Template for waiting line and queues. It facilitates ease in the calculation of waiting time
InstructionsSteady State Queuing Models 26 Oct 2007John O.
McClain [email protected] Graduate School of ManagementSage
Hall, Cornell UniversityIthaca NY 14853This spreadsheet is intended
for teaching purposes. You are welcome to use it in anymanner and
change it as you see fit. This model comes without any guarantee,
and isdistributed free of charge.Note: If the worksheets don't seem
to work properly,Click HereContents:DescriptionsModelsModels
included in this workbookDefinition of Steady StateUsing the
ModelsExponential Service and Interarrival TimesModels with limited
waiting capacity (balking)Finite Queue worksheetModels with
infinite waiting capacity (no balking)Infinite Queue
worksheetGeneral Service and Interarrival Times,
ApproximationInfinite waiting capacity, ApproximationInfinite Queue
Approx. worksheetLimited waiting capacity, SimulationQueue
Simulation worksheetThe ModelsThe Finite Queue model assumes that
there is a limit to the waiting line, and thatcustomers will not
join the queue when that limit is reached. Those customers
arepermanently lost, but the arrival rate of future customers is
not affected.Assumptions: Identical Servers, Poisson arrivals,
Exponential service times.(More)The Infinite Queue model assumes
that there is no limit to the waiting line. That is,customers are
extremely patient and will wait indefinitely.Assumptions: Identical
Servers, Poisson arrivals, Exponential service times,and Arrival
Rate < (Number of Servers)(Service Rate capacity per server)This
model also allows up to 4 priority classes
(non-preemptive).(More)The Infinite Queue Approximation gives a
fairly simple formula that allows you toadjust the CV (coefficient
of variation) for arrival and service times. Output
includesaverages but not probabilities.Assumptions: Identical
Servers, Arrival Rate < (# Servers)(Service Rate)(More)The
Finite Queue Simulation begins with no customers, and simulates
using theGamma distribution for time between arrivals and for
service times. If CV (coefficient of variation) = 1.0, the Gamma is
the same as the Exponential,in which case the simulation results
should converge to the Finite Queue Model. For CV 1, the average
queue should converge to a value similar to theInfinite Queue
Approximation, if the service capacity exceeds the arrival rate,and
if the simulation's queue capacity large enough.Assumptions:
Identical Servers, Gamma inter-arrival times, Gamma service
times.(More)Each of these models is described in more detail below,
and examples are worked out.Steady State, Defined.These models give
"Steady State" results. This has two important implications:The
probability distributions of arrivals and service times do not
change with time.For example, you cannot model variations in the
arrivals at different times of day.The outputs are long run
averages.For example, if the model gives 9% probability that the
queue is empty, it means that9% of the time there will be no one
waiting. But the 9% does not apply, for example, ifyou start with
no one waiting and watch the system for 15 minutes.Using the
ModelsYour inputs always go in the yellow cells, like this:Please
be careful with your time units. Two of the inputs are rates, and
they must have thesame time units. For example, suppose the arrival
rate is 4 customers per hour, and theaverage service time is 10
minutes. To be consistent, the service rate must also be givenin
customers per hour, which would be 60/10 or 6.For the first 3
models, the results are available immediately, as soon as you enter
an input.However for the simulation, once you change the inputs,
you must click a button and waitfor the simulation to finish. The
program then writes new output on the spreadsheet.Finite Queues
(limited waiting line capacity)Assumptions: Identical Servers,
Poisson arrivals, Exponential service times.The model also assumes
that arrivals cease when the queue is full. This is "balking".Your
Inputs: The 4 basic inputs for the finite queuing model are S, M, l
and m.There are S identical servers, and the queue can hold M
customers.Therefore the system can hold up to M+S customers (M in
queue and S in service).The arrival rate of customers is l, and the
service rate is m for each server.Another input looks like this:Use
it to find the probability of a queue exceeding a given length, Q.
For example, to find theprobability of 11 or more customers waiting
for service, type 10 in the yellow box.Example:City Clinic serves a
population that requires an average of 45 visits per 8-hour
day.There are two nurse-practitioners, each capable of serving 25
patients per day.Customers go to another clinic if the waiting room
is full when they arrive.a.If there is no waiting area at all, what
fraction of the patients will leave without service?b.How large
should the waiting area be so that at least 95% of patients will be
served?c.If the waiting area holds 20 patients, how often will more
than 10 be waiting?Solution:a.On the Finite Queue worksheet, put in
S = 2, M = 0, l = 45 and m = 25.Answer: Customers who Balk =
36.65%, so this is how many leave without service.b.Choose larger
values for M until Customers who Balk is below 5%. Answer: M=9.Go
to the Finite Queue Graph sheet to see the entire probability
distribution displayed.c.Put in M=20 and Q=10. Answer:
19.22%Experiments:d.Using M=20 as the capacity of the waiting area,
change the number of servers to 3and watch what happens to the
Finite Queue Graph.e.Change the number of servers to 1 and watch
what happens to the Finite Queue Graph.Note that the queue is never
empty when there is only one server to handle the load.Infinite
Queues (unlimited waiting line capacity)Assumptions: Identical
Servers, Poisson arrivals, Exponential service times.Your Inputs:
The 3 basic inputs for the infinite queuing model are S, l and
m.There are S identical servers, and the queue can hold an
unlimited number of customers.The arrival rate of customers is l,
and the service rate is m for each server.Another input looks like
this:Use it to find the probability of a queue exceeding a given
length, Q. For example, to find theprobability of 11 or more
customers waiting for service, type 10 in the yellow box.Similarly,
this input,gives the probability that a customer will have to wait
0.5 time units* or longer before service,*The time units are the
same as the ones you use for the arrival and service rates.You may
(optionally) specify up to 4 customer categories, each with
different priorities.When there is a waiting line, the highest
priority customers get the next available server.Example:City
Clinic serves a population that requires an average of 45 visits
per 8-hour day.There are two nurse-practitioners, each capable of
serving 25 patients per day.a.What is the average size of the
waiting line, and how long is the average wait?b.What percent of
the time are more than 10 patients are waiting?c.What is the
probability that a patient will have to wait more than one-half of
a day?d.20% of the patients have severe injuries that require
immediate attention. How long dothese "high-priority" patients have
to wait, on average?e.Does the use of a priority system change the
total size of the waiting line?Solution:a.On the Infinite Queue
worksheet, put in S = 2, l = 45 and m = 25.This will cause Nq =
7.674 patients waiting, on average, and Tq = 0.1705 days waiting,on
average. (Tq is in days because the arrival rate is in customers
per day.)b.Put in Q = 10. Answer: 26.76%c.Put in T = 0.5. Answer:
7%d.Put in 0.8 as the fraction of priority 2 customers, and put 0
for priorities 3 and 4.The result is Tq (1) = 0.0208 days for
priority 1 customers.e.No. Adding the waiting lines gives a total
of 7.674, the same as part (a).Approximation: Infinite Queues
(unlimited waiting line capacity)Assumptions: Identical
Servers.Your Inputs:There are S identical servers, and the queue
can hold an unlimited number of customers.The arrival rate of
customers is l, and the service rate is m for each server.CV(s) =
Coefficient of Variation of Service Times:CV(a) = Coefficient of
Variation of Inter-arrival Times (i.e. times between
arrivals):Definition: CV = standard deviation divided by the
mean.The advantage of this formula is that it makes no assumptions
about the distributionsof arrivals and service times. Therefore it
is more general than the "Infinite Queue Model".Example:Computer
Clinic serves a population that generates an average of 45 requests
per 8-hour day.There are two technicians, each capable of serving
25 customers per 8-hour day.a.What is the average service
time?b.The standard deviation of service time is 0.16 hours. What
is its CV?c.What is the average inter-arrival time?d.The standard
deviation of inter-arrival time is 0.1 hours. What is its CV?e.What
is the average size of the waiting line, and how long is the
average wait?Solution:a.To serve 25 customers in 8 hours, service
time must be 8/25 = 0.32 hours.b.CV(s) = Standard Deviation divided
by Average = 0.16/0.32 = 0.5c.If 45 customers arrive in 8 hours,
one arrives every 8/45 = 0.178 hours.d.CV(a) = Standard Deviation
divided by Average = 0.1/0.178 = 0.562e.On the Infinite Queue
Approx. worksheet, put in S = 2, l = 45, m = 25, CV(a) = 0.562and
CV(s) = 0.5. Result: Nq = 2.186 patients waiting, on average, and
Tq = 0.0486 dayswaiting, on average. (Tq is in days because the
arrival rate is in customers per day.)Simulation: Finite Queue
CapacityAssumptions: Identical Servers, Gamma Distributions for
inter-arrival and service times.Your Inputs:There are S identical
servers, and the queue can hold unlimited customers.The arrival
rate of customers is l, and the service rate is m for each
server.CV(s) = Coefficient of Variation of Service Times:CV(a) =
Coefficient of Variation of Inter-arrival Times (i.e. times between
arrivals):Definition: CV = standard deviation divided by the
mean.Simulated time per repetition, RunLength: Time units per
repetition of the simulation.Time Units are defined by Arrival and
Service Rates.If you use customers per hour for the arrival rate,
You MUST use the SAME UNITS for the service rate, and The time
units for the simulation will be "hours".Repetitions (200), nReps =
the number of times the simulation is repeated.Data collection
occurs after each repetition.Number of Repetitions to Ignore,
WarmUp = number of repetitions NOT included inthe summary
statistics. If "Repetitions" = 12 and WarmUp = 3, then the
summarystatistics will cover runs 4 to 12.Example:FarAway Call
Center provides over-the-phone help for computer owners. They
currentlyreceive an average of 45 calls per hour. Their service
area is world-wide, so the callsarrive at a steady rate, 24 hours a
day. Arrivals are quite random. The CV of inter-arrivaltimes is
1.0. Service times also have CV of 1.0, but management hopes to
reduce thatvariation by a combination of training and other tools
for the servers.The phones are answered by an automatic system,
which informs the caller of thenumber of customers that are waiting
and gives an estimate of the waiting time. Experiencehas shown that
people hang up when they hear that the waiting line is 10
customers.a.Find the average number waiting and the probability
that more than 5 are waiting.What is the average waiting time for a
customer?What fraction of customers hang up without receiving
service?b.How do your answers compare to the theoretical values
using the Finite Queue model?c.If the CV of service time is reduced
to 0.3, what is the effect on the answers to part (a)?d.Comment on
the changes that you see between the two results.Solution:a.On the
Queue Simulation worksheet, put in S = 5, M = 10, l = 45 and m =
10.Enter 1.0 for CV(a) and CV(s), and set RunLength = 100, nReps =
12, WarmUp = 2.Then click the "Simulate" Button.Answers: Your
answers will differ because each simulation has different
customers.Average number waiting, Nq = 2.8 P(>5) in queue =
23%Average Waiting Time (Tq) = 0.065 days Fraction who balk =
3.7%b.Virtually the same: 2.73, 21.8%, 0.063 days, and 3.49%,
respectively.c.Same method except CV(s) = 0.3Answers: Your answers
will differ because each simulation has different customers.Average
number waiting = 2.4 P(>5) in queue = 16%Average Waiting Time
(Tq) = 0.054 days Fraction who balk = 1.4%d.Less variablity of
service means that the number of customers in the system
remainscloser to the average. That lowers the probability of the
system being full, which meansless balking. It also lowers the
probability of a long queue.Experiments:e.Change the RunLength to
10 and see what happens.Note that the "Results" become much more
variable. The simulation's accuracy dependson a lot of
observations.
&C&F, &A, page &PClick
Here(More)(More)(More)(More)Models included in this
workbookDefinition of Steady StateUsing the ModelsModels with
limited waiting capacity (balking)Models with infinite waiting
capacity (no balking)Infinite waiting capacity,
ApproximationLimited waiting capacity, SimulationFinite Queue
worksheetInfinite Queue worksheetInfinite Queue Approx.
worksheetQueue Simulation worksheet
Module1Steady-State, Finite Capacity Queues1 Servers, Queue
Capacity = 10, Arrival Rate = 10, Service Rate = 5Basic
Inputs:Number of Servers, S =1Queue Capacity, M =10Arrival Rate, l
=10Service Rate Capacity of each server, m =5Arrivals:Average Rate
Joining System (Lambda-Bar) =4.9987789988Average Rate Leaving
Without Service (balking) =5.0012210012Customers who Balk:
Probability that System is Full (Pfull) =50.01%The Waiting
Line:Average Number Waiting in Queue (Nq) =9.003Average Waiting
Time (Tq) =1.8010747435Q: Probability of more than0customers
waiting= 99.93%Service:Average Utilization of Servers
=99.98%Average Number of Customers Being Served (Ns)
=0.9997557998The Total System (waiting line plus customers being
served):Average Number in the System (N) =10.003Average Time in
System (Tq + Ts) =2.0010747435Probability Distribution:n = total
number of customers in the systemq = number of customers in the
waiting
linenP(n)CumulativeqP(q)Cumulative00.00020.000210.00050.000700.00070.000720.00100.001710.00100.001730.00200.003720.00200.003640.00390.007630.00390.007550.00780.015440.00780.015460.01560.031050.01560.031070.03130.062360.03130.062280.06250.124870.06250.124890.12500.249880.12500.2498100.25010.499990.25010.4998110.50011.0000100.50011.0000
&APage &P&C&F, &A, page &P
Module1
&C&F, &F, page &P1 Servers, Queue Capacity = 10,
Arrival Rate = 10, Service Rate = 5Total Number of Customers in the
System (waiting or being served)Steady-State Probabilities for
Finite Capacity Queue
Finite QueueSteady-State, Infinite Capacity QueuesModel is
OKBasic Inputs:Number of Servers, S =2Arrival Rate, l =45Service
Rate Capacity of each server, m =25The Waiting Line:Average Number
Waiting in Queue (Nq) =7.674Average Waiting Time (Tq)
=0.1705263158Q: Probability of more than20customers waiting=
9.33%T: Probability of more than0.5time-units waiting=
7%Service:Average Utilization of Servers =90.00%Average Number of
Customers Receiving Service (Ns) =1.8The Total System (waiting line
plus customers being served):Average Number in the System (N)
=9.474Average Time in System (Tq + Ts) =0.2105263158An Option:
Multiple Classes of CustomersClassfraction(Ignore)Nq (k)Tq
(k)highest priority =
10.20.820.1870.020820.80.17.4870.2080300.10.0000.0000400.10.0000.0000
&C&F, &A, page &PIgnore this column. It contains
an intermediate calculation that has no physical
interpretation.
Infinite QueueApproximate Formula for Steady-State, Infinite
Capacity QueuesModel is OKBasic Inputs:Number of Servers, S
=2Arrival Rate, l =45Coefficient of Variation of Inter-arrival
time, CV(a) =0.562Service Rate Capacity of each server, m
=25Coefficient of Variation of Service time, CV(s) =0.5The Waiting
Line:Average Number Waiting in Queue (Nq) =2.186
