1 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety

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)


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


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)


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.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Now Let’s Look at the Rest of the System; The Little’s Law Applies Everywhere

Flow time T Ti + Tp


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


Operational Performance Measures

Flow time T = Ti + Tp


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.


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


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


Utilization – Variability - Delay Curve

VariabilityIncreases

Average Time in System T

Utilization (U) U100%

Tp


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


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


Exact Ii for Poisson Arrival Rate, Exponential Service Time


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


=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.


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.


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


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


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


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


Processing Time, Waiting Time; Long Waiting Line

Increasing CapacityPolling

Ra =R= 10/hourTp = 5 minutesInterarrival time PoissonService time exponential

Ra = 10/hour


More Servers + Specialization

Ra/2 = 10/hour

Ra/2 = 10/hour


Polling: Lower Waiting Time, Longer Processing Time (Perhaps)

Ra = 10/hour


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


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


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


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


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


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


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


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


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


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


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%


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)


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


Assignment 2: M/M/1 Performance Evaluation

What if the arrival rate is 11 per hour?


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


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


Suppose the service time is a constant What is the answer of the previous question?

Assignment 5: M/D/c Example

Documents

1 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety