EEE 414 traffic [Compatibility Mode].pdf
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Transcript of EEE 414 traffic [Compatibility Mode].pdf
Traffic Engineering
Traffic EngineeringTraffic Engineering• One billion+ terminals in voice network alone
– Plus data, video, fax, finance, etc.
• Imagine all users want service simultaneously…its not even nearly possible (despite our common intuition)– In practice, the actual amount of equipment provisioned is vastlyIn practice, the actual amount of equipment provisioned is vastly
less than would support all users simultaneously
• And yet, by and large, we get the impression of phone and data networks that work very well!networks that work very well!
• How is this possible?
Traffic theory !!y
Traffic Engineering Traffic Engineering –– TradeTrade--offsoffs• Design number of transmission paths, or radio channels?
– How many required normally?– What if there is an overload?
• Design switching and routing mechanisms– How do we route efficiently?How do we route efficiently? – E.g.
• High-usage trunk groups -Trunk group that is the primary direct route between two switching systems. The group is provided with an alternate route for overflow traffic in order to provide an acceptable level of blocking.
• Overflow trunk groups• Where should traffic flows be combined or kept separate?Where should traffic flows be combined or kept separate?
• Design network topology– Number and sizing of switching nodes and locations
N b d i i f t i i t d l ti– Number and sizing of transmission systems and locations– Survivability
Characterization of Telephone TrafficCharacterization of Telephone Traffic
•• Calling RateCalling Rate () – also called arrival rate, or attempts rate, etc.– Average number of calls initiated per unit time (e.g. attempts per g p ( g p p
hour)– Each call arrival is independent of other calls (we assume)– Call attempt arrivals are random in time– Until otherwise, we assume a “large” calling group or source pool
Tαγ If receive calls from a terminal in time TT:T
If receive calls from mm terminals in time T:
Tαγ g
Group calling rateTm
αγ
Per terminalGroup calling rate Per terminalcalling rate
Characterization of Telephone Traffic (2)Characterization of Telephone Traffic (2)• Calling rate assumption:
– Number of calls in time T is Poisson distributed: e x
– In our case
...2 ,1 ,0!
)(
xx
exp
T
– x – No of Call Arrivals– λ – average call arrival rate
Ti b t ll i “ ti ll ” di t ib t d• Time between calls is “-ve exponentially” distributed:
tetf t 0)( 1
mean
• In probability theory and statistics, the Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate
• Class Question: What do these observations about telephone traffic imply about the nature of the traffic sources?
p gand independently of the time since the last event.
--ve Exponential Holding Timesve Exponential Holding Times•• Implies the “MemoryImplies the “Memory--less” propertyless” property
– Prob. a call last another minute is independent of how long the call has already lasted! Call “forgets” that it has already survived to time T1a eady asted Ca o gets t at t as a eady su ed to t e 1
tTPTTtTTP 11
• Proof:
1
1111
TTP
TTtTTPTTtTTP
1
1 TTP
tTTP
hT
htT
ee
/
/)(
1
1
htetTP /)(
Recall:
hT
hthT
eee/
//
1
1
hte / tTP
etTP )(
•In probability theory and statistics, the exponential distribution (a.k.a. negative exponential distribution) is a family of continuous probability distributions. It describes the time between events in a Poisson process.
Characterization of Telephone Traffic (3)Characterization of Telephone Traffic (3)
•• Holding TimeHolding Time (hh)– Mean length of time a call lasts– Probability of lasting time t or more is also –ve exponential in
nature:0)( / tetTP ht
– Real voice calls fits very closely to the negative exponential form
00)( ttTP
Real voice calls fits very closely to the negative exponential form above
– As non-voice “calls” begin to dominate, more and more calls have a constant holding time characteristic
•• Departure RateDeparture Rate ():
h1
Some Real Holding Time DataSome Real Holding Time Data
Traffic Traffic Volume (V)Volume (V)
hV = # calls in time period T
h h ldi tihV h = mean holding time
V = volume of calls in time period T
• In N. America this is historically usually expressed in terms of “ccsccs”:– Hundred call seconds
“cc” “cc” “ss”
– 1 ccs is volume of traffic equal to:one ci c it b s fo 100 seconds o
Centum Call Seconds
– one circuit busy for 100 seconds, or– two circuits busy for 50 seconds, or– 100 circuits busy for one second, etc.
TrafficTraffic Intensity (A)Intensity (A)• Also called “traffic flowtraffic flow” or simply “traffictraffic”.
= # calls in time period Th = # calls in time period T = # calls in time period TV = # calls in time period T # calls in time period T
h = mean holding time
T = time period of observations
hA
Th
# calls in time period T
h = mean holding time
T = time period of observations
# calls in time period T
h = mean holding time
T = time period of observationsR ll R ll
V
T
R ll
# calls in time period T
h = mean holding time
T = time period of observationspp
= calling rate
p
= calling rate
= departure rateT
Recall:
h1
Recall:
V h
Recall:p
= calling rate
= departure rate
• Units:– “ccs/hourccs/hour”, or
dimensionless (if h and T are in the same units of time)
V = call volume
– dimensionless (if h and T are in the same units of time)
“ErlangErlang” unit
The The ErlangErlang• Dimensionless unit of traffic intensity
• Named after Danish mathematician A. K. Erlang (1878-1929)Named after Danish mathematician A. K. Erlang (1878 1929)
• Usually denoted by symbol EE.
• 1 Erlang is equivalent to traffic intensity that keeps:g q y p– one circuit busy 100% of the time, or– two circuits busy 50% of the time, or– four circuits busy 25% of the time, etc.y ,
• 26 Erlangs is equivalent to traffic intensity that keeps :– 26 circuits busy 100% of the time, or
52 circuits busy 50% of the time or– 52 circuits busy 50% of the time, or– 104 circuits busy 25% of the time, etc.
Class Class • Could 4 E be produced as a traffic intensity by:
– 16 sources? (What is the utilization?)– 4 sources (same)– 1 source?
• What is special about the traffic intensity if it pertains to one source or terminal only?
Erlang (2)Erlang (2)• How does the ErlangErlang unit correspond to ccsccs?
100 call seconds 0.027E100 call seconds1 ccs hour1 hour × 60 min hr × 60 sec min
3600 call seconds
× 60 min hr × 60 sec min
• Percentage of time a terminal is busy is equivalent to the traffic
1E3600 call seconds36 ccs hour1 hour × 60 min hr × 60 sec min
× 60 min hr × 60 sec min
• Percentage of time a terminal is busy is equivalent to the traffic generated by that terminal in Erlangs, or
• Average number of circuits in a group busy at any time
• Typical usages:– residence phone -> 0.02 E– business phone -> 0 15 Ebusiness phone > 0.15 E– interoffice trunk -> 0.70 E
Traffic Offered, Carried, and LostTraffic Offered, Carried, and Lost•• Offered TrafficOffered Traffic (TTO O ) equivalent to Traffic Intensity (AA)
– Takes into account all attempted calls, whether blocked or not, d h d h ldand uses their expected holding times
• Also Carried TrafficCarried Traffic (TTC C ) and Lost TrafficLost Traffic (TTL L )
• Consider a group of 150 terminals each with 10% utilization (or• Consider a group of 150 terminals, each with 10% utilization (or in other words, 0.1 E per source) and dedicated servicededicated service:
11 each terminal has an1
150
1
150
outgoing trunk(i.e. terminal:trunk ratio = 1:1)
150150TO = A = 150 x 0.10 E = 15.0 E
TC = 150 x 0.10 E = 15.0 E
T 0 ETL = 0 E
Traffic Offered, Carried, and Lost (2)Traffic Offered, Carried, and Lost (2)• A = TO = TC + TL
TrafficLost
TrafficIntensity Offered
Traffic
CarriedTraffic
Traffic
• T = T x Prob Blocking (or congestion)• TL = TO x Prob. Blocking (or congestion)
= P(B) x TO = P(B) x A
•• Circuit UtilizationCircuit Utilization () - also called Circuit EfficiencyCircuit Efficiency– proportion of time a circuit is busy, or– average proportion of time each circuit in a group is busy
CT # of Trunks# of Trunks
Grade of Service (gos)Grade of Service (gos)• In general, the term used for some traffic design objective
• Indicative of customer satisfactionIndicative of customer satisfaction
• In systems where blocked calls are cleared, usually use:
L LT T ( )L L
O L C
T T ( )T T + T
P Bgos
• Typical gos objectives:• Typical gos objectives:– in busy hour, range from 0.2% to 5% for local calls, however– generally no more that 1%– long distance calls often slightly higherlong distance calls often slightly higher
• In systems with queuing, gos often defined as the probability of d l di ifi l th f tidelay exceeding a specific length of time
Grade of Service Related TermsGrade of Service Related Terms•• Busy HourBusy Hour
– One hour period during which traffic volume or call attempts is the h h ll d dhighest overall during any given time period
•• Peak (or Daily) Busy HourPeak (or Daily) Busy Hour– Busy hour for each day, usually varies from day to dayy y, y y y
•• Busy SeasonBusy Season– 3 months (not consecutive) with highest average daily busy hour
•• High Day Busy Hour (HDBH)High Day Busy Hour (HDBH)– One hour period during busy season with the highest load
Grade of Service Related Terms (2)Grade of Service Related Terms (2)•• Average Busy Season Busy Hour (ABSBH)Average Busy Season Busy Hour (ABSBH)
– One hour period with highest average daily busy hour during the b
•• Average Busy Season Busy Hour (ABSBH)Average Busy Season Busy Hour (ABSBH)– One hour period with highest average daily busy hour during the
bbusy seasonbusy season– For example, assume days shown below make up the busy season:
1-Apr 2-Apr 3-Apr 4-Apr 5-Apr 6-Apr 7-Apr 8-Apr 9-Apr 10-Apr 11-Apr 12-Apr 13-Apr 14-Apr 15-Apr 16-Apr 17-Apr 18-Apr 19-Apr 20-Apr 21-Apr Mean00:00 to 01:00 1.4 1.4 1.2 1.5 1.1 1.5 1.7 1.5 1.0 1.0 1.8 1.5 1.8 1.6 1.2 1.9 1.8 1.6 1.4 1.5 1.2 1.501 00 t 02 00 1 2 1 8 1 6 1 3 1 0 1 6 1 1 1 1 1 0 1 2 1 7 2 0 2 0 1 8 1 3 1 7 1 4 1 9 1 1 1 4 1 5 1 5
HighestHighestABSBHABSBH
01:00 to 02:00 1.2 1.8 1.6 1.3 1.0 1.6 1.1 1.1 1.0 1.2 1.7 2.0 2.0 1.8 1.3 1.7 1.4 1.9 1.1 1.4 1.5 1.502:00 to 03:00 1.4 1.8 1.5 1.9 1.2 1.0 1.2 1.1 1.1 1.7 1.5 1.5 1.9 1.9 1.3 1.5 1.8 1.1 1.1 1.2 1.5 1.403:00 to 04:00 1.2 1.8 1.7 1.4 1.7 1.1 1.5 1.6 1.1 1.9 1.0 1.0 1.4 1.5 1.6 1.1 1.4 1.9 1.4 1.2 1.1 1.404:00 to 05:00 1.8 1.8 2.3 2.2 2.0 1.7 2.3 1.6 2.2 1.5 2.1 1.6 2.3 2.1 1.7 2.5 1.6 2.0 1.7 1.5 2.3 1.905:00 to 06:00 2.2 2.3 1.9 2.4 2.5 2.0 2.0 1.7 1.8 1.6 2.0 2.0 2.2 2.2 2.1 1.8 1.6 1.7 2.0 2.3 2.1 2.006:00 to 07:00 1.7 2.2 1.7 2.5 2.2 2.1 2.2 2.0 2.3 1.6 2.4 2.2 1.5 2.1 2.2 1.8 1.8 1.7 2.1 2.0 2.1 2.007:00 to 08:00 2.0 2.8 2.2 2.4 2.3 2.4 2.9 2.0 2.4 2.4 2.1 2.9 2.3 2.1 2.9 2.7 2.8 2.3 2.1 2.1 2.7 2.408:00 to 09:00 3.4 3.1 2.8 2.9 2.5 2.7 2.9 3.0 3.4 3.4 3.1 2.9 2.9 2.9 3.3 3.2 3.5 3.1 3.1 3.1 2.5 3.009:00 to 10:00 3.4 3.4 4.0 3.2 3.5 3.4 3.1 3.7 3.3 3.3 3.5 3.9 3.4 4.0 3.7 3.7 3.1 3.4 3.9 3.9 3.4 3.510:00 to 11:00 5.0 4.4 4.8 4.9 4.1 3.0 4.0 4.9 4.2 4.9 4.7 4.2 3.8 3.0 4.6 4.9 4.4 5.0 4.7 3.6 3.8 4.311:00 to 12:00 4.8 5.0 4.7 4.3 4.5 3.8 3.4 4.2 5.0 4.6 5.0 4.7 3.2 3.4 5.0 4.8 4.1 4.3 4.4 3.6 3.7 4.312:00 to 13:00 4.5 4.2 4.1 4.8 4.6 3.8 3.3 4.0 4.2 4.6 4.7 4.0 3.3 3.1 5.0 4.9 4.6 4.1 4.2 3.2 3.6 4.113:00 to 14:00 4.3 4.2 4.7 4.5 4.8 3.2 3.1 4.1 4.5 4.6 4.9 4.7 3.6 3.6 4.8 4.2 4.8 4.9 4.4 3.3 3.0 4.214:00 to 15:00 4.8 4.7 4.5 4.1 4.4 3.6 3.7 4.5 4.3 4.3 4.9 4.5 3.5 3.5 4.3 4.3 4.3 4.5 4.3 3.3 3.2 4.214:00 to 15:00 4.8 4.7 4.5 4.1 4.4 3.6 3.7 4.5 4.3 4.3 4.9 4.5 3.5 3.5 4.3 4.3 4.3 4.5 4.3 3.3 3.2 4.215:00 to 16:00 4.4 4.9 4.4 4.8 4.5 3.8 3.2 4.1 4.8 4.4 4.5 4.2 3.3 3.9 4.3 4.9 4.4 4.3 4.5 3.7 3.3 4.216:00 to 17:00 3.2 3.2 3.8 3.5 3.7 3.1 3.5 3.5 3.2 3.2 3.8 3.4 3.2 4.0 3.3 4.0 3.9 3.0 3.3 3.5 3.3 3.517:00 to 18:00 2.7 2.6 2.7 2.9 3.3 3.1 3.4 2.9 3.2 2.8 2.7 3.0 3.3 3.2 2.5 2.9 2.8 3.4 3.5 2.9 3.2 3.018:00 to 19:00 3.0 2.9 3.0 2.7 2.9 3.4 3.3 3.4 2.7 3.3 3.5 3.5 2.7 3.1 3.1 3.3 3.4 3.1 3.0 3.3 3.3 3.119:00 to 20:00 3.3 3.3 2.6 3.4 3.2 2.7 2.7 3.4 3.4 3.0 3.0 3.4 3.1 2.8 3.2 3.4 3.0 3.4 3.4 3.1 2.9 3.120:00 to 21:00 2.9 2.3 2.1 2.9 2.9 3.0 3.0 2.4 2.3 2.9 3.0 2.1 2.2 2.9 3.0 2.6 2.4 2.5 2.7 2.7 2.6 2.621:00 to 22:00 2 1 1 6 2 3 1 6 2 2 2 1 2 4 1 9 1 6 2 1 2 4 1 7 1 8 2 4 1 8 1 9 2 2 1 9 2 2 2 2 1 6 2 0
Note: Red indicatesdaily busy hour
21:00 to 22:00 2.1 1.6 2.3 1.6 2.2 2.1 2.4 1.9 1.6 2.1 2.4 1.7 1.8 2.4 1.8 1.9 2.2 1.9 2.2 2.2 1.6 2.022:00 to 23:00 1.5 2.1 1.9 1.6 1.7 1.6 2.3 2.5 2.4 1.7 2.1 1.8 2.0 2.4 1.7 1.9 2.2 2.3 1.7 2.4 1.8 2.023:00 to 00:00 1.5 1.0 1.1 1.1 1.5 1.8 1.5 1.4 1.8 1.1 1.9 1.2 1.6 1.9 1.8 1.1 1.5 2.0 1.8 1.6 1.4 1.5
Hourly Traffic VariationsHourly Traffic Variations
Daily Traffic VariationsDaily Traffic Variations
Seasonal Traffic VariationsSeasonal Traffic Variations
Seasonal Traffic Variations (2)Seasonal Traffic Variations (2)
Typical Call Attempts BreakdownTypical Call Attempts Breakdown• Calls Completed - 70.7%
• Called Party No Answer - 12.7%Called Party No Answer 12.7%
• Called Party Busy - 10.1%
• Call Abandoned - 2.6%
• Dialing Error - 1.6%
• Number Changed or Disconnected - 0.4%
• Blockage or Failure - 1.9%
3 Types of Blocking Models3 Types of Blocking Models• Blocked Calls Cleared (BCCBCC)
– Blocked calls leave system and do not return– Good approximation for calls in 1st choice trunk group
• Blocked Calls Held (BCHBCH)– Blocked calls remain in the system for the amount of time it wouldBlocked calls remain in the system for the amount of time it would
have normally stayed for– If a server frees up, the call picks up in the middle and continues– Not a good model of real world behaviour (mathematical g (
approximation only)– Tries to approximate call reattempt efforts
• Blocked Calls Wait (BCWBCW)• Blocked Calls Wait (BCWBCW)– Blocked calls enter a queue until a server is available– When a server becomes available, the call’s holding time begins
Traffic formula selection Decision treeTraffic formula selection Decision tree
Blocked Calls Cleared (BCC)Blocked Calls Cleared (BCC)
10 minutes
2 sources
Source #1
Offered Traffic 1 3
Total Traffic Offered:
Source #2
Offered Traffic 2 4
TO = 0.4 E + 0.3 ETO = 0.7 E
Only one server
Traffic
1st call arrives and is served
2nd call arrives but server already busyTraffic
Carried 1
server already busy
22nd call is cleared
1
3rd call arrives and is served
3 4
Total Traffic Carried:4th call arrives and is served
Total Traffic Carried:TC = 0.5 E
Blocked Calls Held (BCH)Blocked Calls Held (BCH)
10 minutes
2 sources
Source #1
Offered Traffic 1 3
Total Traffic Offered:
Source #2
Offered Traffic 2 4
TO = 0.4 E + 0.3 ETO = 0.7 E
Traffic
Only one server1st call arrives and is served
2nd call arrives but server busy
Traffic
Carried 1 21 2 3 42nd call is served
3rd call arrives and is servedTotal Traffic Carried:
2nd call is held until server free
4th call arrives and is served
Total Traffic Carried:TC = 0.6 E
Blocked Calls Wait (BCW)Blocked Calls Wait (BCW)
10 minutes
2 sources
Source #1
Offered Traffic 1 3
Total Traffic Offered:
Source #2
Offered Traffic 2 4
TO = 0.4 E + 0.3 ETO = 0.7 E
1st ll i d i d
Only one server
Traffic
1st call arrives and is served
2nd call arrives but server busy
2nd call waits until server freeTraffic
Carried 1 2 2nd call served1 2
3rd call arrives, waits, and is served
3
th ll i i d
4
Total Traffic Carried:4th call arrives, waits, andis served
Total Traffic Carried:TC = 0.7 E
Blocking ProbabilitiesBlocking Probabilities• System must be in a Steady StateSteady State
– Also called state of statistical equilibrium–– Arrival RateArrival Rate of new calls equals Departure RateDeparture Rate of
disconnecting calls– Why?
If ll i f h h d ?• If calls arrive faster that they depart?• If calls depart faster than they arrive?
Binomial Distribution ModelBinomial Distribution Model• Assumptions:
–– mm sources–– AA Erlangs of offered traffic
• per source: TO = A/m• probability that a specific source is busy: P(B) = A/m
• Can use Binomial Distribution to give the probability that a certain number (kk) of those m sources is busy:
kmk
mA
mA
km
kP
1)(
kmk
mA
mA
kmkm
1)!(!
!mmk mmkmk )!(!
Binomial Distribution Model (2)Binomial Distribution Model (2)• What does it mean if we only have N serversN servers (N<m)?
– We can have at most N busy sources at a time– What about the probability of blocking?
• All N servers must be busy before we have blocking
)()( NkPBP )(...)1()( mkPNkPNkP
m kmk AAm
kmk
mA
mA
km
kP
1)(
m
Nk mA
mA
km
1
kk
Remember:
1
011
N
k
kmk
mA
mA
km
Binomial Distribution Model (3)Binomial Distribution Model (3)• What does it mean if k>N?
– Impossible to have more sources busy than servers to serve them– Doesn’t accurately represent reality
• In reality, P(k>N) = 0
– In this model, we still assign P(k>N) = A/m – Acts as good model of real behaviour
• Some people call back, some don’t
• Which type of blocking model is the Binomial Distribution?yp g– Blocked Calls Held (BCH)
Time Congestions Time Congestions vs.vs. Call CongestionCall Congestion• Time Congestion
– Proportion of time a system is congested (all servers busy)– Probability of blocking from point of view of servers
• Call Congestion– Probability that an arriving call is blockedProbability that an arriving call is blocked– Probability of blocking from point of view of calls
• Why/How are they different?
Time Congestion:
)()( NkPBP
Call Congestion:
)()( NkPBP )()(
Probability that allservers are busy.
)()(
Probability that there aremore sources wanting serviceth ththan there are servers.
Poisson Traffic ModelPoisson Traffic Model• Poisson approximates Binomial with large mlarge m and small A/msmall A/m
k
!)(
kekP
k = Mean # of
Busy Sources
Note: )(lim BinomialPoissonm
• What is ?• What is ?– Mean number of busy sources
– = A
!)(
kAekP
kA
Poisson Traffic Model (2)Poisson Traffic Model (2)• Now we can calculate probability of blocking:
)()( NkPBP )(...)1()( PNPNP
Remember:
AkA kA Ae
A
kA!
)(kAekP
kA
Nk k!
Nk
Aek!
N kA1A
N
k
k
ekA
1
0 !1
)()( ANPBPExample:
)107(P),()( ANPBP “P” = Poisson “A” = Offered Traffic
)10,7(P
PoissonPoisson P(B) with 10 E10 Eoffered to 7 servers7 servers
“N” = # Serversoffered to 7 servers7 servers
Traffic TablesTraffic Tables• Consider a 1% chance of blocking in a system with N=10 trunks
– How much offered traffic can the system handle?
A
k
k
k
Ak
ekAe
kA
9
010 !1
!01.0
• How do we calculate A?– Very carefully, or
ff bl– Use traffic tables
Traffic Tables (2)Traffic Tables (2)P(B)=P(N,A)P(B)=P(N,A)
NN
AA
Traffic Tables (3)Traffic Tables (3)P(N,A)=0.01P(N,A)=0.01
N=10N=10
A=4.14 EA=4.14 E
If system with N = 10 trunksIf system with N = 10 trunkshas P(B) = 0.01:has P(B) = 0.01:
System can handleSystem can handleSystem can handleSystem can handleOffered traffic (A) = 4.14 EOffered traffic (A) = 4.14 E
Poisson Traffic TablesPoisson Traffic TablesP(N,A)=0.01P(N,A)=0.01
N=10N=10
A=4.14 EA=4.14 E
If system with N = 10 trunksIf system with N = 10 trunkshas P(B) = 0.01:has P(B) = 0.01:
System can handleSystem can handleSystem can handleSystem can handleOffered traffic (A) = 4.14 EOffered traffic (A) = 4.14 E
Efficiency of Large GroupsEfficiency of Large Groups• What if there are N = 100 trunks?
– Will they serve A = 10 x 4.14 E = 41.4 E with same P(B) = 1%?– No!– Traffic tables will show that A = 78.2 E!
• Why will 10 times trunks serve almost 20 times traffic?Why will 10 times trunks serve almost 20 times traffic?– Called efficiency of large groupsefficiency of large groups:
For N = 10 A = 4 14 E efficiency%44114.4
AFor N = 10, A = 4.14 E efficiency %4.4110
N
For N = 100, A = 78.2 E efficiency %2.78100
2.78
NA
100N
The larger the trunk group, the greater the efficiency
TrafCalc SoftwareTrafCalc Software• What if we need to calculate P(N,A) and not in traffic table?
–– TrafCalcTrafCalc: Custom-designed software • Calculates P(B) or A, or• Creates custom traffic tables
TrafCalc Software (2)TrafCalc Software (2)• How do we calculate P(32,20)?
TrafCalc Software (3)TrafCalc Software (3)• How do we calculate A for which P(32,A) = 0.01?
Erlang B ModelErlang B Model• More sophisticated model than Binomial or Poisson
• Blocked Calls Cleared (BCC)Blocked Calls Cleared (BCC)
• Good for calls that can reroute to alternate route if blocked
• No approximation for reattempts if alternate route blocked toopp p
• Derived using birthbirth--death processdeath process– See selected pages from Leonard Kleinrock, Queueing Systems
Volume 1: Theory John Wiley & Sons 1975Volume 1: Theory, John Wiley & Sons, 1975
Erlang B BirthErlang B Birth--Death ProcessDeath Process• Consider infinitesimally small time tt during which only one
arrival or departure (or none) may occur
• Let be the arrival rate from an infinite pool or sources
• Let = 1/h= 1/h be the departure rate per callN t if kk ll i t d t t i kk– Note: if kk calls in system, departure rate is kk
• Steady State Diagram:
Blockage
0 1 2 N-1 N……
2 N(N-1)3
Immediate Service
Erlang B BirthErlang B Birth--Death Process (2)Death Process (2)• Steady State (statistical equilibrium)
– Rate of arrival is the same as rate of departure– Average rate a system enters a given state is equal to the average
rate at which the system leaves that state
Probability of moving
from state 1 to state 2? PP11
0 1 2 N-1 N……P0 P1 P2 PN-1 PN
2 N(N-1)3
Probability of movingProbability of movingfrom state 2 to state 1? 22PP22
Erlang B BirthErlang B Birth--Death Process (3)Death Process (3)
0 1 2 N-1 N……
P0 P1 P2 PN-1 PN
• Set up balance equations: 2 N(N-1)3
0 1P P 0 1P P 1 0P P
0 1P P
1 1 2 02P P P P
2 2 3 12 3P P P P
0 1P P
1 22P P
2 33P P
2 12P P
20
2P
3 3 4 23 4P P P P
3
1k kP k P
3 23P P
30
6P
1 1 2( 1) N N N NN P P N P P
N P P
1k k
P N P
0
!
k
kPPk
1N NN P P 1N NP N P !k
Erlang B BirthErlang B Birth--Death Process (4)Death Process (4)
Rule of Total Probability:
N iN P 1P
0
!
k
kPPk
Recall:
01
N
ii
P
0
0 !
N
i
Pi
0
0
1!
iN
i
P
i
1
k
A h
Recall:
kA
0
!
1!
k iN
i
kP
i
0
!
!
iNk
i
AkP A
i
0 !i i For blocking, must be in state k = N:
( ) ( ) !
NAN( ) ( , ) NP B B N A P
“B” = Erlang B
“N” = # Servers0
!
!
iN
i
NAi
N = # Servers“A” = Offered Traffic
Erlang B Traffic TableErlang B Traffic Table
Example: In a BCC system with m= sources, we can accept a
B(N,A)=0.001B(N,A)=0.001
0.1% chance of blocking in the nominal case of 40E offered traffic. However, in the extreme case of a 20% overload we can accept a
B(N,A)=0.005B(N,A)=0.005
20% overload, we can accept a 0.5% chance of blocking.
How many outgoing trunks do we need? A 40 EA 40 Eneed? A=40 EA=40 EN=59N=59Nominal design: 59 trunks
AA48 E48 EOverload design: 64 trunks
N=64N=64
Requirement: 64 trunks
Example (2)Example (2)P(N,A)=0.01P(N,A)=0.01
N=32N=32
A=20.3 EA=20.3 E
P(N,A) & B(N,A) P(N,A) & B(N,A) -- High BlockingHigh Blocking• We recognize that Poisson and Erlang B models are only
approximations but which is better?– Compare them using a 4-trunk group offered A=10E
Erlang BErlang B
(4 10) 0 64666B
PoissonPoisson
(4 10) 0 98966P(4,10) 0.64666B
(1 ( ))CT A P B 10 (1 0.64666)
3 533T E
(4,10) 0.98966P
(1 ( ))CT A P B 10 (1 0.98966)
0 103T E3.533CT E
3.533 0.884
0.103CT E
0.103 0.0264
4 4
How can 4 trunks handle 10E offeredHow can 4 trunks handle 10E offeredtraffic and be busy only 2.6% of the time?traffic and be busy only 2.6% of the time?
P(N,A) & B(N,A) P(N,A) & B(N,A) -- High Blocking (2)High Blocking (2)• Obviously, the Poisson result is so far off that it is almost
meaningless as an approximation of the example.– 4 servers offered enough traffic to keep 10 servers busy full time
(10E) should result in much higher utilization.
• Erlang B result is more believable.– All 4 trunks are busy most of the time.
• What if we extend the exercise by increasing A?Erlang B result goes to 4E carried traffic– Erlang B result goes to 4E carried traffic
– Poisson result goes to 0E carried
• Illustrates the failure of the Poisson model as valid for situations with high blocking– Poisson only good approximation when low blocking– Use Erlang B if high blocking
Engset Distribution ModelEngset Distribution Model• BCC model with small number of sources (m > N)
= mean departure rate per call mean departure rate per call
= mean arrival rate of a single source
k = arrival rate if in the system is state k Blockagek y
kk = = (m(m--k)k)
m (m-1) (m-2) [M-(N-2)] [m-(N-1)]
Blockage
0 1 2 N-1 N……P0 P1 P2 PN-1 PN
2 N(N-1)3
Immediate Service
Engset Traffic Model (2)Engset Traffic Model (2)• Balance equations give:
k 10
!!( )!
k
kmP P
k m k
and 0
0
1iN
i
Pmi
therefore:k
k iN
mk
Pm
but can show that: Am A
0i
mi
N mA
( ) ( )P B P k N ( , , ) iN
m A NE m N A
mAA i
0i m A i
“E” = Engset
Engset Traffic TableEngset Traffic Table M = 30 sourcesM = 30 sources
# trunks (N)# trunks (N)
Traffic offered (A)Traffic offered (A)
P(B)=E(m,N,A)P(B)=E(m,N,A) N=10N=10
Example: 30 terminals each provide
AA==4.8 E4.8 E
P(B) 0 01P(B) 0 01Example: 30 terminals each provide 0.16 Erlangs to a concentrator with a goal of less than 1% blocking.
How many outgoing trunks do we
P(B)<0.01P(B)<0.01
y g gneed?
A = 30 x 0.16 = 4.8 E
Requirement: N = 10 TrunksN = 10 TrunksCheck m < 10 x N?M=30 < 10 x 10 = 100Requirement: N = 10 TrunksN = 10 Trunks M=30 < 10 x 10 = 100
Erlang C Distribution ModelErlang C Distribution Model• BCW model with infinite sourcesinfinite sources (m) and infinite queue lengthinfinite queue length
= arrival rate of new calls arrival rate of new calls
= mean departure rate per callBlockage
0 1 2 N Q1 Q2…… ……P0 P1 P2 PN PQ1 PQ2
2 NNNN3
Immediate Service
Erlang C Distribution Model (2)Erlang C Distribution Model (2)• Balance equations give:
10 ,
!
k
kA PP k Nk
and0 ,
!
k
k k N
A PP k NN N and 0 1
0
1
! !
N iN
i
PA N AN N A i
• But P(B) = P(kN):
0( )k
k N
A PP B
0k
N
PA 0
kNA AP
but can show that:kA N
A
( )!k N
k N N N !N
k N N N N
00! k
PN N
0k N N A
NA NNA N
0( )!
A NP B PN N A
1
0
!( , )
! !
N iN
i
N N AC N AA N AN N A i
“ ” l 0! !iN N A i “C” = Erlang C
Erlang C Traffic TablesErlang C Traffic Tables N=18N=18
# trunks (N)# trunks (N) P(B)=C(N,A)P(B)=C(N,A)
Traffic offered (A)Traffic offered (A)A 7 EA 7 E
( )( )
Example:
h h b b l f bl k
A=7 EA=7 E C(18,7)=0.0004C(18,7)=0.0004
What is the probability of blocking in an Erlang C system with 18 servers offered 7 Erlangs of traffic?
Delay in Erlang CDelay in Erlang C• Expected number of calls in the queue?
( )k N P
( )kAk N P
kNA AP k
( ) k
k Nk N P
0( )!k N
k Nk N P
N N
0! k
kP k
N N
0NP A A N
( , )A C N A
( )h C N A
!N N A N A
N A
( , )C N A
N A
Mean #Calls DelayedMean Delay over All Calls ( )h C N AT Recall:
yMean Delay over All Calls = Arrival Rate of Calls
( , )C N AN A
T
Mean Delay of Delayed Calls = hMean Delay of Delayed Calls N A
Also:
( ) ( )hT
N AP delay T C N A e
( ) ( , ) N AP delay T C N A e
Comparison of Traffic ModelsComparison of Traffic Models
Erlang C (BCW,Erlang C (BCW, sources)sources)
i i l ( C )i i l ( C )
Poisson (BCH, Poisson (BCH, sources)sources)
Erlang B (BCC, Erlang B (BCC, sources)sources)
Erlang C (BCW, Erlang C (BCW, sources)sources)
P(B)
Binomial (BCH, m sources)Binomial (BCH, m sources)
Engset (BCC, m sources)Engset (BCC, m sources)
Offered Traffic (A)
Efficiency of Large GroupsEfficiency of Large Groups• Already seen that for same P(B), increasing servers results in
more than proportional increase in traffic carried
example 1: (10, 4.14) 0.01P and (100,78.2) 0.01P
l 2example 2: (32, 20.3) 0.01P (33, 20.1) 0.005P and
example 3: (8, 2.05) 0.001B (80,57.8) 0.001B and
• What does this mean?– If it’s possible to collect together several diverse sources, you can
• provide better gos at same cost, or• provide same gos at cheaper cost
Efficiency of Large Groups (2)Efficiency of Large Groups (2)• Two trunk groups offered 5 Erlangs each, and B(N,A)=0.002
N1=?5 E How many trunks total?
From traffic tables, find B(13,5) 0.002 N1=13
N2=?5 E N2=13 Ntotal = 13 + 13 = 26 trunks
Trunk efficiency?
CT 10(1 0.002) 0 384CTN
10(1 0.002) 0.384
26
38.4% utilization 38.4% utilization
Efficiency of Large Groups (3)Efficiency of Large Groups (3)• One trunk group offered 10 Erlangs, and B(N,A)=0.002
How many trunks?
N=?10 E
How many trunks?
From traffic tables, find B(20,10) 0.002 N=20
N 20 kN = 20 trunks
Trunk efficiency?
CTN
10(1 0.002) 0.499
20
49.9% utilization 49.9% utilization
For same gos, we can save 6 trunks!
Efficiency of Large Groups (4)Efficiency of Large Groups (4)
B=0.1B=0.1
B=0.01B=0.01
A B=0.1B=0.1
B=0.01B=0.01
B=0.001B=0.001 B=0.001B=0.001
N N
Sensitivity to OverloadSensitivity to Overload• Consider 2 cases:
Case 1: N = 10 and B(N,A) = 0.01Case 1: N 10 and B(N,A) 0.01
B(10,4.5) 0.01, so can carry 4.5 E
What if 20% overload (5.4 E)? B(10,5.4) 0.03( ) ( )
3 times P(B) with 20% overload
Case 1: N = 30 and B(N,A) = 0.01
B(30,20.3) 0.01, so can carry 20.3 E
What if 20% overload (24.5 E)? B(30,24.5) 0.08
8 times P(B) with 20% overload!
“Trunk Group Splintering”“Trunk Group Splintering”• if high possibility of overloads, small groups may be better
Incremental Traffic Carried by NIncremental Traffic Carried by Nthth TrunkTrunk
• If a trunk group is of size N-1, how much extra traffic can it carry if you add one extra trunk?
– Before, can carry: TC1 = A x [1-(B(N-1,A)]
– After, can carry: TC2 = A x [1-(B(N,A)]
2 1N C CA T T 1 ( , ) 1 ( 1, )A B N A B N A
( 1, ) ( , )A B N A B N A ( ) ( )A N A B N A
• What does this mean?
–– Random HuntingRandom Hunting: Increase in trunk group’s total carried traffic
( ) ( , )NA N A B N A for very low blocking
after adding an Nth trunk
–– Sequential HuntingSequential Hunting: Actual traffic carried by the Nth trunk in the group
Incremental Traffic Carried by NIncremental Traffic Carried by Nthth Trunk (3)Trunk (3)
Fixed B(N A)
AN
Fixed B(N,A)
N
ExampleExample• Individual trunks are only economic if they can carry 0.4 E or
more. A trunk group of size N=10 is offered 6 E. Will all 10 t k b i l?trunks be economical?
( 1, ) ( , )NA A B N A B N A
10 6 (9,6) (10,6)A B B
6 0.07514 0.04314
0.192 E 0.4 E
At least the 10At least the 10thth trunk is not trunk is not economicaleconomical