L2J Server

Hi guys, quick question. Ive got the next specs

Core i53570k
14gb ram
2ssd 120gb (one for server and one for DB)
and, the mos important feature: 5mb simetric. I t means, I download and upload al 700kbps.

With this specs, at a national level, how much players can i host?.

Thanks in advance.

monakian wrote:Hi guys, quick question. Ive got the next specs

Core i53570k
14gb ram
2ssd 120gb (one for server and one for DB)
and, the mos important feature: 5mb simetric. I t means, I download and upload al 700kbps.

With this specs, at a national level, how much players can i host?.

Thanks in advance.

5mbps == ~ 4500 without lag

really? and why the private servers always say 100mbps international bandwith, and when you enter is laggy and there are... 1200 chars online.
Are those 100mbps a lie?

Sacrifice wrote:

monakian wrote:Hi guys, quick question. Ive got the next specs

Core i53570k
14gb ram
2ssd 120gb (one for server and one for DB)
and, the mos important feature: 5mb simetric. I t means, I download and upload al 700kbps.

With this specs, at a national level, how much players can i host?.

Thanks in advance.

5mbps == ~ 4500 without lag

5 mbp/s = 20-30 ppl without lag

UnAfraid wrote:

Sacrifice wrote:

monakian wrote:Hi guys, quick question. Ive got the next specs

Core i53570k
14gb ram
2ssd 120gb (one for server and one for DB)
and, the mos important feature: 5mb simetric. I t means, I download and upload al 700kbps.

With this specs, at a national level, how much players can i host?.

Thanks in advance.

5mbps == ~ 4500 without lag

5 mbp/s = 20-30 ppl without lag

E???... how do you calculate this??. Is a question that I make you UnAfraid... exist any formula??

Formula is:
1Kb average packet * amount_of_unoptimized_l2j_useless_packet_sends * 100packets_avg_per_sec * x10_if_on_siege = total_bandwidth

Example:

1 * 2 * 100 * 1 = 200Kb/sec
1 * 2 * 100 * 10 = 2Mb/sec on siege

Thats all for 1 player.

5Mbps / 200Kbps = 25 online players.
5Mbps / 2Mbps = 2.5 online players on siege.

Nik wrote:Formula is:
1Kb average packet * amount_of_unoptimized_l2j_useless_packet_sends * 100packets_avg_per_sec * x10_if_on_siege = total_bandwidth

Example:

1 * 2 * 100 * 1 = 200Kb/sec
1 * 2 * 100 * 10 = 2Mb/sec on siege

Thats all for 1 player.

5Mbps / 200Kbps = 25 online players.
5Mbps / 2Mbps = 2.5 online players on siege.

I do not agree on the rate of amount_of_unoptimized_l2j_useless_packet_sends, this meaning that on each 100 packets 1 is valid only?... I dont believe... its too high, except if connection is a fatality.

Take in consideration a 10% of BER (bit error rate) and is too high for actual connections... normally is 0.1%

Edit BER <= 1%

So, whats the truth about bandwith?. with 5megabites per second i can host, like, 200 people?. Or im talking nonsense?

Sacrifice wrote:E???... how do you calculate this??. Is a question that I make you UnAfraid... exist any formula??

Traffic monitoring is the formula

Check with wireshark filtering port of gameserver

You could use task manager also, no external tools are required

amount_of_unoptimized_l2j_useless_packet_sends is definitely higher than 2, specially with effects

Nik wrote:Formula is:
1Kb average packet * amount_of_unoptimized_l2j_useless_packet_sends * 100packets_avg_per_sec * x10_if_on_siege = total_bandwidth

Example:

1 * 2 * 100 * 1 = 200Kb/sec
1 * 2 * 100 * 10 = 2Mb/sec on siege

Thats all for 1 player.

5Mbps / 200Kbps = 25 online players.
5Mbps / 2Mbps = 2.5 online players on siege.

We must convert allways all datas inside all formulas in the same units... In Nik's values I fix a minor error, but its sustantially important.

1 * 2 * 0,1 * 1 = 0,2 Kbps per player
1 * 2 * 0,1 * 10 = 2 Kbps per player on siege.

Then... results are very, very diferent... how can to see it.

5000 Kbps / 0,2 Kbps = 25000 players
5000 Kbps (5Mbps) / 2 Kbps = 2500 players

I will give my two cents.

5Mbps upload
= 5 Mb (Megabit) per second
= 5 Mb * 1024 = 5120 Kb (Kilobit)
= 5120 Kb / 8 = 640 KB (Kilobyte)
= 640 KB (Kilobyte) per second

For the amount of packets your best bet it to calculate with the worst case: All players on the server can see each other, are in a full party and are in an area full of npcs. Now you can seperate between private and public updates. Public updates are thoose updates all players have to see. Private updates are thoose only the player itself need to see. The following is by no means "THE 100% SOLUTION". It is an approximation how much packets are send to all players in one second in a worst case scenario.

• Movement: 2 + (sorounding_players * 2) + (sorounding_npcs)
• Hitting & Skill Usage: 5 + (sorounding_players * 5) + (sorounding_npcs * 2.5)
• Effects: 2 + (sorounding_players * 2) + (sorounding_npcs)
• Status: 2 + (sorounding_players / 9 * 2) + (sorounding_players)
• Party Effects: 18 + (sorounding_players / 9 * 18)

This worst case scenario may never be reached in real conditions.

An example with 100 players and 100 npcs
Movement: 2 + (100 * 2) + (100) = 302
Hitting & Skill Usage: 5 + (100 * 5) + (100 * 2.5) = 755
Effects: 2 + (100 * 2) + (100) = 302
Status: 2 + (100 / 9 * 2) + (100) = ~124
Party Effects: 18 + (100 / 9 * 18) = 218
= ~1701 packets per second

When an average packet size is 1 Kilobyte, you need 1701 Kilobyte per second upload on a worst case for 100 players. Because this worst case scenario practically never happens you can half that to about 850 Kilobyte per second upload for 100 players.