1Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety inventory protects against stockouts due to variability of arrival time and processing time. Inventory also permits economies of scale.
Make-to-order (MTO) operations. Each order is specific, cannot be stored in advance. Ex. banks, restaurants, retail checkout counters, airline reservation, hospitals , repair shops, call centres. Production systems also try to follow Dell Computer model. We needs to maintain sufficient capacity to deal with uncertainty in both arrival and processing time. Safety Capacity vs. Safety Inventory.
Make to Stock (MTS) vs. Make to Order (MTO)
2Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
A Call Centre
Sales RepsProcessing
Calls
(Service Process)
Incoming Calls(Customer Arrivals) Calls
on Hold(Service Inventory)
Answered Calls(Customer Departures)
Blocked Calls(Due to busy signal)
Abandoned Calls(Due to long waits)
Calls In Process(Due to long waits)
The Call Centre Process
3Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
The time of the arrival of an order is not known ahead of time. It is a random variable with estimated mean and standard deviation. The time of the next telephone call is not known.
The service time is not known (precisely) ahead of time. It is a random variable with estimated mean and standard deviation. The time a customers spends on the web page of
amazon.com is not precisely known. The time a customer spends speaking with the teller in
the bank is unknown
Characteristics of Waiting Lines (Queuing Systems)
4Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Article: The Psychology of Waiting Lines
Unoccupied time feels longer than occupied time. Pre-process waits feels longer than in-process waits. Anxiety makes waits seem longer. Uncertain waits are longer than known, finite waits. Unexplained waits are longer than explained waits. Unfair waits are longer than equitable waits. The more valuable the service, the longer I will wait. Solo waiting feels longer than group waiting.
5Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Variability in arrival time and service time leads to Idleness of resources Waiting time of customers (orders) to be processed
We are interested in evaluating two measures: Average waiting time of flow units. Average waiting
time in the waiting line and in the system (Waiting line + Processor).
Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).
Let us first look at the Servers or Processors
Characteristics of Queuing Systems
6Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Tp : Processing time. Tp units of time. Ex. on average it takes 5 minutes to
serve a customer. Rp : processing rate. Rp flow units are handled per unit of time. If Tp is 5 minutes. Compute Rp. Rp= 1/5 per minute, or 60/5 = 12 per hour.
AVERAGE Processing Time TpAVERAGE Processing Rate Rp
7Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Tp : processing time. Rp : processing rate. What is the relationship between Rp and Tp?If we have one resource Rp = 1/Tp
What is the relationship between Rp and Tp when we have more than one resource; We have c recourses
Rp = c/Tp
Each customer always spends Tp unites of time with the server
More than One Server; c Servers
8Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Tp = 5 minutes. Processing time is 5 minute. Each customer on average is with the server for 5 minutes.
c = 3, we have three servers.Processing rate of each server is 1/5 customers per
minute, or 12 customer per hour.Rp is the processing rate of all three servers.Rp = c/Tp Rp = 3/5 customers/minute, or 36 customers/hour.
Average Processing Rate of c Servers
9Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Inter-arrival Time (Ta) and Arrival Rate (Ra)
Ta : customer inter-arrival time.
On average each 10 minutes one customer arrives. Ra: customer arrival (inflow) rate.
What is the relationship between Ta and RaTa = every ten minutes one customer arrivesHow many customers in a minute? 1/10; Ra= 1/Ta= 1/10Ra = 1/10 customers per min; 6 customers per hourRa= 1/Ta
10Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Throughput = Min (Ri,Rp)
Ra MUST ALWAYS <= Rp.We will show later that even Ra=Rp is not possible.Incoming rate must be less than processing rate.Throughput = Flow Rate R = Min (Ra, Rp) .Stable Process = Ra< Rp R = RaSafety Capacity Rs = Rp – Ra
11Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
What is the waiting time in the servers (processors)?Throughput?
Buffer (waiting line) and Processors (Servers)
Flow time T = Ti + Tp
Inventory I = Ii + Ip
Ti: waiting time in the inflow bufferIi: number of customers in the inflow buffer
12Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
U = UtilizationU =inflow rate / processing rate U = throughout / process capacityU = R/ Rp < 1Safety Capacity = Rp – R
For example , R = 6 per hour, processing time for a single server is 6 min Rp= 12 per hour,
U = R/ Rp = 6/12 = 0.5Safety Capacity = Rp – R = 12-6 = 6
Utilization is Always Less than 1
13Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Given a single server. And a utilization of U= 0.5How many flow units are in the server ?
Given the Utilization, How Many Flow Units are in the Processor(s)
U = 0.5 means 50% of time there is 1 flow unit in the server 50% of time there is 0 flow unit in the server0.5 1 + 0.5 0 = 0.5Average Inventory in the server is equal to utilizationIp= 1U = U
14Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Given the Utilization, How Many Flow Units are in the Processor(s) Given 2 servers. And a utilization of U = 0.5How many flow units are in the servers ?
U = 0.5 means 50% of time there is 1 flow unit in each server 50% of time there is 0 flow unit in each server0.5 1 + 0.5 0 = 0.5 flow unit in each serverAverage Inventory in the server is equal to utilization timesthe number of servers Ip= 2U = cU
15Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Processing time: Tp, Ex. Tp = 5 minutesNumber of servers: c, Ex. c=3Tp is also waiting time in the server, no mater one server
or c servers. Tp in this example is always 5 min.Processing rate Rp= c/Tp. Ex. Rp =3/5 per min; 36/hrUtilization: U. Ex. U = 0.8 in our exampleNumber of the flow units in all servers, Ip = cUIn our example, Ip = 3 0.8 = 2.4Can we compute R? TR = ITp R = cU R = cU/Tp5 R = 2.4 R = 0.48 flow units per minute or 28.8 / hr We learned it without looking at any formula
What We Have Learned Without Looking for any Formula
16Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Variability in arrival time and service time leads to Idleness of resources Waiting time of customers (orders) to be processed
We are interested in evaluating two measures: Average waiting time of flow units. Average waiting
time in the waiting line and in the system (Waiting line + Processor).
Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).
Characteristics of Queuing Systems
17Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Two key drivers of process performance are Utilization and Variability.
The higher the utilization the longer the waiting line. High capacity utilization U= R/Rp or low safety
capacity Rs =R – Rp, due to High inflow rate R Low processing rate Rp = c / Tp, which may be due
to small-scale c and/or slow speed 1 /Tp The higher the variability, the longer the waiting line.
Utilization and Variability
18Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Variability in the interarrival time and processing time can be measured using standard deviation. Higher standard deviation means greater variability.
Coefficient of Variation: the ratio of the standard deviation of interarrival time (or processing time) to the mean.
Ca = coefficient of variation for interarrival times Cp = coefficient of variation for processing times
Drivers of Process Performance
19Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Now Let’s Look at the Rest of the System; The Little’s Law Applies Everywhere
Flow time T Ti + Tp
Inventory I = Ii + Ip
R
I = R T
R = I/T = Ii/Ti = Ip/Tp
Ii = R Ti Ip = R Tp
We have already learned Rp = c/TpU= R/Rp = Ip/c
20Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Operational Performance Measures
Flow time T = Ti + Tp
Inventory I = Ii + Ip
Ti: waiting time in the inflow buffer = ?
Ii: number of customers waiting in the inflow buffer =?A set of excel sheet are provided at the end of this
lecture. We relay on the approximation formula, given shortly,
and our understanding of the waiting line logics.We also provide an exact table for a special case.
21Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
U= R /Rp,where Rp = c / Tp
Ca and Cp are the Coefficients of Variation
Standard Deviation/Mean of the inter-arrival or processing times (assumed independent)
The Queue Length APPROXIMATION Formula
U1U 1)2(c
Utilization effect U-part
Variability effect
V-part
2
22pa CC
iI
22Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
This part captures the capacity utilization effect. It shows that queue length increases rapidly as U approaches 1.
Factors affecting Queue Length
U1U 1)2(c
iI
This part captures the variability effect. The queue length increases as the variability in interarrival and processing times increases.
Even if the processing capacity is not fully utilized, whenever there is variability in arrival or in processing times, queues will build up and customers will have to wait.
2
22pa CC
23Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Utilization – Variability - Delay Curve
VariabilityIncreases
Average Time in System T
Utilization (U) U100%
Tp
24Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
If inter-arrival and processing times are constant, queues will build up if and only if the arrival rate is greater than the processing rate.
If there is (unsynchronized) variability in inter-arrival and/or processing times, queues will build up even if the average arrival rate is less than the average processing rate.
If variability in interarrival and processing times can be synchronized (correlated), queues and waiting times will be reduced.
Lessons Learned
25Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Coefficient of Variations for Alternative Distributions
Interarrival Time or Processing Time distribution General (G)
Poisson (M)
Exponential (M)
Constant (D)
Mean Interarrival Time or Processiong Time Tp (or Ta) Tp (or Ta) Tp (or Ta) Tp (or Ta)
Standard Deviation of interarrival or Processing Time Sp (or Sa) Tp (or Ta) Tp (or Ta) 0
Coeffi cient of Varriation of Interarrival or Processing Time Sp/(Tp (or Sa/Ta) 1 1 0
26Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Exact Ii for Poisson Arrival Rate, Exponential Service Time
27Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
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.
M/M/1; Poisson arrival rate, Exponential service times, one server, No capacity limit.
M/G/12/23; Poisson arrival rate, General service times, 12 servers, Queue capacity is 23
Terminology and Classification of Waiting Lines
28Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
=AVERAGE () Avg. interarrival time = 6 min.
Ra = 1/6 arrivals / min.
=STDEV() Std. Deviation = 3.94
Ca = 3.94/6 = 0.66
C2a = (0.66)2 = 0.4312
Example; Coefficient of Variation of Interarrival TimeA sample of 10 observations on Interarrival times in
minutes 10,10,2,10,1,3,7,9, 2, 6 minutes.
29Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Tp= 5 minutes
Rp = 1/5 processes/min.
Std. Deviation = 2.83
Cp = 2.83/5 = 0.57
C2p = (0.57)2 = 0.3204
Example: Coefficient of Variation of Processing TimeA sample of 10 observations on Processing times in
minutes 7,1,7 2,8,7,4,8,5, 1 minutes.
30Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Ra=1/6 < RP =1/5 R = Ra
U= R/ RP = (1/6)/(1/5) = 0.83
With c = 1, the average number of passengers in queue is as follows:
Example: Utilization and Safety Capacity
On average 1.56 passengers waiting in line, even though safety capacity is Rs= RP - Ra= 1/5 - 1/6 = 1/30 passenger per minute, or 2 per hour.
Ii = [(0.832)/(1-0.83)] ×[(0.662+0.572)/2] = 1.56
U1U 1)2(c
2
22pa CC
iI 0.831
0.831)2(1
2
57.066.0 22
31Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Waiting time in the line?Ti=Ii/R = (1.56)(6) = 9.4 minutes.Waiting time in the system? T = Ti+TpSince Tp= 5 T = Ti+ Tp= 14.4 minutesTotal number of passengers in the process is: I = RT = (1/6) (14.4) = 2.4Alternatively, 1.56 in the buffer. How many in the process?I = 1.56 + 0.83 = 2. 39
Example: Other Performance Measures
32Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Compute R, Rp and U: Ta= 6 min, Tp = 5 min, c=2R = Ra= 1/6 per minuteProcessing rate for one processor 1/5 for 2 processorsRp = 2/5U = R/Rp = (1/6)/(2/5) = 5/12 = 0.417
Example: Now suppose we have two servers.
On average 0.076 passengers waiting in line.safety capacity is Rs= RP - Ra= 2/5 - 1/6 = 7/30
passenger per minute, or 14 passengers per hour
U1U 1)2(c
2
22pa CC
iI 0.4171
0.4171)2(2
2
57.066.0 22
33Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Other performance measures: Ti=Ii/R = (0.076)(6) = 0.46 minutesCompute T? T = Ti+TpSince TP = 5 T = Ti + Tp= 0.46+5 = 5.46 minutes
Total number of passengers in the process is: I= 0.08 in the buffer and 0.417 in the process. I = 0.076 + 2(0.417) = 0.91
Other Performance Measures for Two Servers
c ρ Rs Ii Ti T I
1 0.83 0.03 1.56 9.38 14.38 2.4
2 0.417 0.23 0.077 0.46 5.46 0.91
34Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Processing Time, Waiting Time; Long Waiting Line
Increasing CapacityPolling
Ra =R= 10/hourTp = 5 minutesInterarrival time PoissonService time exponential
Ra = 10/hour
35Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
More Servers + Specialization
Ra/2 = 10/hour
Ra/2 = 10/hour
36Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Polling: Lower Waiting Time, Longer Processing Time (Perhaps)
Ra = 10/hour
37Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Effect of Pooling
Ra
Server 1
Queue
Server 2
Ra
Server 2Queue 2
Ra/2
Server 1Queue 1
Ra/2
Ra =R= 10/hourTp = 5 minutesInterarrival time PoissonService time exponential
38Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Effect of Pooling : 2M/M/1
Ra
Server 2Queue 2
Ra/2
Server 1Queue 1
Ra/2
Ra/2 = R= 5/hourTp = 5 minutes c = 1 Rp = 12 / hourU= 5/12 = 0.417
3.0417.01
417.0 )11(2
Ii
RTiIi Ti53.0 min 3.6or hour 06.0Ti
min 6.856.3 TpTiT
39Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Comparison of 2M/M/1 with M/M/2
Ra
Server 2Queue 2
Ra/2
Server 1Queue 1
Ra/2
3.0417.01
417.0 )11(2
Ii
min 6.856.3 TpTiT
6.0)3.0(22 Ii
Server 1
Queue
Server 2
Ra
40Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Effect of Pooling: M/M/2
Server 1
Queue
Server 2
Ra
Ra=R= 10/hourTp = 5 minutes c = 2Rp = 24 /hourU= 10/24U= 0.417 AS BEFORE for each processor
2.0417.01
417.0 )12(2
Ii
RTiIi Ti102.0 min 1.2or hour 02.0Ti
min 2.652.1 TpTiT
41Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Under Design A, We have Ra = 10/2 = 5 per hour, and Tp = 5 minutes, c
=1, we arrive at a total flow time of 8.6 minutes Under Design B,
We have Ra =10 per hour, Tp= 5 minutes, c=2, we arrive at a total flow time of 6.2 minutes
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 partially
dependent on each preceding customer’s processing time
Combining queues reduces variability and leads to reduce waiting times
Effect of Pooling
42Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Financial Performance Measures Sales: Throughput Rate Cost: Capacity utilization, Number in queue / in
system Customer service: Waiting Time in queue /in system
Performance Improvement Levers Decrease variability in customer inter-arrival and
processing times. Decrease capacity utilization. Synchronize available capacity with demand.
Performance Measures
43Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
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
44Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
If the capacity utilization can be decreased, there will also be a decrease in delays and queues.
Since U=R/Rp, to decrease capacity utilization there are two options Manage Arrivals: Decrease inflow rate Ra 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
45Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Capacity Adjustment Strategies Personnel shifts, cross training, flexible resources Workforce planning & season variability Synchronizing of inputs and outputs, Better
scheduling
3. Synchronizing Capacity with Demand
46Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
The M/M/c Model EXACT Formulas
Model 4 (M/M/s Queue):Multiple servers, Infinite population, Poisson arrival, FCFS, Exponential service time, Unlimited waiting roomYellow cells need user inputed values
InputsUnit of time hourArrival rate (lambda) 16 customers per hourService rate (mu) 20 customers per hourNumber of identical servers (s) 2 servers
OutputsDirect outputs from inputsMean time between arrivals 0.063 hourMean time per service 0.05 hourTraffic intensity 0.4
Summary measuresAverage utilization rate of server 40.0%Average number of customers waiting in line (Lq) 0.15238 customersAverage number of customers in system (L) 0.95238 customersAverage time waiting in line (Wq) 0.00952 hourAverage time in system (W) 0.05952 hourProbability of no customers in system (P0) 0.42857 (this is the probability of empty system)Probability that all servers are busy 22.9% (this is also the "percentage who wait in queue")Probability that at least one server is idle 77.1% (this is also the "percentage who don't wait in queue")
Distribution of number of customers in systemn (customers) P(n in system)
1 0.342857
Distribution of time in queuet (time in queue) P(wait > t)
0.333333333 0.000077
47Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
The M/G/1 Model EXACT Formulas
Model 3 (M/G/1 Queue):Single server, Infinite Population, Poisson Arrival, FCFS, General Service Time, Unlimited Queue LengthYellow cells need user inputed values
InputsUnit of time minuteArrival rate (lambda) 0.6 customers per minuteAverage service rate (mu) 1.2 customers per minute
Service time standard deviation 0.3 minute
OutputsDirect outputs from inputsMean time between arrivals 1.667 minuteMean time per service 0.833333333 minuteTraffic intensity 0.5
Summary measuresUtilization rate of server 50.0%Average number of customers waiting in line (Lq) 0.406656 customersAverage number of customers in system (L) 0.906656 customersAverage time waiting in line (Wq) 0.67776 minuteAverage time in system (W) 1.511093333 minuteProbability of no customers in system (P0) 50.0%
48Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
The M/G/1 Model EXACT Formulas
Model 5 (M/M/s/b queue):Multiple servers, Infinite population, Poisson arrival, FCFS, Exponential service time, Limited buffer (b)Yellow cells need user inputed values
InputsUnit of time hourArrival rate (lambda) 10 customers per hourService rate (mu) 4 customers per hourNumber of identical servers (s) 4 serversBuffer (waiting room) size 50 customers (The maximum number in system is equal to number of servers plus waiting room size)
OutputsDirect outputs from inputsMean time between arrivals 0.100 hourMean time per service 0.25 hourTraffic intensity 0.625
Summary measuresAverage utilization rate of servers 62.5%Average number of customers waiting in line (Lq) 0.5331 customersAverage number of customers in system (L) 3.0331 customers Average time waiting in line (Wq) 0.0533 hourAverage time in system (W) 0.3033 hourProbability of no customers in system (P0) 0.0737 (Probability of empty system)Probability of rejecting a customer (balking rate) 0.00% (Reject rate)Effective arrival rate 10 (Entering rate)
49Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
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?
Assignment 1: M/M/1 Performance Evaluation
50Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Assignment 2: M/M/1 Performance Evaluation
What if the arrival rate is 11 per hour?
51Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
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 ?
Assignment 3: M/G/c
52Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
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)?
Assignment 4: M/M/c Example
53Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1
Suppose the service time is a constant What is the answer of the previous question?
Assignment 5: M/D/c Example