Javier Garcia - Verdugo Sanchez - Six Sigma Training - W4 Monte Carlo Simulation

Page 1/2008 BB W4 Monte Carlo Simulation 07, D. Szemkus/H. Winkler

Monte Carlo Simulation

Week 4


How we can predict the distribution of the results?

How robust is the system?

– Goal: Estimation of a mean and a standard deviation for comparison with requirements and for further optimization

– Monte Carlo – Assessment of the current variation

– DOE - Analysis and optimization

How we can predict the distribution of the results?

How robust is the system?

– Goal: Estimation of a mean and a standard deviation for comparison with requirements and for further optimization

– Monte Carlo – Assessment of the current variation

– DOE - Analysis and optimization

Output ResponsesSimulationInput factors

controllable

Input factors

Not controllable

Statistical Analysis by Simulation


How does it work?

In the project phase “analysis” you have determined the effect of input factors on results. A model can be described by regressionor by DOE.

It exists an equation for physical or chemical relations.

With Minitab we are able to simulate the known or expected data distribution in order to predict the their importance on the results.

Based on this we can find out which changes of the input factorsare necessary to improve the result. We receive a good foundation to decide at an early stage that redesigns are necessary. At this stage they can be performed at an acceptable cost level.

Statistical Analysis by Simulation


5 stacked metal parts should not exceed a total dimension of 250,4 mm. The actual dimension for the mean of each sheet metal piece is 50 mm with a tolerance of +/- 0,1 mm.

The goal is to find out if the current capability is sufficient. This can be done by statistical tolerancing method but also by simulation.

We generate in Minitab 5 columns with normal distributed data with a mean = 50 and a StDev. = 0,033. In the second step we add the 5 columns and store the sum in the column c6. Now we check the process capability for 250, 4 mm max of the data in column c6.

Based on this results we decide if the specification for the metal parts is adequate in order to get a sufficient process capability.

An Example from Tolerancing


Step 1: Generate Data in Column 1 - 5

MP1 MP2 MP3 MP4 MP5

49,9740 50,0077 49,9886 50,0293 49,9222

49,9779 49,9806 49,9916 49,9520 49,9841

50,0383 49,9811 49,9391 50,0083 49,9832

50,0398 50,0404 49,9941 50,0323 50,0134

49,9718 49,9938 49,9770 49,9855 50,0204

50,0159 50,0278 50,0350 50,0237 50,0039


Calc

>Row Statistics…

Calc

>Row Statistics…

Step 2: Calculation of the Sum

MP1 MP2 MP3 MP4 MP5 Sum49,9740 50,0077 49,9886 50,0293 49,9222 249,92249,9779 49,9806 49,9916 49,9520 49,9841 249,88650,0383 49,9811 49,9391 50,0083 49,9832 249,95050,0398 50,0404 49,9941 50,0323 50,0134 250,12049,9718 49,9938 49,9770 49,9855 50,0204 249,94950,0159 50,0278 50,0350 50,0237 50,0039 250,10650,0141 49,9966 50,0242 49,9473 49,9945 249,977

Calc

>Calculator…

Calc

>Calculator…

or


Step 3: Capability Analysis

Stat

>Quality Tools

>Capability Analysis normal…

Stat

>Quality Tools

>Capability Analysis normal…

Is the long term capability = 1,75 as a prediction sufficient?Is the long term capability = 1,75 as a prediction sufficient?

250,4250,3250,2250,1250,0249,9249,8

USL

LSL *Target *USL 250,4Sample Mean 250Sample N 1000StDev (Within) 0,0766033StDev (O v erall) 0,0759855

Process Data

C p *C PL *C PU 1,74C pk 1,74

Pp *PPL *PPU 1,75Ppk 1,75C pm *

O v erall C apability

Potential (Within) C apability

PPM < LSL *PPM > USL 0,00PPM Total 0,00

O bserv ed PerformancePPM < LSL *PPM > USL 0,09PPM Total 0,09

Exp. Within PerformancePPM < LSL *PPM > USL 0,07PPM Total 0,07

Exp. O v erall Performance

WithinOverall

Process Capability of Summ


R2

R1

R3

R4U1

UoU2

-+

-+

R5

Part 1: Summing unit Part 2: inverting amplifier

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

2

2

1

134 R

U

R

URU

4

50 R

RUU a=

An Electrical Circuit


The voltage drop results from the following equation:

How does the voltage drop looks like?

(Addition of each single distribution)

Assumption:

R3 = R4 = R5 be 10 K ± 10%

R1 = R2 be 1K ± 10%

U1 be 50 mV ± 5 mv

U2 be 100 mV ± 10 mv

R5 R3U1 + R3U2

R4 R1 R2

Uo=

Establish the Equation


Distribution of the Initial Data

11001000900

40

30

20

10

0

R2

Fre

que

ncy

11010090

40

30

20

10

0

U2

Fre

que

ncy

11001000900

40

30

20

10

0

R1

Fre

que

ncy

555045

40

30

20

10

0

U1

Fre

que

ncy


Distribution of the Initial Data

11000100009000

40

30

20

10

0

R3

Fre

que

ncy

11000100009000

40

30

20

10

0

R4

Fre

que

ncy

11000100009000

30

20

10

0

R5

Fre

que

ncy


The mean value of U0 = 1,51 with a standard deviation of 0,18 Sigma; 6 sigma limits : 0.43 <= V0 <= 2.59

What values we get if the resistors have a +/- 5% tolerance?

The mean value of U0 = 1,51 with a standard deviation of 0,18 Sigma; 6 sigma limits : 0.43 <= V0 <= 2.59

What values we get if the resistors have a +/- 5% tolerance?

The Result

2160198018001620144012601080

Median

Mean

15151510150515001495

1st Q uartile 1381,3Median 1498,83rd Q uartile 1626,6Maximum 2194,4

1506,1 1513,2

1494,5 1503,9

177,2 182,1

A -Squared 13,76P-V alue < 0,005

Mean 1509,6StDev 179,6V ariance 32263,5Skewness 0,308720Kurtosis -0,107116N 10000

Minimum 1007,9

A nderson-Darling Normality Test

95% C onfidence Interv al for Mean

95% C onfidence Interv al for Median

95% C onfidence Interv al for StDev95% Confidence Intervals

Summary for Uo


U0 for resistor with 5% tolerance

Mean = 1.49

Standard deviation = 0.10

6 sigma limits:

0.89 <= V0 <= 2.09

U0 for resistor with 10 % tolerance

Mean = 1.51

Standard deviation = 0.18

6 sigma limits:

0.43 <= V0 <= 2.59

The Result, with a 5% Resistor Tolerance


R2

R1

R4

R3

What value has RTotal ?Define the 6 sigma limits

RTotal = 1/(1/ (R1 + R3) + 1/ (R2 + R4) )

R1, R2 Normal distribution: Mean = 100, Stdev = 5

R3, R4 Uniform distribution: lower limit = 90, upper limit = 110

A Similar Problem


The Data Sheet

R1 n (100;5) R2 n (100;5) R3 U 90-110 R4 U 90-110 R total

105,3533548 100,3022008 102,4032289 93,05186376 100,1483757

102,4196231 109,1777291 106,4016145 92,97866104 102,7173859

99,20740871 102,2156768 91,33668526 96,20693558 97,20178174

100,2706008 96,68913464 98,49202587 99,96066105 98,8502832

109,4715529 98,68895419 105,7885608 102,8158035 104,0777187

93,18341572 102,2955652 92,72075957 90,92933067 94,74692828

* * * * *

* * * * *

* * * * *

* * * * *


The 6 sigma limits are 83,85 <= RTotal <= 116,97The 6 sigma limits are 83,85 <= RTotal <= 116,97

The Result

109,2106,4103,6100,898,095,292,489,6

Median

Mean

99,9599,9099,85


99,855 99,960

99,823 99,946

2,641 2,715

A -Squared 0,87P-V alue 0,025

Mean 99,908StDev 2,677V ariance 7,167Skewness 0,016691Kurtosis -0,125713N 10000

Minimum 89,416





Summary for R total


Example: “Distance between screw holes”

0,0044 mm0,4 mmM4

0,0044 mm0,4 mmM2

VarianceToleranceMeasure

What are the tolerances for M0 with the given tolerances (µ, σ) for M2 and M4?

?? Theorem of PythagorasTheorem of Pythagoras


Approach 1: Statistical solution for“Distance between screw holes”

Calculate solution by using Gauss´ error propagation formula:

M0 = 18,03 ± 0,20 mm

assuming that the measurements G1, ..Gn are all uncorrelated.


Approach 2: Get the solutionwith the Monte Carlo Simulation

1. Generate random data for M2 and M4 with σ = 0,0663325

Calc

>Random Data>Normal…

Calc

>Random Data>Normal…

Calc

>Calculator…

Calc

>Calculator…

2. Calculate M0 with Pythagoras:


18,30018,22518,15018,07518,00017,92517,85017,775

Median

Mean

18,0287518,0285018,0282518,0280018,0277518,02750


18,028 18,029

18,028 18,029

0,066 0,067

A -Squared 0,28P-V alue 0,637

Mean 18,028StDev 0,066V ariance 0,004Skewness -0,0041889Kurtosis 0,0011327N 100000

Minimum 17,748





Summary for M0

Approach 2: Solve the problemwith the Monte Carlo Method

3. Analyze M0 with Descriptive StatisticsStat

>Basic Statistics>Display Descriptive Statistics…

Stat

>Basic Statistics>Display Descriptive Statistics…

The 3 sigma limits are 18,03 ± 0,20 mm = 17,83 <= M0 <= 18,23The 3 sigma limits are 18,03 ± 0,20 mm = 17,83 <= M0 <= 18,23

Simulation or calculation: The results are the same!

Engineering

Javier Garcia - Verdugo Sanchez - Six Sigma Training - W4 Monte Carlo Simulation