Chapter 9-Quality Management .ppt

quality

1. CHAPTER 9
QUALITY
MANAGEMENT

2. TOTAL QUALITY
MANAGEMENT (TQM)
TQM refers to “Organization wide effort to
achieve quality”.
TQM is a philosophy about quality that
suggests for involving everyone in the
organization is a quest for Quality

3. SOME BASIC TOOLS OF TQM
Cause-Effect diagram
Control chart

4. CAUSE-EFFECT OR FISH-BONE
DIAGRAM
•Purpose: Identifies root causes of a problem.
•Structure: Problem as the "head" with categories as "bones.“
•Use: Visualizes causes for targeted problem-solving.
•Categories: People, Process, Technology, etc.

5. CAUSE-EFFECT DIAGRAM (OR FISH
BONE DIAGRAM)
Problem:
As an AI engineer you have to figure what went wrong in Apple Electric Car
project?
People
Process
Software
Hardware
Management
Plan
Tail Head

6. CONTROL CHART
Problem Statement: The e-commerce webpage's response time fluctuates,
occasionally exceeding acceptable performance limits, impacting user
experience and sales.

7. Sven basic tools of TQM Control Charts
Errors in Control Charts
Type I error-Producer's risk
Probability of rejecting a good lot
Type II error-Consumer's risk
Probability of accepting a bad lot

8. TAGUCHI LOSS FUNCTION
(QUALITY LOSS FUNCTION)
•Measures Quality Deviation: The Taguchi Loss Function calculates the
cost or negative impact when a product or process moves away from its
ideal target, even if it's still within acceptable limits.
•Formula: It uses a formula showing how sensitive the cost is to the
difference.
•Focus on Consistency: Unlike traditional methods that focus only on
staying within limits, Taguchi encourages minimizing variations to improve
quality and reduce costs over time.

9. TAGUCHI LOSS
FUNCTION

10. PROBLEM 1 ( COST OF COMPLAIN)
Company X receives an average of 10 complaints per
month. In November they received 15
complaints .Management sets an acceptable level at
2 .It costs the company $150 directly per complaint to
correct the problems. Find out the loss associated with
the fall of quality.
Solution:
Here, Nominal (average) value = 10
Real Time Value (y)= 15, Cost (A) = $150, d= 2

11. QUALITY LOSS FOR A PROCESS
We can define a processes average
loss as:
s is process (product) Standard
Deviation
y
bar
is process (product) mean

12. PROBLEM 2
In a metal showpiece manufacturing plant, cost of rework is
$2. Process specification is: 8.5+.05 units. It was found that
the production process mean was = 8.492 and standard
deviation was = 0.016 . Find out the loss due to quality
deviation of the manufacturing process.
Solution:

13. LOSS FUNCTION FORMULA
PERFORMANCE
L(x)= k(x−T)^2
L(x) : The loss or cost associated with the deviation from the target
value. It measures the impact of not achieving the ideal quality.
k: A constant that determines how sensitive the loss is to the deviation.
A higher k means small deviations result in bigger losses.
x: The actual value or performance of the process/product. This is the
measured outcome that may deviate from the target.
T: The target value or ideal performance. This is the desired value the
process or product should achieve for optimal quality.

14. PROBLEM 1: SOFTWARE RESPONSE
TIME OPTIMIZATION
Problem: An e-commerce company has set a target response time
of 2 seconds, but during peak hours, the response time increases to
3 seconds. Find the loss ? Assume k= 100
Solution:
Target (T): 2 seconds
Actual Value (x): 3 seconds
Constant (k): 100
Using the Taguchi Loss Function: L(x) = k (x - T)^2
L(x) = 100 (3 - 2)^2 = 100 x 1 = 100 dollars

15. Six Sigma Distribution

16. SIX SIGMA DISTRIBUTION
PROBLEM STATEMENT
The Department’s monitoring server uptime per day.
Average uptime: 23.7 hours, Standard deviation: 0.15
hours. Calculate Upper and Lower Control Limits (UCL
and LCL) for:
1)Six Sigma Quality,
2)Twelve Sigma Quality.
3)Visualize the control limits and interpret their meaning.

17. SOLUTION
Control Limits Calculation:
1. Six Sigma:
- UCL = µ + 6 * σ = 23.7 + 6 * 0.15 = 24.6 hours
- LCL = µ - 6* σ = 23.7 - 6 * 0.15 = 22.8 hours
2. Twelve Sigma:
- UCL = µ + 12* σ = 23.7 + 12 * 0.15 = 25.5 hours
- LCL = µ - 12 * σ = 23.7 - 12 * 0.15 = 21.9 hours