Transcript Slide 1

ISQS 3344
INTRODUCTION TO
PRODUCTION AND
OPERATIONS MANAGEMENT
SPRING 2014
Quantitative Review III
WAITING LINE THEORY
INFINITE POPULATION, SINGLE-SERVER,
SINGLE LINE, SINGLE PHASE FORMULAE
  lambda  mean arrival rate
  mu  mean service rate
 
L


 average system utilizatio

 
 average number
n
of customers
in system
INFINITE POPULATION, SINGLE-SERVER,
SINGLE LINE, SINGLE PHASE FORMULAE
L Q   L  average number
W 
1
 
 average
W Q   W  average
Pn  1    
n
of customers
time in system
including
time spent waiting
 probabilit
in line
service

in line
y of n customers
in the system
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
 15
 

 0 . 75 or 75 %
 20
 Average Number of Students in the System

15
L

 3 students
   20  15

HELP DESK EXAMPLE
Average Number of Students Waiting in Line
L Q   L  0 . 75 3   2 . 25 students
 Average Time a Student Spends in the System

W 

1
 

1
20  15
 0 . 2 hours or 12 minutes
Average Time a Student Spends Waiting
(Before Service)
W Q   W  0 . 75 0 . 2   0 . 15 hours or 9 minutes
HELP DESK EXAMPLE

Probability of n Students in the System
P0  1      1  0 . 75 1  0 . 25
0
P1  1      1  0 . 75 ( 0 . 75 )  0 . 188
1
P2  1    
2
 1  0 . 75 ( 0 . 75 )  0 . 141
2
P3  1      1  0 . 75 ( 0 . 75 )  0 . 105
3
P4  1    
4
3
 1  0 . 75 ( 0 . 75 )  0 . 079
4
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 = 0.67
P3 = (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 hour
Mean service rate  = 3 minutes per customer
Average time spent waiting in line 𝑊𝑄 = 𝜌𝑊
𝜆
30
30
𝜌= =
=
= 1.5
60
𝜇
20
3
1
1
1
𝑊=
=
=
= −0.10
𝜇−𝜆
20 − 30
−10
𝑊𝑄 = 𝜌𝑊 = 1.5 ∗ −0.10 = −0.15
Waiting time continually increases!
CHAPTER 6 & SUPPLEMENT 6
Quality Control and
Statistical Process Control (SPC)
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 average
k = # 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 3sigma 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
x
15 . 825  16 . 0  15 . 875
3
x  15 . 9
x 
1 . 33
4

1 . 33
 . 665
2
UCL  15 . 9  3 * . 665  17 . 895 Ounces
LCL  15 . 9  3 * . 665  13 . 905 Ounces
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 sample
k = # of samples

When selecting D4 and D3 use number of
observations for n.
FACTOR FOR RANGE (R) CHART
Sample Size (n)
Factors for R-Chart
D3
D4
2
3
4
0.00
0.00
0.00
3.27
2.57
2.28
5
6
0.00
0.00
2.11
2.00
7
8
9
0.08
0.14
0.18
1.92
1.86
1.82
10
11
12
0.22
0.26
0.28
1.78
1.74
1.72
13
14
0.31
0.33
1.69
1.67
15
0.35
1.65
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 PROBLEM
Develop control limits for the sample ranges.
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
Sample Ranges
0.4
0.4
0.3
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) CHART
c = number of defects
c = average number of defects per sample
c
number
of incidents
number
c 
UCL
observed
of samples
(k)
c
c
 c  z c ,
LCL
c
 c  z c
z = standard normal variable or the # of standard
deviations desired to use to develop the control limits
DEFECTIVE (C) CHART PROBLEM
The 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.
_
c
# complaints
# of samples

16
 1.6
10
UCL c  c  z c  1.6  3 1.6  5 . 39
LCL c  c  z c  1.6  3 1.6   2 . 19  0
Week
# of complaints
1
2
2
2
3
1
4
3
5
2
6
0
7
1
8
2
9
2
10
1
Total
16
Note: Lower control limit
can’t be negative.
FRACTION DEFECTIVE (P) CHART
p = proportion of nonconforming items
p = average proportion of nonconforming items
total number
p 
total number

p
of units sampled("
,
k" times" n" )
p (1  p )

UCL
of defects
n
p
 p  z p ,
LCL
p
 p  z
p
z = standard normal variable or the # of standard
deviations desired to use to develop the control limits
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
0.12
2
2
25
0.08
3
4
25
0.16
4
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
0.12
2
2
25
0.08
3
4
25
0.16
4
3
25
0.12
Total
12
100
0.12
# Defectives
p 
Total Inspected
p (1  p )
σp 
n
UCL
LCL
p
p


12
 0 .12
100
(.12)(.88)
 0.065
25
 p  z  σ   .12  3(.065)  .315
 p  z  σ   .12  3(.065)   .075  0
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 
Cp 


product' s design specificat
6 standard
specificat
deviations
ion width
process width
ion range
of the production

system
USL  LSL
6
Cp < 1: process not capable of meeting design specs
Cp ≥ 1: process capable of meeting design specs
PROCESS CAPABILITY INDEXES (Cp & Cpk)

Cpk helps to address a possible lack of
centering of the process.
 USL  μ μ  LSL 
Cpk  min 
,

3σ
3σ


 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.
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
Cp 

USL  LSL
6

16 . 2  15 . 8
6 0 . 1 

0 .4
 0 . 66
0 .6
Cpk
C pk
  LSL 
 USL  
 min 
or

3
3


 16 . 2  15 . 9
15 . 9  15 . 8 


 min 
or
3 0 . 1  
 3 0 . 1 
0 .1 
 0 .3
 min 
or
  min 1 or 0 . 33   0 . 33
0 .3 
 0 .3
CHAPTER 3
Project Management
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
9
0
0
A
1
1
0
Identify
Topic
0
0
1
1
1
B
6
Research
Topic
5
6
6
0
6
Legend
ES ID
C
9
Draft
Paper
3
9
SL
Description
LS
DUR
LF
Edit
Paper
2
11
9
E
10
1
Create
Graphics
10
1
11
9
F
10
10
Group Term Paper
11
9
1
EF
D
References
1
11
11
0
11
G
12
Final
Draft
1
12
FORWARD PASS COMPUTATION
9
D
11
Edit
Paper
Always
start
at 0
EF =
ES+DUR
0
A
1
Identify
Topic
1
EF =
ES+DUR
1
B
6
EF =
ES+DUR
6
Research
Topic
5
C
9
9
Draft
Paper
ES ID
E
10
Create
Graphics
3
1
9
Legend
EF =
ES+DUR
2
F
SL
Description
LS
DUR
LF
1
Group Term Paper
G
10
12
Final
Draft
1
References
EF
11
BACKWARD PASS COMPUTATION
Legend
ES ID
Description
LS
DUR
LF
0
A
1
1
Identify
Topic
0
1
1
LS =
LF - DUR
1
D
11
Edit
Paper
SL
0
9
Group Term Paper
EF
B
6
6
Research
Topic
5
LS =
LF - DUR
6
C
9
9
2
11
9
E
10
Draft
Paper
6
3
LS =
LF - DUR
9
11
Create
Graphics
10
9
1
10
1
F
11
10
References
1
11
G
12
Final
Draft
11
1
LS =
LF - DUR
12
EF=LF
DETERMINING SLACK
9
SL =
LS – ES
or
LF - EF
SL =
LS – ES
or
LF - EF
0
A
1
1
0
Identify
Topic
0
0
1
1
1
B
6
Research
Topic
5
6
SL =
LS – ES
or
LF - EF
6
0
6
Legend
ES ID
C
9
Draft
Paper
3
9
0
SL
Description
LS
DUR
LF
Edit
Paper
2
11
9
E
10
1
Create
Graphics
10
1
11
9
F
10
10
Group Term Paper
11
9
1
EF
D
11
0
11
G
12
Final
Draft
1
12
References
1
11
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
1
2
3
4
5
6
7
8
9
10
Description
Pour Foundation
Erect Frame
Roof
Windows
Doors
Electrical
Rough-in Frame
Door Opener
Paint
Clean-up
Predecessor
None
1
2
2
2
2
3, 4, 5, 6
5, 6
7, 8
9
Time(Days)
3
4
4
1
1
3
2
1
2
1
GARAGE PROBLEM
3
Roof
Project Duration: ______ days
Critical Path: _______________
4
4
1
Pour
Foundation
3
2
Erect
Frame
Windows
1
7
Rough-in
Frame
2
4
9
10
Paint
Clean-up
2
5
8
Doors
Door
Opener
1
1
1
Legend
6
Electrical
3
ES ID EF
SL
Description
LS
DUR
LF
GARAGE PROBLEM
0
0
0
1
3
Pour
Foundation
3
3
3
2
7
0
Erect
Frame
3
4
7
3 11
0
Roof
7
4 11
7
4
3
8
Windows
10 1 11
Project Duration: ___16___ days
Critical Path: __1237910__
11 7 13
0
Rough-in
Frame
11 2 13
7
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.
13 9 15
15 10 16
0
0
Paint
13 2 15
7
5
8
3
Doors
10 1 11
15 1 16
10 8 11
2
Clean-up
Door
Opener
12 1 13
Legend
ES ID EF
7
6 10
1
Electrical
SL
Description
8
3 11
LS
DUR
LF
FINAL EXAM
 When:
 Friday
May 9, 2014
 7.30p.m. – 10.00p.m.
 Where:
 TBA
THANKS FOR A GREAT SEMESTER!