# Production Yield

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- This topic has 4 replies, 5 voices, and was last updated 10 years, 10 months ago by Sap.

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- January 19, 2006 at 7:01 pm #24466

ConfusedParticipant@Confused**Include @Confused in your post and this person will**

be notified via email.Hello all

I was wondering if someone can stir me the right direction. I working for a company at the moment and they want me to calculate the overall yield. Currently the formula being used is output / input. I was looking at the roll throughout yield it says to multiply the subprocess yields together… i’m getting confused at the moment let say i’ve got 5 station and the input / output is .8, .9, .8, .9, .8 which gives me rty of .46 but the average of the 5 stations is .86. what is this information telling me, no matter what my sub process yields are my overall process is still going to be less than .5 or is that different.

0January 19, 2006 at 10:41 pm #58705

J CastagnaParticipant@J-Castagna**Include @J-Castagna in your post and this person will**

be notified via email.Hi Confussed

Yes its depressing isnt it!

This isnt my area of expertise, but I think it has a few great lessons.

Firstly, it shows the value in looking at the entire system rather than the individual process in isolation. Starting with 100 widgets Only 46 will actually get through unscathed with out rework or scrapping. Most people seeing the process working at full tilt not the rework being done to fix prior problems.

Secondly it shows the value in having 6 sigma (or 99.99966%) as a process goal.

Starting with a 100 widgets ending with a 100 widgets!

Thirdly, most people naturally use arithmetic means. They dont use geometric means naturally as they generally find them difficult to understand and use. But when they start seeing things from a new perspective they generally prefer them.

Cheers

Justin0January 19, 2006 at 10:48 pm #587061) You can’t average percentages.

2) Think about your process using numbers.

At station one you input 100 widgets and at a throughput of .8 you would have 80 widgets that move on to station two.

At station two you input 80 widgets and at a throughput of .9 you would have 72 widgets that move on to station three.

At station three you input 72 widgets and at a throughput of .8 you would have 57 widgets that move on to station four.

At staion four you input 57 widgets and at a throughput of .9 you would have 51 widgets that move on to station five.

At station five you input 51 widgets and at a throughput of .8 you would have 40 widgets that made it through all 5 stations.

When I multiply 0.8*0.9*0.8*0.9*0.8 I get 0.417. Since I truncated instead of rounded at each station the math works.

Make sense?0January 23, 2006 at 11:48 pm #58712

thevillageidiotMember@thevillageidiot**Include @thevillageidiot in your post and this person will**

be notified via email.This might prove useful…..Calculate the defects and the number of units at each process. Now calculate the defects per unit (DPU) for each process. You can simply add the DPU for all processes now and use the inverse natural log of the negative DPU to give you RTY for the system as a whole…..Beautifully simple and easy to understand.

Example:

10 defects out of 100 units = .1 DPU

20 defects out of 100 units = .2 DPU

30 defects out of 100 units .3 DPU

.1 + .2 +.3 = .6 DPU

Inv (ln) -.6 = .54 * 100 = 54% RTY

Good luck and verify.

0March 18, 2009 at 8:30 am #59571Hi Doug,

In case we have 2 different inputs, then we ned to calculate the yield for them separately?

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