea it was anywhere near as much as that!
the stripping of the description tags isn't inc in this formula, so you can add a bunch of extra users.
Stress test, 407 users on connection with 150 kbps upload, no TAGs (September 2004)
this was a test from ppk with stripped tags on ptokax.
the formula gives 224 users for that upload.
plop