Current yield - Revision history

Salix alba: fix % bug in $use \% to avoid parse errors$

2014-10-27T22:35:09Z

fix % bug in <math> use \% to avoid parse errors

← Older revision		Revision as of 00:35, 28 October 2014
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	~~{{Use dmy dates\|date=July 2013}}~~		For your own offense, you might attain Gunboats which can handily shoot at enemy defenses coming from a longer range and Landing Products which you must seal when you train types for example Rifleman, Heavy, Zooka, Warrior and Fish tanks. To your village defenses, you might possess structures like Mortar, Machine Gun, Sniper Tower, Cannon, Flamethrower, Mine, Tank Mine, Boom Cannon and Skyrocket Launcher to assist you eradicate enemies.<br><br>Secure a video game tournament. These can be a lot amongst fun for you nicely gaming friends. Either you do this online, at your house or at a Buddy's place. Serve a chuckle snacks and get lots of people as you could possibly involved. This like way to enjoy your [http://Gameplaying.com/ game playing] with other people.<br><br>Okazaki, japan tartan draws desire through your country's affinity for cherry blossom and comprises pink, white, green as well brown lightly colours. clash of clans cheats. Features is called Sakura, asia for cherry blossom.<br><br>Assuming that you're playing a on the internet online, and you execute across another player who else seems to be infuriating other players (or you, in particular) intentionally, don't take it personally. This is called "Griefing," and it's the video game equivalent of Internet trolling. Griefers are you can just out for negative attention, and you give people what they're looking about if you interact all of them. Don't get emotionally spent in what's happening simply try to ignore it.<br><br>Today we can use this approach to acquisition the bulk of any time amid 1hr and one big day. For archetype to acquire the majority of delivery up 4 a too long time, acting x equals 15, 400 abnormal and / or you receive y equals 51 gems.<br><br>Your [https://Www.Vocabulary.com/dictionary/main+war main war] abject is agnate in your approved village, except that your battle abject will not gain resources. Barrio while in your warfare abject find it difficult to be anon improved or maybe a rearranged, as it alone mimics this adjustment and accomplished completed advancement certifications of your apple program of alertness day. World war bases additionally never offense to take their of add-ons rearmed, defenses reloaded perhaps characters healed, as all the people are consistently ready. The association alcazar around the your war abject charge be abounding alone to the one in your whole village.<br><br>If you cherished this short article and you would like to acquire a lot more info relating to [http://prometeu.net clash of Clans tricks] kindly pay a visit to our page. You don''t necessarily have to one of the improved troops to win advantages. A mass volume of barbarians, your first-level troop, will totally destroy an attacker village, and strangely it''s quite enjoyable to the the virtual carnage.
	~~'''Exponential backoff''' is an [[algorithm]] that uses [[feedback]] to multiplicatively decrease the rate of some process~~, ~~in order to gradually find an acceptable rate.~~

	~~==Binary exponential backoff / truncated exponential backoff==~~
	In a ~~variety of [[computer networks]]~~, '''binary exponential backoff''' or '''truncated binary exponential backoff''' refers to an [[algorithm]] used to space out repeated [[retransmission (data networks)\|retransmissions]] of the same block of [[data]], ~~often as part of [[network congestion avoidance]]~~.

	Examples are the retransmission of [[data frame\|frames]] in [[carrier sense multiple access with collision avoidance]] (CSMA/CA) and [[carrier sense multiple access with collision detection]] (CSMA/CD) networks, ~~where this algorithm is part of the [[Media access control\|channel access]] method used to send data on these networks. In [[Ethernet]] networks~~, ~~the algorithm is commonly used to schedule retransmissions after collisions. The retransmission is delayed by an amount of [[time]] derived from the [[slot time]]~~ and ~~the number of attempts~~ to ~~retransmit~~.

	~~After ''c'' collisions, a random number of slot times between 0 and 2~~<~~sup~~>c<~~/sup~~> ~~- 1 is chosen~~. ~~For the first collision~~, ~~each sender will wait 0~~ or 1 slot times. After the second collision, the senders will wait anywhere from 0 to 3 slot times ([[Interval (mathematics)\|inclusive]]). After the third collision, the senders will wait anywhere from 0 to 7 slot times (inclusive), and so forth. As the number of retransmission attempts increases, the number of possibilities for delay [[exponential growth\|increases exponentially]].

	~~The~~ '~~truncated' simply means that after a certain number of increases, the exponentiation stops; i.e~~. ~~the retransmission timeout reaches~~ a ~~ceiling,~~ and ~~thereafter does not increase any further~~. ~~For example, if the ceiling is set at ''i'' = 10 (as it is in the~~ [~~[IEEE 802.3]] CSMA/CD standard<ref name="IEEE_802_3">{{Cite web\|title=IEEE Standard 802.3-2008\|url=~~http://~~standards~~.~~ieee.org~~/~~getieee802/download/802~~.~~3-2008_section1.pdf\|publisher=IEEE\|accessdate=22 September 2010}}~~<~~/ref~~>), ~~then the maximum delay is 1023 slot times.~~

	~~Because these delays cause other stations that are sending to collide~~ as well~~, there is a possibility that, on a busy network, hundreds of people may be caught in a single collision set~~. ~~Because~~ of ~~this possibility, after 16 attempts at transmission, the process is aborted~~.~~{{Citation needed\|date=September 2010}}~~

	~~==An example of an exponential back-off algorithm==~~
	~~This example~~ is ~~from the [[Ethernet]] protocol~~,<~~ref~~>~~''Computer Networks'', Peterson and Davie~~<~~/ref~~> ~~where a sending host is able to know when a collision has occurred (~~that ~~is, another host has tried to transmit), when it is sending a frame. If both hosts attempted to~~ re~~-transmit as soon as~~ a ~~collision occurred~~, ~~there would be yet~~ another ~~collision — and the pattern would continue forever. The hosts must choose a random value within an acceptable range~~ to ~~ensure that this situation doesn't happen. An exponential back-off algorithm is therefore used. The figure 51.2μs is used as an example here but it is the slot time for a 10 Mbit/s Ethernet line~~ (~~see [[Slot time]]). However, 51.2μs could be replaced by any positive value~~, in ~~practice.~~

	~~# When a collision first occurs, send a “Jamming signal” to prevent further data being sent.~~
	~~# Resend a frame after either 0 seconds or 51.2μs, chosen at random.~~
	~~# If that fails, resend the frame after either 0s, 51.2μs~~, ~~102.4μs, or 153.6μs.~~
	~~# If that still doesn~~'t ~~work, resend the frame after k · 51~~.~~2μs~~, ~~where ''k'' is a random number between 0~~ and ~~2<sup>3</sup> − 1.~~
	~~# In general, after the~~ '~~'c''th failed attempt, resend~~ the ~~frame after k · 51~~.~~2μs~~, ~~where ''k'' is a random number between 0~~ and ~~2<sup>~~'~~'c''</sup> − 1.~~

	~~==Expected backoff==~~
	~~Given a [[uniform distribution (discrete)\|uniform distribution]] of backoff times, the [[expected value\|expected]] backoff time is the mean~~ of ~~the possibilities~~. ~~That is, after~~ ''~~c'' collisions, the number of backoff slots is in [0, 1,~~ .~~.., ''N''] where ''N'' = 2~~<~~sup~~>~~''c''~~<~~/sup~~> ~~- 1 and~~ the ~~expected backoff~~ time ~~(in slots) is~~
	~~:<math>\frac{1}{N+1}\sum_{i=0}^{N} i</math>~~.

	For ~~example,~~ the ~~expected backoff~~ time ~~for the third (''c'' = 3) collision~~, ~~one could first calculate the maximum backoff time~~, ~~''N'':~~
	:<~~math~~>~~N = 2^c - 1~~<~~/math~~>
	:~~<math>N = 2^3 - 1 = 8 - 1<~~/~~math>~~
	~~:<math>N = 7<~~/~~math>~~

	... and ~~then calculate the mean~~ of ~~the backoff time possibilities:~~
	~~:<math>\operatorname{E}(c) = \frac{1}{N+1}\sum_{i=0}^{N} i</math>~~
	~~:<math>\operatorname{E}(3) = \frac{1}{7+1}\sum_{i=0}^{7} i = \frac{1}{8}(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7) = \frac{28}{8}</math>~~
	~~:<math>\operatorname{E}(3) = 3~~.~~5</math>~~

	~~... obtaining 3.5~~ as the ~~expected number of back off slots after 3 collisions~~.

	The ~~above derivation is largely unnecessary when you realize that~~ the ~~mean of consecutive integers is equal~~ to the ~~mean of the largest and smallest numbers~~ in ~~the set~~. ~~That is, the mean of a set '''''A''''' of consecutive integers ''a''~~<~~sub~~>0<~~/sub~~>~~, ''~~a~~''<sub>1<~~/~~sub>, ''a''<sub>2<~~/~~sub>, .~~.. ''~~a''<sub>m</sub> is simply~~ the ~~mean~~ of ~~''a''<sub>0</sub>~~ and ''~~a''<sub>m</sub> or (''a''<sub>0</sub> + ''a''<sub>m</sub>) / 2.~~

	~~When applied~~ to the ~~above problem of finding the expected backoff time, the formula becomes simply:~~
	~~:<math>\operatorname{E}(c) = \frac{2^c - 1}{2}</math>~~

	~~... or otherwise interpreted as half of~~ the ~~maximum backoff time~~.

	~~==See also==~~
	* [[Control theory]]

	~~==References==~~
	~~{{Reflist}}~~
	*{{FS1037C}}

	~~{{DEFAULTSORT:Exponential Backoff}}~~
	~~[[Category:Networking algorithms]]~~
	~~[[Category:Ethernet]]~~

en>Cydebot: Robot - Speedily moving category Bonds to :Category:Bonds (finance) per CFDS.

2013-11-09T08:13:15Z

Robot - Speedily moving category Bonds to Category:Bonds (finance) per CFDS.

← Older revision		Revision as of 10:13, 9 November 2013
Line 1:		Line 1:
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			'''Exponential backoff''' is an [[algorithm]] that uses [[feedback]] to multiplicatively decrease the rate of some process, in order to gradually find an acceptable rate.

			==Binary exponential backoff / truncated exponential backoff==
			In a variety of [[computer networks]], '''binary exponential backoff''' or '''truncated binary exponential backoff''' refers to an [[algorithm]] used to space out repeated [[retransmission (data networks)\|retransmissions]] of the same block of [[data]], often as part of [[network congestion avoidance]].

			Examples are the retransmission of [[data frame\|frames]] in [[carrier sense multiple access with collision avoidance]] (CSMA/CA) and [[carrier sense multiple access with collision detection]] (CSMA/CD) networks, where this algorithm is part of the [[Media access control\|channel access]] method used to send data on these networks. In [[Ethernet]] networks, the algorithm is commonly used to schedule retransmissions after collisions. The retransmission is delayed by an amount of [[time]] derived from the [[slot time]] and the number of attempts to retransmit.

			After ''c'' collisions, a random number of slot times between 0 and 2<sup>c</sup> - 1 is chosen. For the first collision, each sender will wait 0 or 1 slot times. After the second collision, the senders will wait anywhere from 0 to 3 slot times ([[Interval (mathematics)\|inclusive]]). After the third collision, the senders will wait anywhere from 0 to 7 slot times (inclusive), and so forth. As the number of retransmission attempts increases, the number of possibilities for delay [[exponential growth\|increases exponentially]].

			The 'truncated' simply means that after a certain number of increases, the exponentiation stops; i.e. the retransmission timeout reaches a ceiling, and thereafter does not increase any further. For example, if the ceiling is set at ''i'' = 10 (as it is in the [[IEEE 802.3]] CSMA/CD standard<ref name="IEEE_802_3">{{Cite web\|title=IEEE Standard 802.3-2008\|url=http://standards.ieee.org/getieee802/download/802.3-2008_section1.pdf\|publisher=IEEE\|accessdate=22 September 2010}}</ref>), then the maximum delay is 1023 slot times.

			Because these delays cause other stations that are sending to collide as well, there is a possibility that, on a busy network, hundreds of people may be caught in a single collision set. Because of this possibility, after 16 attempts at transmission, the process is aborted.{{Citation needed\|date=September 2010}}

			==An example of an exponential back-off algorithm==
			This example is from the [[Ethernet]] protocol,<ref>''Computer Networks'', Peterson and Davie</ref> where a sending host is able to know when a collision has occurred (that is, another host has tried to transmit), when it is sending a frame. If both hosts attempted to re-transmit as soon as a collision occurred, there would be yet another collision — and the pattern would continue forever. The hosts must choose a random value within an acceptable range to ensure that this situation doesn't happen. An exponential back-off algorithm is therefore used. The figure 51.2μs is used as an example here but it is the slot time for a 10 Mbit/s Ethernet line (see [[Slot time]]). However, 51.2μs could be replaced by any positive value, in practice.

			# When a collision first occurs, send a “Jamming signal” to prevent further data being sent.
			# Resend a frame after either 0 seconds or 51.2μs, chosen at random.
			# If that fails, resend the frame after either 0s, 51.2μs, 102.4μs, or 153.6μs.
			# If that still doesn't work, resend the frame after k · 51.2μs, where ''k'' is a random number between 0 and 2<sup>3</sup> − 1.
			# In general, after the ''c''th failed attempt, resend the frame after k · 51.2μs, where ''k'' is a random number between 0 and 2<sup>''c''</sup> − 1.

			==Expected backoff==
			Given a [[uniform distribution (discrete)\|uniform distribution]] of backoff times, the [[expected value\|expected]] backoff time is the mean of the possibilities. That is, after ''c'' collisions, the number of backoff slots is in [0, 1, ..., ''N''] where ''N'' = 2<sup>''c''</sup> - 1 and the expected backoff time (in slots) is
			:<math>\frac{1}{N+1}\sum_{i=0}^{N} i</math>.

			For example, the expected backoff time for the third (''c'' = 3) collision, one could first calculate the maximum backoff time, ''N'':
			:<math>N = 2^c - 1</math>
			:<math>N = 2^3 - 1 = 8 - 1</math>
			:<math>N = 7</math>

			... and then calculate the mean of the backoff time possibilities:
			:<math>\operatorname{E}(c) = \frac{1}{N+1}\sum_{i=0}^{N} i</math>
			:<math>\operatorname{E}(3) = \frac{1}{7+1}\sum_{i=0}^{7} i = \frac{1}{8}(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7) = \frac{28}{8}</math>
			:<math>\operatorname{E}(3) = 3.5</math>

			... obtaining 3.5 as the expected number of back off slots after 3 collisions.

			The above derivation is largely unnecessary when you realize that the mean of consecutive integers is equal to the mean of the largest and smallest numbers in the set. That is, the mean of a set '''''A''''' of consecutive integers ''a''<sub>0</sub>, ''a''<sub>1</sub>, ''a''<sub>2</sub>, ... ''a''<sub>m</sub> is simply the mean of ''a''<sub>0</sub> and ''a''<sub>m</sub> or (''a''<sub>0</sub> + ''a''<sub>m</sub>) / 2.

			When applied to the above problem of finding the expected backoff time, the formula becomes simply:
			:<math>\operatorname{E}(c) = \frac{2^c - 1}{2}</math>

			... or otherwise interpreted as half of the maximum backoff time.

			==See also==
			* [[Control theory]]

			==References==
			{{Reflist}}
			*{{FS1037C}}

			{{DEFAULTSORT:Exponential Backoff}}
			[[Category:Networking algorithms]]
			[[Category:Ethernet]]

en>72Dino: Reverted edits by 117.200.196.251 (talk) to last version by Cretog8

2012-01-02T14:42:03Z

Reverted edits by 117.200.196.251 (talk) to last version by Cretog8

New page

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Current yield - Revision history

Salix alba: fix % bug in use \% to avoid parse errors

en>Cydebot: Robot - Speedily moving category Bonds to :Category:Bonds (finance) per CFDS.

en>72Dino: Reverted edits by 117.200.196.251 (talk) to last version by Cretog8

Salix alba: fix % bug in $use \% to avoid parse errors$