waiting line systems

7/29/2019 Waiting Line Systems

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Waiting Line Systems

Queuing Models


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Examples of Waiting Lines

At Petrol Pumps, Restaurants, Malls,

banks, post offices

Planes awaiting clearance from controltowers, trucks waiting to load or unloadcargo, cabs at airports, buses enteringterminals

Employees waiting to swipe/punch cardsfor entry/exit


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DehradunExp

FerozPurJanta Exp

PaschimExpress

FrontierMail

JammuTawi Exp

Any Train Booking at All Booking Counters Shorter Queue and Less Idle Servers

Specific Service CountersIdle Server at some counter,Longer Queue at some counter

Indian Railways

25 YearsAgo


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Bulk Cash> 50K

PowerAdvantage

AccountHolderIn Person

ThirdParty

Premier

Segregation of Service and Service PriorityHongKong Bank (HSBC)

Nationalized Banks Security is a Warm FeelingService is a .. Feeling

Receipts and PaymentsAt different counters.Idle time is more

Longer queues


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EnterExit

CashierChitle Bandhu,Sadashiv Peth, Pune

CustomerCounter (server)

Every customer gets a numbered token. Customerspends. Less time at each counter. Purchases are

updated at counter. Waiting Time at cashier reduced


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Service Exit

ProcessingOrder

Waiting LineArrivals

CallingPopulation

Major Elements of Waiting Line Systems


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Goals of a Waiting Line System

Queuing Analysis for the design of Systemcapacity

Waiting line models are predictive models ofexpected behavior of a system in which waiting

line forms Balance cost of providing customer service vs

cost of customers waiting for service Design or redesign to satisfy desired

specifications (Bank Manager wants maximumno of waiting customers = 7 in a bank, henceestimate the no of tellers required)


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Calling Population

Infinite or unlimited

Probability of arrival isnot significantly

influenced by the factthat some customersare waiting

Open to general

public (theatres,banks, post offices,restaurants, etc..)

Finite or limited

Systems have limitedaccess for service

Limited no ofmachines, limited noof trucks


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Customer Arrivals

Discrete people arriving, trucks arriving,telephone calls

Arrivals in single units or batches (bus load of

passengers arriving at fast food joints, Theatreor cinema hall patrons)

Distribution of arrivals follow the PoissonDistribution (ex : 4 cars per hour)

The average time in between arrivals follows thenegative exponential distribution. Ex Timebetween arrivals is 15 mins


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Exit

Single Channel/Server, Single Line, Single Phase

Exit

Exit

Single Channel/Server, Single Line, Multiple Phase

Payment of College Fees


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Exit

Multiple Channel/Server, Single Line, Single Phase

Exit

Exit

Multiple Channel/Server, Multiple Line, Single Phase

Exit

Exit

Exit

Customer Service at Banks

Hospitals Out Patient Department


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Exit

Multiple Channel/Server, Single Line, Multiple Phase

Exit

Exit

Multiple Channel/Server, Multiple Line, Multiple Phase

Exit

Exit

Exit


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Poisson Distribution Assumptions

Probability of occurrence of an event (arrival) ina given interval does not affect the probability ofoccurrence of an event in another nonoverlapping interval (afternoon & night show)

The expected no of occurrences of an event(arrival) in an interval is proportional to the sizeof the interval (15 mins and 30 mins).

Probability of occurrence of an event (arrival) inone interval is equal to the probability ofoccurrence of the event in another equal-sizeinterval/


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Probability in a Poisson Distribution

P(X) = e-u uX / X!

u = expected mean number of occurrences within agiven interval

e = Eulers constant = 2.71828

X = number of occurrences of an event in a given timeinterval

P(X) = Probability of occurrence of event X

Excel Formula = POISSON(X,u,True) for Cummulative

Excel Formula = POISSON(X,u,False) for NonCummulative


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Poisson Distribution Problem

Machines arrive at a repair shop at the rate of 3per 20 minute period. (u = 3)

What is the probability that there will be 6

machines arriving in a 20 minute period?P(X = 6)

What is the probability that there will be 12machines arriving in 1 hour? (u = 9 per hour),

P(X = 12) What is the probability that there will be less

than 3 machines arriving in a 20 minute period?(u = 3), P(X< 3) = P(X = 0) + P(X = 1) + P(X = 2)


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Types of Customers

Arriving customers mayrefuse to enter thesystem as there is a longwaiting line

Customers may arrive,wait for some time, andleave without beingserved

Customers may switchlines in order to reducethe waiting time

Balking

Reneging

Jockeying


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Processing Order (Priority)

First come, First served

Assign priority and process waitingcustomers as per priority order

High and Low Priority Customers


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Service

Number of servers

Number of steps or

phases

Distribution of processingor service time

Single or multiple

Single phase or multiple

phase

Negative exponentialdistribution

Most customers requireshort service time, while afew require more servicetime


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Probability in a Exponential

Distribution

Probability density function f(t) = e- t for

t, >= 0 and f(t) = 0 elsewhere

= mean no of occurrences of a particularevent per time unit

Mean u = Std Dev S = 1/

P(t=b) = e- b = 1- EXPONDIST(b, ,True)

P(a


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Exponential Distribution Problem

At a grocery store, time between arrivals is

an exponential distribution. On anaverage, 2 customers arrive every four

minutes. arr = 2/4 = 0.5 per min

Service time is also exponentiallydistributed with a mean service rate of 40

customers per hour. ser = 40/60 = 0.67per min


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Exponential Distribution Problem

Probability that no more than 4mins elapse between successivearrivals of customers

Probability that more than 6 minselapse between successive arrivalsof customers

Probability that between 4 and 6

mins elapse between successivearrivals of customers Expected time (u) between

successive arrivals and standarddeviation (s) of time betweenarrivals of customers

Expected time of service Std deviation of service time Probability that service time is 1

min or less Probability that service time is 2

min or more Probability that service time is

between 1 min and 2 min

P(t=6) = e- (0.5)(6) =0.049787

P(4


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In the WaitingLine

BeingServed

In the System

AverageNumber

Lq /u L=Lq +/u

AverageTime

Wq =Lq/u 1/u W=Wq +1/u

-------------System -----------

Average Number Waiting and Average Waiting and Service Time

Where = mean arrival rate and u = mean service rate


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Lq The average number waiting for service

L The average number in the system (waiting forservice or being served)

P0The probability of zero units in the system

p The system utilization (percentage of time serversare busy serving customers)

WqThe average time customers wait for service

WfThe average time customers spend in the system(waiting for service and service time)

M The expected maximum number waiting forservice for a given level of confidence

Measures of System Performance


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Waiting Line Problem A car wash franchise estimates for a new proposed facility: car

arrival rate = 20 cars per hour and car service rate u = 25 cars perhour. Service time is variable as cars are hand washed. Car arrivalsfollow Poisson distribution, while Car service time followsexponential distribution. Cars are processed one at a time. (Singleline, single server (s = 1)) Determine the following:

Average no of cars being washed ( =/u = 20/25 = 0.80 car)

Average no of cars in the system (cars being washed or waiting tobe washed, where average number waiting in line Lq = 3.2) (L=Lq+/u= 3.2 + 0.80 = 4.0 cars)

Average time in the line (avg time cars wait to get washed) (=Lq/u =3.2/20 = 0.16 hour = 0.16*60 = 9.6 mins)

Average time cars spend in the system (waiting in line and beingwashed) (= Wq +1/u = 9.6 + 1/25 = 0.20 hour = 0.20*60 = 12 mins)

System utilization ( =/(su) = 20/(1*25) = 0.80 or 80%) where (s =no of servers = 1)


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Basic Single Channel (M/M/1) Model

M/M/1 = Markovian (Poisson) arrivals,Markovial (negative exponential) service, and1 (single) server

One server or channel

A Poisson arrival rate

A negative exponential service time

First-come, First-served processing order

An infinite calling population

No limit on queue length


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(/u)Wa + (1 - /u)0(/u) + (1 - /u)

= (/u)WaWq =

Wq = Weighted Average of waiting time of ActualWaiting and Non waiting

Wa = Average Waiting TimeProportion of Customers Waiting = (/u)

Proportion of Customers who do not wait = (1 - /u)

Average Waiting Time


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Performance Measure Notation Formulae

System utilization p /u

Average number in line Lq 2/u(u- )

Average number in line L Lq + /u

Average time in Line Wq Lq / = /(u(-u) = (/u)Wa

Average time in System W Wq + 1/u

Probability of Zero units in System P0 (1 - /u)

Probability of n units in System Pn P0(

/u)

n

Probability that waiting line wont exceed kunits

Pn


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M/M/1 Problem

At a ticket counter, mean arrival rate of

customers = 3 per minute. Mean servicerate is 4 customers per minute. Calculate

all the performance measures.

Consider n = 2 customers in the systemand k = maximum length of the queue = 5


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System utilization p /u = = 0.75 or 75%

Average number in line Lq 2/u(u- ) = 2.25 customers

Average number in line L Lq + /u = 2.25 + = 3 customers

Average time in Line Wq Lq / = 2.25/3 = 0.75 minute

Average time in System W Wq + 1/u = 0.75 + = 1.0 minute

Probability of Zero units in System P0 (1 - /u) = (1 0.75) = 0.25

Probability of n units in System Pn P0(/u)n = 0.25(0.75)2 = 0.1406

Probability that waiting line wontexceed k units

Pn


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Basic Multiple Channel (M/M/S) Model

M/M/S = Markovian (Poisson) arrivals,

Markovial (negative exponential) service, andS (multiple) server (S > 1)

More than One server or channel

A Poisson arrival rate A negative exponential service time


An infinite calling population No upper limit on queue length

The same mean service rate for all servers


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Multiple Server, Priority Servicing

Model

Basic Multiple Server Model but there is

priority servicing.

Within each class (or priority), waiting unitsare processed in the order they arrive.

Unit with low priority will have a longwaiting line


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No Priority

Priority 1

Priority 2

Priority 3

New arrivals

Multiple Server, Priority Servicing Model


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Psychology of Waiting

Perception of waitingof the customer

Desirable waiting timeexploited

Magazines providedat doctors or dentistwaiting rooms

Music, in-flight movies Fill up forms

Mirrors near lifts

Impulse purchases atsupermarkets

Additional spending


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Value of Waiting Line Models

Assumptions are criticized as :

Often service times are not negativeexponential

System is not in steady-state, but tends tobe dynamic

Service is difficult to define because

service requirements can varyconsiderably from customer to customer


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OTHER WAITING LINEMODELS


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First Come First Served processing

in M/M/S Model

Customers wait in a single line (post offices).

Others systems record order of arrival (busyrestaurants) or have customers take a number

on arrival (bakeries, customer service in banks) Supermarket checkouts do not fall in multiple

servers (though they have multiple servers) ascustomers do not form a single line)

Condition su > must be satisfied

B i M l i l S S M/M/S


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System utilization p /su

Average number in line Lq u(/u)sP0/((s-1)!(su- )2)

Average number in line L Lq + /u

Average time in Line Wq Lq / = /(u(-u) = (/u)Wa

Average time in System W Wq + 1/u

Probability of Zero units in System P0 1/((/u)n/n!) + (/u)s/(s!(1-(/su))

from n = 0 to s-1

Probability of n units in System Pn P0(/u)n/(n!) for n s

Probability that an arrival will have to wait forservice

Pw (/u)s P0/ (s!(1- /su))

Average waiting time for an arrival not servedimmediately

Wa 1/(su- )

Basic Multiple Server System M/M/S


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/u s Lq P0

0.25 1 0.083 0.750

2 0.004 0.778

0.50 1 0.5 0.52 0.033 0.6

3 0.003 0.606

0.75 1 2.25 0.25

2 0.123 0.455

3 0.015 0.471

1.0 2 0.333 0.333

3 0.045 0.364

4 0.007 0.367

Infinite Source values for Lq and Po given /u and s

And so on Refer Excel Worksheet


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Determining Maximum Length of

Waiting Lines

Amount of space required toaccommodate waiting customers

Number of customers waiting will not

exceed a specified percentage of time Determine the line length (n) that probably

will not exceed 95% or 99% of the time

pn

= K where p = /su K = (1 probability)/Lq (1- p)

n = LogK/Logp


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Problem

Determine the maximum number of customers waiting inline for probabilities of both 95% and 99%. s = 1, = 4per hour, u = 5 per hour

p = /su = 4/(1)(5) = 0.80

For /u = 0.80, s = 1, Lq = 3.2

For 95%, K = (1-0.95)/3.2(1-0.80) = 0.078

n = log 0.078/log0.80 = 11.43 or 12

For 99%, K = (1-0.99)/3.2(1-0.80) = 0.0156

n = log 0.0156/log0.80 = 18.64 or 19


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During noon hour at a bank, the arroval rate is 8 customers per minute,and the service rate is 2 customers per minute. Thus /u = 8/2 = 4.

How many servers would be required so that the line does not exceed 4 waiting

Customers? 6 servers at 95%, 7 servers at 99%

Servers(s)

Lqp = /su n95 n99

5 2.216 0.80 10 17

6 0.570 0.67 4 8

7 0.180 0.57 1 4

8 0.059 0.50 0


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Cost Considerations

Minimize Total Cost = Customer Waiting

Cost + Capacity (Service) Cost

Iterative procedure : Increase Capacity (orchannels) by one. And compute the totalcost.

Total cost will initially decrease with

increase in number of channels and thenincrease.


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Trucks arrive at a warehouse at a rate of 15 per hour during business hoursCrew can unload at a rate of 5 per hour. Crew wage rates have increased andCrew size is to be determined. Crew cost is Rs 100/hour, while driver cost isRs 120 per hour. Thus /u = 15/5 = 3. Values of Lq are as per calculations

for given s and /u

Crew Size (s) Service Cost

Rs 100 x s

Avg No inSystem L = Lq

+ /u

Waiting Cost= L x Rs 120/-

Total Cost =Service Cost

+ Waiting

Cost4 Rs 400 1.528 + 3 =

4.5284.528 x 120 =

Rs 543.26Rs 943.26

5 Rs 500 0.354 + 3 =3.354

3.354 x 120 =Rs 402.58

Rs 902.58(minimum)

6 Rs 600 0.099 + 3 =3.099

3.099 x 120 =Rs 371.88

Rs 971.88

7 Rs 700 0.028 + 3 =3.028

3.028 x 120 =Rs 363.36

Rs 1063.36


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Other Queuing Models

Poisson Arrival Rate with anyservice Distribution

Poisson Arrival Rate and ConstantService Time

Finite Queue Length

Finite Calling Population

Multiple Server, Priority Servicing

Model

M/G/1 (single Server)

M/D/1 (single server,Deterministic service time


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Basic Single Channel (M/G/1) Model

M/M/1 = Markovian (Poisson) arrivals,General service distribution, and 1 (single)server


A Poisson arrival rate Service time follows any distribution

Estimate of variance is necessary





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PerformanceMeasure Formula

System Utilization p /su (where s = 1)

Average number

waiting in line

Lq ((/u)2 +2v)/2(1-/u)

where v = variance

Average number inSystem

L Lq + /u

Average time waitingin line

Wq Lq /

Average time in thesystem

W Wq + 1/u

M/G/1 Poisson Arrival rate with any Service Distribution


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Problem Any Service Distribution

Customer arrival rate is Poisson = 0.25/min

Service time has mean = 2 min (u = = 0.5)and standard deviation = 0.9 min (variance =

0.81). Compute performance measures


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Performance Measure Formula

System Utilization p /su (where s = 1) = 0.25/0.50 = 0.50

Average number

waiting in line

Lq ((/u)2 +2v)/2(1-/u) where v = variance

= (0.25/0.50)2

+ 0,252

(0.81)/2(1-(0.25/0.5)) =0.301 customers


L Lq + /u = 0.301 + 0.5 = 0.801 customers

Average time waiting in

line

Wq Lq / = 0.301/0.25 = 1.203 min


W Wq + 1/u = 1.203 + 1/0.5 = 3.203 min

M/G/1 Poisson Arrival rate with any Service Distribution


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Basic Single Channel (M/D/1) Model

M/M/1 = Markovian (Poisson) arrivals,Constant Service time, and 1 (single) server


A Poisson arrival rate

Service time is constant

Variance = zero





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System Utilization p /su (where s = 1)

Average number

waiting in line

Lq (/u)2/2(1-/u) = ()2/2u(u-)


L Lq + /u

Average time waiting inline

Wq Lq /


W Wq + 1/u

M/D/1 Poisson Arrival rate with Constant Service Time


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Problem Constant Service Time

Customer arrival rate is Poisson = 12/hour

Service time has mean = 2 min (u = = 0.5)and standard deviation = 0.9 min (variance =

0.81). Compute performance measures


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System Utilization p /su (where s = 1) = 12/20 = 0.60

Average number

waiting in line

Lq 2/2u(u-) = 122/2(20)(20-12) = 0.45 customer


L Lq + /u = 0.45 + 0.6 = 1.05 customers

Average time waiting inline

Wq Lq / = 0.45/12 = 0.0375 hour = 2.25 min


W Wq + 1/u = 2.25 + 60(1/20) = 5.25 min

M/G/1 Poisson Arrival rate with Constant Service Time

B i Si l Ch l ith Fi it Q


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Basic Single Channel with Finite QueueLength Model

M/M/1 = Markovian (Poisson) arrivals,Markovian Exponential Distribution Servicetime, and 1 (single) server


A Poisson arrival rate Exponential Distribution Service rate

Limit on Maximum length of Queue First-come, First-served processing order

An infinite calling population New customers not allowed when queue ismaximum (parking lots)

Basic Single Server System M/M/1


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System utilization p /u

Average number in line Lq L (1-P0)

Average number in system L (/u)/(1- /u) (m+1)(/u)m+1/(1- (/u)m+1)

Average time in System W Lq /((1-Pm)) + 1/u

Average time in Line Wq W- 1/u

Probability of Zero units in System P0 (1 - /u)/(1-(/u)m+1)where m = max length of queue

Probability of n units in System Pn P0(/u)n

Basic Single Server System M/M/1


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Problem Finite Length

Arrival rate = 9 per hour

Service rate = 15 per hour

Max length of Queue = 5 customers

Calculate performance measures

Basic Single Server System M/M/1 with Finite Queue Length


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System utilization p /u = 9/15 = 0.60

Average number in line Lq L (1-P0) = 1.206 (1-0.420) = 0.63

Average number in system L (/u)/(1- /u) (m+1)(/u)m+1/(1- (/u)m+1)

= 0.6/(1-0.6) (5+1)(0.6)5+1)/(1-0.65+1) =1.206

Average time in System W Lq /((1-Pm)) + 1/u = 0.63/(9(1 0.033)) +1/15 = 0.139 hr

Average time in Line Wq W- 1/u = 0.139 1/15 = 0. 072 hour

Probability of Zero units in System P0

(1 - /u)/(1-(/u)m+1) = (1-0.6)/(1-0.65+1) = 0.420

Probability of n units in System Pn P0(/u)n = 0.420(0.6)5 = 0.033

Maximum length of Queue m Given = 5

Basic Single Server System M/M/1 with Finite Queue Length

Basic Single Channel with Finite Calling


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Basic Single Channel with Finite CallingPopulation Model

M/M/1 = Markovian (Poisson) arrivals,Markovian Exponential Distribution Servicetime, and 1 (single) server


A Poisson arrival rate Exponential Distribution Service rate


A Finite calling population

Machine Operator responsible for loadingand unloading five machines

waiting line systems

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