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utdallas.edu/~metin 1 Waiting Times Chapter 7 These slides are based in part on slides that come with Cachon & Terwiesch book Matching Supply with Demand http://cachon-terwiesch.net/3e/. If you want to use these in your course, you may have to adopt the book as a textbook or obtain permission from the authors Cachon & Terwiesch.
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Page 1: Waiting Times - The University of Texas at Dallasmetin/Or6302/Folios/omqueue.pdf · We are interested in the waiting times in the queue and the queue length. utdallas.edu/~metin 5

utdallas.edu/~metin

1

Waiting Times

Chapter 7

These slides are based in part on slides that come with Cachon & Terwiesch

book Matching Supply with Demand http://cachon-terwiesch.net/3e/. If you

want to use these in your course, you may have to adopt the book as a textbook

or obtain permission from the authors Cachon & Terwiesch.

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utdallas.edu/~metin2

Learning Objectives

Interarrival and Service Times and their variability

Obtaining the average time spent in the queue

Pooling of server capacities

Priority rules

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utdallas.edu/~metin3

Where are the queues?

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utdallas.edu/~metin

Input:Passengers in an airport

Customers at a bank

Patients at a hospital

Callers at a call center

Arrival rate

Resources:

Check-in clerks at an airport

Tellers at a bank

Nurses at a hospital

Customer service representatives

(CRS) at a call center

Capacity

BufferProcessing

A Queue is made of a server and a queue in front

Queues

We are interested in the waiting times in the queue and the queue length.

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Blocked calls

(busy signal)Abandoned calls

(tired of waiting)

Calls

on Hold

Reps

processing

calls

Answered

CallsIncoming

calls

Call center

Financial consequences

Lost throughput Holding cost

Lost goodwill

Lost throughput (abandoned)

$$$ Revenue $$$Cost of capacity

Cost per customer

At peak, 80% of calls dialed received a busy signal.

Customers getting through had to wait on average 10 minutes

Extra phone line expense per day for waiting was $25,000.

An Example of a Simple Queuing System

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utdallas.edu/~metin67:00 7:10 7:20 7:30 7:40 7:50 8:00

Patient

Arrival

Time

Service

Time

1

2

3

4

5

6

7

8

9

10

11

12

0

5

10

15

20

25

30

35

40

45

50

55

4

4

4

4

4

4

4

4

4

4

4

4

A Somewhat Odd Service ProcessConstant Arrival Rate (0.2/min) and Service Times (4 min)

Arrival rate 0.2/min = 1/(4 mins) = 1 every five minutes, which implies interarrival time of 5 minutes.

Units of arrival rate 1/min whereas units of interarrival time is min.

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utdallas.edu/~metin 7

Where is Variability?

There certainly is significant (actually infinite) amount of waiting

when the arrival rate is greater than the service rate

– Equivalently, the processing capacity is less than the arrival rate

More interestingly, variability can cause long waiting times.

Variability in

– Arrival process

– Processing times

– Availability of resources; Absent, sick, broken or vacationing servers.

– Types of customers; Priority versus regular customers.

– Routing of flow units; Recall the Resume Validation Example.

– Response of customers to waiting for a while; Wait more or abandon

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Input:• Unpredicted Volume swings

• Random arrivals (randomness is the rule,

not the exception)

• Incoming quality

• Product Mix

Resources:

• Breakdowns / Maintenance

• Operator absence

• Set-up times

Tasks:

• Inherent variation

• Lack of Standard Operating Procedures

• Quality (scrap / rework)

Routes:

• Variable routing

• Dedicated machines

Buffer Processing

Especially relevant in service operations

(what is different in service industries?):

• emergency room

• air-line check in

• call center

• check-outs at cashier

Variability: Where does it come from? Examples

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PatientArrival

Time

Service

Time

1

2

3

4

5

6

7

8

9

10

11

12

0

7

9

12

18

22

25

30

36

45

51

55

5

6

7

6

5

2

4

3

4

2

2

2

Time

7:10 7:20 7:30 7:40 7:50 8:007:00

Patient 1 Patient 3 Patient 5 Patient 7 Patient 9 Patient 11

Patient 2 Patient 4 Patient 6 Patient 8 Patient 10 Patient 12

0

1

2

3

2 min. 3 min. 4 min. 5 min. 6 min. 7 min.

Service times

Nu

mb

er

of ca

se

s

Random Arrival Rate and Service Times

Averages 5 4Interarrival

time

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Inventory

(Patients at lab)

5

4

3

2

1

0

7:00 7:10 7:20 7:30 7:40 7:50 8:00

Wait time

Service timePatient

Arrival

Time

Service

Time

1

2

3

4

5

6

7

8

9

10

11

12

0

7

9

12

18

22

25

30

36

45

51

55

5

6

7

6

5

2

4

3

4

2

2

2

Variability Leads to Waiting Time

Average Arrival Rate (0.2/min) and Service Times (4 min)

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0:1

5

2:0

0

3:4

5

5:3

0

7:1

5

9:0

0

10

:45

12

:30

14

:15

16

:00

17

:45

19

:30

21

:15

23

:00

0

20

40

60

80

100

120

140

160

Time0

20

40

60

80

100

120

140

160

Time

Number of customers

Per 15 minutes

Is the incoming call rate stationary?

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0

6:00:00 7:00:00 8:00:00 9:00:00 10:00:00

0

100

200

300

400

500

600

700

Cumulative

Number of

Customers

6:00:00 7:00:00 8:00:00 9:00:00 10:00:006:00:00 7:00:00 8:00:00 9:00:00 10:00:00

Time

Expected arrivals

if stationary

Actual, cumulative

arrivals

0

7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00

0

10

20

30

40

50

60

70

7:15:00 7:18:00 7:21:00 7:24:00 7:27:00 7:30:00

Time

Cumulative

Number of

Customers

How to test for stationary?

Not stationary over a day, try over an hour or over 30 minutes.

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0

0.2

0.4

0.6

0.8

1

Pro

ba

bil

ity{I

nte

rarr

iva

lti

me

t}

Perc

en

tag

e o

f call

s w

ith

{In

tera

rriv

alti

me

t}t time

0

10

20

30

40

50

60

70

80

90

100

Nu

mb

er

of

ca

lls

wit

h in

tera

rriv

alti

me

t

Duration t

Exponential distribution for Interarrival times

Prob 𝐼𝐴 ≤ 𝑡 = 1 − exp −𝑡

𝑎, IA interarrival time

E 𝐼𝐴 = 𝑎, expected time between two arrivals in a row

StDev 𝐼𝐴 = 𝑎, expected and StDev are the same for exponential distribution

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0

0.2

0.4

0.6

0.8

1

0:00:00 0:00:09 0:00:17 0:00:26 0:00:35 0:00:43 0:00:52 0:01:00 0:01:09

-0

0.2

0.4

0.6

0.8

1

0:00:00 0:00:09 0:00:17 0:00:26 0:00:35 0:00:43 0:00:52 0:01:00 0:01:09

Distribution Function

-

Interarrival time

Empirical distribution

Exponential distribution

Comparing empirical and theoretical distributions

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Analyzing the Arrival Process

CVa=StDev(IA)/E(IA)Stationary

Arrivals?

YES NO

Break arrival process up

into smaller time

intervals

YES

Exponentially distributed

inter-arrival times?

NO

• Compute a: average interarrival time

• CVa=1

• All results of chapters 8 and 9 apply

• Compute a: average interarrival time

• CVa= St.dev. of interarrival times / a

• Results of chapter 8 or 9 do not apply,

require simulation or more complicated models

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A

800

600

400

200

0

Std. Dev = 141.46

Mean = 127.2

N = 2061.00Call durations

[seconds]

Frequency

2.5

2

1.5

1

0.5

0

0:00 23:00

Time of the

day

Call duration

[minutes] Week-end averages

Week-day averages

Analyzing the Service Times

Seasonality and Variability

CVp: Coefficient of variation

of service times

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Inflow Outflow

Inventory

waiting Iq

Entry to

system

DepartureBegin

Service

Waiting Time Tq Service Time p

Flow Time T=Tq+p

Inventory

in service Ip

Computing the expected waiting time Tq

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Utilization

a

p

p

a

Capacity

FlowRateunUtilizatio

/1

/1

Example: Average Activity time=p=90 seconds

Average Interarrival time=a=300 seconds

Utilization=90/300=0.3=30%

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Waiting Time Formula for Exponential Arrivals

Service time factor

Utilization factor

Variability factor

21

22

pa CVCV

nutilizatio

nutilizatioTimeActivity queue in Time

The Waiting Time Formula

EX: Average Activity time=p=90 seconds; Average Interarrival time=a=300 seconds; CVa=1 and CVp=1.333

sec57.532

333.11

3.01

3.090queuein timeAverage

22

Waiting Time Formula above is a restatement of Pollaczek-Khinchin (PK) Formula:

PK Formula: 𝑇𝑞 =1

𝑎

1

1−𝑝/𝑎

𝑆𝑒𝑐𝑜𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑡𝑖𝑚𝑒

2

= 𝑝𝑝/𝑎

1−𝑝/𝑎

1

𝑝2𝑝2+Variance of activity time

2

= 𝑝𝑝/𝑎

1−𝑝/𝑎

1+𝐶𝑉𝑝2

2= 𝑝

𝑢

1−𝑢

1+𝐶𝑉𝑝2

2

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Bank Teller Example

An average of 10 customers per hour come to a bank teller who serves each customer in 4 minutes on average. Assume exponentially distributed interarrival and service times.

(a) What is the teller’s utilization?

(b) What is the average time spent in the queue waiting?

(c) How many customers would be waiting for this teller or would be serviced by this teller on average?

(d) On average, how many customers are served per hour?

Answer: p=4 mins; a=6 mins=60/10

!!!input Average hour per 100*)3/1(15*(2/3)output Average

hour.per 0 outputs 1/3, wpidle is teller If

hour.per 15 outputs 2/3, busy wp is teller If d)

248*6

1pT*

1I c)

82

11

66.01

66.04T b)

66.06

4u a)

q

22

q

a

a

p

4 minutes 2 mins 4 minutes 2 mins

Arrive Arrive

Depart Depart

Deterministic Counterpart

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utdallas.edu/~metin21

Average

flow

time T

Theoretical Flow Time

Utilization 100%

Increasing

Variability

The Flow Time Increase Exponentially in Utilization

2

1 TimeActivity

queuein

22

pa CVCV

nutilizatio

nutilizatiop

Timep

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Inflow Outflow

Inventory

waiting Iq

Entry to system DepartureBegin Service

Waiting Time Tq Service Time p

Flow Time T=Tq+p

Inventory

in service Ip

Inventory in the system I=Iq+ Ip

Computing Tq with m Parallel Servers

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Utilization with m servers

ma

p

pm

a

Capacity

FlowRateunUtilizatio

)/1(*

/1

Example: Average Activity time=p=90 seconds

Average Interarrival time=a=11.39 secs over 8-8:15

m=10 servers

Utilization=90/(10x11.39)=0.79=79%

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21

timequeuein Time

221)1(2pa

m CVCV

nutilizatio

nutilizatio

m

Activity

Approximate Waiting Time Formula for Multiple (m) Servers

Waiting Time Formula for Parallel Resources

Example: Average Activity time=p=90 seconds

Average Interarrival time=a=11.39 seconds

m=10 servers

CVa=1 and CVp=1.333

min916.1sec94.1149094.24

sec94.242

333.11

79.01

79.0

10

90queuein Time

221)110(2

pTT q

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utdallas.edu/~metin25

Customers send emails to a help desk of an online retailer every 2

minutes, on average, and the standard deviation of the inter-arrival time

is also 2 minutes. The online retailer has three employees answering

emails. It takes on average 4 minutes to write a response email. The

standard deviation of the service times is 2 minutes.

(a) Estimate the average customer wait before being served.

(b) How many emails would there be -- on average -- that have been

submitted to the online retailer, but not yet answered?

Answer: a=2 mins; CVa=1; m=3; p=4 mins; CVp=0.5

a) Find Tq. b) Find Iq=(1/a)Tq

Online Retailer Example

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• Target Wait Time (TWT)

• Service Level = Probability{Waiting TimeTWT}; needs distribution of waiting time

• Example: Deutsche Bundesbahn Call Center

- now (2003): 30% of calls answered within 20 seconds

- target: 80% of calls answered within 20 seconds

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

Waiting time [seconds]

Fraction of

customers who

have to wait x

seconds or less Waiting times for those customers

who do not get served immediately

Fraction of customers who get served

without waiting at all

90% of calls had to

wait 25 seconds or

less

Service Levels in Waiting Systems

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Bank of America’s Service Measures

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• Variability is the norm, not the exception

- understand where it comes from and eliminate what you can

- accommodate the rest

• Variability leads to waiting times although utilization<100%

• Use the Waiting Time Formula to

- get a quantitative feeling of the system

- analyze specific recommendations / scenarios

• Adding capacity is expensive, although some safety capacity is necessary

• Next case:

- application to call center

- careful in interpreting March / April call volume

Waiting Lines: Points to Remember

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1. Collect the following data:

- number of servers, m

- activity time, p

- interarrival time, a

- coefficient of variation for interarrival (CVa ) and processing time (CVp )

2. Compute utilization: u= ma

p

3. Compute expected waiting time

Tq

21

221)1(2pa

m CVCV

nutilizatio

nutilizatio

m

timeActivity

4. Based on Tq , we can compute the remaining performance measures as

Flow time T=Tq+p

Inventory in service Ip=m*u

Inventory in the queue=Iq=Tq/a

Inventory in the system I=Ip+Iq

Summary of the formulas

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Staffing levels

Cost of direct labor per serviced unit

Ex: $10/hour wage for each CSR; m=10

Activity time=p=90 secs; Interarrival time=11.39 secs

1-800 number line charge $0.05 per minute

u

pm per time) (wages x

1/a

per time) (wages x

rate Arrivalrate Flow

per time wagesTotalLaborDirect ofCost

call/0958.0$05.0x916.1callper charge line ofCost 1.916;T Recall

cents/call 64.3179.0

cents/min) (16.66 x min/call 1.5LaborDirect ofCost

79.039.11x10

90

a x m

punUtilizatio

Because ma=p/u

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Cost of direct labor per serviced unit

Another example with 9 servers

Ex: m=9. $10/hour wage for each CSR; Activity time=p=90 secs;

Interarrival time=11.39 secs; 1-800 number line charge $0.05 per minute

cents/call 474.28878.0

cents/min) (16.66 x min/call 1.5LaborDirect ofCost

878.039.11x9

90

a x m

punUtilizatio

min709.2sec54.1629054.72

sec54.7239.1*218.5*102

333.11

878.01

878.0

9

90queuein Time

221)19(2

pTT q

call/1354.0$05.0*709.2callper charge line ofCost 2.709;TWith

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Cost of Staffing Levels

m u Labor cost

per call

Line cost

charge per call

Total cost

per call

8 1.3458

9 0.878 0.2847 0.1354 0.4201

10 0.790 0.3164 0.0958 0.4122

11 0.4323

12 0.4593

13 0.4887

14 0.5193

15 0.5503

The optimal staffing level m=10

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Number of

CSRs

17161514131211109876543210

20

40

60

80

100

120

140

160

0:1

5

2:0

0

3:4

5

5:3

0

7:1

5

9:0

0

10:4

5

12:3

0

14:1

5

16:0

0

17:4

5

19:3

0

21:1

5

23:0

0

Time

Number of customers

Per 15 minutes

0

20

40

60

80

100

120

140

160

0:1

5

2:0

0

3:4

5

5:3

0

7:1

5

9:0

0

10:4

5

12:3

0

14:1

5

16:0

0

17:4

5

19:3

0

21:1

5

23:0

0

Time

Number of customers

Per 15 minutes

Staffing Levels under various Interarrival Times

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utdallas.edu/~metin

Independent Resources

2x(m=1)

Pooled Resources

(m=2)0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

60% 65%

m=1

m=2

m=5

m=10

70% 75% 80% 85% 90% 95%

Waiting

Time Tq

Utilization u

Implications:

+ balanced utilization

+ Shorter waiting time (pooled safety capacity)

- Change-overs / set-ups

The Power of Pooling

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Service times:

A: 9 minutes

B: 10 minutes

C: 4 minutes

D: 8 minutes

D

A

C

B9 min.

19 min.

23 min.

Total wait time: 9+19+23=51min

B

D

A

C

4 min.

12 min.

21 min.

Total wait time: 4+12+21=37 min

• Flow units are sequenced in the waiting area (triage step)

• Shortest Processing Time (SPT) Rule

- Minimizes average waiting time

• Do you want to wait for a short process or a long one?

Service-Time-Dependent Priority Rules

• Problem of having “true” processing times

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•Sequence based on importance

- emergency cases; identifying profitable flow units

•First-Come-First-Serve

- easy to implement; perceived fairness

•The order in which customers are served does Not affect the average waiting time.

•W(t): Work in the system

•An arrival at t waits until the work W(t) is completed

Service-Time-Independent Priority Rules

Time t

W(t)

First

arrivalSecond

arrivalLast

departure

p1

p2

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Service-Time-Independent Order does not

affect the waiting time

Time t

W(t)

Time t

W(t)

First finishes Second finishes

Work is conserved even when the processing order changes.

No matter what the order is, the third arrival finds the same amount of work W(t).

Order: 1-1-2 Order: 1-2-1

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Summary

Interarrival and Service Times and their variability

Obtaining the average time spent in the queue

Pooling of server capacities

Priority rules


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