Poisson distribution in Queueing Theory

1. Gary Spencer 14/15 UG EECS Project
Queen Mary, University of London.
Page 1 of 2
Poisson Distribution
Investigating the effectiveness of smart resource allocation strategies
If an eventoccursat randomin time,thenthere isameannumberof occurrencesina giventime
interval.We definethe meannumberof occurrencesaslambdaƛand � isour random variable.
Speakingwithateachingassistant(TA),teachingprocedural programminginthe QueenMary
InformaticsTeachingLaboratory,aconservative estimate onarelativelywell attendeddayisthateach
TA can expecttoserve 6 studentsin30 minutes.GiventhataTA is serving6 studentsin30 minutes,
whatis the probability thatfouror fewerstudentsneedassistance inthe same timeframe?
6 = ƛ; “≤4 students” = �
p(0;6) + p(1;6) + p(2;6) + p(3;6) + p(4;6)
=
60
𝑒−6
0!
+
61
𝑒−6
1!
+
62
𝑒−6
2!
+
63
𝑒−6
3!
+
64
𝑒−6
4!
≈ 0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339 = 0.2851
≈ 0.285
≈> 28.5%
Unlike binomialdistribution,eventsinpoissondistributedformare endless.Giventhe eventthataTA
servesfouror lessthanfourstudentsin30 minutes hasan approximate probabilityof successat28.5%.
Giventhata TA attends6 studentsin30 minutes,whatisthe probabilitythata TA attendsmore than6
studentsin30 minutes?
6 = ƛ; “>6 students”= �
= 1-P(≤6 students)
= 1-[p(0;6) + p(1;6) + p(2;6) + p(3;6) + p(4;6) + p(5;6) + p(6;6)]
= 1 − 𝑒(
6
0
𝑒
−6
0!
+6
1
𝑒
−6
1!
+6
2
𝑒
−6
2!
+6
3
𝑒
−6
3!
+
6
4
𝑒
−6
4!
+
6
5
𝑒
−6
5!
+
6
6
𝑒
−6
6!
)
≈ 1-(0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339 + 0.1606 + 0.1606)
= 0.3937
≈ 0.394
≈> 39.4%

2. Gary Spencer 14/15 UG EECS Project
Queen Mary, University of London.
Page 2 of 2
The probabilityof successthata TA servesmore than6 studentsin30 minutesis39.4%.Runningat
100% utilisation,aqueue with6people,servicetime couldnotexceed5minutestonotexceed100%
utilisationandeachstudentwouldneedtobe servicedimmediatelyafterthe last.
Problems in the poisson model
For a queue withmore than6 students,servicetime wouldneedtobe lessthan5 minutestoprevent
utilisationtakinglongerthanthe 30 minute time frame,asinour example. Real systemsare notlike
this,theyhave substantial variability.A studentrequestinghelpfromaTA isunscheduledandwe don’t
knowhowlongit will take toservice astudent.If we can reduce variabilityinservice time,we willsee
the queue move quicker,however,we have limitedcontrol overhow peoplearrive andwhatissuesthey
bringto the attentionof a TA. The onlythingwe can do to increase utilisationistohave the abilityto
respondondemandandtry to affordthe eventspeople bringtothe queue.
Maximising utilisation
Utilisationisthe rate at whichpeople enterthe queueingsystem, tothe rate theycan be serviced.How
utilisedthe queue iscantell howbusythe queue isata pointintime.If there isa TA ina lab,and there
are more studentsrequiringattentionthancanbe served(giventreatmenttimes),aqueue will be
formedandwill growovertime.Evenif the TA systemcankeepupwithdemandinthe lab,studentswill
still findthemselveswaiting.
I can maximise throughputinthe queue bychangingthe mechanismsthatIhave control over.Those
mechanisms are servers andthe orderat whichpeople enterthe queue.
If we have twoidentical,separateTA’s(serversinthe queue),pooledinone lab,thenthe systemspeed
shouldincrease byafactor of 2. In the eventof twice the arrivals,the systemisrunning attwice the
speedandso average waitingtime inthe queue shoulddropbyhalf.Generallyitseemspoolingisa
goodidea,thoughthere’sahiddenriskif the labisheavilyloaded,andgiventhe nature of how labs
needtobe assessed,arounddue datesformarkssurgeswill happen.