Waiting Lines Supplement B - PowerPoint Presentation

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Waiting Lines Supplement B - PPT Presentation

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Slide1

Waiting LinesSupplement B

B - 0

Slide2

What are waiting lines and why do they form?

B - 02

Waiting lineOne or more customers waiting for service.Waiting Lines form due to a temporary imbalance between the demand for service and the capacity of the system to provide the service.

Slide3

Structure of Waiting-Line Problems

An input, or customer population, that generates potential customersA waiting line of customersThe service facility, consisting of a person (or crew), a machine (or group of machines), or both necessary to perform the service for the customerA priority rule, which selects the next customer to be served by the service facility

B - 0

Slide4

Customer Population

B - 04

Customer population

Service system

Waiting line

Priority rule

Service facilities

Served customers

Slide5

The Service System

Number of LinesSingle or MultipleArrangement of Service FacilitiesChannels and Phases

B - 0

Slide6

Service facilities

Waiting Line Arrangements

Single Line

Multiple Lines

B- 0

Slide7

Service Facility Arrangements

Service facility

Single

channel, single phase

Single

channel, multiple phase

Service facility 1

Service facility 2

B- 0

Slide8

Multiple channel, single phase

Service facility 1

Service facility 2

Service Facility Arrangements

Service facility 3

Service facility 4

Service facility 1

Service facility 2

Multiple

channel,

multiple

phase

B- 0

Slide9

Service Facility Arrangements

Routing for :

1–2–4Routing for : 2–4–3Routing for : 3–2–1–4

Mixed

arrangement

Service facility 1

Service facility 4

Service facility 3

Service facility 2

B- 0

Slide10

Priority Rules

First-come, first-served (FCFS) - most commonEarliest due date (EDD)Shortest processing time (SPT)Preemptive discipline

B -

Slide11

Probability Distributions

Arrival distribution

B - 11

n = e-T for n = 0, 1, 2,…

(T)nn!

where

= Probability of n arrivals in

time periods

 = Average numbers of customer arrivals per period

= 2.7183

Slide12

Example B.1

Management is redesigning the customer service process in a large department store. Accommodating four customers is important. Customers arrive at the desk at the rate of two customers per hour. What is the probability that four customers will arrive during any hour?

In this case customers per hour, T = 1 hour, and n = 4 customers. The probability that four customers will arrive in any hour is

P4 =

e–2 = 0.090

[2(1)]

–2(1)

B -

Slide13

P(t ≤ T) = 1 – e-T

where μ = average number of customers completing service per period t = service time of the customer T = target service time

Probability Distributions

B - 13

Service Time distribution

Slide14

Example B.2

The management of the large department store in Example B.1 must determine whether more training is needed for the customer service clerk. The clerk at the customer service desk can serve an average of three customers per hour. What is the probability that a customer will require less than 10 minutes of service?

Because  = 3 customers per hour, we convert minutes of time to hours, or T = 10 minutes = 10/60 hour = 0.167 hour.

P(t ≤ T) = 1 – e–T

P(t ≤ 0.167 hour) = 1 – e–3(0.167) = 1 – 0.61 = 0.39

B - 14

Slide15

Using Waiting-Line Models

Balance costs against benefits Operating characteristics Line lengthNumber of customers in systemWaiting time in lineTotal time in systemService facility utilization

B -

Slide16

Single-Server Model

Single-server, single line of customers, and only one phaseAssumptions are:Customer population is infinite and patientCustomers arrive according to a Poisson distribution, with a mean arrival rate of Service distribution is exponential with a mean service rate of Mean service rate exceeds mean arrival rateCustomers are served FCFSThe length of the waiting line is unlimited

B - 16

Slide17

 = Average utilization of the system =

= Average number of customers in the service system =

m – l

Lq = Average number of customers in the waiting line =  L

W = Average time spent in the system, including service =

m – l

Wq = Average waiting time in line = W

Rn = Probability that n customers are in the system = (1 – r )r n

Single-Server Model

B- 17

Slide18

Example B.3

The manager of a grocery store in the retirement community of Sunnyville is interested in providing good service to the senior citizens who shop in her store. Currently, the store has a separate checkout counter for senior citizens. On average, 30 senior citizens per hour arrive at the counter, according to a Poisson distribution, and are served at an average rate of 35 customers per hour, with exponential service times characteristics:

a. Probability of zero customers in the systemb. Average utilization of the checkout clerkc. Average number of customers in the systemd. Average number of customers in linee. Average time spent in the systemf. Average waiting time in line

B - 18

Slide19

Example B.3

The checkout counter can be modeled as a single-channel, single-phase system. The results from the Waiting-Lines Solver from OM Explorer are below:

B -

Slide20

Application B.1

Customers arrive at a checkout counter at an average 20 per hour, according to a Poisson distribution. They are served at an average rate of 25 per hour, with exponential service times. Use the single-server model to estimate the operating characteristics of this system.

= 20 customer arrival rate per hour = 25 customer service rate per hour

B - 20

Slide21

Application B.1

2. Average number of customers in the service system

= 4

25 – 20

m – l

Lq = L

W =

m – l

Wq = W

3. Average number of customers in the waiting line

= 0.8(4) = 3.2

4. Average time spent in the system, including service

= 0.2

25 – 20

Average waiting time in line

= 0.8(0.2) = 0.16

B - 21

1. Average utilization of system

 =

= =

0.8

Slide22

Example B.4

The manager of the Sunnyville grocery in Example B.3 wants answers to the following questions:

What service rate would be required so that customers average only 8 minutes in the system?For that service rate, what is the probability of having more than four customers in the system?c. What service rate would be required to have only a 10 percent chance of exceeding four customers in the system?

B - 22

Slide23

Example B.4

a. We use the equation for the average time in the system and solve for 

 – 

8 minutes = 0.133 hour =



– 30

0.133

 – 0.133(30) = 1

 = 37.52 customers/hour

B - 23

Slide24

Example B.4

b. The probability of more than four customers in the system equals 1 minus the probability of four or fewer customers in the system.

= 1 – Pn

n = 0

= 1

– (1 – ) n

n = 0



= 0.80

37.52

and

Then,

= 1 – 0.2(1 + 0.8 + 0.82 + 0.83 + 0.84) = 1 – 0.672 = 0.328

Therefore, there is a nearly 33 percent chance that more than four customers will be in the system.

B - 24

Slide25

Example B.4

c. We use the same logic as in part (b), except that  is now a decision variable. The easiest way to proceed is to find the correct average utilization first, and then solve for the service rate.

P = 1 – (1 –  )(1 +  +  2 +  3 +  4)

= 1 – (1 –  )(1 +  +  2 +  3 +  4) +  (1 +  +  2 +  3 +  4)

= 1 – 1 –  –  2 –  3 –  4 +  +  2 +  3 +  4 +  5 =  5

 = P1/5

If P = 0.10

 = (0.10)1/5 = 0.63

B - 25

Slide26

Example B.4

Therefore, for a utilization rate of 63 percent, the probability of more than four customers in the system is 10 percent. For  = 30, the mean service rate must be

 = 47.62 customers/hour

0.63



B -

Slide27

Application B.2

From the checkout counter example in Application B.1, what service rate is required to have customers average only 10 minutes in the system?

W =

m – l

= 0.17 hr (or 10 minutes)

0.17(

 – ) = 1, where  = 20 customers arrival rate per hour

 =

= 25.88 or about 26 customers per hour

1 + 0.17(20)

0.17

B -

Slide28

Multiple-Server Model

Service system has only one phase, multiple-channelsAssumptions (in addition to single-server model)There are s identical serversThe service distribution for each server is exponentialThe mean service time is 1/s should always exceed 

B - 28

Slide29

Example B.5

The management of the American Parcel Service terminal in Verona, Wisconsin, is concerned about the amount of time the company’s trucks are idle (not delivering on the road), which the company defines as waiting to be unloaded and being unloaded at the terminal. The terminal operates with four unloading bays. Each bay requires a crew of two employees, and each crew costs $30 per hour. The estimated cost of an idle truck is $50 per hour. Trucks arrive at an average rate of three per hour, according to a Poisson distribution. On average, a crew can unload a semitrailer rig in one hour, with exponential service times. What is the total hourly cost of operating the system?

B - 29

Slide30

Example B.5

Calculate the average number of trucks in the system.

B - 30

Labor cost:

$30(s) = $30(4)=$120.00Idle truck cost:$50(L) = $50(4.53)=226.50Total hourly cost=$346.50

Slide31

Application B.3

Suppose the manager of the checkout system in Application B.2 decides to add another counter. The arrival rate is still 20 customers per hour, but now each checkout counter will be designed to service customers at the rate of 12.5 per hour. What is the waiting time in line of the new system?

s = 2,  = 12.5 customers per hour,  = 20 customers per hour

B -

Slide32

Application B.3

B - 32

Slide33

Application B.3

B - 33

Slide34

Little’s Law

A fundamental law that relates the number of customers in a waiting-line system to the arrival rate and waiting time of customers.

B - 34

Average time

in the facility

W =

customers customer/hour

Work-in-process =

L = W

 =

arrival rate

Slide35

Finite-Source Model

AssumptionsFollows the assumption of the single-server, except that the customer population is finiteOnly N potential customersIf N > 30, then the single-server model with the assumption of infinite customer population is adequate

B - 35

Slide36

Example B.6

The Worthington Gear Company installed a bank of 10 robots about 3 years ago. The robots greatly increased the firm’s labor productivity, but recently attention has focused on maintenance. The firm does no preventive maintenance on the robots because of the variability in the breakdown distribution. Each machine has an exponential breakdown (or interarrival) distribution with an average time between failures of 200 hours. Each machine hour lost to downtime costs $30, which means that the firm has to react quickly to machine failure. The firm employs one maintenance person, who needs 10 hours on average to fix a robot. Actual maintenance times are exponentially distributed. The wage rate is $10 per hour for the maintenance person, who can be put to work productively elsewhere when not fixing robots. Determine the daily cost of labor and robot downtime.

B - 36

Slide37

Example B.6

 = 1/200, or 0.005 break-down per hour, and  = 1/10 = 0.10 robot per hour.

B - 37

The daily cost of labor and robot downtime is

Labor cost:($19/hour)(8 hours/day)(0.462 utilization)=$36.96Idle robot cost:(0.76 robot)($30/robot hour)(8 hours/day)=182.40Total daily cost=$219.36

Slide38

Simulation models can be used analyze a problem when the relationship between variables is nonlinear or when the situation involves too many variables or constraints to handle with optimizing approachesUsed to conduct experiments without disrupting real systems Used to obtain operating characteristic estimates in much less time (time compression) Simulation is useful in sharpening managerial decision-making skills through gaming

Waiting Lines and Simulation

B -

Slide39

SimQuick Software

Easy-to-use package that is simply an Excel spreadsheet with some macrosModels can be created for a variety of simple processesA first step with SimQuick is to draw a flowchart of the process using SimQuick’s building blocksInformation describing each building block is entered into SimQuick tables

B -

Slide40

Buffer

Done

Passenger Security Process

Workst.

Add. Insp. 2

Workst.

Add. Insp. 1

Buffer

Sec. Line 2

Dec. Pt. DP

Workst.

Insp. 1

Workst.

Insp. 2

Buffer

Sec. Line 1

Entrance Arrivals

B -

Slide41

Simulation Results

Element

Types

Element

Names

Statistics

Overall

Means

Entrance(s)

Door

Objects entering process

237.23

Buffer(s)

Line 1

Mean Inventory

Mean cycle time

5.97

3.12

Line 2

Mean Inventory

Mean cycle time

0.10

0.53

Done

Final Inventory

224.57

B -

Slide42

Application B.4

DBT Bank has 8 copy machines located in various offices throughout the building. Each machine is used continuously and has an average time between failures of 50 hours. Once failed, it takes 4 hours for the service company to send a repair person to have it fixed. What is the average number of copy machines in repair or waiting to be repaired?

= 1/50 = 0.02 copiers per hour = 1/4 = 0.25 copiers per hour

B -

Slide43

Application B.4

B- 43

Slide44

Application B.4

B- 44

Slide45

Decision Areas for Management

Arrival ratesNumber of service facilitiesNumber of phasesNumber of servers per facilityServer efficiencyPriority ruleLine arrangement

B - 45

Slide46

Application B.5

The Hilltop Produce store is staffed by one checkout clerk. The average checkout time is exponentially distributed around an average of two minutes per customer. An average of 20 customers arrive per hour.What is the average utilization rate?

B - 46

 =



= .667

Slide47

Application B.5

What is the probability that three or more customers will be in the checkout area?First calculate 0, 1, and 2 customers will be in the checkout area:

Rn = (1 – r )r 0

= (0.333)(0.667)0 = 0.333

Rn = (1 – r )r 1

= (0.333)(0.667)1 = 0.222

Rn = (1 – r )r 2

= (0.333)(0.667)2 = 0.148

Then calculate 3 or more customers will be in the checkout area:

1 – R0 – R1 – R3 =

1 - 0.333 – 0.222 – 0.148 = 0.297

B -

Slide48

Application B.5

What is the average number of customers in the waiting line?

– l

q = L = 

= 1.333

What is the average time customers spend in the store?

– l

= 0.1

hr  60 min/hr = 6 minutes

B -

Slide49

Solved Problem

A photographer takes passport pictures at an average rate of 20 pictures per hour. The photographer must wait until the customer smiles, so the time to take a picture is exponentially distributed. Customers arrive at a Poisson-distributed average rate of 19 customers per hour.

What is the utilization of the photographer?b. How much time will the average customer spend with the photographer?

B - 49

Slide50

Solved Problem

b. The average customer time spent with the photographer is

W =

m – l

= 1 hour

20 – 19

B -

The assumptions in the problem statement are consistent with a single-server model. Utilization is

 =

= 0.95

Slide51

B - 51

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.

Printed in the United States of America.

Waiting Lines Supplement B - PowerPoint Presentation

Waiting Lines Supplement B - PPT Presentation

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