ISQS 3344 INTRODUCTION TO PRODUCTION AND

OPERATIONS MANAGEMENT

SPRING 2014Quantitative Review III

WAITING LINE THEORY

INFINITE POPULATION, SINGLE-SERVER,SINGLE LINE, SINGLE PHASE FORMULAE

systemincustomersofnumberaverage

nutilizatiosystemaverage

rateservicemeanratearrivalmean

L

mulambda

INFINITE POPULATION, SINGLE-SERVER,SINGLE LINE, SINGLE PHASE FORMULAE

systemtheincustomersofyprobabilit1

linein waitingspenttimeaverage

serviceincludingsystemintimeaverage1

lineincustomersofnumberaverage

nP

WW

W

LL

nn

Q

Q

HELP DESK EXAMPLE A help desk in the computer lab

serves students on a first-come, first served basis. On average, 15 students need help every hour. The help desk can serve an average of 20 students per hour.

HELP DESK EXAMPLE Based on the description, we know:

= 20 (Exponential distribution) = 15 (Poisson distribution)

Average System Utilization

Average Number of Students in the System

%75or75.02015

students31520

15

L

HELP DESK EXAMPLE Average Number of Students Waiting in

Line

Average Time a Student Spends in the System

Average Time a Student Spends Waiting (Before Service)

students25.2375.0 LLQ

minutes 12or hours 2.01520

11

W

minutes9orhours15.02.075.0 WWQ

HELP DESK EXAMPLE Probability of n Students in the System

079.0)75.0(75.011

105.0)75.0(75.011

141.0)75.0(75.011

188.0)75.0(75.011

25.0175.011

444

333

222

11

00

P

P

P

P

P

WAITING LINE PROBLEM Consider a single-line, single-server waiting

line system. The arrival rate is 100 people per hour, and the service rate µ is 150 people per hour. What is the probability of having 3 people in the system?

Pn = (1 - ) n

= / = 100/150 = 0.67P3 = (1 - ) 3 = (1 – 0.67)(0.67)3 = 0.0992

WAITING LINE PROBLEM Consider a single-line, single-server

waiting line system. Suppose that customers arrive according to a Poisson distribution at an average rate of 60 per hour, and the average (exponentially distributed) service time is 50 seconds per customer. What is the average number of customers in the system?

L = / ( - ) = 60 / (72 – 60) = 5

WAITING LINE PROBLEM

Consider a single-server queuing system.If the arrival rate is 30 customers per hour and it takes 3 minutes on average to serve a customer, what is the average waiting time in the waiting line in minutes?

WAITING LINE PROBLEM (SOLUTION)

Mean arrival rate = 30 customers per hourMean service rate = 3 minutes per customer

Average time spent waiting in line

Waiting time continually increases!

Quality Control andStatistical Process Control (SPC)

CHAPTER 6 & SUPPLEMENT 6

CONTROL CHARTS For Variable Data:

Mean ( ) Chart Range (R) Chart

For Non-variable Data: Defective (C) Chart Fraction Defective (P) Chart

CONTROL CHARTS FOR VARIABLE DATA Mean ( ) Chart:

Tracks the central tendency (the average or mean value observed) over time

Range (R) Chart:Tracks the spread of the distribution over time (estimates the observed variation)

MEAN ( ) CHART

= sample averagek = # of samples

n = # of observations in each sample

z = standard normal variable or the # of standard deviations desired to use to develop the control limits

MEAN ( ) CHART PROBLEM

Assume the standard deviation of the process is given as 1.33 ounces. Management wants a 3-sigma chart (only 0.26% chance of alpha error). Observed values shown in the table are in ounces. Calculate the UCL and LCL.Sample 1 Sample 2 Sample 3

Observation 1

15.6 16.1 16.0

Observation 2

16.0 16.2 15.9

Observation 3

15.8 15.8 15.9

Observation 4

15.9 15.9 15.7

Sample Means

15.825 16.0 15.875

MEAN ( ) CHART PROBLEM

Ounces 905.13665.*39.15Ounces 895.17665.*39.15

665.233.1

433.19.15

3875.150.16825.15

LCLUCL

x

x

x

RANGE (R) CHART Range chart measures the dispersion or

variance of the process while the mean chart measures the central tendency of the process.R = range of each samplek = # of samples

When selecting D4 and D3 use number of observations for n.

FACTOR FOR RANGE (R) CHART

D3 D42 0.00 3.273 0.00 2.574 0.00 2.285 0.00 2.116 0.00 2.007 0.08 1.928 0.14 1.869 0.18 1.82

10 0.22 1.7811 0.26 1.7412 0.28 1.7213 0.31 1.6914 0.33 1.6715 0.35 1.65

Factors for R-ChartSample Size (n)

RANGE (R) CHART PROBLEM

Four samples of 10 observations each have been taken form a Soft drink bottling plant in order to test for volume dispersion in the bottling process. The average sample range was found to be .4 ounces. Develop control limits for the sample range.

RANGE (R) CHART PROBLEMSample 1 Sample 2 Sample 3

Observation 1

15.6 16.1 16.0

Observation 2

16.0 16.2 15.9

Observation 3

15.8 15.8 15.9

Observation 4

15.9 15.9 15.7

Sample Means

15.825 16.0 15.875

Sample Ranges

0.4 0.4 0.3

Develop control limits for the sample ranges.

CONTROL CHARTS FOR NON-VARIABLE DATA

Defective (C) Chart: Used when looking at # of defects

Fraction Defective (P) Chart: Used for yes or no type judgments (acceptable/not acceptable, works/doesn’t work, on time/late, etc.)

DEFECTIVE (C) CHARTc = number of defects = average number of defects per sample

z = standard normal variable or the # of standard deviations desired to use to develop the control limits

c

cccc

c

zczc

c

c

LCL,UCL

(k) samples ofnumber observed incidents ofnumber

DEFECTIVE (C) CHART PROBLEMThe number of weekly customer complaints are monitored in a large hotel using a C-chart. Develop three sigma (z=3) control limits using the data table.

Week # of complaints

1 22 23 14 35 26 07 18 29 210 1

Total 16

019.21.631.6ccLCL

39.51.631.6ccUCL

1.61016

samples of #complaints#c

c

c

_

z

zNote: Lower control limit can’t be negative.

FRACTION DEFECTIVE (P) CHARTp = proportion of nonconforming items = average proportion of nonconforming items

z = standard normal variable or the # of standard deviations desired to use to develop the control limits

p

pppp

p

zpzpnpp

p

LCL,UCL

)1(

,)n"times"k"sampled(" units ofnumber total

defects ofnumber total

FRACTION DEFECTIVE (P) CHART PROBLEM

A Production manager for a tire company has inspected the number of defective tires in four random samples with 25 tires in each sample. The table shows the number of defective tires in each sample of 25 tires. z=3. Calculate the control limits.Sample # of

defective tires

# of tires in each sample

Proportion defective

1 3 25 0.122 2 25 0.083 4 25 0.164 3 25 0.12

Total 12 100 0.12

FRACTION DEFECTIVE (P) CHART PROBLEM

Sample # of defective

tires

# of tires in each sample

Proportion defective

1 3 25 0.122 2 25 0.083 4 25 0.164 3 25 0.12

Total 12 100 0.12

0.0753(.065).12σzpLCL

.3153(.065).12σzpUCL

0.06525

(.12)(.88)n

)p(1pσ

.12010012

Inspected TotalDefectives#p

p

p

p

Note: LCL can’t be negative.

PROCESS CAPABILITY Can a process or system meet its

requirements? Product Specifications

Preset product or service dimensions, tolerances, e.g., bottle fill might be 16 oz. ± .2 oz.

Based on how product is to be used or what the customer expects

Process Capability Assessing capability involves evaluating process

variability relative to preset product or service specifications

PROCESS CAPABILITY INDEXES (Cp & Cpk) Cp assumes that the process is centered

in the specification range.

Cp < 1: process not capable of meeting design specs Cp ≥ 1: process capable of meeting design specs

6 widthprocession widthspecificat

system production theof deviations standard 6rangeion specificatdesign sproduct'

LSLUSLC

C

p

p

PROCESS CAPABILITY INDEXES (Cp & Cpk) Cpk helps to address a possible lack of

centering of the process.

min = minimum of the two = mu or mean of the process Cpk < 1: process not capable or not centered Cpk ≥ 1: process capable or centered

Cp=Cpk when process is centered.

3σLSLμ,

3σμUSLminCpk

PROCESS CAPABILITY EXAMPLE Design specifications call for a target

value of 16.0 +/- 0.2 ounces (USL = 16.2 & LSL = 15.8)

Observed process output has a mean of 15.9 and a standard deviation of 0.1 ounces

PROCESS CAPABILITY EXAMPLE Cp

Cpk 66.0

6.04.0

1.068.152.16

6

LSLUSLC p

33.033.0or 1min3.01.0or

3.03.0min

1.038.159.15or

1.039.152.16min

3or

3min

LSLUSLC pk

Project Management

CHAPTER 3

Reference:Erik Larson and Clifford Gray, 2011, Project Management: The Managerial Process, McGraw Hill.

NETWORK COMPUTATION PROCESS Forward Pass – Earliest Times

Early Start (ES) – How soon can the activity start? Early Finish (EF) – How soon can the activity finish?

Backward Pass – Latest Times Late Start (LS) – How late can the activity start? Late Finish (LF) – How late can the activity finish?

Slack (SL) – How long can the activity be delayed?

Critical Path (CP)– The longest path in the network which, when delayed, will delay the project

FORWARD PASS COMPUTATION You add activity times along each path in

the network (ES + Duration = EF). You carry the early finish (EF) to the next

activity where it becomes its early start (ES), unless

The next succeeding activity is a merge activity. In this case, you select the largest early finish number (EF) of all its immediate predecessor activities.

BACKWARD PASS COMPUTATION You subtract activity times along each path

starting with the project end activity (LF - Duration = LS).

You carry the late start (LS) to the next preceding activity to establish its late finish (LF), unless

The next preceding activity is a burst activity. In this case, you select the smallest late start number (LS) of all its immediate successor activities to establish its late finish (LF).

DETERMINING SLACK Slack for an activity is simply the difference

between the LS and ES (LS – ES) or between LF and EF (LF – EF).

Slack tells us the amount of time an activity can be delayed and yet not delay the project.

When the LF = EF for the end project activity, the critical path can be identified as those activities that also have LF = EF or a slack of zero (LF – EF = 0 or LS – ES = 0).

ACTIVITY-ON-NODE NETWORK

CRITICAL PATH METHOD NETWORK

Description

Legend

ID EF

LF

ES

SL

LS DUR

CreateGraphics

9 E 10

11110

1

EditPaper

11D9

1129

0

DraftPaper

9C6

0

6 3 9

ResearchTopic

1 B 6

0

1 5 6

IdentifyTopic

110

0

0 A 1

References

9 F 10

11110

1

FinalDraft

11 G 12

12111

0

Group Term Paper

FORWARD PASS COMPUTATION

Description

Legend

ID EF

LF

ES

SL

LS DUR

CreateGraphics

9 E 10

1

EditPaper

11D9

2

DraftPaper

9C6

3

ResearchTopic

1 B 6

5

IdentifyTopic

1

0 A 1

References

9 F 10

1

FinalDraft

11 G 12

1

Group Term Paper

Always

start at 0

EF =ES+DU

R

EF =ES+DU

R

EF =ES+DU

R

EF =ES+DU

R

BACKWARD PASS COMPUTATION

Description

Legend

ID EF

LF

ES

SL

LS DUR

CreateGraphics

9 E 10

11110

EditPaper

11D9

1129

DraftPaper

9C6

6 3 9

ResearchTopic

1 B 6

0

1 5 6

IdentifyTopic

110

0 A 1

References

9 F 10

11110

1

FinalDraft

11 G 12

12111

Group Term Paper

EF=LF

LS =LF - DUR

LS =LF - DUR

LS =LF - DUR

LS =LF - DUR

DETERMINING SLACK

Description

Legend

ID EF

LF

ES

SL

LS DUR

CreateGraphics

9 E 10

11110

1

EditPaper

11D9

1129

0

DraftPaper

9C6

0

6 3 9

ResearchTopic

1 B 6

0

1 5 6

IdentifyTopic

110

0

0 A 1

References

9 F 10

11110

1

FinalDraft

11 G 12

12111

0

Group Term Paper

SL =LS – ES

orLF - EF

SL =LS – ES

orLF - EF

SL =LS – ES

orLF - EF

CRITICAL

PATH?

GARAGE PROBLEM Compute the early, late, and slack activity times Determine the planned project duration Identify the critical path What should you do if the Doors activity is going

to take two extra days?ID Description Predecessor Time(Days)

1 Pour Foundation None 3 2 Erect Frame 1 4 3 Roof 2 4 4 Windows 2 1 5 Doors 2 1 6 Electrical 2 3 7 Rough-in Frame 3, 4, 5, 6 2 8 Door Opener 5, 6 1 9 Paint 7, 8 2

10 Clean-up 9 1

GARAGE PROBLEM

Description

LegendID EF

LF

ESSLLS DUR

Windows

4

1

Roof

3

4

Electrical

6

3

ErectFrame

2

4

PourFoundation

3

1

Doors

5

1

Paint

9

2

Rough-inFrame

7

2

DoorOpener

8

1

Clean-up

10

1

Project Duration: ______ daysCritical Path: _______________

GARAGE PROBLEM

Description

LegendID EF

LF

ESSLLS DUR

Windows

7 4 8

111103

Roof

1137

11470

Electrical

106718 3 11

ErectFrame

3 2 703 4 7

PourFoundation

33000 1 3

Doors

7 5 8

111103

Paint

13 9 15

152130

Rough-inFrame

11 7 13

132110

DoorOpener

10 8 11

131122

Clean-up

15 10 16

161150

Project Duration: ___16___ daysCritical Path:

__1237910__

If activity 5 is going to take two extra days, you probably do not have to do any thing because this activity has three days of slack – no effect on project duration.

FINAL EXAM When:

Friday May 9, 2014 7.30p.m. – 10.00p.m.

Where: TBA

THANKS FOR A GREAT SEMESTER!