A Basic Queueing System

clashjudiciousΗλεκτρονική - Συσκευές

8 Νοε 2013 (πριν από 3 χρόνια και 9 μήνες)

99 εμφανίσεις

1

A Basic Queueing System

Customers
Queue
Served Customers
Queueing System
Service
facility
S
S
S
S
C
C
C
C
C C C C C C C
Served Customers
2

Herr Cutter’s Barber Shop


Herr Cutter is a German barber who runs a one
-
man barber shop.


Herr Cutter opens his shop at 8:00 A.M.


The table shows his queueing system in action over a typical morning.

Customer

Time of

Arrival

Haicut

Begins

Duration

of Haircut

Haircut

Ends

1

8:03

8:03

17 minutes

8:20

2

8:15

8:20

21 minutes

8:41

3

8:25

8:41

19 minutes

9:00

4

8:30

9:00

15 minutes

9:15

5

9:05

9:15

20 minutes

9:35

6

9:43







3

Arrivals


The time between consecutive arrivals to a queueing system are called the
interarrival times
.


The expected number of arrivals per unit time is referred to as the
mean
arrival rate.


The symbol used for the mean arrival rate is


l

=
Mean arrival rate

for customers coming to the queueing system

where
l

is the Greek letter lambda.


The mean of the probability distribution of interarrival times is



1 /
l

=
Mean interarrival time


Most queueing models assume that the
form

of the probability distribution of
interarrival times is an
exponential distribution
.

4

Evolution of the Number of Customers

20
40
60
80
1
2
3
4
Number of
Customers
in the
System
0
Time (in minutes)
100
5

The Exponential Distribution for Interarrival Times

M
e
a
n
T
i
m
e
0
6

Properties of the Exponential Distribution


There is a high likelihood of small interarrival times, but a small chance of a
very large interarrival time. This is characteristic of interarrival times in
practice.


For most queueing systems, the servers have no control over when customers
will arrive. Customers generally arrive
randomly
.


Having
random arrivals

means that interarrival times are completely
unpredictable, in the sense that the chance of an arrival in the next minute is
always just the same.


The only probability distribution with this property of random arrivals is the
exponential distribution.


The fact that the probability of an arrival in the next minute is completely
uninfluenced by when the last arrival occurred is called the
lack
-
of
-
memory
property
.

7

The Queue


The
number of customers in the queue

(or
queue size
) is the number of
customers waiting for service to begin.


The
number of customers in the system

is the number in the queue
plus

the
number currently being served.


The
queue capacity

is the maximum number of customers that can be held in
the queue.


An
infinite queue

is one in which, for all practical purposes, an unlimited
number of customers can be held there.


When the capacity is small enough that it needs to be taken into account, then
the queue is called a
finite queue
.


The
queue discipline

refers to the order in which members of the queue are
selected to begin service.


The most common is
first
-
come, first
-
served

(FCFS).


Other possibilities include
random selection
, some
priority procedure
, or even
last
-
come, first
-
served
.

8

Service


When a customer enters service, the elapsed time from the beginning to the
end of the service is referred to as the
service time
.


Basic queueing models assume that the service time has a particular
probability distribution.


The symbol used for the
mean

of the service time distribution is



1 /
m

=
Mean service time


where
m

is the Greek letter mu.


The interpretation of
m

itself is the
mean service rate
.



m

= Expected service completions per unit time for a single busy server

9

Some Service
-
Time Distributions


Exponential Distribution


The most popular choice.


Much

easier to analyze than any other.


Although it provides a good fit for interarrival times, this is much less true for
service times
.


Provides a better fit when the service provided is random than if it involves a fixed
set of tasks.


Standard deviation:
s

= Mean


Constant Service Times


A better fit for systems that involve a fixed set of tasks.


Standard deviation:
s

= 0.


Erlang Distribution


Fills the middle ground between the exponential distribution and constant.


Has a
shape parameter
,
k
that determines the standard deviation.


In particular,
s

= mean / (

k
)

10

Standard Deviation and Mean for Distributions

Distribution

Standard Deviation

Exponential

mean

Degenerate (constant)

0

Erlang, any
k

(1 /

k)
(Mean)

Erlang,
k

= 2

(1 /

2
)
(Mean)

Erlang,
k

= 4

(1 /
2
)
(Mean)

Erlang,
k

= 8

(1 / 2

2
)
(Mean)

Erlang,
k

= 16

(1 /
4
)
(Mean)

11

Labels for Queueing Models

To identify which probability distribution is being assumed for service times (and
for interarrival times), a queueing model conventionally is labeled as follows:






Distribution of service times







/


/



Number of Servers




Distribution of interarrival times

The symbols used for the possible distributions are


M

= Exponential distribution (Markovian)


D

= Degenerate distribution (constant times)


E
k

= Erlang distribution (shape parameter =
k
)


GI

= General independent interarrival
-
time distribution (any distribution)


G

= General service
-
time distribution (any arbitrary distribution)

12

Summary of Usual Model Assumptions

1.
Interarrival
times are
independent

and identically distributed according to a
specified probability distribution.

2.
All arriving customers enter the queueing system and
remain there

until
service has been completed.

3.
The queueing system has a single
infinite queue
, so that the queue will hold
an unlimited number of customers (for all practical purposes).

4.
The queue discipline is
first
-
come, first
-
served
.

5.
The queueing system has a
specified number of servers
, where each server is
capable of serving any of the customers.

6.
Each
customer is served individually

by any one of the servers.

7.
Service

times are
independent

and identically distributed according to a
specified probability distribution.

13

Examples of Commercial Service Systems

That Are Queueing Systems

Type of System

Customers

Server(s)

Barber shop

People

Barber

Bank teller services

People

Teller

ATM machine service

People

ATM machine

Checkout at a store

People

Checkout clerk

Plumbing services

Clogged pipes

Plumber

Ticket window at a movie theater

People

Cashier

Check
-
in counter at an airport

People

Airline agent

Brokerage service

People

Stock broker

Gas station

Cars

Pump

Call center for ordering goods

People

Telephone agent

Call center for technical assistance

People

Technical representative

Travel agency

People

Travel agent

Automobile repair shop

Car owners

Mechanic

Vending services

People

Vending machine

Dental services

People

Dentist

Roofing Services

Roofs

Roofer

14

Examples of Internal Service Systems

That Are Queueing Systems

Type of System

Customers

Server(s)

Secretarial services

Employees

Secretary

Copying services

Employees

Copy machine

Computer programming services

Employees

Programmer

Mainframe computer

Employees

Computer

First
-
aid center

Employees

Nurse

Faxing services

Employees

Fax machine

Materials
-
handling system

Loads

Materials
-
handling unit

Maintenance system

Machines

Repair crew

Inspection station

Items

Inspector

Production system

Jobs

Machine

Semiautomatic machines

Machines

Operator

Tool crib

Machine operators

Clerk

15

Examples of Transportation Service Systems

That Are Queueing Systems

Type of System

Customers

Server(s)

Highway tollbooth

Cars

Cashier

Truck loading dock

Trucks

Loading crew

Port unloading area

Ships

Unloading crew

Airplanes waiting to take off

Airplanes

Runway

Airplanes waiting to land

Airplanes

Runway

Airline service

People

Airplane

Taxicab service

People

Taxicab

Elevator service

People

Elevator

Fire department

Fires

Fire truck

Parking lot

Cars

Parking space

Ambulance service

People

Ambulance

16

Choosing a Measure of Performance


Managers who oversee queueing systems are mainly concerned with two
measures of performance:


How many customers typically are waiting in the queueing system?


How long do these customers typically have to wait?


When customers are internal to the organization, the first measure tends to be
more important.


Having such customers wait causes
lost productivity
.


Commercial service systems tend to place greater importance on the second
measure.


Outside customers are typically more concerned with how long
they

have to wait
than with how many customers are there.

17

Defining the Measures of Performance

L

=

Expected
number of customers in the system
, including those being
served (the symbol
L

comes from
L
ine
L
ength).

L
q

=

Expected
number of customers in the queue
, which excludes customers
being served.

W

=

Expected
waiting time in the system

(including service time) for an
individual customer (the symbol
W

comes from
W
aiting time).

W
q

=

Expected
waiting time in the queue

(excludes service time) for an
individual customer.


These definitions assume that the queueing system is in a
steady
-
state condition
.

18

Relationship between
L
,
W
,
L
q
, and
W
q


Little’s formula states that


L

=
l
W

and


L
q

=
l
W
q


Since 1/
m

is the expected service time


W

=
W
q

+ 1/
m


Combining the above relationships leads to


L

=
L
q

+
l/m



19

Using Probabilities as Measures of Performance


In addition to knowing what happens
on the average
, we may also be
interested in
worst
-
case scenarios
.


What will be the
maximum

number of customers in the system? (Exceeded no more
than, say, 5% of the time.)


What will be the
maximum

waiting time of customers in the system? (Exceeded no
more than, say, 5% of the time.)


Statistics that are helpful to answer these types of questions are available for
some queueing systems:


P
n

=
Steady
-
state probability of having exactly
n

customers in the system.


P
(
W


t
) = Probability the time spent in the system will be no more than
t
.


P
(
W
q

≤ t
) = Probability the wait time will be no more than
t
.


Examples of common goals:


No more than three customers 95% of the time:
P
0

+
P
1

+
P
2

+
P
3

≥ 0.95


No more than 5% of customers wait more than 2 hours:
P
(
W

≤ 2 hours) ≥ 0.95

20

The Dupit Corp. Problem


The Dupit Corporation is a longtime leader in the office photocopier
marketplace.


Dupit’s service division is responsible for providing support to the customers
by promptly repairing the machines when needed. This is done by the
company’s
service technical representatives
, or
tech reps
.


Current policy:

Each tech rep’s territory is assigned enough machines so that
the tech rep will be active repairing machines (or traveling to the site) 75% of
the time.


A repair call averages 2 hours, so this corresponds to 3 repair calls per day.


Machines average 50 workdays between repairs, so assign 150 machines per rep.


Proposed New Service Standard:

The average waiting time before a tech rep
begins the trip to the customer site should not exceed two hours.

21

Alternative Approaches to the Problem


Approach Suggested by John Phixitt:

Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing
machines.


Approach Suggested by the Vice President for Engineering:

Provide new
equipment to tech reps that would reduce the time required for repairs.


Approach Suggested by the Chief Financial Officer:

Replace the current
one
-
person tech rep territories by larger territories served by multiple tech
reps.


Approach Suggested by the Vice President for Marketing:

Give owners of
the new printer
-
copier priority for receiving repairs over the company’s other
customers.

22

The Queueing System for Each Tech Rep


The customers:

The machines needing repair.


Customer arrivals:

The calls to the tech rep requesting repairs.


The queue:

The machines waiting for repair to begin at their sites.


The server:

The tech rep.


Service time:

The total time the tech rep is tied up with a machine, either
traveling to the machine site or repairing the machine. (Thus, a machine is
viewed as leaving the queue and entering service when the tech rep begins the
trip to the machine site.)

23

Notation for Single
-
Server Queueing Models



l

=
Mean arrival rate

for customers


= Expected number of arrivals per unit time


1/
l

= expected interarrival time




m

=
Mean service rate

(for a continuously busy server)


= Expected number of service completions per unit time


1/
m

= expected service time




r

= the
utilization

factor


= the average fraction of time that a server is busy serving customers


=
l

/
m


24

The
M/M/
1 Model


Assumptions

1.
Interarrival times

have an exponential distribution with a mean of 1/
l
.

2.
Service times

have an exponential distribution with a mean of 1/
m
.

3.
The queueing system has one server.


The
expected number of customers in the system

is



L

=
r

/ (
1


r
)

=
l

/ (
m



l
)


The
expected waiting time in the system

is



W

= (1 /
l
)
L

= 1 / (
m



l
)


The
expected waiting time in the queue

is



W
q

=
W



1/
m

=
l

/ [
m
(
m



l
)]


The
expected number of customers in the queue

is



L
q

=
l
W
q

=
l
2

/ [
m
(
m



l
)]

25

The
M
/
M
/1 Model


The

probability of having exactly
n

customers in the system is



P
n

= (1


r
)
r
n

Thus,


P
0

= 1


r


P
1

= (1


r
)
r


P
2

= (1


r
)
r
2



:



:


The probability that the
waiting time in the system

exceeds
t

is




P
(
W

>
t
) =
e

m
(1

r
)
t

for
t

≥ 0


The probability that the
waiting time in the queue

exceeds
t

is



P
(
W
q

>
t
) =
r
e

m
(1

r
)t

for
t

≥ 0

26

M/M/
1 Queueing Model for the Dupit’s Current Policy

Data
Results

3
(mean arrival rate)
L =
3

4
(mean service rate)
L
q
=
2.25
s =
1
(# servers)
W =
1
Pr(W > t) =
0.368
W
q
=
0.75
when t =
1

0.75
Prob(W
q
> t) =
0.276
when t =
1
n
P
n
0
0.25
1
0.1875
2
0.1406
3
0.1055
4
0.0791
5
0.0593
6
0.0445
7
0.0334
8
0.0250
9
0.0188
10
0.0141
27

John Phixitt’s Approach (Reduce Machines/Rep)


The proposed new service standard is that the average waiting time before
service begins be two hours (i.e.,
W
q


1
/
4

day).


John Phixitt’s suggested approach is to lower the tech rep’s utilization factor
sufficiently to meet the new service requirement.



Lower
r

=
l

/
m
, until
W
q


1
/
4

day,

where


l

= (Number of machines assigned to tech rep) / 50.


28

M/M/
1

Model for John Phixitt’s Suggested Approach

(Reduce Machines/Rep)

3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Data
Results
l 
2
(mean arrival rate)
L =
1
m 
4
(mean service rate)
L
q
=
0.5
s =
1
(# servers)
W =
0.5
Pr(W > t) =
0.135
W
q
=
0.25
when t =
1
r 
0.5
Prob(W
q
> t) =
0.068
when t =
1
n
P
n
0
0.5
1
0.25
2
0.1250
3
0.0625
4
0.0313
5
0.0156
6
0.0078
7
0.0039
8
0.0020
9
0.0010
10
0.0005
29

The
M/G/
1 Model


Assumptions

1.
Interarrival times

have an exponential distribution with a mean of 1/
l
.

2.
Service times

(T) can have
any
probability distribution.


E(T) = 1/
m

Ⱐ噡爨吩r㴠
s
2
.

3. The queueing system has one server.


The
probability of zero customers in the system

is



P
0

= 1


r


The
expected number of customers in the queue

is



L
q

=
l
2
[Var(T
)
+䔨吩
2
] / [2(1


l
䔨吩⥝


The
expected number of customers in the system

is



L

=
L
q

+
l
/
m

The
expected waiting time in the queue

is



W
q

=
L
q

/
l


The
expected waiting time in the system

is



W

=
W
q

+ 1/
m

30

The Values of
s

and
L
q

for the
M/G/
1 Model

with Various Service
-
Time Distributions

Distribution

Mean

s

Mo摥d

L
q

Exponential

1/
m

1/
m

M/M/
1

r
2

/ (1


r
)

Degenerate (constant)

1/
m

0

M/D/
1

(
1
/
2
) [
r
2

/ (1


r
)]

Erlang, with shape parameter
k

1/
m

(1/

k) (1/
m
)

M/E
k
/
1

(
k
+1)/(2
k
) [
r
2

/ (1


r
)]


The
expected number of customers in the queue

is



L
q

=
l
2
[Var(T
)
+䔨吩
2
] / [2(1


l
䔨吩⥝E

[l
2
s
2

+
r
2
] / [2(1


r


31

VP for Engineering Approach (New Equipment)


The proposed new service standard is that the average waiting time before
service begins be two hours (i.e.,
W
q


1
/
4

day).


The Vice President for Engineering has suggested providing tech reps with
new state
-
of
-
the
-
art equipment that would reduce the time required for the
longer repairs.


After gathering more information, they estimate the new equipment would
have the following effect on the service
-
time distribution:


Decrease the mean from
1
/
4

day to
1
/
5

day.


Decrease the standard deviation from
1
/
4

day to
1
/
10

day.

32

M/G/
1 Model for the VP of Engineering Approach

(New Equipment)

3
4
5
6
7
8
9
10
11
12
B
C
D
E
F
G
Data
Results
l 
3
(mean arrival rate)
L =
1.163
1/m 
0.2
(expected service time)
L
q
=
0.563
s
0.1
(standard deviation)
s =
1
(# servers)
W =
0.388
W
q
=
0.188
r 
0.6
P
0
=
0.4
33

The
M/M/s

Model


Assumptions

1.
Interarrival times

have an exponential distribution with a mean of 1/
l
.

2.
Service times

have an exponential distribution with a mean of 1/
m.

3.
Any number of servers (denoted by
s
).


With multiple servers, the formula for the
utilization factor

becomes



r

=
l

/
s
m


but still represents that average fraction of time that individual servers are
busy.



34

Values of
L

for the
M/M/s

Model for Various Values of
s

s = 1
s = 2
s = 3
s = 4
s = 5
s = 7
s = 10
s = 15
s = 20
s = 25
100
10
0.5
0.1
0.2
0
0.1
0.3
0.5
0.7
0.9
1.0
Utilization factor
r  l
s
m
Steady-state expected number of customers in the queueing system
35

CFO Suggested Approach (Combine Into Teams)


The proposed new service standard is that the average waiting time before
service begins be two hours (i.e.,
W
q


1
/
4

day).


The Chief Financial Officer has suggested combining the current one
-
person
tech rep territories into larger territories that would be served jointly by
multiple tech reps.


A territory with two tech reps:


Number of machines = 300

(versus 150 before)


Mean arrival rate =
l

= 6

(versus
l

= 3 before)


Mean service rate =
m

= 4

(as before)


Number of servers =
s

= 2

(versus
s

= 1 before)


Utilization factor =
r

=
l
/
s
m

= 0.75

(as before)

36

M/M/s

Model for the CFO’s Suggested Approach

(Combine Into Teams of Two)

3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Data
Results
l 
6
(mean arrival rate)
L =
3.4286
m 
4
(mean service rate)
L
q
=
1.9286
s =
2
(# servers)
W =
0.5714
Pr(W > t) =
0.169
W
q
=
0.3214
when t =
1
r 
0.75
Prob(W
q
> t) =
0.087
when t =
1
n
P
n
0
0.1429
1
0.2143
2
0.1607
3
0.1205
4
0.0904
5
0.0678
6
0.0509
7
0.0381
8
0.0286
9
0.0215
10
0.0161
37

CFO Suggested Approach (Teams of Three)


The Chief Financial Officer has suggested combining the current one
-
person
tech rep territories into larger territories that would be served jointly by
multiple tech reps.


A territory with three tech reps:


Number of machines = 450

(versus 150 before)


Mean arrival rate =
l

= 9

(versus
l

= 3 before)


Mean service rate =
m

= 4

(as before)


Number of servers =
s

= 3

(versus
s

= 1 before)


Utilization factor =
r

=
l
/
s
m

= 0.75

(as before)

38

M/M/s

Model for the CFO’s Suggested Approach

(Combine Into Teams of Three)

3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
B
C
D
E
G
H
Data
Results
l 
9
(mean arrival rate)
L =
3.9533
m 
4
(mean service rate)
L
q
=
1.7033
s =
3
(# servers)
W =
0.4393
Pr(W > t) =
0.090
W
q
=
0.1893
when t =
1
r 
0.75
Prob(W
q
> t) =
0.028
when t =
1
n
P
n
0
0.0748
1
0.1682
2
0.1893
3
0.1419
4
0.1065
5
0.0798
6
0.0599
7
0.0449
8
0.0337
9
0.0253
10
0.0189
39

Comparison of
W
q

with Territories of Different Sizes

Number of

Tech Reps

Number of

Machines

l

m

s

r

W
q

1

150

3

4

1

0.75

0.75 workday (6 hours)

2

300

6

4

2

0.75

0.321 workday (2.57 hours)

3

450

9

4

3

0.75

0.189 workday (1.51 hours)

40

Values of
L

for the
M/D/s

Model for Various Values of
s

100
0.1
1.0
10
0
0.1
0.3
0.5
0.9
1.0
s = 2
s = 3
s = 4
s = 5
s = 7
s = 10
s = 15
s = 20
s = 25
s = 1
0.7
Utilization factor
Steady-state expected number of customers in the queueing system
r  l
s
m
41

Values of
L

for the
M/E
k
/
2 Model for Various Values of
k

100
10
1.0
0.1
0
0.2
0.4
0.6
0.8
1.0
k = 1
k = 2
k = 8
Utilization factor
r  l
s
m
Steady-state expected number of customers in the queueing system
42

The Four Approaches Under Considerations

Proposer

Proposal

Additional Cost

John Phixitt

Maintain one
-
person territories, but
reduce number of machines assigned
to each from 150 to 100

$300 million per year

VP for Engineering

Keep current one
-
person territories,
but provide new state
-
of
-
the
-
art
equipment to the tech
-
reps

One
-
time cost of $500
million

Chief Financial Officer

Change to three
-
person territories

None, except
disadvantages of larger
territories

Decision:

Adopt the third proposal

43

Some Insights About Designing Queueing Systems

1.
When designing a single
-
server queueing system, beware that giving a
relatively high utilization factor (workload) to the server provides surprisingly
poor performance for the system.

2.
Decreasing the
variability

of service times (without any change in the mean)
improves the performance of a queueing system substantially.

3.
Multiple
-
server queueing systems can perform satisfactorily with somewhat
higher utilization factors than can single
-
server queueing systems. For
example,
pooling servers

by combining separate single
-
server queueing
systems into one multiple
-
server queueing system greatly improves the
measures of performance.

4.
Applying
priorities

when selecting customers to begin service can greatly
improve the measures of performance for high
-
priority customers.

44

Effect of High
-
Utilization Factors (Insight 1)

3
4
5
B
C
D
E
G
H
Data
Results
l 
0.5
(mean arrival rate)
L =
1
m 
1
(mean service rate)
L
q
=
0.5
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
A
B
C
D
E
Data Table Demonstrating the Effect of
Increasing
r
on Lq and L for M/M/1
l  r
L
q
L
1
0.5
1
0
0.01
0.0001
0.0101
0
0.25
0.0833
0.3333
0
0.5
0.5
1
0
0.6
0.9
1.5
0
0.7
1.6333
2.3333
0
0.75
2.25
3
0
0.8
3.2
4
0
0.85
4.8167
5.6667
0
0.9
8.1
9
0
0.95
18.05
19
0
0.99
98.01
99
0
0.999
998.001
999
0
20
40
60
80
100
0
0.2
0.4
0.6
0.8
1
System Utilization (r)
Average Line Length (L)
45

Effect of Decreasing
s

(Insight 2)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
A
B
C
D
E
F
G
H
Template for the M/G/1 Queueing Model
Data
Results
l 
0.5
(mean arrival rate)
L =
0.8125
1/m 
1
(expected service time)
L
q
=
0.3125
s
0.5
(standard deviation)
s =
1
(# servers)
W =
1.625
W
q
=
0.625
r 
0.5
P
0
=
0.5
Data Table Demonstrating the Effect of Decreasing s on Lq for M/G/1
Body of Table Shows L
q
Values
s
0.3125
1
0.5
0
0.5
0.500
0.313
0.250
r (l)
0.75
2.250
1.406
1.125
0.9
8.100
5.063
4.050
0.99
98.010
61.256
49.005
46

Economic Analysis of the Number of Servers to Provide


In many cases, the consequences of making customers wait can be expressed
as a
waiting cost
.


The manager is interested in minimizing the total cost.


TC = Expected total cost per unit time


SC = Expected service cost per unit time


WC = Expected waiting cost per unit time

The objective is then to choose the number of servers so as to


Minimize TC = SC + WC


When each server costs the same (
C
s

= cost of server per unit time),


SC =
C
s

s


When the waiting cost is proportional to the amount of waiting (
C
w

= waiting
cost per unit time for each customer),


WC =
C
w

L

47

Acme Machine Shop


The Acme Machine Shop has a tool crib for storing tool required by shop
mechanics.


Two clerks run the tool crib.


The estimates of the mean arrival rate
l

and the mean service rate (per server)
m

are


l

= 120 customers per hour


m

= 80 customers per hour


The total cost to the company of each tool crib clerk is $20/hour, so
C
s

= $20.


While mechanics are busy, their value to Acme is $48/hour, so
C
w

=
$48.


Choose
s

so as to Minimize TC = $20
s

+ $48
L
.

48

Excel Template for Choosing the Number of Servers

3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
C
D
E
F
G
Data
Results
l 
120
(mean arrival rate)
L =
1.736842105
m 
80
(mean service rate)
L
q
=
0.236842105
s =
3
(# servers)
W =
0.014473684
Pr(W > t) =
0.02581732
W
q
=
0.001973684
when t =
0.05
r 
0.5
Prob(W
q
> t) =
0.00058707
when t =
0.05
n
P
n
0
0.210526316
1
0.315789474
Cs =
$20.00
(cost / server / unit time)
2
0.236842105
Cw =
$48.00
(waiting cost / unit time)
3
0.118421053
4
0.059210526
Cost of Service
$60.00
5
0.029605263
Cost of Waiting
$83.37
6
0.014802632
Total Cost
$143.37
7
0.007401316
Economic Analysis:
49

Comparing Expected Cost vs. Number of Clerks

1
2
3
4
5
6
7
8
9
10
H
I
J
K
L
M
N
Data Table for Expected Total Cost of Alternatives
Cost of
Cost of
Total
s
r
L
Service
Waiting
Cost
0.50
1.74
$60.00
$83.37
$143.37
1
1.50
#N/A
$20.00
#N/A
#N/A
2
0.75
3.43
$40.00
$164.57
$204.57
3
0.50
1.74
$60.00
$83.37
$143.37
4
0.38
1.54
$80.00
$74.15
$154.15
5
0.30
1.51
$100.00
$72.41
$172.41
$0
$50
$100
$150
$200
$250
0
1
2
3
4
5
Number of Servers (s)
Cost ($/hour)
Cost of
Service
Cost of
Waiting
Total Cost