CONTROL CHART FOR QUALITY CONTROL

_

X-R CHART

_ _X-R chart is a pair of chart consisting of a average chart (X chart) and a range chart (R chart). The X chart is used to control the mean value of the characteristic and R chart is used to restrict the range of variations in the values. This chart drawn when the characteristics are measurable. Each chart consist of three values : a central line ; upper control limit & lower control limit. The central line represents the arithmetic average of sample means which equal to the mean of the population. The upper & lower control limit are located at three standard deviation on either side of the central line.

PREPAERING X-R CHART• The arithmetic mean of samples are distributed according to

normal distribution with sample mean X equal to the mean of the population µ and std deviation of the sample mean σ equal to √σ/n where σ is std deviation of the population and n is the sample size.

The statistical limit of average X-chart in different situation is _ _ _

(i) X and σx known : _ _ _

Central line = CLx = X

_ _ _ _

Upper control line = UCLx = X + 3σx

_ _ _ _

Lower control line = LCLx = X – 3σx

CONTINUED…………….• (ii) µ and σ known : _ _ _

Central line = CLx = X

_

Upper control line = UCLx = µ + 3σ/√n

_

Lower control line = LCLx = µ - 3σ/√n

_

(iii) X and σ known :

_ _ _

Central line = CLx = X

_ _ _

Upper control line = UCLx = X + 3σ/√n

_ _ _

Lower control line = LCLx = X - 3σ/√n

CONTROL RANGE CHART (R-CHART) _ _ _ • Control limit = X + 3.σx

_ _• = X ± 3.σ/√n

_ _ _

= X ± 3.d`/√n * R

Now, 3.d`/√n expressed as a constant sample size and expressed as A2

_ _ _ _

Upper control line = UCLx = X + A2.R

_ _ _ _

Lower control line = LCLx = X – A2.R

_ _

Central line = R ± 3. σR

PROBLEMS• Individual measurement of dia (mm)

_• Plot X & R chart and find out which of the given expression out of control.

• Given constant A2 = 0.5768 & D3= 0 and D4 = 2.114

SAMPLE NO.

1 2 3 4 5

1 25.00 25.01 25.00 25.03 25.01

2 25.00 25.03 25.00 25.04 25.03

3 25.01 25.02 25.02 25.03 25.02

4 25.01 25.02 25.02 25.01 25.04

5 25.02 25.02 25.03 25.03 25.00

6 25.06 25.03 25.02 25.00 24.99

7 24.99 24.98 25.02 25.02 24.99

8 25.02 25.01 25.01 24.99 25.02

9 25.03 25.01 24.97 25.01 25.03

10 25.02 24.99 24.99 24.98 24.97

SOLUTION _• THE mean of sample 1, mean X1

( X11 + X12 + X13 + X14 + X15 ) / 5

=( 25 + 25.01 + 25 + 25.03 + 25.01 )/ 5

= 125.05 / 5

= 25.01

For sample 1 highest measurement is = 25.03

lowest measurement is = 25.00

Range for sample1 = 25.03 – 25.00

= 0.03

Arithmetic mean & range of sample table SAMPLE NUMBER ARITHMETIC MEAN RANGE

1 25.01 0.03

2 25.02 0.04

3 25.02 0.02

4 25.02 0.03

5 25.02 0.03

6 25.02 0.07

7 25.00 0.04

8 25.01 0.03

9 25.01 0.06

10 24.98 0.05

CALCULATION OF GRAND AVERAGE AND MEAN RANGE

• GRAND AVERAGE ( X DOUBLE BAR)

_ _• X = (25.01+25.02+25.02+25.02+25.02+25.02+25.00

+25.01 + 24.98 ) / 10

= 25.01 _

Now, mean range R = (0.03+0.04+0.02+0.03+0.03+0.07+0.04

+0.03+0.06+0.05 ) / 10

= 0.04

Set up control chart for the average _ _ _• 1. central line = CLx = X =25.01_ _ _ _• 2. upper control line = UCLx = X + A2.R

=25.01 + 0.5768 * 0.04

= 25.01 + 0.0231

= 25.03

_ _ _ _

3. lower control limit =LCLx = X - A2.R

= 25.01 - .0231

= 24.99

Therefore average chart ( X - chart )

UCL

CL

LCL

Since the tenth average fallen outside the limit, so we can conclude process is not under control so, tenth average should be deleted and again test the control limits.

Control limit with the remaining data is:

_ _

• X = ( 25.01+25.02+25.02+25.02+25.02+25.02+25.00

+25.01+25.01 ) / 9

= 25.01

_

R = (0.03+0.04+0.02+0.03+0.03+0.07+0.04+0.03+0.06) / 9

= .039

_ _

Now, revised control parameters of X-R chart :

_ _ _

Central line = CLX = X = 25.01

_ _ _ _

Upper control line= UCLX = X + A2.R

= 25.014 + 0.5768 * 0.039

= 25.014 + 0.022 = 25.036 = 25.04

_ _ _ _

Lower control limit = LCLX = X + A2.R

= 25.014 – 0.022

= 24.99

Revised average chart ( X – chart )UCL

CL

LCL

Since all the process within the control limit so, the X – chart represents the process under control.

RANGE CHART ( R-CHART) _• Central line CLR = R = 0.039 = 0.04 _• Upper control limit = UCLR =_ D3.R = 2.115 * 0.039 =0.08• Lower control limit = LCLR = D3.R = 0 * 0.039 = 0

The values of R for the test data have been plotted into the chart within the control limit, hence variability is within under control.

UCL

CL

LCL

CONTROL CHART FOR ATTRIBUTE(p-CHART)

• Another form of inspection is where the items comprising the sample are classified into two types : ‘acceptable’ & ‘non-acceptable’ such as inspection is known as “inspection by attributes”.

• Inspection by attribute covers those situations where either :

(i) quantitative measurement are not possible as with the inspection of damages, scratches, paint runs, matching of colour against std. shade etc.

(ii) quantitative measurement consumes too much time.

The control chart which do this kinds of inspection is known as

control chart for fraction defectives ( p-Chart)

Basic theory underlying p-Charts• The central line of fraction is determined by average _

percentage defectives ( average fraction defective )= p

_ _ _ • Therefore control limit = p ± √ [p (1-p)] / n

• So, p is = no. of defectives / no. of pieces inspected

_• Therefore, p = ∑p / m

[ sample wise percentage defective / no. of sample ]

_ _ _

Upper Control Limit = p + 3.√ [p (1-p)] / n

_ _ _

Lower Control Limit = p - 3.√ [p (1-p)] / n

PROBLEM

• 10 Sample each of size 50, of a pipe were inspected in

pressure testing. The results of the inspection are given below:

• Draw a p-chart and state your conclusion.

Sample no. 1 2 3 4 5 6 7 8 9 10

No. of defectives

2 3 2 0 2 3 2 1 2 3

SOLUTION………..Sample no.

(1)No. of defectives

(2)Fraction defectives(p)(3) = (2) / sample size

1 2 0.04

2 3 0.06

3 2 0.04

4 0 0.00

5 2 0.04

6 3 0.06

7 2 0.04

8 1 0.02

9 2 0.04

10 3 0.060.40

Sum of fraction defectives = 0.40 _Therefore, Average fraction defectives = 0.40 / 10 = 0.04 ( p )

_ • (i) The central line = p = 0.04

_ _ _ • (ii) Upper control limit = p + 3. √ [p (1-p)] / n

= 0.04 + 3. √[0.04(1-0.04)]/ 50

= 0.04 + 0.083

= 0.123

_ _ _• (iii) Lower control limit = p - 3.√ [p (1-p)] / n

=0.04 – 3. √[0.04 (1-0.04)]/50

=0.04 – 0.083

= -0.043 = 0

(since ve− control limit taken as 0)

CL

LCL

UCL