Chapter 16
Capacity Planning and
Queuing Models
Terminology
Capacity
 is the ability to deliver service over a particular time
period.
 is determined by the resources available to the
organization in the form of facilities, equipment and labor.
Capacity Planning
 is the process of determining the types and amounts of
resources that are required to implement an
organization’s strategic business plan.
 is a challenge for service firms because of the open
system nature of service operations.
Keynotes




The capacity planning decision involves a trade-off
between the cost of providing a service and the cost of
inconvenience of customer waiting.
The cost of service capacity is determined by the
number of servers on duty, whereas customer
inconvenience is measured by waiting time.
The lack of control over customer demands for service
and the presence of the customer in the process
complicate capacity planning.
For services, it is necessary to predict the degree of
customer waiting associated with different levels of
capacity.
Capacity Planning Challenges





Inability to create a steady flow of demand to fully utilize
capacity
Idle capacity always a reality for services.
Customer arrivals fluctuate and service demands also
vary.
Customers are participants in the service and the level of
congestion impacts on perceived quality.
Inability to control demand results in capacity measured in
terms of inputs (e.g. number of hotel rooms rather than
guest nights).
14-4
Strategic Role of Capacity Decisions
Capacity decision in services have strategic importance
based on the time horizon.
 Lack of short-term capacity planning can generate
customers for competition (e.g. restaurant staffing)
 Capacity decisions that must be balanced against the
costs of lost sales if capacity is inadequate or against
operating losses if demand does not reach expectations.
 Strategy of building ahead of demand is often taken to
avoid losing customers.

14-5
Queuing System Cost Tradeoff
Let: Cw = Cost of one customer waiting in
queue for an hour
Cs = Hourly cost per server
C = Number of servers
Total Cost/hour = Hourly Service Cost +
Hourly Customer Waiting Cost
Total Cost/hour = Cs C + Cw Lq
Note: Only consider systems where C    

16-6
Analytical Queuing Models
A popular system classifies parallel-server queuing models
using A / B / C.

A represents the distribution of time between arrivals.

B represents the distribution of service times

C represents the number of parallel servers.
The descriptive symbols used for the arrival and service
distributions include
M= exponential interarrival or service time distribution (or
Poisson distribution of arrival or service rate)
D= deterministic or constant interarrival or service time
Ek = Earlang distribution with shape parameter k (if k=1 then
exponential; if k= ∞ then deterministic)
G= general distribution with mean and variance (normal,
uniform or any empirical distribution)
M / M / 1 = a single server queuing model with Poisson
arrival rate and exponential service time distribution.
Classification of Queuing Models
Queuing Models
Poisson Arrivals
Standard
(Infinite Queue)
Exponential
Service Times
I
Single
Server
M / M /1
II
Multiple
Servers
M / M /c
Finite Queue
General
Service Times
III
Single
Server
M / G /1
IV
Self
Service
M / G /∞
Exponential
Service Times
V
Single
Server
M / M /1
VI
Multiple
Servers
M / M /c
Notation in Equations













n= number of customers in the systems
λ = mean arrival rate
μ = mean service rate per busy server
ρ = (λ / μ) mean number of customers in service
N = max number of customers allowed in the system
c = number of servers
Pn = probability of exactly n customers in the system
Ls = mean number of customers in the system
Lq = mean number of customers in queue
Lb = mean number of customers in queue for a busy system
Ws = mean number of customers in the system
Wq = mean number of customers in queue
Wb = mean number of customers in queue for a busy system
Standard M / M / 1 Model Assumptions
Calling Population: An infinite or very large population of
callers arriving. The callers are independent of each other
and not influenced by the queuing system.
 Arrival process: Negative exponential distribution of
interarrival times or Poisson distribution of arrival rate.
 Queue configuration: Single waiting line with no restrictions
on length and no balking or reneging.
 Queue discipline: First come first serve (FCFS)
 Service Process: One server with negative exponential
distribution of service times.

Queuing Formulas
Single Server Model with Poisson Arrival and Service Rates: M/M/1
1. Mean arrival rate: 

2. Mean service rate:


3. Mean number in service:

4. Probability of exactly “n” customers in the system:
5. Probability of “k” or more customers in the system:
6. Mean number of customers in the system:
Ls 
7. Mean number of customers in queue:
8. Mean time in system:
9. Mean time in queue:
Lq 
1
Ws 

Wq 


14-12
Pn   n (1  )
P( n  k )   k




Queuing Formulas (cont.)
Single Server General Service Distribution Model: M/G/1
 2  2 2
Lq 
2(1   )
Mean number of customers in queue for two servers: M/M/2
3
Lq 
4  2
Relationships among system characteristics:
Ls  Lq  
1
Ws  Wq 
Ws 
Wq 
1

1


Ls
Lq
14-13
Congestion as   10
.
100
10

Then:
Ls 

8
0
0.2
0.5
0.8
0.9
0.99
6
4
2
0
0


With:

1 
Ls
0
0.25
1
4
9
99
1.0
16-14
General Queuing Observations
1. Variability in arrivals and service times contribute equally to
congestion as measured by Lq.
2. Service capacity must exceed demand.
3. Servers must be idle some of the time.
4. Single queue preferred to multiple queue unless jockeying
is permitted.
5. Large single server (team) preferred to multiple-servers if
minimizing mean time in system, WS.
6. Multiple-servers preferred to single large server (team) if
minimizing mean time in queue, WQ.
16-15
Appendix D
Equations for selected queuing models
1.
Standard M/M/1 Model (0<ρ<1.0)
2.
Standard M/M/c Model (0<ρ<c)
3.
Standard M/G/1 Model (V(t)=service time
variance)
4.
Self-service M/G/∞ Model
5.
Finite-Queue M/M/1 Model
6.
Finite-Queue M/M/c Model
1. Standard M/M/1 Model (0<ρ<1.0)
1.
Calling population. An infinite or very large population of callers
arriving. The callers are independent of each other and not
influenced by the queuing system.
2.
Arrival process. Poisson distribution of arrival rate.
3.
Queue configuration. Single waiting line with no restrictions on
length and no balking or reneging.
4.
Queue discipline. FIFO
5.
Service process. One server with exponential distribution of
service times.
Example 16.2.



λ=6 boats per hour (poisson)
μ=6 minutes per boat (10 boat per hour)
(exponential)
M/M/1 model (infinite population, no queue
length restrictions, no balking or reneging, and
FCFS queue discipline)
Ls
λ
Single server
μ
Lq

ρ=λ/μ=6/10= 0.60

Probability that the system is busy and an
arriving customer waits:
P(n  k )   k
P(n≥1)= ρ1=0.601=0.60

Probability of finding the ramp idle :
P0=1- ρ=1-0.60=0.40

Mean number of boats in the system:

6
Ls 

 1.5 boats
   10  6

Ls 


Mean number of boats in queue:

Lq 

 (0.60)(6)
Lq 

 0.90 boat

10  6


Mean time in the system:
Ws 
1

1
1
Ws 

 0.25 hour (15 min.)
   10  6

Mean time in queue:
Wq 


0.60
Wq 

 0.15 hour (9 min.)
   10  6

The boat ramp is busy 60 % of the time.
◦ Thus, arrivals can expect immediate access to the ramp without delay
40% of the time.
The mean time in the system of 15
minutes is the sum of the mean time in
queue of 9 minutes and the mean service
time of 6 minutes.
 Arrivals an expect to find the number in
the system to be 1.5 boats and the
expected number in queue to be 0.9 boat.


The number of customers in the system can
be used to identify system states.
◦ For example, when n=0, the system is idle.
◦ When n=1, the server is busy but no queue
exists.
◦ When n=2, the server is busy and a queue of 1
has formed.
◦ The probability distribution for n can be very
uaeful in determining the proper size of a waiting
room (i.e., the number of chairs) to
accommodate arriving customers with a certain
probability of assurance that each will find a
vacant chair.

For the boot ramp example, determine the
number of parking spaces needed to ensure
that 90 % of the time, a person arriving at
the boot ramp will find a space to park
while waiting to launch.
Pn  n (1  )
n
Pn
P (number of customers ≤ n)
0
(0.6)0(0.4)=0.40
0.40
1
(0.6)1(0.4)=0.24
0.64
2
(0.6)2(0.4)=0.144
0.784
3
(0.6)3(0.4)=0.0864
0.8704
4
(0.6)4(0.4)=0.05184
0.92224
Repeatedly using the probability distribution for
system states for increasing values of n, we
accumulate the system state probabilities until
90 % assurance is exceeded.
 A system state of n=4 or less will occur 92% of
the time.
 This suggests that room for 4 boat trailers
should be provided because 92% of the time
arrivals will find 3 or fewer people waiting in
queue to launch.

EXAMPLE



A Social Security Administration branch is considering the following two
options for processing applications for social security cards:
◦ Option 1: Three clerks process applications in parallel from a single
queue. Each clerk fills out the form for the application in the presence of
the applicant. Processing time is exponential with a mean of 15 minutes.
Interarrival times are exponential.
◦ Option 2: Each applicant first fills out an application without the clerk’s
help. The time to accomplish this is exponentially distributed, with a
mean of 65 minutes. When the applicant has filled out the form, he or
she joins a single line to wait for one of the three clerks to check the
form. It takes a clerk an average of 4 minutes (exponentially distributed)
to review an application.
The interarrival time of applicants is exponential, and an average of 4.8
applicants arrive each hour.
Which option will get applicants out of the more quickly?
For Option 1

Option 1 is an M/M/c system with λ = 4.8
applicants/hr. and µ = 4 applicants/hour.

c=3 and ρ = 4.8/4 = 1.2 , from table Lq = 0.094
applicants

Ls=Lq + ρ = 0.094 + 1.2 = 1.294 applicants

Ws = Ls/λ = 1.294/4.8 = 0 .27 hours
EXAMPLE

Last National Bank is concerned about the level
of service at its single drive-in window. A study
of customer arrivals during the window’s busy
period revealed that, on average, 20 customers
per hour arrive, with a Poisson distribution, and
they are given FCFS service, requiring an
average of 2 minutes, with service times having
an exponential distribution.