Poisson distribution
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The document discusses using a Poisson distribution to model the number of students seeking assistance from a teaching assistant (TA) in a 30 minute period. It calculates that the probability a TA will help 4 or fewer students in 30 minutes is 28.5% and the probability a TA will help more than 6 students is 39.4%. It notes limitations in using this model for a real queueing system due to variability in student arrival times and assistance needs. Maximizing throughput in the queue by adding additional TAs or changing the arrival order is proposed to reduce average waiting times.
![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 occurrencesaslambdaand 鐃 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 = ; 4 students = 鐃
p(0;6) + p(1;6) + p(2;6) + p(3;6) + p(4;6)
=
60
6
0!
+
61
6
1!
+
62
6
2!
+
63
6
3!
+
64
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 = ; >6 students= 鐃
= 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 (
6
0
6
0!
+6
1
6
1!
+6
2
6
2!
+6
3
6
3!
+
6
4
6
4!
+
6
5
6
5!
+
6
6
6
6!
)
1-(0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339 + 0.1606 + 0.1606)
= 0.3937
0.394
> 39.4%](https://image.slidesharecdn.com/kcoib5qpqtvfo5ouggtq-signature-10a26006b5b4ef787ca065fc4b1f17da893c39fddd58e60695add0c4f19c7f17-poli-150626232500-lva1-app6892/85/Poisson-distribution-1-320.jpg)
Poisson distribution
- 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 occurrencesaslambdaand 鐃 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 = ; 4 students = 鐃 p(0;6) + p(1;6) + p(2;6) + p(3;6) + p(4;6) = 60 6 0! + 61 6 1! + 62 6 2! + 63 6 3! + 64 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 = ; >6 students= 鐃 = 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 ( 6 0 6 0! +6 1 6 1! +6 2 6 2! +6 3 6 3! + 6 4 6 4! + 6 5 6 5! + 6 6 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 dont 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,separateTAs(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,thoughtheresahiddenriskif the labisheavilyloaded,andgiventhe nature of how labs needtobe assessed,arounddue datesformarkssurgeswill happen.