Waiting Line

Ch 12. Waiting Line Models
Contents
1. Structure of Waiting Line System
2. Single-Channel Waiting Line Model with Poisson Arrivals and
Exponential Service Times
3. Multiple-Channel Waiting Line Model with Poisson Arrivals and
Exponential Service Times
4. Economic Analysis of Waiting Lines
5. Other Waiting Line Models
6. Single-Channel Waiting Line Model with Poisson Arrivals and
Arbitrary Service Times
7. Multiple-Channel Model with Poisson Arrivals, Arbitrary Service
Times and No Waiting Line
8. Waiting Line Model with Finite Calling Population

9. Estimations of Arrival Process and Service Time Distribution
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Waiting Line Models
 Waiting line or Queue



Model is developed to help manager make to better decision for the operation of waiting line.
Erlang (a Danish Telephone engineer) began a study of congestion and waiting times in the completion of telephone calls.  Operating Characteristic (performance Measure) for a waiting Line Model





Probability that no units are in the system
Probability that an arriving unit has to wait for service
Average Number of units in waiting line or system
Average Time a unit spends in waiting line or system

 Make a decision that balance desirable service level against the cost of providing the service

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1. Structure of Waiting Line System
 Single-channel Waiting Line System
Service Rule

Server

Costumer
Arrives

Customer
Leaves
Waiting Line

Service System

 Elements for Waiting Line System





Population of arrivals and their arrival process
Capacity of waiting Line
Service discipline and service facility structure
Service process
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1.1 Population of Arrivals
 Infinite Population of arrival





Population of arrival to system is finite in most cases.
Customers to watch the movie ‘Abata’ may live in the Busan metropolitan area.
For the convenience of analysis, we assume that it is an infinite population.  Finite Population of arrival


When the size of population of arrivals to system is small and the probability of costumer’s arrival is dependent of its size, we assume the population of arrivals is finite.



A factory operates 10 machines. An arrival process of the breakdowned machines to repair-shop is dependent on the number of machines break-downed.

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1.2 Arrival Process
 Arrival process of customers to system
 The arrivals occur randomly and independently.
 The Inter-arrival time is described either by historical distribution or theoretical distribution.
 Mean arrival rate is the number of customers per unit time. 





When the mean arrival rate is 3 persons/hour, the mean interarrival time is 20min/person.
The mean arrival rate is a reciprocal of the mean inter-arrival time. Generally in waiting line theory, the Poisson arrival process provides a good description of the arrival pattern.

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1.3 Poisson Arrival Process
Assumptions for Poisson arrival process
(1) During an arbitrary time period t, the probability that customers arrive to system is proportional to time period t.
(2) For an infinitely small period, the probability that customers more than 1 arrive to system is 0.
(3) The number of costumers that arrive to system for a time period is independent of that for mutually not overlapped period.

Only Poisson distribution satisfies the above 3 assumptions. 6

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1.4 Poisson distribution
Poisson distribution







Mean arrival rate = 
Mean number of arrival for an interval (0, t) or (s, s+ t) = m  t
Number of arrivals for (0, t) or (s, s +t) = x e = 2.71828
The number of arrivals for (0, t) follows Poisson distribution. ( t ) e p( x)  x! x

7

 t

, x  0,1,2,3, 

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Example 1
Burger Dome’s costumer arrival data
 Mean arrival rate =   0.75






Mean number of arrivals for an interval (0, 1) = m  (0.75)(1)  0.75 x = Number of costumer arrivals during a one-minute period Then Poisson distribution will be

( ) x e   p ( x) 
, x  0,1,2,3,  x! 8

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Poisson probability for the number of costumer arrivals for one minute period
Number of Arrivals
0
1
2
3
4
5 or over

Probability
0.4724
0.3543
0.1329
0.0332
0.0062
0.0010

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Example 2
 Mean arrival rate :   1.5 person / min
 x = number of cars that arrive at toll gate for 2 min
 m  t  3

( t ) x e   t
3 x e 3 f ( x) 

, x  0,1,2,3,  x! x!
 Poisson Distribution x f(x)

x

f(x)

f(x)
0.5

0
1
2
3
4
5

0.0498
0.1494
0.2240
0.2240
0.1680
0.1008

6
7
8
9
10
11

0.0504
0.0216
0.0081
0.0027
0.0008
0.0002

0.4
0.3
0.2
0.1
0.0

0 1 2 3 4 5 6 7 8 9 10 11 12
10

x
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1.5 Service System
Waiting line capacity



Finite waiting line
Infinite waiting line

Structure of service facility



Single-phase service structure
Multi-phase service structure

Number of servers



Single-channel system
Multiple-channel system
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Service Discipline
Service discipline of server


Order that costumers in waiting line receive service



FCFS (first come first out) or FIFO (first in first out)



LCFO or LIFO (last come first out or last in first out)



Served with priority or Preemption

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Service Time
 Service Time





Deterministic or Probabilistic service time
Mean service rate is the number of costumers served in unit time.
 When 3 costumers are served for an hour, the mean service rate is 3 person/hour, and mean service time is 20 minute
/person.
 The mean service rate is the reciprocal of the mean service time. Usually in waiting line analysis, typically we assume the exponential service time distribution.

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Service Time distribution
 Service Time distribution





Deterministic or Probabilistic service time
When service time distribution is an exponential distribution, formulas is available for providing useful information.
Exponential distribution with the mean service rate 

g (t )  e  t , t  0


the probability that service time is less than t is

P(servicetime  t )  1  eut


where e = 2.71828
When the mean service rate = 1.0 costumer/ 1 min

P(servicetime  1.0 min)  1  e1(1.0)  1  0.3679  0.6321
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1.6 System Performance Measure
Performance measure
 utilization factor: average served time ratio per server  mean waiting time in Queue
 average queue length
 time in system or response time
 average number of customers in system
Improving service level results in reducing the average queue length (or mean waiting time in
Queue), and utilization factor of server.
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1.7 Relationship between Poisson Distribution and Exponential Distribution
 If the number of costumers (=X) served in time period (0, t) follows Poisson distribution, then

( t ) 0 e  t
P( x  0) 
 e  t
0!
 P ( No costumer will be served until time t )
= P( service time T is greater than t)
 That is, P(X=0) = P( T>t )

P(T  t )  1  P(T  t )  e  t
 Cumulative distribution of T

P(T  t )  1  P(T  t )  1  e  t

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Probability density function of service time
 The probability density function of service time is a derivative of cumulative density function. So we have

dF (t ) d d  f (t ) 
 P(T  t )  (1  e  t )  e  t dt dt dt  This result means that If the number of costumers (=X)

served in time period (0, t) follows Poisson distribution with mean of t , then the service time T should be the exponential distribution with mean service time of 1 /  .

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Steady State Operation
Transient state






The beginning or start-up period is referred to as a transient state.
The transient state is dependent on the initial condition. Transient state ends when the system reaches a steady state.

Waiting line models describes the steady-state operating characteristics of a waiting line.

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2. Single-Channel Waiting Line Model with
Poisson Arrivals and Exponential Service Times
 Single-Channel Waiting Line Model



Poisson Arrival process with the mean arrival rate 
Exponential Service Times with the service rate 

 Utilization Factor:    / 




This value provides that an arriving unit has to wait because the service facility is in use.
The value of utilization factor should be less than 1.

 Little’s Law:


average number in system = (average arrival rate) (average waiting time in system)

L  W ;

Lq  Wq
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Operating characteristics
1. Probability that no units are in the system:
P  1
0


 1 


2. Probability that an arriving unit has to wait for service:


Pw   
3. Average number of units in the waiting line:

2
Lq 
(  )
4. Average numberofunits in the system:



L  Lq  
  
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Operating characteristics
5. Average waiting time in the system:
1
L  W  W 
 

6. Average waiting time in the waiting line:
1
1

Wq  W  
 
     (   )
1

 or; Lq  Wq  Wq 
(   )
7. Probability of n units in the system:

 n n Pn  ( ) P0   P0

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Operating characteristics for the Burger
Dome Problem
1. Probability that no units are in the system:

P  1
 1  0.75  0.25
0

2. Probability that an arriving unit has to wait for service:
Pw 


 0.75


2
0.752
Lq number of
3. Average   (    )  units .in )  2waiting line: the .25customers
1(1  0 75



0.75
L 

 in the system:
4. Average number of units 3 cos tumers
   (1  0.75)

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Operating characteristics for the Burger
Dome Problem
5. Average waiting time in the system:
1
W 
 4 min
 

6. Average waiting time in the waiting line:
Wq  W 

1



 4  1  3 min

7. Probability of n units in the system:

Pn   n P0  (0.75) n (0.25)

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Manager’s Use of waiting line model
 Operating characteristics are unsatisfactory.
 The average waiting time of 3 minutes before beginning to order appears somewhat long for a business based on fast food.
 The average number of costumers waiting in line (2.25) indicates that something should be done to improve the waiting line operation.
 The probability of seven or more people in system is
0.1335.
Number of Costumers
Probability

0

1

2

3

4

5

7

0.25 0.19 0.14 0.11 0.08 0.06 0.04

7 or more
0.13

 Manager considers alternative designs for improving the waiting line operation.
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Improving the waiting line operation
Possible service rate improvement




Increase the mean service rate by making a creative design or using new technology.
Add service channels so that more costumers can be served simultaneously.

Design alternatives



Increase the mean service rate to 1.25 costumers/min
The second alternative is a topic of the next section.

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Operating Characteristics with the mean service rate of 1.25
Operating Characteristics

Values

Probability of no customers in system
Average Number of costumers in queue
Average Number of costumers in system
Average time in queue
Average time in system
Probability that an arriving costumers has to wait
Probability that seven or more costumers are in system

0.4
0.9
1.5
1.2 min
2.0 min
0.6

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0.028

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3. Multiple-Channel Waiting Line Model with
Poisson Arrivals and Exponential Service Times
 Multiple-Channel Waiting Line Model
 Number of service channels (=k)


Poisson Arrival process with the mean arrival rate 



Exponential Service Times with the service rate 



Total service rate of k channels: k



Utilization factor:    / k  1

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3.1 structure of Two-channel waiting line
 Two-channel waiting line
Server A
Service
Rule

Waiting Line
Costumer
Arrives

Server B

Customer
Leaves

Costumer goes to Next open
Channel

Service System

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Operating characteristics
1. Probability that no units are in the system:
1  n 1  k k
P0  { [ ( ) ]  ( ) (
)}1
k!  k   n  0 n!  k 1

2. Probability that an arriving unit has to wait for service:

Pw  1  P0

3. Average number of units in the waiting line:

( /  ) k 
Lq 
P
2 0
4. Average numberk  units in the system:
(k  1)!( of  )


L  Lq 

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Operating characteristics
5. Average waiting time in the system:
W 

L

;W  Wq 

1



6. Average waiting time in the waiting line:
Lq
1
Wq  W  or ;Wq 



7. Probability of n units in the system:
( /  ) n
Pn 

Pn  k

n!
( /  ) n k! k

( nk )

P0

for n  k

P0

for n  k
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Operating characteristics for the Burger
Dome Problem
1. Probability that no units are in the system:

P  0.4545
0
2. Probability that an arriving unit has to wait for service:

Pw 0 2045
3. Average numberof .units in the waiting line:

Lq  0.1227customers
4. Average number of units in the system:

0.75
L  Lq   0.127 
 0.8727 cos tumers

1
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Operating characteristics
5. Average waiting time in the system:
L

0.8727
W  
 1.1636

0.75

6. Average waiting time in the waiting line:
Lq

0.1227
Wq 

 0.1636 min

0.75

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Comparison 2-channel system with a single channel system
 Operating characteristics
 The average waiting time in system:
W (one)=4 min is reduced to W (two)=1.6336 min.
 The average number of costumers waiting in line:
Lq (one)=2.25 is reduced to Lq (two)=0.1227.
 The average waiting time in queue:
Wq (one)=3 min is reduced to W (two)=0.1636 min.
 The probability that costumer has to wait for service:
Pw (one)=0.75 is reduced to Pw (two)=0.2045
 The probability of n costumers in system
Number of Costumers
Probability

0

1

2

3

4

0.455 0.341 0.128 0.048 0.018

33

5 or more
0.011

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4. Economic Analysis of waiting Line
 Evaluation on operating characteristics of waiting line Model




Waiting line performance goals.
Cost of operating waiting line
Total cost model including costs of waiting and service
Total Cost

Total
Cost
per hour Service Cost

Waiting Cost
Number of Channels (k)
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4.1 Total cost per time period
Definition of Notation






The
The
The
The

waiting cost per time period for each unit = CW service cost per time period for each channel = C S average number of units in system = L number of channels = k

The total cost per time period = TC

TC  CW L  CS k

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4.2 Burger Dome Example
Specify the waiting cost and service cost






The waiting cost of costumer is a direct cost and usually difficult to evaluate.
L
The service cost is easier to determine. It includes the server’s wage, benefits, and any other indirect cost.
Burger Dome estimates the waiting cost and service cost to be $10 and $7 per hour, respectively.

The total cost per time period




Single-channel system
TC  CW L  CS k  10(3)  7(1)  $37 / hour
Two-channel system
TC  CW L  CS k  10(0.8727)  7(2)  $22.73 / hour
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5. Other Waiting Line Model
 Kendall suggested a notation for classifying the wide variety of different waiting line models:





A = the probability distribution for the arrivals
B = the probability distribution for the service time k = the number of channels
Three-symbol Notation: A/B/k

 Notation for the probability distribution




M = Poisson distribution for arrivals and Exponential distribution for service time
D = arrivals or service time for deterministic distribution
G = arrivals or service time for general distribution with known mean and variance

 M/M/1 Model denotes a single-channel waiting line model with Poisson arrivals and exponential service time.
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6. Single-Channel Waiting Line Model with
Poisson Arrivals and Arbitrary Service Times
 Operating characteristics for M/G/1 Model
 the mean arrival rate = 
 the mean service rate = 
 Standard deviation of the service time =



1. Probability that no units are in the system:


P  1
 1 
0

2. Probability that an arriving unit has to wait for service:


Pw   

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Operating characteristics
3. Average number of units in the waiting line:

2 2  ( /  ) 2
Lq 
2(1   /  )

4. Average number of units in the system:

L  Lq 

5. Average waiting time in the system:
W 

L



6. Average waiting time in the waiting line:
Wq  W 

1


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6.1 Hartlage’s Seafood Supply Example
 Operating characteristics for M/G/1
 the mean arrival rate = 0.35 costumes per min
 the mean service rate = 0.5 costumers per min
 Standard deviation of the service time = 1.2 min
L
 1. Probability that no units are in the system:


0.35
P0  1   1 
 0.30

0.50
2. Probability that an arriving unit has to wait for service:


Pw   0.70

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Operating characteristics
3. Average number of units in the waiting line:
2 2  ( /  ) 2 (0.35) 2 (1.2) 2  (0.35 / 0.50) 2
Lq 

 1.1107customers
2(1   /  )
2(1  0.35 / 0.50)

4. Average number of units in the system:

0.35
L  Lq   1.1107 
 1.8170

0.50

5. Average waiting time in the system:
L

1.8170
W  
 5.1733 min

0.35

6. Average waiting time in the waiting line:
Wq  W 

1



 5.1733 

1
 3.1733 min
0.50
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6.2 Constant Service Times
 Operating characteristics for M/D/1
 the mean arrival rate =

 the constant service time:
 0
 Production and manufacturing environment where machine controlled service time is constant.

Average number of units in the waiting line:


In M/G/1, the standard deviation is equal to 0.

( /  ) 2
Lq 
2(1   /  )


The other operating characteristics can be easily found by use of the expressions in M/G/1 model.
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7. Multiple-Channel Model with Poisson Arrivals,
Arbitrary Service Times and No Waiting Line
 System descriptions







The system has k channels.
The arrivals follow a Poisson distribution with the mean service rate 
The service time for each channel may have any probability distribution. The mean service rate  is the same for each channel.
An arrival enters the system only if at least one channel is available.
An arrival when all channels are busy, the arrival is allowed to enter system (blocked).

 Design of telephone and communication system



Arrivals are calls and channels are communication lines available.
When channels are busy, additional calls receive busy signal and are denied access to the system.

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7.1 Operating characteristics for M/G/k Model with Blocked Costumers Cleared
 Operating characteristics
 the mean arrival rate = 
 the mean service rate = 
 The number of channels = k
 The probability that j of the k channels are busy for j =0,
1,2, …, k:

Pj 

( /  ) j / j! k ( / ) i / i!

i 0



What is the probability of arrivals that are blocked and denied access to the system ?

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7.2 Micro-data Software Example
 System Description









Telephone ordering system for its computer software products
Callers place orders with Micro-data by using the company’s 800 telephone number.
Calls to this telephone arrive at average rate 12 calls per hour.
Each Micro-data sales representative handles an average rate 6
L
calls per hour. the company’s 800 telephone number has 3 channels, each operated be a separated representative.
Calls received are automatically transferred to an open line if available. If all the lines are busy, callers are denied to enter the system.
Denied calls are considered as lost sales.

 How many lines are necessary to provide sufficient capacity to handle 90% of calls?
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7.3 Operating characteristics for Micro-data

Software, Inc.
 The probability that j of the k channels are busy:
( /  ) j / j!
Pj  k
( / ) i / i!

i 0

(12 / 6)3 / 3!
 P3 
 0.2105
0
1
2
3
(12 / 6) / 0!(12 / 6) / 1!(12 / 6) / 2!(12 / 6) / 3!

 P4  0.0925
 Average number in the system with k=4:


12
L  (1  Pk )  L  (1  0.0952)  1.8095

6

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7.4 Economic Analysis for Micro-data
Software, Inc.
 Probabilities of Busy Lines for Micro-data 4-Line system
Number of busy Lines

Probability

0
1
2
3
4

0.1429
0.2857
0.2857
0.1905
0.0925

 With 9.52% of the calls blocked and 12 calls per hour, an
8-hour day will have an average of 8(12)(0.0952)=9.1 blocked calls.
 The cost of additional line and additional sales representative should be balanced against the cost of the blocked calls.
47

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8. Waiting Line Model with Finite Calling
Population
 Finite Calling Population



Maximum number of units that seek service is finite.
Depending on the number of units in the waiting line, the mean arrival rate for the system changes.

 Assumptions






The arrivals for each unit follow a Poisson distribution with the mean service rate  .
The service time for each channel has an exponential distribution with the mean service rate  .
The population of units that seek service is finite

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8.1 Operating characteristics for the M/M/1
Model with a finite Calling Population
 Operating characteristics
 the mean arrival rate for each unit = 
 the mean service rate = 
 the size of population = N
1. Probability that no units are in the system:

N!
 n 1
P0  [
( ) ] n 0 ( N  n)! 
N

2. Probability that an arriving unit has to wait for service:

Pw  1  P0
3. Average number of units in the waiting line:


Lq  N 
(1  P0 )

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Operating characteristics
4. Average number of units in the system:

L  Lq  (1  P0 )

5. Average waiting time in the system:
W  Wq 

1


6. Average waiting time in the waiting line:
Wq 

Lq
( N  L )

7. The probability of n units in the system
N!
 n
Pn 
( ) P0
( N  n)! 

n  0,1,, N
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8.2 Kolkmeyer Manufacturing Company
Example
 System Description









This company has 6 identical machines (finite population).
Each machine operates an average of 20 hours between breakdowns. So Mean arrival rate for repair service is 0.05 per hour.
Machine breakdowns is described as a Poisson arrival process.
One repairman provides repair service.
Exponential repair service time with the mean service rate of 0.5 machine per hour.
Calls received are automatically transferred to an open line if available.  One of the primary applications of this model is a machine repair problem.

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Operating characteristics
 mean arrival rate for each unit = 0.05
 mean service rate = 0.50
 size of population = 6
1. Probability that no units are in the system:
N

6!
0.05 n 1
P0  [
(
) ]  0.4845 n 0 (6  n)! 0.5

2. Probability that an arriving unit has to wait for service:

Pw  1  0.4845  0.5155
3. Average number of units in the waiting line:

0.05  0.5
Lq  6 
(1  0.4845)  0.3297machine
0.05
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Operating characteristics
4. Average number of units in the system:

L  0.3295  (1  0.4845)  0.8451machine

5. Average waiting time in the system:
1
W  1.279  
 3.279
0.5

6. Average waiting time in the waiting line:
0.3297
Wq 
 1.279hours
(6  0.8451)(0.05)

7. The probability of n units in the system
6!
0.05 n
Pn 
(
) (0.4845)
(6  n)! 0.5
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7.3 Economic Analysis for Kolkmeyer
Manufacturing Company
 Operating characteristics with 1 and 2 repairmen
One
Po
Pw
L
Lq
W
Wq

Two

0.4858
0.5142
0.8451 machine
0.3297 machine
3.279 hour
1.279 hour

0.5602
0.1036
0.5661 machine
0.0227 machine
2.0834 hour
0.0834 hour

 Compare the cost of machine downtime with the cost of the repair personnel.
 Management can determine whether the improved service of the 2 channel system is cost effective or not.
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9. Estimations of Arrival Process and Service
Time Distribution
 Number of trucks for one hour
Number of trucks

Frequency

Probability

0
1
2
3
4
5
6
7
8

1
7
14
20
20
16
12
7
3

0.01
0.07
0.14
0.20
0.20
0.16
0.12
0.07
0.03

Total

100

1.00

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9.1 Estimation for Poisson Arrival
Distribution

ˆ
   (0 1  1 7  2 14    8  3) / 100  4trucks / hour
 Probability of Poisson Distribution
Number of trucks for 1 hour (X)

Theoretical
Probability

Observed
Probability

0
1
2
3
4
5
6
7
8
9
10

0.0183
0.0733
0.1465
0.1954
0.1953
0.1563
0.1042
0.0596
0.0297
0.0133
0.0053

0.01
0.07
0.14
0.20
0.20
0.16
0.12
0.07
0.03
0.00
0.00

Total

1.00

1.00
56

4 x e 4 f ( x) 
,
x! x  0,1,2,3, 

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Fitness Test for Poisson Distribution
A test how well an observed distribution is fitted to an assumed theoretical distribution is referred to as a fitness test.


Use

 2 test or Kolmogorov-Smirnov test.

f(x)
0.2

Observed
Distribution

0.15

Theoretical
Distribution

0.1

0.05

X

0.0

0 1 2 3 4 5 6 7 8 9 10
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Fitness Test of Distribution
 Fitness of Distribution
For a discrete case, if difference between the observed probability and theoretical probability at each point is less than a prescribed value, then the observed distribution is well fitted to the theoretical distribution.
For a continuous case, if difference between the observed probability and theoretical probability for each pre-determined interval is less than a prescribed value, then the observed distribution is well fitted to the theoretical distribution.





 

2

k

Test:

2 

( N i  nP ) 2

i i 1

nP i   2 ( k  1)

where n= total number of data
Ni = number of data included in interval i
Pi = theoretical probability for interval i

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9.2 Estimation for service Time
Distribution
 Service time has various distributions.
 We may use the observed distribution directly or theoretical distribution to describe the service process.

 In waiting line analysis, we apply the theoretical distribution to describe the service process. For convenience of analysis, an exponential service time distribution is assumed.
 For instance, a mean service time is 12 min, the mean service rate is 5 units per hour and exponential service time distribution is f (t )  5e 5t.

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Assignment
Solve Problems 10, 14, 22, and 31

Presentation



Case Problem 1: Regional Airlines
Case Problem 2: Office Equipment, Inc.

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61

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