This document discusses calculating buffer credit requirements for long distance fibre channel links. It provides an overview of fibre channel frame structure and sizes. It then shows calculations for link distances and speeds to determine the number of frames and bits that could be in transit simultaneously. This is used to calculate the required number of buffer-to-buffer credits. It proposes a rule of thumb that for a 4Gbps link, 1.33 buffer credits are needed per kilometer. It notes to add an additional 20% credits as recommended by best practices. The document seeks feedback on the proposed 1.33 factor for 4Gbps links.

1. B2B FC CREDITSPERFORMANCEDEADLOCKS
Jan 31, 2013 3:17 PM (
So thiswasabout gettingthe link"up"onboth ends,butwhataboutactuallysendingdataaccross those
links?Longerlinkscanactuallycontainmultiple fibrechannel frames!!The speedof lightisn'tinfinate,
so ittakestime (microsecondstosub milli seconds) foraframe to actuallytravel fromTx (transmitter)
to Rx (receiver).
If a fibre opticpathisan FCISL, theninknowingitsline bit-length,we cancalculate the maximum
numberof framesthat can simultaneouslyfit(source todestinationandback) onthat physical link.That
numberisthe exactanmountof buffertobufferscreditsyourISLswitchportswill require.
Why?
To safelystore-and-forwardframesatthe source end,while theyare still intransittothe destination
port,and acknowledgedashavingbeenreceived.Sothe sendingswitchwill keepthatframe ina buffer
while the ACKhasn'tbeenreceived.
To provide a“successfullysent”I/Oacknowledgmenttothe sendingportor HBA soit doesnothave to
waitfor the frame to reach the otherside of the link,andthe acknowledgmentthathasto come back.
So if youupgrade a port'sspeedfrom2 to 4 Gbps for example,the amountof framesthatcan be on the
linksimultaniouslydoubledandthe amountof B2B creditshasto be doubledaswell.
Please note thatalthoughthe linkbetweensitescanbe fastenoughandhave enoughbuffercreditsto
facilitate linerate communication,the receivingarraycan still be a bottleneckwhichcancause the
sendingside toslowdown.Considerforexample MirrorView orSRDF.The primarystorage array is
writtentoby a numberof hostsand the the back endof the primaryarray can handle the amountof
IOps.All I/Osthatwere writtenare alsosentto the secondaryarray to be replicated.If the receivingend
cannot handle thatamountof (write) IOpsbecause the numberof disksistoolow orthere'snot enough
cache available,the receiving(busy) arrayisprocessingthe incomingI/Osnotasfastas the ISLbetween
the sitescan provide andsoacknowledgementsaren'tsentbackto the primaryarray instantaniously,
but a little bitslower.Sointhe endthe sendingapplication receivesthisacknowledgementsomewhat
laterand we all knowthat comminucationisbuiltonsend/receive andACK,right?If the ACKcomesin
slow,the conversationisslowerthanyou'dlike ittobe.

2. So configuringthe linkbetweensitesisone thing,sizingthe recveivingarrayisanother(important)
thing!
Before we beginanexample of howtocalculate bufferrequirements,itisimportanttoknow the
numerical definitionof aFibre Channel Gigabit,aswell astounderstandthe structure of a Fibre Channel
Frame.
In the Fibre Channel world,one gigabitisdefinedtobe 1,062,500,000 bits(whichisnot
1024x1024x1024). Other Fibre Channel Gigabitvaluesare thenderivedfromthisreference definition.
For example,two(Fibre Channel) Gb=2 x 1,062,500,000 bits= 2,125,000,000 bits.To avoidconfusion
withthe traditional (nonFibre Channel) definitionof aGb, throughoutthisdocumentIwill use the
symbol Gbfcto mean“1,062,500,000 bits”,or 1 Fibre Channel Gb.
In summary:
1 Gbfc = 1,062,500,000
2 Gbfc = 2,125,000,000
4 Gbfc = 4,250,000,000
8 Gbfc = 8,500,000,000
10 Gbfc = 10,625,000,000
16 Gbfc = 17,000,000,000
Next,we showthe anatomyof a Fibre Channel Frame withnotes.
Start of Frame:4 bytesor 32 bits
StandardFrame Header:24 bytesor192 bits
Data (payload):[0– 2,112] bytesor [0 – 16,896] bits
CRC: 4 bytesor32 bits
End of Frame: 4 bytesor 32 bits
TOTAL (Nbrbits/frame):[36– 2,148] bytesor 288 – 17,184 bits
Notes:
The term byte usedhere means8 bits(notthe 10 bitsthat resultfrom 8/10 bitencoding).
The maximumFibre Channel frame sizeis2,148 bytes.
The final frame size mustbe a multiple of 4bytes.Thusthe Data (payload) segmentwill,asnecessary,
be paddedwith1 to 3 “fill-bytes”toachieve anoverall 4byte frame alignment.

3. The standard Frame Headersize is24 bytes.However,upto64 additional bytes(foratotal of an 88 byte
header) canbe includedforapplicationsthatneedextensive control information.Since the total frame
size cannotexceedthe maximumof 2,148 bytes,these additional Headerbyteswillsubtractfromthe
Data segmentsize byas muchas 64 bytes(perframe).Thisiswhythe maximumData(payload) size is
2,112 (because [2,112 – 64] = 2,048, whichisexactly2K-bytesof data).
The final frame,once constructed,ispassedthroughthe 8 byte to 10 byte conversionprocess.
In the FC world,1 Word = 4 x 8/10 bit encodedbytes(40bits).
Thenwe have the speedof light:
299,792.458 km/ s == 0.00000333564095 s / km, so that's 3.33564095 microsecondsperkilometer,
that's 3.3 millionthof asecondfor eachkm "traveled".
NOTE: thisisin vacuum!I'll adjustthe outcome later.The speedof lightinfiberisapproximately
200,000 kmper secondinsteadof almost300,000.
Nextwe take the lineardistance betweenthe twositesanddetermine:
How manysecondsittakesfor 1-bitto travel the one-waydistance linearbetweenthe twosites;thatis,
expressthe “distance”betweenthe twositesinseconds.Thisisdeterminedbythe speedof light.
Havingdetermined the one-waydistance insecondsbetweenthe twosites(afixednumber),we can
nowdetermine the maximumnumberof bitsthatcan exist,intransit,betweenthe twositesatanyone
time.Inotherwords,we can calculate the “distance”betweenthe twosites inbits(asopposedto
miles).Thisisdeterminedbythe speedof light,aswell asbythe rate at whichthe transmitting
equipment(e.g.afibre channel port) cancreate electrical variationsontothe medium(i.e.fibre optic
line).Inotherwords,howfastcan itpush bitsontoone endof the fibre opticline (usually1Gbfc,2
Gbfc, 4 Gbfc or 10 Gbfc).
Let say,for the purposesof discussion,thatwe have redundant(i.e.two) fibre opticpathsbetweena
primaryand secondarysite.
The lineardistance (asopposedtodisplacement)is91.732608 Km (or about57 miles) forone path,and
43.452288 km(or about27 miles) forthe otherpath.
Suppose the linkspeedis2Gbfc:
Distance (Km) Distance (secs) # of Bits(1Way) 8/10 FC bytes
91.732608 (57mi) 0.000305987044 650,222.468 65,022.2468
43.452288 (27mi) 0.000144941231 308,000.116 30,800.0116
Again:thisisin vacuuminsteadof fiber!Butstaywithme,I'll fix itlateron!!

4. Calculationsnotesforthe tablesabove:
Column2 indicates the amountof time (inseconds) ittakesone OPTICALVARIATION (howevermany
numberof bitsthat optical variationhappenstorepresent) totravel the one-waydistance specifiedin
columnone.Itcomesfrom multiplyingthe numberof secondsittakeslight totravel 1 km (i.e.
0.00000333564095 s / km),bythe numberof km traveledwhich,isspecifiedinthe firstColumn.
Column3 is the productof column2 (i.e.the line distance inseconds) andthe FCbitinsertionrate
(2,125,000,000). This columneffectivelyrepresentsthe amountof additional I/Othatcouldhave been
processedatthe host,had the response tothat I/Obeeninstantaneous;(actuallytwice thatamount,
since acknowledgmenttime forthose I/O’s,comingbackthe otherway,have to be accounted foras
well).
Column4 is Column3 dividedby10.This isdone to groupsingle bitsintransitintoequivalent8/10bit
“byte”quantities.The Fibre Channel protocol convertsevery8bitbyte intoa 10 bit equivalent(viathe
8/10 bit encodingalgorithm) before transmittingit.Sothe value inthisColumn4will determine the
numberof buffersneededforanISL switchport.
Frame Length8 (inkm) = (299,792.458 km/s) x (Seconds-Between- Inserted-Bitssecs/bit) x (Number-Of-
Bits-Per-Framebits)
Frame Length10 (inkm) = (299,792.458 km/s) x (Seconds-Between- Inserted-Bitssecs/bit) x (Number-Of-
Bits-Per-Framebits) x 10/8
Where:
Seconds-Between-Inserted-Bits=1/1,062,500,000 (for1Gbps FC) = 1/2,125,000,000 (for2Gbps FC)
Number-Of-Bits-Per-Frame=Variable dependingonDatapayloadand/orHeadersize.See table and
notesabove.Inmostcases the Headerwill be the Standardsize of 24 bytes.
Numberof bits/ frame
(after8/10 bitencoding)
Frame Length(km) @ 2 Gbps FC
(.000141078803764705 km/bit)
Numberof in-transitframes(1-
way) for 91.732608 km leg.
17,184 / 21,480 3.0303 km (2112 PL-bytes) 30.2170 frames(buffers)
Andagain: thisisvacuum,not infiber!!!Staywithme a little bitlonger....
Notes:
Column1 representsthe total numberof 8/10 bitsin a Fibre Channel frame (withastandardheadersize
of 24 bytes) atvaryingData payloads(PL).

5. Column2 representsthe productof the 10 bitvalue inColumn1, and (1/2,125,000,000). Thus,this
Column(Column2) essentiallyrepresentsthe lineardistance that1 (one) single frame consumesforthe
specifiedpayload(PL).
Column3 representsthe quotientderivedbydividingthe longest(worstcase) DWDMline distance (in
kilometers),bythe numberof kilometersperframe (calculatedinColumn2).Thus,thiscolumn
essentiallyindicateshowmanyadditionalONE-WAYframesworthof datacouldhave beenprocessedby
the host/application,hadthe response tothe firstframe beeninstantaneous.Inotherwords,thisishow
manyONE-WAY(notround trip) switch buffersyouwouldneedtoallow non-stoptransmission.Double
(i.e.roundtrip) the valuesinthiscolumn3to yieldthe numberof buffersrequiredof yourISLswitch
ports.
So inour example forrunning2longlinksusing2 Gbfc speed(invacuum):
the numberof B2B creditsneededforthe 91.7 km legwill be 2 x 30.2 = 60.4, soroundingupto 61 B2B
credits.
the numberof B2B creditsneededforthe 43.4 km legwill be 2 x (43.4/91.7) = 2 x 14.3 = 28.6, so
roundingupto 29 B2B credits.
Can't we make a rule of thumbout of this?
I admit,the numbersIusedare takenfroman example documentIhave lyingaroundsoI didn'tneedto
the math myself ,butmakinga ROT out of itbackfiresatme.
The easiestwayto calculate the rule of thumbisgettingbackto 1 km, soat 2 Gbfc speedthe numberof
B2B creditsneededfora1 km linkis2 x 30.2170 / 91.732608 = 0.6588 IN VACUUM.
Otherexampleswe cannowuse to more easilyrememberthese"nice"numbers:
2 Gbfc overa 2 km link= 2 x 0.6588 = ± 1.32
4 Gbfc overa 2 km link= 4 x 0.6588 = ± 2.64
4 Gbfc overa 20 km link= 40 x 0.6588 = ± 26.4
So whatwouldbe a nice numberto remember?4Gbfc requires1.3x the numberof km? 1.33 x the
numberof km? I guesswe have justcreateda rule of thumb!1.33 can be rememberedveryeasily!
Right?
So runninga 4Gb linkover44 km requires1.33 x 44 = 58.52 := 59 B2B credits.

6. A 2 Gbfc linkusingthe same distance (44km) will needhalf of that,so30 B2B credits.Sohalf the speed,
half the buffercredits,doublethe speed,doublethe buffercredits.
HOWEVER: the speedof lightinvacuumneedstobe devidedbythe refractionindexof fiber,whichis
about1.5.
BuffersRule of Thumb(takingthe actual speedof lightintoaccount):
4 Gbfc needs1.33 times1.5 timesthe distance inkilometer=1.995 buffersperkilometer
Thisextra1.5 is because of the refractionindex of the carrierused.Mostfibershave anindex of
approximately1.5,whichwill "slowdown"the speedof lighttoabout200,000 kilometerspersecond.
A longdistance connectionrunningat4 Gbfc needs2buffersperkilometer
A longdistance connectionrunningat4 Gbfc needs3.22 bufferspermile
(fromkm to mile youneedtomultiplythe outcome inkmby1.609, whichisnotthat easyto remember,
but youcan figure outa numberthatworksfor you,right?)
EMC as well asBrocade recommendaddinganextra20% B2B creditsto the amountyou justcalculated.
Message waseditedby:RRR Added20% extrasimilartothe bestppracticesbyEMC and Brocade.
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16. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov20, 2012 3:17 AM(inresponse toRRR)
If anyone thinksImade a mistake here,please informme,sowe can setthisstraight.
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17. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
Vincze Nov20, 2012 4:48 AM(inresponse toRRR)

7. Nice breakdown,RRR
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18. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov20, 2012 6:01 AM(inresponse toVincze)
Thanks.AndI almostlostmy post!!WhenI was done Iaccidentlyswitchedtothe advanced editorand
nothinghappened....aaaargh!FortunatelyIcopiedthe html texttonotepadonlyaminute before,soI
was able torestore mypost. Tookme an hourand a half or so to type this,so I wasgladit wasn't lostin
the end.
So whatabout this1.33 factor for a 4Gb link?Doyouthinkit's correct?I can't findanyerrorsin my
calculations
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19. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
JonKlaus Nov20, 2012 7:32 AM(inresponse toMark)
AndI was afraidmy postwas long...yikes!
I originallycalculatedall buffercreditsbackfrom2gbit,whichis/was1creditper km.4Gbit wouldneed
2 creditsperkm,etc.
Don't forgetthat toomuch creditscan't harm performance (assumingyou're notrunningshortonother
links),andthatyoucan alwaysmonitorcreditusage onthe switch.CiscoDevice Managerhasa counter
that showsthe slackyou've got:

8. In thiscase it's an ISL that's maybe a couple of hundredmeterslong,soitcan cope with1 or 2 credits
dependingonactual speedanddistance.The portisneverthelessassigned32credits,of which32 are
still available(seebottom2numbers,CurrRxBbCreditsandCurrTxBbCredits.
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20. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov22, 2012 1:43 AM(inresponse toMark)
Tonight(Europe time) we plannedtodothe TweetChat,butyesterday SeanThulin mentionedtousover
TwitterthatThursday Novthe 22nd isThanksgivinganditmightnotbe a goodday for the TweetChat.
So aftersome careful considerationswe decidedtomove the TweetChattonextMonday,November
the 26th. Same time,same place,newday!
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21. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
JonKlaus Nov23, 2012 4:26 AM(inresponse toRRR)
Happy Thanksgivingeveryone inthe USand aroundthe world!I hope the turkeytastedgood!
So..A lotof people are readingthistopic,butthe repliesare abiton the low side.Isthere no-one else
that wantsto share theirexperienceswithlongdistance links?Are longdistancelinksthateasy?

9. Please join the conversation;we won'tbite!
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22. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov28, 2012 2:49 AM(inresponse toJonKlaus)
So we talkedaboutlight,multi mode andsingle modeaswell asattenuation.Sothe maximumdistance
isdefinedbyhowsensitive the Rx side is.
Alsowe explainedwhyBuffertoBufferCreditsare neededandhow youcalculate how manyof these
youneed.
So the nextthingwe'dlike todiscussisa wayto make optimal use of the expensivefiberlinkyoumight
have.A way to do thisisby usingcolors.Since lasersare monochromaticanywayusingasort of prismto
bundle colorscomingfromdifferentlasersintoasingle multicoloredbeamisthe solution!
TechniqueslikeCWDM,DWDM andEWDM were developedtofacilitateasmanycolorsas possible into
a single beam.
WDM standsfor wavelength-divisionmultiplexing.Multiplexingsaysitall:itcombinesmultiple source
intoone.
The bestknownWDM productis CWDM: Course WDM. In 2002 the standardwas setfor CWDM as using
wavelengthsbetween1270 and1610 nm as well ashaving20 nm spacingsbetweenthe "colors".The
mostcommonlyusedCWDMMultiplexers(the prisms) provide 8connectionsso8 sendinglongwave
ports can use the same physical darkfiber.Andalthougheachof these "colors"isnamedaftera visible
color(like brown,yellow,blue andsoon) the actual wavelengthsare farbeyondthe visibilityof the
humaneye.
The wavelengthsbetween1270 and 1470 are consideredunusablebecause of the increased
attenuation,soonly1470 to 1610 issupported.

10. The "colors"usedfor CWDM are:
1470 nm (gray)
1490 nm (violet)
1510 nm (blue)
1530 nm (green)
1550 nm (yellow)
1570 nm (orange)
1590 nm (red)
1610 nm (brown)
The mentioned colorsare the namesCiscocallsthese differentwavelength.Byusingcoloredtagson
each SFPthe humaneye can noweasilyidentifyeachwavelength.
Needlesslytosaythat to connecta networkonsite A usingforexample the 1530 nmwavelength,the
same wavelengthof 1530 nm is neededonsite B.

11. CWDM isunaware of the protocol usedtotransport data accross,so you can easilycombine tcp/ip,fibre
channel andevenanalogue tvsignalsaslongas eachsignal usesa differentwavelength.
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23. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
JonKlaus Nov23, 2012 8:22 AM(inresponse toRRR)
So we knowhowto getthe signal acrosslongdistancesandwe know how to efficientlyloadthe link
usingbuffercredits.If we needmore bandwidththanone linkcandeliver,we cansenddifferent
wavelengthsoverone linkusingCWDMandDWDM to make thingsevenmore efficient.Butthere'sone
thingwe can't influence (much):latency.
Whenwe sendinformationbetweentwodistanceswe use light.The speedof lightis299,792,458m/s.
Thisis ina vacuum; infiberit'sa bit lower,somewherearound200,000,000m/s. Puttingitdifferently,
that's 200,000km/s or 200km/ms. This meansthatif we wantto cover200km of distance,we'll have
1ms of latencybetweenputtinglightonthe fiberandthe otherendseeingit.
To performreliable communications,we can'tsignal uni-directionallyandleave itatthat.We wantto
knowif the data arrivedat the other endintact.Thismeansthat we needtouse the bi-directional
latency,alsocalledround-triptime.Our200km linkisnow costingus 2ms of latency.Nottoo much
right?Well...helloSCSI!
FC data transportusesSCSIcommands.If we assume Iwill be the sendingstorage arrayonthe source
endand RRR will be the receiving storage onthe otherend,our(SCSI) conversationwouldbe something
alongthe linesof:
RTT direction1:HeyRRR,I wantto sendyou some data.
RTT direction2:Sure Jon,goahead.
RTT direction1:Okay,here isthe data.
RTT direction2:ThanksJon,gotit ingoodshape.
Or, since a picture saysmore than a thousandwords,a picture fromthe Brocade whitepapers:

12. Thismeansthat insteadof 1ms latency,our200km linkisactuallycostingus4ms of latency.Andthat's
onlyforthe linkitself.
If we send32KB of data,we also needtoput that onthe line.IgnoringFCframe overhead,at1Gbit/sa
32KB I/Otakes0,244ms to load.You can compare thisto a truck enteringthe highway:once the front
fendercrossesthe startingline,the backof itis still not"loaded"ontothe highway.
If we alsohave protocol convertersinbetweensource anddestination(because forexampleyouswitch
fromFC to FCIPand at the otherendback fromFCIPto FC) youwill needtoaddanother0,5ms for each
conversion.With4round-trips,you'll endupwith4msof protocol conversiontime.
So we've got4ms of delaydue tothe speedof light.We have 4ms due to protocol conversionsor
routers.Andwe have 0,244ms of latencydue tothe fact we can onlyloadthe data on the line at
1Gbit/s.Addit all up and at 200km youhave an additional latencyof 8,244ms.
Consideringthataunmirroredwrite I/Othatishittingwrite cache onlycostsus somethinginthe
neighborhoodof 0,5ms,thisisa lot.Switchon mirroringoverthatsecondarylinkandyourwrite latency
isnow 0,5 + 8,244 + 0,5ms = 9,244ms. If the secondarystorage systemisalsorunningoutof write cache
and needstoforce flush,you're ina worldof pain...
Moral of the story:be careful of synchronousreplicationoverlongdistances!
Andhave a goodweekend...
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13. 24. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
JonKlaus Nov26, 2012 10:07 AM (inresponse toJonKlaus)
Andwe're off withthe tweetchat!Joinusontwitterwithhashtag#EMCATE or go the the following
website:http://tweetchat.com/room/EMCATE
Hope to see youthere for questions,commentsorjust a casual chat!
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25. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov26, 2012 12:13 PM (inresponse toJonKlaus)
We talkedaboutFCIP,buffervcredits,inorderdelivery,portgroups,sharedbandwidth,butalsoSONET
OC-48 to coveralmost1,000 miles.Prettyimpressive!
Lots of questions,lotsof answer,butstill alotof unansweredones.Needmore time.......
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26. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
RRR Nov27, 2012 3:50 AM(inresponse toRRR)
I fixedit.Itotallyforgotaboutthe speedof light,whichislowerinfibercomparedtovacuum.It'sabout
1.5 timesslower,sothe speedof lightinfiberisapproximately200kkm/s.
Thismeansmore frameswill live onthe link,somore buffersare needed.
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27. Re:Ask The Expert:Discussingthe challengesof LongDistance Links

14. merlijnNov27,2012 4:59 AM(inresponse toJonKlaus)
Hi Jon,
Can the counteractuallydropto 0, or is1 the minimum?
Because there’salwaysatleast1 BBC for a port.
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28. Re:Ask The Expert:Discussingthe challengesof LongDistance Links
JonKlaus Nov27, 2012 5:46 AM(inresponse tomerlijn)
Hi Merlijn,
Yes,the countercan actuallydropto 0. At that point,transmissionof new frameswillhaltuntill atleast
1 ACKisback and the counteris incrementedagain.
If you run the followingCLIcommandagainstan interface,forexample fc1/1:
sh intfc1/1 countersdetails|grepcredit
You will see somethinglike:
0 waitsdue to lackof transmitcredits
14576233333 transitionsof tx BB creditout of zero state
14 transitionsof rx BB creditto zero state
I had to google abit to make sense of it..apparently"transitionsof tx BBcreditout of zero state" does
not indicate aproblem.Ihave resetthe countersandit keepsincrementing;the othertwovalues
remain0. Thisis a 1km linkwhichhas32 credits,so ithardlyeverdipsunder31.
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29. Re:Ask The Expert:Discussingthe challengesof LongDistance Links

15. RRR Nov27, 2012 7:22 AM(inresponse toJonKlaus)
0 waitsdue to lackof transmitcredits
14576233333 transitionsof tx BB creditout of zero state
14 transitionsof rx BB creditto zero state
Is this"waitsdue to lackof transmitcredits"cumulativeorsimplyacounterfora certainpointintime to
presentthe value atTHAT moment?Ibet that if it'scumulative,youcaneasilytrackbuffershortages
and act uponthat.
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