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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Performance Measure Notation Formulae
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
First-come, First-served processing order
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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Performance Measure Notation Formulae
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
One server or channel
A Poisson arrival rate Service time follows any distribution
Estimate of variance is necessary
First-come, First-served processing order
An infinite calling population
No limit on queue length
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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
Average number inSystem
L Lq + /u = 0.301 + 0.5 = 0.801 customers
Average time waiting in
line
Wq Lq / = 0.301/0.25 = 1.203 min
Average time in thesystem
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
One server or channel
A Poisson arrival rate
Service time is constant
Variance = zero
First-come, First-served processing order
An infinite calling population
No limit on queue length
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Performance Measure Formula
System Utilization p /su (where s = 1)
Average number
waiting in line
Lq (/u)2/2(1-/u) = ()2/2u(u-)
Average number inSystem
L Lq + /u
Average time waiting inline
Wq Lq /
Average time in thesystem
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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Performance Measure Formula
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
Average number inSystem
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
Average time in thesystem
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
One server or channel
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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Performance Measure Notation Formulae
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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Performance Measure Notation Formulae
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
One server or channel
A Poisson arrival rate Exponential Distribution Service rate
First-come, First-served processing order
A Finite calling population
Machine Operator responsible for loadingand unloading five machines