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

2. Topics Principles of queuing theory Extensions Key points
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

3. “Customer arrives every 10 minutes, service process takes 8 minutes”
100%
100%
90%
90%
80%
80%
“Customer arrives every 10 minutes, service process takes 8 minutes”
Ideal situation
70%
70%
60%
60%
Probability
50%
50%
40%
40%
30%
30%
20%
20%
10%
10% 0% 0% 15 30 45 60 75 90
105
120
135
150
165
180
195
Reality 0% 5%
10%
15%
20%
25%
30% 10 20 30 40 50 60 70 80 90
100
110
120
130
140
150
160
170
180
190
Yli
40%
50%
60%
70%
80%
90%
100%
Arrival rate
Probability 0% 5%
10%
15%
20%
25% 10 20 30 40 50 60 70 80 90
100
110
120
130
140
150
160
170
180
190
Yli
Probability
40%
60%
80%
90%
Service rate
100%

4. What is queuing theory about?
Average number of jobs
in queue (WIPq )
Average waiting time
in queue (CTq )
Service
Arrival rate (ra
Service rate (re
Average number of jobs in the system (WIP )
Average time in the system (CT )

5. Cost Trade-Off: Capacity Utilization
Total costs
Low 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 of
operations
Cost
Cost of
queuing
0 %
Utilization
100 %

6. Cost Trade-Off: Capacity Level
Total costs
Cost of
operations
Optimum
Cost of
queuing
Capacity

7. Kendall's Classification 1/2/3/4/5/6
Queue
Server
1 = the nature of arrival process
2 = the nature of service process
3 = the number of (parallel) machines
4 = queue discipline
5 = the maximum allowable number of jobs
6 = the size of population

8. 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

9. Inter-arrival and Service Times Exponential Distribution
1.0
0.9
0.8
0.7
Exp(1)
0.6
Exp(3)
0.5
Exp(5)
0.4
0.3
0.2
0.1
0.0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Time (t)

10. Inter-arrival and Service Times Poisson Distribution
0.40
0.35
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.
0.30
0.25
0.20
0.15
Poisson(1)
0.10
Poisson(3)
0.05
Poisson(5)
0.00 2 4 6 8 10 12 14 16 18 20
Arrivals per hr

11. M/M/1 Queue - Relevant Equations
Utilization rate u
Average waiting time CTq
Average queue length WIPq
Average time of jobs in system CT
Average number of jobs in system WIP
e.g. Krajewski & Ritzman

12. M/M/m Queue - Relevant Equations
Utilization rate u
Average queue length WIPq
Average waiting time CTq
Average time of jobs in system CT
Average number of jobs in system WIP
Probability for being idle p0
e.g. Krajewski & Ritzman

14. 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
Note! This VUT equation is also called Kingman’s Equation.
(VUT = Variability x Utilization x Time)
Hopp and Spearman 2000, 270

15. 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
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
Server
Queue
50% ?

16. Extensions to the Basic Models

17. 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,

18. M/M/1/b Queue Model of Station 2 1 2 B Infinite raw materials
There is room for b=B+2 jobs in system, B in the buffer and one at each station B
Infinite raw
materials 1 2
Model of Station 2
Formulas valid for u1
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
Hopp and Spearman 2000,

19. Illustrating the Effect of Blocking
Example 1
Illustrating the Effect of Blocking
te1=21
te2=20 1 2
Conclusion:
M/M/1/b system has less WIP and less TH than M/M/1 system!
18 % less TH
90 % less WIP
Hopp and Spearman 2000,

20. We need to be able to analyze different types of processes…
Single Stage
Multi-Stage
ATM
Car wash
Single
Channel
Queue
Server
Queue
Queue
Server
Server
Bank teller’s windows
Hospital admissions
Multi-
Channel
Queue
Queue
Queue
Servers
Servers
Servers 6

21. Serial and Network Queuing Systems
Serial (or tandem) queuing systems
Servers arranged in series
e.g. repetitive flow process 1 3 2
Queuing networks
Some servers arranged in parallel and some in series
E.g. cellular manufacturing, job shops 1a 3a 2a 1b 2b 2c 3b

22. Example 2
Serial Queuing System 1 3 2
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.

23. Serial Queuing System u = ra / re = 5/6 = 0.83
Example 2 1 3 2
Serial Queuing System
Using the M/M/1 formulas for the 1st stage
u = ra / re = 5/6 = 0.83
WIPq = ra2/ [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 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. 1
re=5/day
CT=1 day
ra=6/day
WIPq=4.17 3 2
CT=1 day
ra=6/day
WIPq=4.17
WIPq=4.17

24. Serial Queuing System Demand increases from 5.0 to 5.7 orders per day
Example 2 1 3 2
Serial Queuing System
Demand increases from 5.0 to 5.7 orders per day
u = ra / re = 5.7/6 = 0.95
WIPq = ra2/ [re(re-ra)] = 5.72 / 6(6-5.7) = orders
CTs = 1/(re -ra) = 1/(6-5.7) = 3.3 days
In front of the 1st stage there will be an average of 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 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. 1
re=5.7/day
CT=3.33 d
ra=6/day
M/M/1
WIPq= 18.05 3 2
ra=6/day
CT=3.33 d
WIPq= 18.05
M/M/1

25. Serial Queuing System No randomness in processing times
Example 2 1 3 2
Serial Queuing System
No randomness in processing times
Using the M/D/1 formulas for the first stage
WIPq = u2 / [2(1-u)] = / 2(1-0.95) = orders
CTq = WIPq / ra = / 5.7 = days
CT = CTq +1/re = /6 = 1.75 days
re=5.7/day 1
CT=1.75 d
WIPq=9.025
M/D/1
ra=6/day 2
CT=0.167 d
D/D/1
ra=6/day 3
CT=0.167 d
D/D/1
ra=6/day
WIPq=0
WIPq=0
No queues develop in front of stages 2 and 3, and so the average time in the stage is the processing time days per job.
On average, there will be a total of jobs in inventory in the system. The average throughput time is 2.08 days.
Inventories are reduced by 83% and throughput time decreases by nearly 80%.

26. M/D/1 Queue: Relevant Equations
Average waiting time CTq
Average queue length WIPq
Average # of jobs in system WIP
Probability for being idle p0

27. Serial Queuing System Process and Resources
Example 3
Serial 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 A B
Stage 1
Stage 2

28. Serial Queuing System Questions
Stage 2
Stage 1
Example 3
Serial Queuing System Questions
What is the long run utilization rate for server 1 and for server 2?
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.
Same as b) except type B jobs have non-preemptive priority over type A jobs.
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.
Inter-arrival and service time distributions are exponential.

29. Serial Queuing System Summary Statistics of a Simulation
Stage 2
Stage 1
Example 3
Serial Queuing System Summary Statistics of a Simulation
QL=queue length

30. Serial Queuing System Simulation Results (Questions b, c and d)
Stage 2
Stage 1
Example 3
Serial Queuing System Simulation Results (Questions b, c and d)

31. Serial Queuing System Simulation Results (Question e)
Stage 2
Stage 1
Example 3
Serial Queuing System Simulation Results (Question e)

32. Queuing Network Precedence constraints as illustrated
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= 7
n2= 2
a4= 2
b4= 4
n4= 1
a1= 3
b1= 9
n1= 3
a3= 3
b3= 5
n3= 3
Uniform[0,7]

33. Queuing Network Simulation Setting and Questions1 3 2 4
Start
End
Example 4
Queuing 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?

34. Queuing Network Summary Statistics1 3 2 4
Start
End
Example 4
Queuing Network Summary Statistics

35. Queuing Network Simulation Results1 3 2 4
Start
End
Example 4
Queuing Network Simulation Results

36. Single Channel, Multiple Priority
Average waiting time
in queue (CTq )
Service
Arrival rate (ra
Average number of jobs
in queue (WIPq )
Service rate (re
New orders are assigned a priority upon arrival
Decision on the processing order

37. 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

38. Multiple Priorities Model
Example 5
Multiple Priorities Model 4 4 3 3 3 2 1 1 1
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

39. Multiple Priorities Model
Example 5 1 2 3
Multiple Priorities Model
ra =  rak = ra1 + ra2 + ra3 = 5 per hr
re = 1 job per hr
m = 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) 3 3 3 2 1 1 1
Average time in system = CTq+ 1/re = CTqk+ 1

40. Multiple Priorities Model
Example 5 1 2 3
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

41. Priority Queuing Network Process and Resources
Example 6
Priority Queuing Network Process and Resources
Two processing resources
Six kinds of tasks each with own processing time distribution
All distributions exponential
Deterministic arrival of type A jobs: one every 4.5 hrs
Station 1
Deterministic arrival of type B jobs: one every 5.0 hrs
MPT 2.0
MPT 1.9
MPT 0.1
Station 2
MPT 2.0
MPT 1.9
MPT 0.1

42. Priority Queuing Network Simulation and Questions
Example 6
Priority Queuing Network Simulation and Questions
MPT 2.0
MPT 1.9
MPT 0.1
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 tested
1: FCFS
2: Type B jobs are given priority at station 1, and type A jobs are prioritized at station 2, otherwise FCFS applied
3: 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?

43. Priority Queuing Network Simulation Results
Example 6
Priority Queuing Network Simulation Results
MPT 2.0
MPT 1.9
MPT 0.1

44. Priority Queuing Network Simulation Results
MPT 2.0
MPT 1.9
MPT 0.1
Example 6
Priority Queuing Network Simulation Results

45. 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.

46. Queuing Notation and Measures
ra = 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 system
WIP = 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
SCV = squared coefficient of variation
MPT = mean processing time
MTPT = mean throughput time
Hopp and Spearman 2000, 265