Sessions 19 and 20 -Quality Control

Operations Management-104 1

Recap of Previous Session

Understand QualityCosts of QualityImpact of Quality on OperationsImpact of Quality on Supply Chain

Operations Management-104

Quality Control

Debabrata Ghosh

*Images available in the public domain


The ANZ Door Manufacturing Case

The COO of ANZ Door Manufacturing Company has recently received complaints from the sales headthat several of his team members (Salespersons) have received customer complaints over the past few months. The complaints vary from higher costs, the door not meeting the customers required specifications or door not having the required quality standards in terms of safety, durability and ease of operation.

The COO decided to form an internal Quality Assurance Team (QAT) to study the concerns and suggest any improvement changes, if required.

Approach:To begin with, the QAT team ran a survey of 1000 past and present customers, asking them to rate their experiences with each of the following aspects of ANZ Door's products and services:Cost of purchasing and maintaining a doorResponse time from ordering a door until its deliveryDegree of door customization permitted in accommodating individual preferencesAfter sales service qualityDoor quality in terms of its fit and finish; ease of operation and durability



Step 1: Based on the survey feedback, the QAT team lists the number of complaints received :

Type of Quality Number of Complaints

Cost 10

Response Time 5

Customization 4

Service Quality 15

Door Quality 25

Step 2: Based on the above table, QAT team applies the Pareto Chart to prioritize customer complaints

0 10 20 30

Door Quality

Service Quality

Cost

Response Time

Customization

Number of Complaints

Number of Complaints



Day

Time/Day

1 2 3 4 5 6 7 8 9 10

9 a.m. 81 82 80 74 75 81 83 86 88 82

11 a.m. 73 87 83 81 86 86 82 83 79 84

1 p.m. 85 88 76 91 82 83 76 82 86 89

3 p.m. 90 78 84 75 84 88 77 79 84 84

5 p.m. 80 84 82 83 75 81 78 85 85 80

Day

Time/Day

11 12 13 14 15 16 17 18 19 20

9 a.m. 86 86 88 72 84 76 74 85 82 89

11 a.m. 84 83 79 86 85 82 86 85 84 80

1 p.m. 81 78 83 80 81 83 83 82 83 90

3 p.m. 81 80 83 79 88 84 89 77 92 83

5 p.m. 87 83 82 87 81 79 83 77 84 77

Refer to Tab: Histogram in Histogram of Door Weights.xls


Assignable variation is caused

by factors that can be clearly

identified and possibly

managed

Common variation is inherent

in the production process

Example: A poorly trained employee that creates variation in finished product output.

Example: The door manufacturing process is inherently variable

Basic forms of Variation

6


An Understanding of standard deviation

(2) 0.9772, or Pr(x + 2) 0.9772.

In the ANZ Door Manufacturing Case, the QAT team would expect 99.73% of the door weights would be within +- 3 standard deviations from the mean (between 69.9 and 95.1 kg).

=0 1 2 3-1-2-3

95%

99.74%



Step 6: Control ChartsA generic control chart : LCL = - zUCL = + z

Average (or X-bar) and Range (or R) Control Charts:

X-bar - Process central tendency based on estimated process mean

RA - x = LCL

RA + x = UCL

2

2

Limits ControlChart x

x = The average of the means of the samples

j = Sample Numberm = Total number of samples

i = ith observationn = Number of observations in each sub-group

=

Refer to Tab: X-bar and R in Histogram of Door Weights.xls



R chart - Process variability based on estimated process range

Rj = Difference between the highest and lowest measurement in the sample

n = Number of observations in each sub-group

RD = LCL

RD = UCL

3

4

Limits ControlChart R

R= Average of the measurement differences R for all samples =

n A2 D3 D4

2 1.88 0 3.27

3 1.02 0 2.57

4 0.73 0 2.28

5 0.58 0 2.11

6 0.48 0 2.00

7 0.42 0.08 1.92

8 0.37 0.14 1.86

9 0.34 0.18 1.82

10 0.31 0.22 1.78

11 0.29 0.26 1.74

9

Source: Exhibit 9A.6, Operations and Supply Management, Chase, Shankar, Jacobs and Aquilano, McGraw Hill 12th Ed., 2010



UCL = 82.5 + 0.58* 10.05 = 82.5 + 5.829 = 88.33LCL = 82.5 0.58*10.05 = 76.67

RA - x = LCL

RA + x = UCL

2

2

Limits ControlChart x

RD = LCL

RD = UCL

3

4

Limits ControlChart R

UCL = 2.11* 10.05 = 21.21 LCL = 0

10


Normal Behavior

Possible problem, investigate

UCL

LCL

Samples

over time

1 2 3 4 5 6

UCL

LCL

Samples

over time

1 2 3 4 5 6

Control Chart Patterns

11


Constructing p-charts

p is the proportion of non-conforming items found in a sample. p-chart is often called fraction non-conforming chart.Assume, that k samples each of size n are collected.Let y, represent the number non-conforming in a particular sample, then proportion non-conforming is y/n.Let, pi be the fraction non-conforming in the i th sample, the average fraction non-conforming for group of k samples is

=

An estimate of the s.d. Is given by:

Thus, UCL =LCL =

+ z

- z

12

Q: The operators of automated sorting machines in a post office must read the ZIP code on letters and divert the letters to the proper carrier routes. Over a months study, 25 samples each of 100 letters were chosen and the number of errors were recorded. The data presented is the following:

Refer to Tab: p-chart in Histogram of Door Weights.xls


Constructing c-charts

13

A non-conforming item may have more than one non-conformance factor. For e.g. A customers order may have several errors such as wrong item, wrong quantity, wrong price etc. To monitor the number of non-conformances per unit, a c-chart is used.

For a c-chart,UCL =

LCL =

where, sc =

z

z

Refer to Tab: c-chart in Histogram of Door Weights.xls


Summary of Control Charts

14

Type of Measurement

Type of Attribute

p-chart c-chart

X-bar and R chart

Discrete Continuous

Proportion Non-conforming

Non-conformances

per unit


Process Capability

15

Process capability is the ability of the process to meet customer specifications

Process Control Process Capability

Looks at internal stability Customer centric focus

Analysis based on samples Each individual unit matters

Pertains to production process Pertains to Design specifications

The ANZ Door Manufacturing Case : If the design specifications given by the design team for customer use varies between 75-85 Kg, process capability studies the variability of the process with respect to these design/customer specifications.


Process Capability

16

(b) Design specifications and process variation the same; process is capable of meeting specifications most of the time.

Design Specifications

Process

(a) Process variation exceeds design specifications; process is not capable of meeting specifications all the time.


Process


(c) Design specifications greater than process variation; process is capable of always conforming to specifications.


Process

Process Capability

17


Fraction of Output within specifications

Q: The ANZ Door Manufacturing Case : Given, design specifications of the doors as 75-85 kg with 75 being the lower specification limit (LSL) and 85 being the upper specification limit (USL), what is the fraction of door output that meets the door specifications (given, production process is normal with mean = 82.5 kg and s.d. = 4.2 kg)

18

Ans:Prob(75


Process Capability Measures

Cp =tolerance range

process range

upper spec limit - lower spec limit

6=

19

Process Capability Ratio

Cpk = minimum

x - lower specification limit

3

=

upper specification limit - x

3

=

,

Process Capability Index

Q: The ANZ Door Manufacturing Case : Whats the process capability?

Refer to Tab: ProcessCapability in ProcessCapability of ANZ Door Manf Company.xls


Understanding Process capability ratio, Defects (ppm) and sigma measure

A 3-sigma process: Cp = = 1

A 6-sigma process: Cp = = 2

6

6

6

12

A 3-sigma process: 0.9973 of the output falls within specifications, which means 0.0027 defects => 2700 defects in parts per million (ppm).

A 6-sigma process: 0.999999998 of the output falls within specifications, which means 1.97318 *10^-9 defects => 0.002 defects in parts per million (ppm)

Cp Sigma level ()Area under

the probability density function

Process yieldProcess fallout (in

terms of PPM)

0.33 1 0.6826894921 68.27% 317311

0.67 2 0.9544997361 95.45% 45500

1.00 3 0.9973002039 99.73% 2700

1.33 4 0.9999366575 99.99% 63

1.67 5 0.9999994267 99.9999% 1

2.00 6 0.9999999980 99.9999998% 0.002 500

60

42

16


Mean shift under a Six-sigma process

Motorola allows for a 1.5 sigma shift in mean which results in a process producing on an average 3.4 defective units per million.

4.5 sigma process. Percentage yield is 99.99966%. Therefore, percent defective is 0.00034%. Thus, Defects in Parts per million = 3.4


Process Capability Improvement


Six-sigma Examples

Mumbai Dabbawalas pick up, deliver and return tiffin lunch boxes from 200,000 homes and apartments to 80,000 office locations that are situated over 40 miles away, hardly using any fuel or modern technology. The distribution process is an ingenious combination of coding, aggregating and sorting boxes and moving them in crates through public trains, push carts and bicycles from each household to a correct office destination and back to home. The error rate in delivery is 1 in more than million trips with a cost of about $10 a month.

Harley-Davidson, Bank of America, HSBC, Hospital Chains


Process Capability Improvement

The ANZ Door Manufacturing Case :

1) Mean Shift: If the average door weight could be brought down to 80 kg thus shifting the entire distribution.

Ans:Prob(75


Munchies snack food company packages potato chips in a process with a process mean of 8.8 oz and a s.d. of 0.12 oz. The packages are designed for 9 oz of chips with a tolerance of 0.5 oz. The company wants to determine the capability of the current process.

Question:

The process mean is off the center. And the capability of the process is

Cpk = Min[ (8.8-8.5)/(3*0.12); (9.5-8.8)/(3*0.12)]= Min[0.83,1.94]= 0.83

Cp = (9.5-8.5)/(6*0.12) = 1.389

What percentage of packages produced are defective?

1- Prob(8.5

Documents

Sessions 19 and 20 -Quality Control