© The McGraw-Hill Companies, Inc., 2004 1. 2 Technical Note 6 Waiting Line Management.

©The McGraw-Hill Companies, Inc., 2004

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2

Technical Note 6

Waiting Line Management


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• Waiting Line Characteristics

• Suggestions for Managing Queues

• Examples (Models 1, 2, 3, and 4)

OBJECTIVES


4

Components of the Queuing System

CustomerArrivals

Servers

Waiting Line

Servicing System

Exit

Queue or


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Customer Service Population Sources

Population Source

Finite Infinite

Example: Number of machines needing repair when a company only has three machines.

Example: Number of machines needing repair when a company only has three machines.

Example: The number of people who could wait in a line for gasoline.

Example: The number of people who could wait in a line for gasoline.


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Service Pattern

ServicePattern

Constant Variable

Example: Items coming down an automated assembly line.

Example: Items coming down an automated assembly line.

Example: People spending time shopping.

Example: People spending time shopping.


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The Queuing System

Queue Discipline

Length

Number of Lines &Line Structures

Service Time Distribution

Queuing System


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Examples of Line Structures

Single Channel

Multichannel

SinglePhase Multiphase

One-personbarber shop

Car wash

Hospitaladmissions

Bank tellers’windows


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Degree of Patience

No Way!

BALK

No Way!

RENEG


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Suggestions for Managing Queues

1. Determine an acceptable waiting time for your customers

2. Try to divert your customer’s attention when waiting

3. Inform your customers of what to expect

4. Keep employees not serving the customers out of sight

5. Segment customers


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Suggestions for Managing Queues (Continued)

6. Train your servers to be friendly

7. Encourage customers to come during the slack periods

8. Take a long-term perspective toward getting rid of the queues


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Waiting Line Models

Model LayoutSourcePopulation Service Pattern

1 Single channel Infinite Exponential

2 Single channel Infinite Constant

3 Multichannel Infinite Exponential

4 Single or Multi Finite Exponential

These four models share the following characteristics: Single phase Poisson arrival FCFS Unlimited queue length


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Notation: Infinite Queuing: Models 1-3

linein tingnumber wai Average

server single afor

rate sevice torate arrival totalof Ratio = =

arrivalsbetween timeAverage

timeservice Average

rate Service =

rate Arrival =

1

1

Lg

linein tingnumber wai Average

server single afor

rate sevice torate arrival totalof Ratio = =

arrivalsbetween timeAverage

timeservice Average

rate Service =

rate Arrival =

1

1

Lg


14Infinite Queuing Models 1-3 (Continued)

linein waitingofy Probabilit

systemin units exactly ofy Probabilit

channels service identical ofNumber =

system in the units ofNumber

served) be to time(including

systemin time totalAverage

linein waiting timeAverage = g

served) being those(including

systemin number Average = s

Pw

nPn

S

n

Ws

W

L

linein waitingofy Probabilit

systemin units exactly ofy Probabilit

channels service identical ofNumber =

system in the units ofNumber

served) be to time(including

systemin time totalAverage

linein waiting timeAverage = g

served) being those(including

systemin number Average = s

Pw

nPn

S

n

Ws

W

L


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Example: Model 1Assume a drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.

Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?

Determine:A) What is the average utilization of the employee?B) What is the average number of customers in line?C) What is the average number of customers in the system?D) What is the average waiting time in line?E) What is the average waiting time in the system?F) What is the probability that exactly two cars will be in the system?


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= 25 cust / hr

= 1 customer

2 mins (1hr / 60 mins) = 30 cust / hr

= = 25 cust / hr

30 cust / hr = .8333

= 25 cust / hr

= 1 customer

2 mins (1hr / 60 mins) = 30 cust / hr

= = 25 cust / hr

30 cust / hr = .8333

Example: Model 1

A) What is the average utilization of the employee?


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Example: Model 1

B) What is the average number of customers in line?

4.167 = 25)-30(30

(25) =

) - ( =

22

Lg 4.167 = 25)-30(30

(25) =

) - ( =

22

Lg

C) What is the average number of customers in the system?

5 = 25)-(30

25 =

- =

Ls 5 = 25)-(30

25 =

- =

Ls


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Example: Model 1

D) What is the average waiting time in line?

mins 10 = hrs .1667 =

=

LgWg mins 10 = hrs .1667 =

=

LgWg

E) What is the average waiting time in the system?

mins 12 = hrs .2 = =Ls

Ws mins 12 = hrs .2 = =Ls

Ws


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Example: Model 1

F) What is the probability that exactly two cars will be in the system (one being served and the other waiting in line)?

p = (1-n

n

)( )p = (1-n

n

)( )

p = (1- = 2

225

30

25

30)( ) .1157p = (1- =

2

225

30

25

30)( ) .1157


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Example: Model 2

An automated pizza vending machine heats and dispenses a slice of pizza in 4 minutes.

Customers arrive at a rate of one every 6 minutes with the arrival rate exhibiting a Poisson distribution.

Determine:

A) The average number of customers in line.B) The average total waiting time in the system.

Determine:

A) The average number of customers in line.B) The average total waiting time in the system.


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Example: Model 2

A) The average number of customers in line.

.6667 = 10)-(2)(15)(15

(10) =

) - (2 =

22

Lg .6667 = 10)-(2)(15)(15

(10) =

) - (2 =

22

Lg

B) The average total waiting time in the system.

mins 4 = hrs .06667 = 10)-51)(15(2

10 =

) - (2 =

Wg mins 4 = hrs .06667 = 10)-51)(15(2

10 =

) - (2 =

Wg

mins 8 = hrs .1333 = 15/hr

1 + hrs .06667 =

1 + =

WgWs mins 8 = hrs .1333 = 15/hr

1 + hrs .06667 =

1 + =

WgWs


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Example: Model 3Recall the Model 1 example:Drive-up window at a fast food restaurant.Customers arrive at the rate of 25 per hour.The employee can serve one customer every two minutes.Assume Poisson arrival and exponential service rates.

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?

If an identical window (and an identically trained server) were added, what would the effects be on the average number of cars in the system and the total time customers wait before being served?


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Example: Model 3Average number of cars in the system

ion)interpolatlinear -using-TN6.10(Exhibit

1760= .Lgion)interpolatlinear -using-TN6.10(Exhibit

1760= .Lg

1.009 = 30

25 + .176 = + =

LgLs 1.009 = 30

25 + .176 = + =

LgLs

Total time customers wait before being served

)( = mincustomers/ 25

customers .176 = = Wait! No

LgWg mins .007

)( =

mincustomers/ 25

customers .176 = = Wait! No

LgWg mins .007


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Notation: Finite Queuing: Model 4

channels service ofNumber

linein units ofnumber Average

)( system

queuingin thoseless source Population =

served being units ofnumber Average

linein wait tohaving

ofeffect theof measure a factor, Efficiency

linein must wait arrivalan y that Probabilit =

S

L

n-N

J

H

F

D

channels service ofNumber

linein units ofnumber Average

)( system

queuingin thoseless source Population =

served being units ofnumber Average

linein wait tohaving

ofeffect theof measure a factor, Efficiency

linein must wait arrivalan y that Probabilit =

S

L

n-N

J

H

F

D


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Finite Queuing: Model 4 (Continued)

required timeservice of proportionor factor, Service

linein time waitingAverage

tsrequiremen servicecustomer between timeAverage

service theperform to timeAverage =

system queuingin units exactly ofy Probabilit

source populationin units ofNumber

served) being one the(including

system queuingin units ofnumber Average =

X

W

U

T

nPn

N

n

required timeservice of proportionor factor, Service

linein time waitingAverage

tsrequiremen servicecustomer between timeAverage

service theperform to timeAverage =

system queuingin units exactly ofy Probabilit

source populationin units ofNumber

served) being one the(including

system queuingin units ofnumber Average =

X

W

U

T

nPn

N

n


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Example: Model 4

The copy center of an electronics firm has four copymachines that are all serviced by a single technician.

Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.

Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?

The copy center of an electronics firm has four copymachines that are all serviced by a single technician.

Every two hours, on average, the machines require adjustment. The technician spends an average of 10minutes per machine when adjustment is required.

Assuming Poisson arrivals and exponential service, how many machines are “down” (on average)?


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Example: Model 4N, the number of machines in the population = 4M, the number of repair people = 1T, the time required to service a machine = 10 minutesU, the average time between service = 2 hours

X =T

T + U

10 min

10 min + 120 min= .077X =

T

T + U

10 min

10 min + 120 min= .077

From Table TN6.11, F = .980 (Interpolation)From Table TN6.11, F = .980 (Interpolation)

L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines

L, the number of machines waiting to be serviced = N(1-F) = 4(1-.980) = .08 machines

H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines

H, the number of machines being serviced = FNX = .980(4)(.077) = .302 machines

Number of machines down = L + H = .382 machinesNumber of machines down = L + H = .382 machines


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Queuing Approximation• This approximation is quick way to analyze a queuing situation. Now, both

interarrival time and service time distributions are allowed to be general.• In general, average performance measures (waiting time in queue, number

in queue, etc) can be very well approximated by mean and variance of the distribution (distribution shape not very important).

• This is very good news for managers: all you need is mean and standard deviation, to compute average waiting time

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Define:

Standard deviation of Xcoefficient of variation for r.v. X =

Mean of XVariance

squared coefficient of variation (scv) = mean

x

x x

C

C C


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Queue Approximation

2( 1) 2 2

1 2

Sa s

q

C CL

s qL L S

Compute S

2 2,a sC CInputs: S, , ,

(Alternatively: S, , , variances of interarrival and service time distributions)

as before, , and q sq s

L LW W


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Approximation Example• Consider a manufacturing process (for example making plastic parts) consisting of a single stage with

five machines. Processing times have a mean of 5.4 days and standard deviation of 4 days. The firm operates make-to-order. Management has collected date on customer orders, and verified that the time between orders has a mean of 1.2 days and variance of 0.72 days. What is the average time that an order waits before being worked on?

Using our “Waiting Line Approximation” spreadsheet we get:Lq = 3.154 Expected number of orders waiting to be completed.

Wq = 3.78 Expected number of days order waits.

Ρ = 0.9 Expected machine utilization.


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End of Technical Note 6

© The McGraw-Hill Companies, Inc., 2004 1. 2 Technical Note 6 Waiting Line Management.

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