1Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Terminology: The characteristics of a queuing system is captured by five parameters: Arrival pattern Service pattern Number of server Restriction on queue capacity The queue discipline
Terminology and Classification of Waiting Lines
2Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
M/M/1 Exponential interarrival times Exponential service times There is one server. No capacity limit
M/G/12/23 Exponential interarrival times General service times 12 servers Queue capacity is 23
Terminology and Classification of Waiting Lines
3Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Coefficient of Variations
Interarrival Time or Processing Time distribution General Poisson Exponential Constant
Mean Interarrival Time or Processiong Time m m m m
Standard Deviation of interarrival or Processing Time s m m 0
Coeffi cient of Varriation of Interarrival or Processing Time s/m 1 1 0
4Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processor (with the
server)? On average how many customers are with the server? On average how many customers are in the system? On average how long a customer stay in the system?
Problem 1: M/M/1 Performance Evaluation
5Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
R = 6 customers per hourRp =1/5 customer per minute, or 60(1/5) = 12/hour r = R/Rp = 6/12 = 0.5On average how many customers are in the waiting
line?
M/M/1 Performance Evaluation
21
22)1(2pi
c
i
CCI
2
11
5.01
22)11(25.0
iI 5.05.0
5.0 2
iI
6Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
GAP Example
ii IRT 5.06 iT hour 6/5.0iT
minutes 5 )6/5.0(60 iT
How long a customer stays in the line?
How long a customer stays in the processor (with the server)?
minutes 5 pT
On average how many customers are with the server?
?pI 5.0 ely,Alternativ pI5.0)10/1)(5( TpRIp
7Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
M/M/1 Performance Evaluation
min 1055 pi TTT
?I 15.05.0 pi III
On average how many customers are in the system?
On average how long a customer stay in the system?
8Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Problem 2: M/M/1 Performance Evaluation
What if the arrival rate is 11 per hour?
08.10
1211
1
1211
1
2
2
iI
ii IRT 08.1011 iT
min 55or hours 91667.011/08.10 iT
12
11
Rp
R
9Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
As the utilization rate increases to 1 (100%) the number of customers in line (system) and the waiting time in line (in system) is increasing exponentially.
M/M/1 Performance Evaluation
c R Rp Ii Ti (min) Tp T Ip I1 6 12 0.50 0.50 5 5 10 0.50 11 7 12 0.58 0.82 7 5 12 0.58 1.41 8 12 0.67 1.33 10 5 15 0.67 21 9 12 0.75 2.25 15 5 20 0.75 31 10 12 0.83 4.17 25 5 30 0.83 51 11 12 0.92 10.08 55 5 60 0.92 111 11.1 12 0.93 11.41 61.7 5 66.7 0.93 12.331 11.2 12 0.93 13.07 70 5 75 0.93 141 11.3 12 0.94 15.20 80.7 5 85.7 0.94 16.141 11.4 12 0.95 18.05 95 5 100 0.95 191 11.5 12 0.96 22.04 115 5 120 0.96 231 11.6 12 0.97 28.03 145 5 150 0.97 291 11.7 12 0.98 38.03 195 5 200 0.98 391 11.9 12 0.99 118.01 595 5 600 0.99 119
10Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute.
On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processors (with the
servers)? On average how many customers are with the servers? On average how many customers are in the system ? On average how long a customer stay in the system ?
Problem 3: M/G/c
11Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Arrival rate: R = 10 per hourAverage interarrival time: Ta = 1/R = 1/10 hr = 6 minStandard deviation of interarrival time: SaService rate per server: 12 per hourAverage service time: Tp = 1/12 hours = 5 minStandard deviation of service time: Sp = 1 min
Coefficient of variation for interarrivals : Ci= Sa /Ta = 1
Coefficient of variation for services: Cp = Sp /Tp = 1/5 =0.2
Number of servers: c =2Rp = c/Tp = 2/(1/12) = 24 per hourρ = R/Rp = 10/24 = 0.42
The Key Information
12Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
M/G/2
107.02
2.01
42.01
42.0
21
22)12(222)1(2
pic
i
CCI
ii IRT 107.010 iT
minutes 0.6hour 0107.0 iT
On average how many customers are in the waiting line?
How long a customer stays in the line?
How long a customer stays in the processors (with the servers)?
Average service time: Tp = 1/12 hours = 5 min
13Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
M/G/2
?T min 6.556.0 T
On average how many customers are with the servers?
?pI
84.0)42.0(2 ely,Alternativ cI p
84.0)10)(12/1( TpRIp
On average how many customers are in the system ?
?I 95.084.0107.0 pi III
On average how long a customer stay in the system ?
14Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Approximation formula gives exact answers for M/M/1 system.
Approximation formula provide good approximations for M/M/2 system.
Comment on General Formula
15Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)?
Problem 4: M/M/c Example
91.0220
200 1iC 1pC
16Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
M/M/c Example
21
22)1(2pi
c
i
CCI
89.62
11
91.01
91.0 )12(2
iI
ii RTI iT20089.6
min 2.1or hour 0345.0iT
17Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Suppose the service time is a constant What is the answer of the previous question?
In this case
Problem 5: M/D/c Example
0pC
21
22)1(2pi
c
i
CCI
45.3
2
01
91.01
91.0 12*2
iI
ii RTI iT20045.3
min 1.03or hour 017.0iT
18Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Effect of Pooling
Ri
Server 1
Queue
Server 2
Ri
Server 2Queue 2
Ri/2
Server 1Queue 1
Ri/2
Ri =R= 10/minTp = 5 secs Interarrival time PoissonService time exponential
19Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Effect of Pooling : 2M/M/1
Ri
Server 2Queue 2
Ri/2
Server 1Queue 1
Ri/2
Ri /2 = R= 5/minTp = 5 secs C = 1 Rp = 12 /min= 5/12= 0.417
3.0417.01
417.0 )11(2
iI
ii RTI iT53.0 sec 3.6or min 06.0iT
sec 6.856.3 pi TTT
20Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Comparison of 2M/M/1 with M/M/2
Ri
Server 2Queue 2
Ri/2
Server 1Queue 1
Ri/2
3.0417.01
417.0 )11(2
iI
sec 6.856.3 pi TTT
6.0)3.0(22 iI
Server 1
Queue
Server 2
Ri
21Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Effect of Pooling: M/M/2
Server 1
Queue
Server 2
Ri
Ri =R= 10/minTp = 5 secs C = 2Rp = 24 /min= 10/24= 0.417 AS BEFORE for each processor
2.0417.01
417.0 )12(2
iI
ii RTI iT102.0 sec 1.2or min 02.0iT
sec 2.652.1 pi TTT
22Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Under Design A, We have Ri = 10/2 = 5 per minute, and TP= 5 seconds, c
=1, we arrive at a total flow time of 8.6 seconds Under Design B,
We have Ri =10 per minute, TP= 5 seconds, c=2, we arrive at a total flow time of 6.2 seconds
So why is Design B better than A? Design A the waiting time of customer is dependent on
the processing time of those ahead in the queue Design B, the waiting time of customer is only partially
dependent on each preceding customer’s processing time
Combining queues reduces variability and leads to reduce waiting times
Effect of Pooling
23Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
1. Decrease variability in customer inter-arrival and processing times.
2. Decrease capacity utilization.3. Synchronize available capacity with demand.
Performance Improvement Levers
24Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
Customers arrival are hard to control Scheduling, reservations, appointments, etc….
Variability in processing time Increased training and standardization processes Lower employee turnover rate more experienced work
force Limit product variety, increase commonality of parts
1. Variability Reduction Levers
25Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines3
If the capacity utilization can be decreased, there will also be a decrease in delays and queues.
Since ρ=R/Rp, to decrease capacity utilization there are two options Manage Arrivals: Decrease inflow rate Ri
Manage Capacity: Increase processing rate Rp
Managing Arrivals Better scheduling, price differentials, alternative
services Managing Capacity
Increase scale of the process (the number of servers) Increase speed of the process (lower processing time)
2. Capacity Utilization Levers