Waiting Lines Queues. Queuing Theory Managers use queuing models to be more efficient in providing...
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Queuing Theory
• Managers use queuing models to be more efficient in providing customer service.
• Models measure average waiting times and average length of waiting lines.
Historical Roots
• Agner Krarup Erlang, a Danish engineer who worked for the Copenhagen Telephone Exchange, published the first paper on queueing theory in 1909.
• David G. Kendall introduced an A/B/C queueing notation in 1953.
Three queuing disciplines used in Telephone Networks
• First In First Out – This principle states that customers are served one at a time and that the customer that has been waiting the longest is served first.[5]
• Last In First Out – This principle also serves customers one at a time, however the customer with the shortest waiting time will be served first.[5]
• Processor Sharing – Customers are served equally. Network capacity is shared between customers and they all effectively experience the same delay
Source: Wikipedia.org
LIFO“Last in First Out”
Elevators are a circumstance where this occurs.
Single-server Single-stage Queue
Service Facility
CustomersIn queue
Arrival Stream
Waiting for the Newest ????????
Single-server Multiple-stage Queue
Service Facility
CustomersIn queue
Pharmacy Conveyor System >>>>>
Little's Theorem
• Little's theorem: L = / • The average number of customers (N) can
be determined from the following equation:
•
• Here lambda () is the average customer arrival rate and mu () is the average service time for a customer.
Queuing system state probabilities(Basic Model)
P0
= Pn
= 1 -
1, 2, 3,…
Probability distribution for the number of customers in the system
( )
nP0
n =
=
Queuing Formulas(Basic Model)
-
=
Average # of Customers in the system
Average Customer Time spent in the system
Average # of Customers waiting (length of line)
Average Customers waiting time
Server Utilization Factor
L =
W =
Wq =Lq
L = -
( - )
=
Lq = ( - )
2
1
Phlebotomy Room Example
• A queuing system for blood draws.
• An average of 25 patients arrive for a blood draw each hour.
• One full-time (very experienced) phlebotomist can take one patient every two minutes, thus 30 draws per hour can be done.
Queuing Formulas(Basic Model)
-
=
Average # of Patients in the system:
Average Patient Time spent in the system:
Average # of Patients waiting:
Average Patients waiting time:
Server Utilization Factor:
L =
W =
Wq =
Lq
-
( - )
=
Lq = ( - )
2
1
= 25 Pts per hour = 30 Pts per hour
25
30 - 25=
1
=
30 - 25
30 (30 - 25)
(25)2
= =25
6=
1
64
30 (30 - 25)
25=
1
=
5
1
6=
hour
hour
customers
= 5 customers
=25
30=
5
6
= The phlebotomist is busy five-sixths of the time.
The system state probabilities
P0
P1
= 1 -
( )
1P0 =
= 1 - 2530
P2( )
2P0 =
.1667=
=
=
2530 (.1667)( )
2530( )
2(.1667)
= .1389
= .1158
This formula provides the probability that n (0, 1, 2, 3, …) patients will be in the blood drawing room. If you add the individual probabilities for values of n cumulatively you would find 54 in the number of patients where all probabilities of n total 1.
Multiple server models
• Uses same notation as basic model but different formulas.
• Formulas are based on FIFO discipline.
• The customer at the head to waiting line proceeds to the first server.
• S = Number of service channels
Queuing system state probabilities(Multiple Servers)
P0
Pn
= 1 ( /)n
n!
P0
If n > S=
[S-1
n=0+
( /)s
S! ( )11 - /S
If 0 < n < S( /)n
n!
( /)n
S!Sn-s
Phlebotomy Room Examplewith a second Phlebotomist
Multiple Servers
• A 2 service channel queuing system for blood draws.
• An average of 50 patients arrive for a blood draw each hour.
• Two full-time phlebotomists can take one patient each every two minutes, thus 60 draws per hour can be done.
The Probability that there are no patients in the system.
P0 = 1 ( /)0
0! +( /)1
1! ( )11 - /2+
( /)2
2!
= 1 (50 /60)0
0! +(50 /60)1
1! ( )11 - 50/2(60)+
(50 /60)2
2!
= 50 Pts per hour = 60 Pts per hourS = 2 service channels
= 1 1 + .833 ( )11 - .416+
(.833)2
2! ]
[ ]
][
[= 1 /[1 + .833 + .594] = 1 / 2.427 = .412
Queuing Formulas(Multiple Servers)
S
Average # of Patients in the system:
Average Patient Time spent in the system:
Average Patients waiting time:
Server Utilization Factor:
L =
W =
Wq =Lq
=
1
Average # of Patients waiting: Lq =S! (1 - /S)2
(/)2 (/S)
Wq +
Lq +
P0
= 50 Pts per hour = 60 Pts per hourS = 2 service channels
Queuing Formulas(Multiple Servers)
S
Average # of Patients in the system:
Average Patient Time spent in the system:
Average Patients waiting time:
Server Utilization Factor:
L =
W =
Wq =Lq
=
1
Average # of Patients waiting: Lq =2! (1 - 50/2(60))2
(50/60)2 (50/2(60))
Wq +
Lq +
(.412) = .175
= 50 Pts per hour = 60 Pts per hourS = 2 service channels
.175
50 = .0035 = 12.6 seconds=
= .0035 + 1/60 = .0035+.016 = .0195
= .0195 hour = 1.17 minutes
= .175 + 50/60 = .175 + .833 = 1.008 pts
= 50/ 2(60) = 0.416
Two Fax machines example
• An organization is considering renting 2 fax machines.• The 2006 model can send 100 faxes per minute.• However, loading the originals and entering the receiving
phone number slows the process. The vendor indicates the effective service rate is .5 job per minute.
• The demand for fax service in the organization is projected at 3 jobs every 5 minutes (.6 job per minute)
• S = 2 service channels = .5 job per minute = .6 job per minute
The Probability that there are no patients in the system.
P0 = 1 ( /)0
0! +( /)1
1! ( )11 - /2+
( /)2
2!
= 1 (.6 /.5)0
0! +(.6 /.5)1
1! ( )11 - .6/2(.5)+
(.6 /.5)2
2!
= .5 job per minute = .6 job per minuteS = 2 service channels
= 1 1 + 1.2 ( )1 1 - .6+
(1.2)2
2! ]
[ ]
][
[= 1 /[1 + 1.2 + 1.8] = 1 / 4 = . 25
Queuing Formulas(Multiple Servers)
Average # of jobs in the fax room:
Average Job Time spent in the fax room:
Average job waiting time per job:
L =
W =
Wq =Lq
1
Average # of Jobs waiting: Lq =S! (1 - /S)2
(/)2 (/S)
Wq +
Lq +
P0
= .5 job per minute = .6 job per minuteS = 2 service channels
=2! [1 - .6/2(.5)]2
(.6/.5)2 [.6/2(.5)] (. 25) = .68 job
.68
.6 = 1.13 minutes=
= 1.13 +1
.5
1= 3.13 minutes
= .68 +.5
.6= 1.88 jobs
Establishing a queuing system costConsider the average hourly cost of operating two rented fax machines. Each job is personally processed by the user. The average hourly payroll cost is $10. Machine rental is a straight $.05 per copy, and an average job involves 12 copies. The average number of jobs per hour is:
.6 X 60 = 36 jobsEach employee spends an average of W = 3.13 minutes:
3.13/60 = .0522 hourAverage cost of labor lost making copies:
$10 X 36 X .0522 = $18.79Hourly rental cost :
$.05 X 12 X 36 = $21.60Total hourly average cost of operating two machines:
$18.79 (labor cost) + $21.60 (equipment rental) = $40.39
Two compared to one fax that is twice as fast.
=
Average # of jobs waiting:
Average Patients waiting time: Wq =
Lq
( - )
Lq = ( - )
2
1(1 - .6)
(.6)2
= =.36
.4= .9
.6
.9= 1.5= Minutes (2008 model)
Job (2008 model)
One would would think that one server twice as fast would produce identical results to two servers.THIS IS NOT TRUE.If so we would not need a different model for multi-channel queues.If a 2008 model fax is twice as fast as the 2006 model is there a difference?If = 1 job per minute.
Average job time spent in fax room: W = -
1 1
=
1 - .6 = 2.5 Minutes (2008 model)
Results in a smaller hourly labor cost. $10 X 36 X = $15.002.560
The new model might rent for a little more than the older model, but would still be cheaper than two 2006 models.
Single Server Model w/a finite queueA waiting line of limited length is called a finite queue.e.g., Hospital Emergency room with a limited number of beds.If the number of patients reaches a given point additional patients are diverted.The patient (customer) who does not enter the system does not return. There are not limits on the number of patients waiting for service
P0 = 1 - /1 - ( /)M + 1
Pn = ( /)n P0 for 1 < n < M
Probabilities for # of patients in the system
Average # of patients in the system1 - /
/ L =
1 - ( /)M+1
(M + 1)( / )M+1
Average length of waiting line Lq = L – (1 – Po)
Average patient waiting times Wq =(1 – PM)
LqW =
(1 – PM)
L