How to calculate the sigma level of a process and Relationship of DPMO

[jackylpt]

Hi there,

Now the six sigma is buzzword in many area. My question is how to calculate process sigma level.

For exmaple, suppose there is no shift in process, our process means is 100, taget is also 100,process sigma(by R/d2 calculating) is 1.8, specification is 90--110,

Question1: how to calculate sigma level?
is it (110-90)/6*1.8?
Question2: now I am also confusing how to calculate the cpk of this ?

thanks

michael

[Bob Ablondi]

The calculation of a Sigma level, is based on the number of defects per million opportunities (DPMO).

In order to calculate the DPMO, three distinct pieces of information are required:
a) the number of units produced
b) the number of defect opportunities per unit
c) the number of defects

The actual formula is:

DPMO = (Number of Defects X 1,000,000)

((Number of Defect Opportunities/Unit) x Number of Units)

Example:

A manufacturer of computer hard drives wants to measure their Six Sigma level.
Over a given period of time, the manufacturer creates 83,934 hard drives.
The manufacturer performs 8 individual checks to test quality of the drives.
During testing 3,432 are rejected.


Defects 3432 DPMO 5111.158768
Opportunities 83934 Sigma Level 4.1
Defect Opportunities per unit 8


Six Sigma Table: 1 690,000
2 308,000
3 66,800
4 6,210
5 320
6 3.4

[Marc]

Thanks for the help on this one, Bob!

[jackylpt]

Originally Posted by Bob Ablondi

The calculation of a Sigma level, is based on the number of defects per million opportunities (DPMO).

In order to calculate the DPMO, three distinct pieces of information are required:
a) the number of units produced
b) the number of defect opportunities per unit
c) the number of defects

The actual formula is:

DPMO = (Number of Defects X 1,000,000)

((Number of Defect Opportunities/Unit) x Number of Units)

Example:

A manufacturer of computer hard drives wants to measure their Six Sigma level.
Over a given period of time, the manufacturer creates 83,934 hard drives.
The manufacturer performs 8 individual checks to test quality of the drives.
During testing 3,432 are rejected.


Defects 3432 DPMO 5111.158768
Opportunities 83934 Sigma Level 4.1
Defect Opportunities per unit 8


Six Sigma Table: 1 690,000
2 308,000
3 66,800
4 6,210
5 320
6 3.4

Dear Bob,

thank you so much for your infomation. I have anther way to calcuate it, but the result is different. Would you please give me some suggestion?

DPU: 3432/83934=0.0408
Y=FPY=e(-DPU)=e(-0.0408)=0.96
probability of a defect {P(d)}=1-Y=1-0.96=0.04
using Z value table is 1.75 sigma

so it is different.

thanks

Michael

[Bob Ablondi]

Originally Posted by jackylpt

Hi there,

Now the six sigma is buzzword in many area. My question is how to calculate process sigma level.

For exmaple, suppose there is no shift in process, our process means is 100, taget is also 100,process sigma(by R/d2 calculating) is 1.8, specification is 90--110,

Question1: how to calculate sigma level?
is it (110-90)/6*1.8?
Question2: now I am also confusing how to calculate the cpk of this ?

thanks

michael

The method shown is with a plus/minus 1.5sigma shift based on Motorola's allowance for process variation in manufacturing.

Bob

[Tim Folkerts]

That pesky 1.5 sigma shift sure makes life difficult! And the question of what an "opportunity" is!

There is a "Sigma calculator" you can use at
http://www.isixsigma.com/sixsigma/si...asp?m=advanced

If you have a situation with...
* 83934 units
* 3432 defective units
* each unit is 1 opportunity

Then you have 4.1% defective = 41,000 DPMO. The calculator above says that if the process is centered then this is a 1.74 sigma process. Using the "standard" 1.5 sigma shift says this is a 3.24 sigma process. (these numbers aren't too hard to come up with - they assume a normal distribution that would produce a similar number of defects at the closer spec limit.)

If you considered each test as an opportunity, then you would have 8 times as many opportunities, and a correspondingly better sigma value.


For the example you gave, with x-bar=100, limits at 90 & 100, and sigma = 1.8, you are currently (110-100)/1.8 = 5.56 sigma from the closest spec limit. However, since the "standard" calculation assumes that the process might drift 1.5 sigma, i.e the center might drift by 1.8*1.5 = 2.7. You would then be only (110 - 102.7)/1.8 = 4.06 sigma from the closer spec limit. This would, I believe, be the "official" sigma level for the process.


Tim F

[Arvind]

http://elsmar.com/Forums/showthread....131#post114131

[CarlosDelgado]

Originally Posted by Tim Folkerts

That pesky 1.5 sigma shift sure makes life difficult! And the question of what an "opportunity" is!

There is a "Sigma calculator" you can use at
http://www.isixsigma.com/sixsigma/si...asp?m=advanced

If you have a situation with...
* 83934 units
* 3432 defective units
* each unit is 1 opportunity

Then you have 4.1% defective = 41,000 DPMO. The calculator above says that if the process is centered then this is a 1.74 sigma process. Using the "standard" 1.5 sigma shift says this is a 3.24 sigma process. (these numbers aren't too hard to come up with - they assume a normal distribution that would produce a similar number of defects at the closer spec limit.)

If you considered each test as an opportunity, then you would have 8 times as many opportunities, and a correspondingly better sigma value.


For the example you gave, with x-bar=100, limits at 90 & 100, and sigma = 1.8, you are currently (110-100)/1.8 = 5.56 sigma from the closest spec limit. However, since the "standard" calculation assumes that the process might drift 1.5 sigma, i.e the center might drift by 1.8*1.5 = 2.7. You would then be only (110 - 102.7)/1.8 = 4.06 sigma from the closer spec limit. This would, I believe, be the "official" sigma level for the process.


Tim F

I cant accept your apreciation as simple as you say

If is a continuous process, How do you calculate DPMO ..... its depen of the kind of the testing, testing frecuency ........... for me yield must be equal to sigma levels in cetered, normal distribution variable ................... if that we can use the same formula said (Upper Level less media)/sigma.......... we must say exactly what we got real in our process and it work for us ........ theory is not required please...... I dont know if you get my email but is carlos.delgado@ca.sabmiller.com and cdelgadonieto@yahoo.com........... whenever you want to dicussed it please send it ....... if you know about 6sigma and know your process we can learn join

Best regards
Carlos Delgado