1. Queuing Models Basic Concepts
Module C7
Queuing Models
Basic Concepts

2. QUEUING MODELS Queuing theory is the analysis of waiting lines
It can be used to:
Determine the # checkout stands to have open at a store
Determine the type of line to have at a bank
Determine the seating procedures at a restaurant
Determine the scheduling of patients at a clinic
Determine landing procedures at an airport
Determine the flow through a production process
Determine the # toll booths to have open on a bridge

3. COMPONENTS OF QUEUING MODELS
Arrival Process
Waiting in Line
Service/Departure Process
Queue -- The waiting line itself
System -- All customers in the queuing area
Those in the queue
Those being served

4. 7 customers in the system
The Queuing Process
1. Customers arrive according to some arrival pattern
The Queue
4 customers in the queue
3. Customers are served according to some service distribution and depart
2. Customers may have to wait in a queue
The System
7 customers in the system

5. ARRIVAL PROCESS Deterministic or Probabilistic (how?)
Determined by # customers in system/balking?
Single or batch arrivals
Priority or homogeneous customers

6. THE WAITING LINE One long line or several smaller lines
Jockeying allowed?
Finite or infinite line length
Customers leave line before service?
Single or tandem queues

7. THE SERVICE PROCESS Single or multiple servers
Deterministic of probabilistic (how?)
All servers serve at same rate?
Speed of service depends on line length?
FIFO/LIFO or some other service priority

8. OBJECTIVE To design systems that optimize some criteria
Maximizing total profit
Minimizing average wait time for customers
Meeting a desired service level

9. TYPICAL SERVICE MEASURES
Average Number of customers in the system -- L
Average Number of customers in the queue -- Lq
Average customer time in the system -- W
Average customer waiting time in the queue -- Wq
Probability there are n customers in the system -- pn
Average number of busy servers (utilization rate) -- 

10. POISSON ARRIVAL PROCESS
REQUIRED CONDITIONS
Orderliness
at most one customer will arrive in any small time interval of t
Stationarity
for time intervals of equal length, the probability of n arrivals in the interval is constant
Independence
the time to the next arrival is independent of when the last arrival occurred

11. NUMBER OF ARRIVALS IN TIME t
Assume  = the average number of arrivals per hour (THE ARRIVAL RATE)
For a Poisson process, the probability of k arrivals in t hours has the following Poisson distribution:

12. Time Between Arrivals The average time between arrivals is 1/
For a Poisson process, the time between arrivals in hours has the following exponential distribution:
f(x) = e-t
This means:
P(next arrival occurs > t hours from now) = e-t
P(next arrival occurs within the next t hours) = 1- e-t

13. POISSON SERVICE PROCESS
REQUIRED CONDITIONS
Orderliness
at most one customer will depart in any small time interval of t
Stationarity
for time intervals of equal length, the probability of completing n potential services in the interval is constant
Independence
the time to the completion of a service is independent of when it started
IS THIS A GOOD ASSUMPTION?

14. NUMBER OF POTENTIAL SERVICES IN TIME t
Unlike the arrival process, there must be customers in the system to have services
Assume  = the average number of potential services per hour (SERVICE RATE)
For a Poisson process, the probability of k potential services in t hours has the following Poisson distribution:

15. THE SERVICE TIME The average service time is 1/
For a Poisson process, the service time has the following exponential distribution:
f(x) = e-t
This means:
P(the service will take t additional hours) = e-t
P(the remaining service will take longer than t hours) = 1- e-  t

16. TRANSIENT vs. STEADY STATE
Steady state is the condition that exists after the system has been operational for a while and wild fluctuations have been “smoothed out”
Until steady state occurs the system is in a transient state -- transiting to steady state
It is the long run steady state behavior that we will measure

17. CONDITIONS FOR STEADY STATE
For any queuing system to be stable the overall arrival rate must be less than the overall potential service rate, i.e.
For one server:  < 
For k servers with the same service rate:  < k
For k servers with different service rates:
 < 1 + 2 + 3 + …+ k

18. STEADY STATE PERFORMANCE MEASURES
We’ve mentioned these before:
Average Number of customers in the system -- L
Average Number of customers in the queue -- Lq
Average customer time in the system -- W
Average customer waiting time in the queue -- Wq
Probability there are n customers in the system -- pn
Average number of busy servers (utilization rate) - 

19. Little’s Laws and Other Relationships
Little’s Laws relate L to W and Lq to Wq by:
L = W
Lq = Wq
Also, (# in Sys) = (# in queue) + (# being served)
Thus
E(# in Sys) = E(# in queue) + E(# being served)
L = Lq 
Thus knowing one of L, W, Lq and Wq allows us to find the other values.

20. CLASSIFICATION OF QUEUING SYSTEMS
Queuing systems are typically classified using a three symbol designation:
(Arrival Dist.)/(Service Dist.)/(# servers)
Designations for Arrival/Service distributions include:
M = Markovian (Poisson process)
D = Deterministic (Constant)
G = General
Sometimes the designation is extended to 4 or 5 symbols to indicate Max queue length and # in population

21. M/M/1
M = Customers arrive according to a Poisson process at an average rate of  / hr.
M = Service times have an exponential distribution with an average service time = 1/ hours
1 = one server
Simplest system -- like EOQ for inventory -- a good starting point

22. M/M/1 PERFORMANCE MEASURES
Average Number of customers in the system -- L = /(- )
Average Number of customers in the queue -- Lq = L - /
Average customer time in the system -- W = L/ 
Average customer waiting time in the queue -- Wq = Lq/ 
Probability 0 customers in the system -- p = 1-/
Probability n customers in the system -- pn =(/)n p0
Average number of busy servers (utilization rate) or
Average number customers being served =  = /

23. EXAMPLE -- Mary’s Shoes
Customers arrive according to a Poisson Process about once every 12 minutes
 = (60min./hr)/(12 min./customer) = 60/12 = 5/hr.
Service times are exponential and average 8 min.
 (service rate) = (60min/hr)/(8min./customer) = 7.5/hr.
One server
This is an M/M/1 system
Will steady state be reached?
 = 5 <  = 7.5/hr YES

24. MARY’S SHOES PERFORMANCE MEASURES
Avg # of busy servers (utilization rate) or
Avg # customers being served =  = / =(5/7.5) = 2/3
Average # in the system -- L = /(- ) = 5/(7.5-5) = 2
Average # in the queue -- Lq = L - / = 2 - (2/3) = 4/3
Avg. customer time in the system -- W = L/  = 2/5 hrs.
Avg cust.time in the queue - Wq = Lq/  = (4/3)/5 = 4/15 hrs.
Prob.0 customers in the system -- p0 = 1-/ 1-(2/3) = 1/3
Prob. n customers in the system -- pn=(/)n p0 =(2/3) n(1/3)

25. COMPUTER SOLUTION
The formulas for an M/M/1 are very simple, but those for other models can be quite complex
Use M/M/k worksheet in Queue Template
Results are in the row corresponding to 1 server

26. Input  and 
Steady State Results
Pn’s

27. M/M/k SYSTEMS
M = Customers arrive according to a Poisson process at an average rate of  / hr.
M = Service times have an exponential distribution with an average service time = 1/ hours regardless of the server
k = k IDENTICAL servers

28. M/M/k PERFORMANCE MEASURES

29. EXAMPLE LITTLETOWN POST OFFICE
Between 9AM and 1PM on Saturdays:
Average of 100 cust. per hour arrive according to a Poisson process --  = 100/hr.
Service times exponential; average service time = 1.5 min. --  = 60/1.5 = 40/hr.
3 clerks; k = 3
This is an M/M/3 system
 = 100/hr < 3( = 40/hr.) i.e. 100 < 120
STEADY STATE will be reached

30. Solution Using the formulas, with  = 100,  = 40, k = 3
Prob.0 customers in the system -- p0 =
Average # in the system -- L =
Average # in the queue -- Lq =
Avg. customer time in the system -- W = hrs.
Avg cust.time in the queue - Wq = .0351hrs.
Avg # of busy servers =  = / =(100/40) = 2.5
Average system utilization rate = /k = 100/120=.83

31. Input  and 
Performance Measures
for 3 servers
Pn’s

32. M/G/1 Systems
M = Customers arrive according to a Poisson process at an average rate of  / hr.
G = Service times have a general distribution with an average service time = 1/ hours and standard deviation of  hours (1/ and  in same units)
1 = one server
Cannot get formulas for pn but can get performance measures

33. Example -- Ted’s TV Repair
Customers arrive according to a Poisson process once every 2.5 hours --
 = 1/2.5 = .4/hr.
Repair times average 2.25 hours with a standard deviation of 45 minutes
 = 1/2.25 = .4444/hr.
 = 45/60 = .75 hrs.
Ted is the only repairman: k= 1
THIS IS AN M/G/1 SYSTEM

34. There are no formulas for the pn’s.
Performance Measures
P0 = 1-/ = 1-(.4/.4444) = .0991
L = (()2 + (/ )2)/(2(1-/ )) + /
= ((.4)(.75)2 + (.4/.4444)2)/(2(.0991)) + (.4/.4444)
= 5.405
Lq = L - / = = 4.504
W = L/  = 5.405/.4 = hrs.
Wq = Lq/  = 4.504/.4 = hrs.
There are no formulas for the pn’s.

35. Input  (in customers/hr.)  (in customers/hr.)  (in hours)
Performance
Measures
Select MG1 Worksheet

36. FINITE QUEUES (M/M/k/F)
Frequently there are systems that have limits to the maximum number of customers in the system F
Thus with probability pF the system is FULL and an arriving customer cannot join the queue-- i.e. we lose pF portion of the potential customers
The effective arrival rate is then e = (1 - pF)
Use e to calculate L, Lq, W, and Wq
Steady state will always be achieved regardless of 
and since the queue cannot build up indefinitely!!

37. Example -- Ryan’s Roofing
Customer calls average 10/hr.
1 server -- average service time -- 3 minutes
 = 60/3 = 20/hr.
3 lines so 2 can be on hold -- they will hold until they are served
Arriving phone call when all 3 lines busy will not join the system

38. Input , , k, F
Performance Measures
pn’s
W = L/ e
e = ( )(10) =
Select MMkF Worksheet

39. M/M/1 QUEUES WITH FINITE CALLING POPULATIONS (M/M/1//m)
Maximum m school buses at repair facility, or m assigned customers to a salesman, etc.
1/ = average time between repeat visits for each of the m customers
 = average number of arrivals of each customer per time period (day, week, mo. etc.)
1/ = average service time
 = average service rate in same time units as 

40. Example -- Pacesetter Homes
4 projects m = 4
Average 1 work stoppage every 20 days/project
“arrival” rate of work stoppages,  = 1/20 = .05/day
Average 2 days to resolve work stoppage dispute
“service” rate,  = 1/2 = .5/day

41. Input , , m
Performance Measures
pn’s
Select MM1 m Worksheet

42. Module C7 Review Components of a queuing system
Arrivals, Queue, Services
Assumptions for Poisson (Markovian) distribution
Requirements for Steady State
Overall service rate > Overall arrival rate
Steady State Performance Measures
L, Lq, W, Wq, pn’s, 
Queuing Systems
M/M/1, M/M/k, M/G/1, M/M/k/F, M/M/1//m
Use of the Queue Template