jpm tqm course mat-4 t-3 imba 2013

7/30/2019 Jpm Tqm Course Mat-4 T-3 Imba 2013

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Control ChartProblem-1For the quality measurement under Inspection process, if the sub-group Sample size is

14, no. of sub-group is 20, sum of measurements of the sample is 357.50 and Sum of

the sample-ranges is 9.90, (i) Indicate Center Line, UCL & LCL for X-chart & R- Chart,

(ii) Comment on the Process Capability and Whether the process is under control, if

specified dimension is 17.5 0.5. (iii) Suggest possible ways for improving thesituation.

[Given A2 = 0.18, d2 = 3.735, D3 = 0.41, D4 = 1.59]

Solution : = Process Mean of measurements, X = X N = 357.50 / 20 = 17.875

_Mean Range, R = R N = 9.90 / 20 = 0.495

(i) For X-chart, = = _Central Line = X =17.875 , UCL = X + A2R= 17.875 + 0.18 x0.495 = 17.9641

LCL=17.875 0.18 x0.495 = 17.7859

For R-chart _ _Central Line = R =0.495 , UCL = D4 x R = 1.59 x 0.495 = 0.78705

_LCL = D3 x R = 0.41 x 0.495 = 0.20295

_(ii) Standard Deviation of Universe, =R / d2 = 0.495 / 3.735 = 0.13253

Natural Spread (Band) = 6 = 6 x 0.13253 = 0.79518

and 3 = 0.39759


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Solution -1(contd.) Control Chart=Upper Natural Tolerance Limit, UNTL = X + 3 = 17.875 + 0.39759 = 18.2726

=Lower Natural Tolerance Limit, LNTL = X - 3 = 17.875 - 0.39759 = 17.4774

=Center Line of natural Spread , X = 17.875,

Specified dimension =17.5 0.5Upper Specified Limit, USL = 17.5 + 0.5 = 18.00Lower Specified Limit, LSL = 17.5 - 0.5 = 17.00

Specified Spread= 18 17 =1.0Center line of Specified Spread = (18 + 17)/2 =17.5

Process Capability Ratio, Cp = Specified Spread = 1.0 . > 1 , but < 26 0.79518

so the process is marginally capable, and capableonly ifthe Process CenterLine is placed on or very close to the Center Line of Specified Spread. But there is

slight off-set between the center line of specified spread (17.5)and Process Center Line

(17.875). Now,UNTL > USL , UNTL is not within the specified limit,so considerable rejections are

expected which

indicates that the process is not under control.LNTL > LSL , LNTL is within the specified limit,so no rejections are expected which

indicates that

the process is under control.

(iii) Suggestion :1) Process is marginally capable, so process improvement is recommended2) Designed Process Center-line must be set very close to Center Line of

the specified spread3) Specified Spread is to be relaxed to some extent, if possible.


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Control Chart

Problem-2

For the quality measurement under Inspection process, if the sub-group Sample size is

4, no. of sub-group is 20, sum of measurement means of the sample is 412.83 and Sum

of the sample-ranges is 3.39, (i) Indicate Center Line, UCL & LCL for X-chart & R- Chart,

(ii) Comment on the Process Capability and Whether the process is under control, if

specified dimension is 20 0.2. (iii) Suggest possible ways for improving thesituation. [Given A2 = 0.73, d2 = 2.059, D3 = 0, D4 = 2.28.]

Solution-2 : = Process Mean of measurements, X = X N = 412.83 / 20 = 20.6415

_Process Mean of Range, R = R N = 3.39 / 20 = 0.169

_

Standard Deviation of Universe, =R / d2 = 0.169 / 2.059 = 0.082Natural Spread (Natural Tolerance range) = 6 = 6 x 0.082 = 0.492Specified Spread = 0.2 (-0.2) = 0.4

Process Capability Ratio, Cp = Specified Spread = 0.4 .


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Control Chart

Problem-3

For the quality measurement under Inspection process, if the sub-group

Sample size is 10, no. of sub-group is 18, sum of measurements of the sample

is 595.8 and Sum of the std. Deviations is 8.24, Estimate

(i) Control Line, UCL & LCL for X-chart & s - Chart.

(ii) Process Capability and whether the process is under control if specifieddimension is 32 2,

(iii) Suggest possible ways for improving the situation.

[Given

A3 = 0.98, A2 = 0.31, C2 = 0.9727, d2 = 3.078, B3 = 0.28, B4 = 1.72 D3 = 0.22, D4 = 1.78]

Solution : =

Process Mean of measurements, X = X N = 595.8 / 18 = 33.10_Mean Std. Deviation, S= S N = 8.24 / 18 = 0.4577

_Population Standard Deviation, =s / C2 = 0.4577 / 0.9727 = 0.496

Process Capability (Natural Tolerance range) = 6 = 6 x 0.496 = 2.976

For X-chart,= = _

Control Line = X =33.10, UCL = X +A3s= 33.1 + 0.98 x 0.4577 = 33.5485LCL = 33.1 0.98 x 0.4577 = 32.6515

ForS -chart_ _

Control Line = s = 0.4577 , UCL = B4 x s = 1.72 x 0.4577= 0.7872_

LCL = B3 x s = 0.28 x 0.4577 = 0.12815

jpm tqm course mat-4 t-3 imba 2013

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