Queuing Models in Operations 35E00100 Service Operations and Strategy #4 Fall 2015.

Queuing Models in Operations

35E00100 Service Operations and Strategy#4 Fall 2015

35E00100 Service Operations and Strategy #4 Aalto/BIZ Logistics2

Topics

Principles of queuing theory Queue parameters Basic queuing models

M/M/1 and M/M/s

G/G/1 and G/G/s

Extensions Systems with finite buffers Serial (tandem) queuing system Queuing network Multiple priority system Priority queuing network

Key points

Useful material in the textbook:Hopp, W. & Spearman, M. (2000) Factory Physics, Chapter 8.6


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Pro

bab

ilit

y

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Service rate

“Customer arrives every 10 minutes, service process

takes 8 minutes”

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Arrival rate

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bab

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Ideal

situatio

n

Reality


What is queuing theory about?

Service

Average number of jobs in the system (WIP )

Average time in the system (CT )

Average number of jobsin queue (WIPq )

Average waiting timein queue (CTq )

Arrival rate (raService rate (re


Utilization

Cost Trade-Off:Capacity Utilization

Cost

Cost ofqueuing

Cost ofoperations

100 % 0 %

Total costsLow utilization levels (u < 0.6 ) provide Better service levels Lower waiting costs (e.g., lost business) Greater flexibility

High utilization levels (u > 0.9 ) provide Better equipment and employee utilization Fewer idle periods Lower production/service costs


Cost Trade-Off:Capacity Level

Capacity

Cost

Total costs

Cost ofoperations

Cost ofqueuing

Optimum


Kendall's Classification 1/2/3/4/5/6

1 = the nature of arrival process2 = the nature of service process3 = the number of (parallel) machines4 = queue discipline5 = the maximum allowable number of jobs6 = the size of population

1

2

3Queue Server


System Characteristics

Inter-arrival and service times M = exponential (Markovian) distribution

Memoryless property

Common in service models G = completely general distribution (with mean and variance

specified) Memoryless property is lost performance approximations

Manufacturing environment

Typical queuing disciplines FCFS, SPT, EDD, LCFS, etc. Priority classes and within class FCFS


0.0

0.1

0.2

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0.5

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0.7

0.8

0.9

1.0

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

Time (t)

Exp(1)

Exp(3)

Exp(5)

Inter-arrival and Service Times

Exponential Distribution


0.00

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0 2 4 6 8 10 12 14 16 18 20

Arrivals per hr

Poisson(1)

Poisson(3)

Poisson(5)

Inter-arrival and Service Times

Poisson Distribution

If the times between arrivals are independent and exponentially distributed, then the number of customer that arrive within a given period of time is Poisson distributed. These are often referred to as “Poisson arrivals”.

For instance, if inter-arrival times are exponentially distributed with mean (1 / ra) = 0.25 hrs, then the number of customer arrivals in a one hour period has a Poisson distribution with mean ra = 4.

0.05


2

1

u

u

or ( )

a

e e a

r

r r r

M/M/1 Queue - Relevant Equations

Average waiting time CTq

Average queue length WIPq

Average number of jobs in system WIP

Average time of jobs in system CT

Utilization rate ua

e

r

r

1et

u

1 eut

u

or a qr CT

or q eCT t

1

u

ue.g. Krajewski & Ritzman


aq

e

rWIP

r

0

2

( )

!(1 )

ma

e

rp ur

m u2( 1) 1

(1 )

m

eu

tm u

a

e

r

mr

M/M/m Queue - Relevant Equations

Average waiting time CTq

Average queue length WIPq

Average number of jobs in system WIP

Average time of jobs in system CT

Utilization rate u

Probability for being idle p0

1q e q

e

CT t CTr

11

0

( / ) ( )

! !(1 )

m n ma e a e

n

r r r r

n m u

e.g. Krajewski & Ritzman


General Inter-arrival & Process Times

G/G/1 queue

Single machine workstations Flow variability, process

variability, or both can combine to inflate queue time

Variability causes congestion

G/G/m queue

Multi-machine workstations (parallel machines)

Fast and accurate Extremely general

eea

q

tu

ucc

tUV

12

CT

22

e

mea

q

tum

ucc

tUV

)1(2

CT

1)1(222

Hopp and Spearman 2000, 270

Note! This VUT equation is also called Kingman’s Equation.

(VUT = Variability x Utilization x Time)


Little’s law Lower utilization, less waiting Change in throughput can be avoided by decreasing variability

Resource pooling Economies of scale achieved by joining multiple queues into a single

queue with multiple machines Problems in one machine do not delay all orders

Some points to be remembered

Arrival rate is rarely stationary Long term average versus ”worst case” estimate

A heuristic to decide how many servers to provide Specify service level by defining the probability for waiting in line Calculate the number of servers

1a a

e e

r rm z

r r

Server

Queue

Server

Queue

50%

50% ?


Extensions to the Basic Models


Effects of Blocking

VUT equation versus real world systems? Blocking models

The principle is to estimate WIP and TH for given set of rates and buffer sizes.

Typically much more complex than non-blocking (open) models. Often simulation is needed to evaluate realistic systems.

Differences of blocking models compared to basic queuing models:

Arrival rate is only the rate of potential arrivals Utilization can equal or exceed 100 % Balking rate to be defined (rate at which arrivals are rejected)

Hopp and Spearman 2000, 273 - 279


1

( / /1/ ) 1

( / /1/ ) 1

( / /1/ ) 2( / /1/ )

( / /1/ ) 1

( 1)

1 1

1

1

, here

b

M M b b

b

M M b ab

M M b eM M b

M M b e

b uuWIP

u u

uTH r

uWIP t

CT w uTH t

M/M/1/b Queue

BInfinite rawmaterials

Model of Station 2

1 2

There is room for b=B+2 jobs in system, B in the buffer and one at each station

Goes to u/(1-u) as b Always less than WIP(M/M/1)

Goes to ra as b Always less than TH(M/M/1)

Little’s law

Formulas valid for u1



1 2

Illustrating the Effect of Blocking

te1=21 te2=20

Conclusion:M/M/1/b system has

less WIP and less TH than M/M/1 system!

18 % less TH

90 % less WIP

2

1

( / /1)

( / /1)1

4

1 1 5

1 5

( / /1/ ) 1

200.9524

21

20 jobs1

1 10.0476 job/min

21

1 1 0.9524 10.039 job/min

211 1 0.9524

( 1) 5(0.9524 )20

1 1 1 0.9524

e

e

M M

M M aeb

(M/M/ /b) ab

b

M M b b

tu

t

uWIP

u

TH rt

- uTH r

- u

b uuWIP

u u

51.8954 job

Example 1


35E00100 Service Operations and Strategy #4 Aalto/BIZ Logistics

We need to be able to analyze different types of processes…

Single Channe

l

Multi-Channel

Single Stage Multi-Stage

Servers

Queue

Servers

Queue

Servers

Queue

ServerQueueServer

QueueServer

Queue

Hospital admissions

Car wash

Bank teller’s windows

ATM


Queuing networks Some servers arranged in parallel and some in series E.g. cellular manufacturing, job shops

Serial and Network Queuing Systems

Serial (or tandem) queuing systems Servers arranged in series e.g. repetitive flow process

1 32

1a 3a

2a

1b

2b

2c

3b


Example 2

Serial Queuing System

Suppose a company produces a variety of similar but customized products using a three stage batch flow process. Customer orders are received randomly during the day according to a Poisson process at the rate of five orders per day. Because of differences in order (batch) sizes, machine setup times, and operating speeds, and because of machine breakdowns quality problems, and worker fatigue, the time to process a customer order at each production stage is exponentially distributed, with an average processing time of 1/6 day.

The 1st stage of the system is then an M/M/1 queue, where orders are the customers and the 1st stage is the server, ra =5 orders/day and re =6 orders/day.

It can be proved that in steady state, the pattern of served customers leaving an M/M/1 queuing system will approximate a Poisson process with an average rate of ra. So, the seriel queuing system can be treated as a series of 3 identical M/M/1 queues.

1 32


Serial Queuing System Using the M/M/1 formulas for the 1st stage u = ra / re = 5/6 = 0.83 WIPq = ra

2/ [re(re-ra)] = 52 / 6(6-5) = 4.17 orders CTs = 1/(re -ra) = 1/(6-5) = 1 day In front of the 1st stage there will be an average of 4.17 customer orders waiting

in the queue to be processed. It will take an order an average of 1 day to make its way through the 1st stage (from receipt of the order until processing is completed at stage 1).

The other two stages are identical, so in front of each stage there will be an average of 4.17 orders in queue, and it will take 1 day to get through each stage. So, on average, there will be a total of 12.51 jobs in inventory in front of the three processing stages. The average throughput time is 3 days, even though the time spent processing an order is one-half of a day.

WIPq=4.171

re=5/dayCT=1 day

ra=6/dayWIPq=4.17

32

CT=1 day CT=1 day

ra=6/day ra=6/dayWIPq=4.17

Example 21 32



Demand increases from 5.0 to 5.7 orders per day u = ra / re = 5.7/6 = 0.95 WIPq = ra

2/ [re(re-ra)] = 5.72 / 6(6-5.7) = 18.05 orders CTs = 1/(re -ra) = 1/(6-5.7) = 3.3 days

In front of the 1st stage there will be an average of 18.05 customer orders waiting in the queue to be processed. It will take an order an average of 3.33 days to make its way through the 1st stage.

On average, there will be a total of 54.15 jobs in inventory in front of the three processing stages. The average throughput time is 10 days, even though the time spent processing an order is one-half of a day.

32

ra=6/day ra=6/day

CT=3.33 d CT=3.33 d

WIPq= 18.05

M/M/1 M/M/1

WIPq= 18.05

1re=5.7/day

CT=3.33 d

ra=6/day

M/M/1

WIPq= 18.05

Example 21 32


No queues develop in front of stages 2 and 3, and so the average time in the stage is the processing time 0.167 days per job.On average, there will be a total of 9.025 jobs in inventory in the system. The average throughput time is 2.08 days.

Example 21 32


No randomness in processing times Using the M/D/1 formulas for the first stage WIPq = u2 / [2(1-u)] = 0.952 / 2(1-0.95) = 9.025 orders CTq = WIPq / ra = 9.025 / 5.7 = 1.583 days CT = CTq +1/re = 1.583 + 1/6 = 1.75 days

WIPq=0

re=5.7/day1

CT=1.75 d

WIPq=9.025

M/D/1

ra=6/dayWIPq=0

2

CT=0.167 d

D/D/1

ra=6/day

3

CT=0.167 d

D/D/1

ra=6/day

Inventories are reduced by 83% and throughput time decreases by nearly 80%.


BA

Example 3Serial Queuing System

Process and Resources

Two single-server work stations Three types of tasks to perform 2 job types (A and B)

Inter-arrival times for A and B Uniform[0,40]

Type A makes two passes

Type B jobs make a single pass through the system Service times

Stage 1: Uniform[0,12]

Stage 2: Uniform[0,10]

Individual service times not known No changeover cost

Can change from one task type to another without penalty No preemption

Processing of jobs cannot be interrupted

Stage 2

Stage 1



Questions

a) What is the long run utilization rate for server 1 and for server 2?

b) What are key findings of the following simulation? The length of the simulation is 120,000 hrs.

First 20,000 hrs are discarded.

Summary statistics of the remaining 100,000 hrs are used. 95% confidence interval for the mean throughput time

(MTPT) of each class = MTPT estimate +/- 10% FCFS discipline is applied.

c) Same as b) except type B jobs have non-preemptive priority over type A jobs.

d) Same as b) except two identical servers at each station and an inter-arrival time distribution of each customer type that is uniformly distributed between 0 and 20 hrs.

e) Inter-arrival and service time distributions are exponential.

Stage 2

Stage 1



Summary Statistics of a Simulation

b) c) d) e)MTPT of A 64,85 76,05 44,7 156,2MTPT of B 32,86 17,26 22,13 78,46

QL - station 1 2,65 2,38 2,77 7,84QL - station 2 0,65 0,63 0,63 2,25

u - station 1 0,897 0,897 0,898 0,901u - station 2 0,749 0,75 0,754 0,753QL=queue length

Stage 2

Stage 1



Simulation Results (Questions b, c and d)

Throughput times for type A (d)

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120-140

140-160

160-180

Throughput times for type B (d)

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0-10 10-20 20-30 20-40 40-50 50-60 60-70 70-80 80-90 90-100

Throughput times for type A (c)

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Throughput times for type B (c)

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Throughput times for type A (b)

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Throughput times for type B (b)

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0,6

0-10 10-20 20-30 20-40 40-50 50-60 60-70 70-80 80-90 90-100

Stage 2

Stage 1



Simulation Results (Question e)

Type A Type B0-50 0,177 0,4356

50-100 0,286 0,3194100-150 0,18 0,1643150-200 0,124 0,0469200-250 0,105 0,0104250-300 0,059 0,0067300-350 0,027 0,0051350-400 0,009 0,0069400-450 0,005 0,0047450-500 0,019 0

Throughput times for type A (e)

0

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0,2

0,25

0,3

0,35

0-50 50-100

100-150

150-200

200-250

250-300

300-350

350-400

400-450

450-500

Throughput times for type B (e)

0

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0,2

0,3

0,4

0,5

0-50 50-100

100-150

150-200

200-250

250-300

300-350

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400-450

450-500

Stage 2

Stage 1


Example 4

Queuing Network

Precedence constraints as illustrated n “servers” are identical Service times per stage j UNIFORM[aj,bj] Many statistically independent replications of the project to be

completed over time Inter-arrival times uniformly distributed between 0 and 7 days

1 3

2 4

Start End

a2= 3 b2= 7n2= 2

a4= 2 b4= 4n4= 1

a1= 3 b1= 9n1= 3

a3= 3 b3= 5n3= 3

Uniform[0,7]


Example 4Queuing Network

Simulation Setting and Questions

Simulation for 60,000 days First 5,000 days are discarded Statistics of the remaining 55,000 days are used as the base case

This data represents the “steady state” behaviour of the system.

Potential improvements One improvement at a time (effect evaluated in isolation) If improvement achieved, the whole task time distribution will be

shifted by one day.

Which of the four improvements is the best for improving the system performance? Why?

1 3

2 4

Start End

Station a' b' mean1 2 8 52 2 6 43 2 4 34 1 3 2



Summary Statistics

MTPT95% confid.

level Utilization of stations

Base case 12.8 (12.4, 13.2) (0.58, 0.72, 0.38, 0.86)Reduction at station 1 12.5 (12.1, 12.9) (0.48, 0.72, 0.38, 0.86)Reduction at station 2 12.1 (11.7, 12.5) (0.58, 0.58, 0.38, 0.86)Reduction at station 3 12.3 (11.9, 12.7) (0.58, 0.72, 0.29, 0.86)Reduction at station 4 11.1 (11.0, 11.2) (0.58, 0.72, 0.38, 0.58)

1 3

2 4

Start End



Simulation Results

Base case

0

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0,3

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0-6 6-8 8-10 10-12

12-14

14-16

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18-20

20-22

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24-100

Reduction at station 1

0

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12-14

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18-20

20-22

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24-100


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0-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-100


0

0,1

0,2

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0-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-100


0

0,1

0,2

0,3

0,4

0-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-100

TPT Base caseReduction

at st 1Reduction

at st 2Reduction

at st 3Reduction

at st 4

0-6 0 0,0031 0 0,0037 06-8 0,0406 0,0648 0,0701 0,086 0,07098-10 0,2153 0,2491 0,2802 0,2706 0,3105

10-12 0,3228 0,3143 0,2932 0,2773 0,374612-14 0,1995 0,1503 0,1852 0,1474 0,179714-16 0,0871 0,0839 0,0679 0,0814 0,04316-18 0,0504 0,05 0,039 0,0494 0,011618-20 0,0319 0,0318 0,0244 0,0318 0,005820-22 0,0197 0,0202 0,0155 0,0198 0,001822-24 0,0135 0,0134 0,0097 0,0134 0,001424-100 0,0192 0,0191 0,0148 0,0192 0,0007

1 3

2 4

Start End


Service

Single Channel, Multiple Priority

Arrival rate (raAverage number of jobs

in queue (WIPq )

Average waiting timein queue (CTq )

Service rate (re

Decision on the processing order

New orders are assigned a priority upon arrival


Performance Measures for Multiple Priority Queuing Model

System utilization

Average waiting time in line for units in kth priority class

Average waiting time in system for units in kth priority class

Average number waiting in line for units in kth priority class

q ek qkWIP r CT

a

e

ru

mra akr r

1 1

1where and 1

(1 )

ka ac

qk kk k q ec

r rCT A B

A B B u WIP mr

q qk eCT CT t


Example 5

Multiple Priorities Model

Arriving jobs are assigned a priority and based on it placed into one of several priority classes. Jobs are then processed by class, highest class first. Within each class, on FCFS basis. Non - preemptive system assumed.

A machine shop handles tool repairs in a large company. As each job arrives in the shop, it is assigned a priority based on urgency of the need for that tool.

Requests for repair can be described by a Poisson distribution. Arrival rates are: ra1 = 2 per hour, ra2 = 2 per hour , and ra3 = 1 per hour. The service rate is one

tool per hour for each server, and there are six servers in the shop.

Determine: System utilization The average time a tool in each of the priority classes will wait for service The average time spends in the system for each priority class The average number of tools waiting for repair in each class

111233344


Multiple Priorities Modelra = rak = ra1 + ra2 + ra3 = 5 per hr

re= 1 job per hrm = 6 servers

System utilization u = 5 / 6*1 = 0.833

Average waiting times per each priority class For ra 5 and m=6, WIPq=2.938 (given)

1112333

19.10938.2)833.01(

5

)1(

q

a

WIPu

rA

111

0

k

c e

ac

mr

rB

667.01*6

211 B

333.01*6

2212

B

167.01*6

12213

B

hrsBBA

CT 765.1)167.0)(333.0(19.10

11

32

3

hrBBA

CT 442.0)333.0)(667.0(19.10

11

21

2

hrBBA

CT 147.0)667.0)(1(19.10

11

10

1

Average time in system = CTq+ 1/re= CTqk+ 1

Example 51

1123


Multiple Priorities Model

Average times in the system per each priority class Average time in system = CTqk+ 1/ re = CTqk+ 1

Average number of units waiting in each priority class WIPqk= rak * CTk

Class CTqk CT (hrs)

1 0.147 1.1472 0.442 1.4423 1.765 2.765

Class rak*CTk WIPqk (units)

1 2(0.147) 0.2942 2(0.442) 0.8843 1(1.765) 1.765

Example 51

1123


Deterministic arrival of type B jobs: one every

5.0 hrs

Deterministic arrival of type A jobs: one

every 4.5 hrs

Example 6Priority Queuing Network

Process and Resources

Two processing resources Six kinds of tasks each with own processing time distribution

All distributions exponential

Station 1

Station 2

MPT 2.0 MPT 1.9 MPT 0.1

MPT 2.0MPT 1.9MPT 0.1



Simulation and Questions Simulation for 120,000 hrs

First 20,000 hours discarded The statistics of the remaining 100,000 hours = Base case

The data represents the “steady state” behavior of the system

Disciplines tested1: FCFS2: Type B jobs are given priority at station 1, and type A jobs are

prioritized at station 2, otherwise FCFS applied3: Type B jobs are given priority at station 1, and type A jobs are

prioritized at station 2, otherwise second priority at each station given to jobs which are waiting for their final operation (otherwise processed on FCFS basis)

Questions What is the long run utilization rate? Can any principles of network scheduling be deduced from the

summary statistics?

MPT 2.0

MPT 1.9

MPT 0.1

MPT 2.0

MPT 1.9

MPT 0.1



Simulation Results

MTPT of A95% Conf

IntervMTPT of B

95% Conf Interv

Utilization of Stations

FCFS 19.1 (18.9, 19.2) 17.0 (16.9, 17.1) (0.887, 0.807)Priority 1 15.7 (14.0, 16.9) 11.2 (11.1, 11.3) (0.887, 0.807)Priority 2 14.6 (14.5, 14.7) 10.8 (10.7, 10.9) (0.887, 0.807)

FCFS

0

0,05

0,1

0,15

0,2

0,25

0,3

0-3 3-6 6-9 9-12 12-15

15-18

18-21

21-24

24-27

27-30

30-33

33-36

36-39

39-42

42-45

45-100

A-FCFS

B-FCFS

MPT 2.0

MPT 1.9

MPT 0.1

MPT 2.0

MPT 1.9

MPT 0.1


Priority Queuing Network

Simulation ResultsExample 6

Priority 2

0

0,05

0,1

0,15

0,2

0,25

0,3

0-3 3-6 6-9 9-12 12-15

15-18

18-21

21-24

24-27

27-30

30-33

33-36

36-39

39-42

42-45

45-100

A-Priority 2

B-Priority 2

Priority 1

0

0,05

0,1

0,15

0,2

0,25

0,3

0-3 3-6 6-9 9-12 12-15

15-18

18-21

21-24

24-27

27-30

30-33

33-36

36-39

39-42

42-45

45-100

A-Priority 1

B-Priority 1

MPT 2.0

MPT 1.9

MPT 0.1

MPT 2.0

MPT 1.9

MPT 0.1


Key Points

Cost trade-offs include the cost of waiting, lost sales, and cost of capacity.

The VUT equation considers the impact of variability and utilization on system performance.

High variability lead to long queues and long waits. Reduction in throughput reduces waiting. Reducing variability improves performance. Pooling servers improves performance.

Different types of systems exists A variety of models is needed in practice. Simple ones easy to solve; simulation can be used to gain

insights in more complex ones.


SCV = squared coefficient of variationMPT = mean processing time MTPT = mean throughput time

Queuing Notation and Measuresra = arrival rate

ta = mean inter-arrival time (=1/ ra)

re = mean service rate for a single server (=m/ te)

u = system utilization (=ra / mre)

te= mean process time (1/re)

m = the number of parallel machines in the systemWIP = average number of jobs in the system (incl. those being served; steady state)

WIPq = average number of jobs in the queue (steady state)

CT = average time spent in the system (including service; steady state)

CTq = average time spent in the queue (steady state)

p0 = the probability that there are no jobs in the system

pn = the probability that there are n jobs in the system

b = buffer size e.g. between machines

ca2 = SCV of the inter-arrival time

ce2 = SCV of the effective process time

Hopp and Spearman 2000, 265

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Queuing Models in Operations 35E00100 Service Operations and Strategy #4 Fall 2015.

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Transcript of Queuing Models in Operations 35E00100 Service Operations and Strategy #4 Fall 2015.