how to calculate this RTY?
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November 4, 2004 at 7:05 am #37440
How to calculate the RTY?
Suppose we have an order to complete 100 of product A. There will be 3 steps to complete. The output is as below:
1st step: 10 scrap, 10 need rework.
2nd step: in order to fulfill the order, we add 10. But still 5 scrap, 5 need rework.
3rd step: in order to fulfill the order, we add 5. But still 8 need rework.0November 4, 2004 at 10:54 am #110247
arvind pathakParticipant@arvindpathak Include @arvindpathak in your post and this person will
be notified via email.My understanding of FPY is that by definition it excludes scrap and rework at whatever stage it comes.
Step 1 :
(10010scrap10 rework)/100=80/100 = 0.800
Step 2 :
((80+10 additional)5 rework 5 scrap)/((80+10 additional)=80/90 = 0.889 >Here the assumption is that 10 additional products have already passed through step 1 at some point of timewith zero defect..
Step 3 :
((80+5 additional)8 rework)/ (80+5 additional)=77/85=0.906>Here the assumption is that 5 additional products have already passed through step 1 & 2 at some point of time with zero defect.
Now RTY (Unit based) can be calculated as below.
RTY = FPY of Step1 x FPY of Step2 x FPY of Step 3
= 0.800 x 0.889 x 0.906
= 0.644 (64.4%)
Whatever we follow we should be consistent as I have seen some messages where if rework is done say at step 2 the same is added into numerator and focus is only on scrap, but I am dead against it! let’s assume there is no facility of rework now think how many good units(nondefective) are deliverable to the customer?Any views!!
0November 4, 2004 at 12:21 pm #110252Hi Sir,
Thanks for your quick response. But your answer doesn’t convince me. And it is difficult for me to give my opinion against your answer. Based on your knowledge, in such a case, how is the whole process FTY (you call this as FPY)?
In this example, it is totally different from the others illustrated in this forum. And different people around me will give me different answers. So, I am confused.0November 4, 2004 at 12:46 pm #110253Using the following formulas, I get:
Throughput yield of the single processes
Ytp1 = exp(dpu1) = exp((scraps+reworks)/units)=exp((10+10)/100)=0.8187
Ytp2 = exp(dpu2) = exp((5+5)/110)=0.9131
Ytp3 = exp(dpu3) = exp((8+0)/115)=0.9328
Rolled throughput yield:
RTY = Ytp1*Ytp2*Ytp3 = 69.73%
Or, it is the same:
RTY = exp(dputot) = exp(((10+10)/100+(5+5)/110+(8+0)/115)) = 69.73%
0November 4, 2004 at 1:11 pm #110254Both answers you have been given are wrong. You want to argue that it doesn’t match your opinion, but you did not tell anyone what your opinion is. What’s up with that?
You have given us information on defectives, not defects. RTY assumes defect info. The best answer that can be given is RTY = Y1xY2xY3. In your case, this is .8 x .9 x .92 = .66.
This assumes that your rework was 100% successful and the “added” units had no defects in the steps before they were added.
0November 4, 2004 at 1:15 pm #110255
arvind pathakParticipant@arvindpathak Include @arvindpathak in your post and this person will
be notified via email.Fer,
Your calculation is relevant for Defect based RTY.Here Michael has asked for Unit (Defective) based RTY.
However , a defective unit may have a single or multiple defects. If no. of defects are being counted in Units, then definitely it makes sense to calculate DPU and exp(DPU)!
Correct me Michael if I am wrong. Don’t get confused. Discussion is on. Difference of opinion would be in putting numerator and denominator!!0November 4, 2004 at 1:25 pm #110257
arvind pathakParticipant@arvindpathak Include @arvindpathak in your post and this person will
be notified via email.Stan,
Why do you include reworked unit in Numerator and Denominator? Imagine what happens if rework is not being done.How many good units can be passed to the customer first time?How did you arrive at 0.9 & .92? Could you specify N/D with some explanation. That would bring in more clarity and learning to all of us!0November 4, 2004 at 1:49 pm #110259I wrongly considered the units, but I don’t understand why we can’t use the formula RTY=exp(total dpu)
The dpu for each process is:
dpu1=0.2
dpu2=0.1
dpu3=0.08
So I should have:
RTY=exp(0.38)=0.684
Where am I wrong?
0November 4, 2004 at 1:51 pm #110260Again, not enough information has really been given, but the assumprtion is that we are adding units to keep the quantity at 100. If so, the rework is also being done. The denominator at every step is 100 because of the added units.
0November 4, 2004 at 1:55 pm #110262Because you only know defectives, not defects. Defects are greater than or equal to defectives, but you don’t know.
0November 4, 2004 at 3:05 pm #110266If you are not careful you will spend more time calculating the RTY than actually solving the problems.
0November 4, 2004 at 3:58 pm #110273
Mike CarnellParticipant@MikeCarnell Include @MikeCarnell in your post and this person will
be notified via email.Michael,
You do seem to be confused.
Nobody has the job of convincing you. If you won’t provide your opinion then you need to figure it out for yourself.
Good luck0November 4, 2004 at 10:28 pm #110289
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.Let me take a shot:
Step 1: 100 go in, 20 are bad with 10 scrapped and 10 reworked. According to the RTY calculations the yield of this step is 80/100 =.8
Step 2: 90 go in (80 good from step 1 and 10 reworked), 10 are bad with 5 scrapped and 5 reworked. The RTY yield at that step is 80/90=.89
Step 3: 85 go in (80 from step 2 plus 5 reworked), Assume no scrap but 8 rework. The RTY yield at that step is 77/85=.906
Final output is 85. Traditional yield would say 85/100=.85
RTY is .8 x .906 x .85 = .61608.Everybody good with this?0November 4, 2004 at 10:58 pm #110293
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.The underlying concept of RTY is twofold. First, the yield per step is based on how much comes in and how much is produced correctly the first time. Thus if 100 came in but 10 were bad then the yield is .90. But, if there is rework then we must be realistic and allow it to flow to the next step. So we adjust the denominator to reflect the good plus rework coming out of the previous step. But only count the good stuff of the next step over that denominator. This way we give credit for the rework…they are now good, but penalize the yield of that step for the rework. It is a compromise of the unrealistic nature of first pass yield and the deception of traditional yield. In your example, as others pointed out, dpu isn’t applicable since you are dealing with they were either good or bad, that is defectives. Hope this clarifies.
0November 5, 2004 at 1:12 am #110300Thanks for all of your inputs. The question was asked during an internal training class, and I can not convince them. In fact, this scenario in the example happens everyday in our plant.
My understanding is:
0.8x(90/100)x(92/100)=0.66
I think it is more realistic, because:
In real situation, we may not know whether each step adds additional parts in order to keep the final output (in our example, it is 100piece).
Second, lets say, the 3 steps were done in 3 different plants. When you collect the data, what figure will they give you?
Thirdly, if theres many scraps, we also can have the hint from inventory control to drive us to solve the real problem.
I hope more people can join this discussion, then we may have an universal sense. Discussion makes us learn from each other.0November 5, 2004 at 1:21 am #110301I am good with it.
Question: He asked about FPY to start with. In my mind, FPY = RTY. Would all agree?0November 5, 2004 at 9:37 am #110325Darth,
theres a point that is still not clear to me
You wrote:
RTY = prod (Yi)
Ok, but whats the right definition of Yi?
You have used this definition: Yi = 1(scraps+reworks)/units
And you get: RTY = .800 x .906 x .850 = 0.61608
I was taught that the Yi that should be used is throughput yield, based on DPU:
Yi = exp(dpu) = exp((scraps+reworks)/units)
The calculation of defects is the same (scraps+reworks), but, according to the above definition, I get
RTY = 0.8187*0.8948*0.9102 = 0.6668
Can you help me clarify this point?Thanks a lot0November 5, 2004 at 12:15 pm #110327The thing that is wrong with your logic is you assume that all defectives come from the original population of 100.
Think about it – Original 100 yield 61% and added 15 yield 100%.
Odds are that several of the defectives at step 2 & 3 come from the added units.
The best information we have (limited sample of 100 at every step) is that step 1 yields .8, step 2 yields .9, step 3 yields .92 for a RTY of .66.0November 5, 2004 at 1:24 pm #110332
Arvind Kumar PathakParticipant@ArvindKumarPathak Include @ArvindKumarPathak in your post and this person will
be notified via email.Darth,
That’s exactly what I calculated in my reply (see second thread). Concept wise we are on the same page.But probably by mistake you typed “.8 x .906 x .85=.61608 ” instead of “0.8 x 0.89 x0.906 =0.645”. I would like to ask Stan if he is convinced by us (Darth & me).
We should get the purest value of RTY where rework should not be considered. Michael, I repeat don’t get confused with exp(DPU). Many of us has clarified this is not applicable for your problem which is defective based, not defects based.
Expecting a polite reply from Stan.
0November 5, 2004 at 1:38 pm #110334Nope – you are wrong.
How can you assume all failures in step 2 and step 3 are from the original units?
It is a bad assumption and odds are some of the failures are from the added units.
This ain’t rocket science. We have incomplete information, so you either get better information, or go with the most likely scenario.0November 5, 2004 at 2:14 pm #110335I agree with Stan.
RTY=66%
FTY=92/100=92%.(not 100% as Stan gives)
Because the calculating is simpler, and it will simplify many of our statistical job, more importantly, it can be suited for many scenarios without thinking whether there is scrap or new launch of material.
If we follow Arvind’s rule, probably it will be a time & labor consuming job as we usually have more than 10 steps to finish a product. But I am not saying Arvind’s calculating is wrong or right.0November 5, 2004 at 3:03 pm #110336What’s the RTY if starting from 100 units we have a bad situation like this?:
1st step: 50 scrap, 50 need rework.
2nd step: in order to fulfill the order, we add 50. But still 50 scrap, 50 need rework.
3rd step: in order to fulfill the order, we add 50. But still 50 scrap, 50 need rework.
Let’s assume the rework is 100% successful0November 5, 2004 at 3:10 pm #110337
K. SubbiahParticipant@K.Subbiah Include @K.Subbiah in your post and this person will
be notified via email.Gents:
Here are the key points in this analysis: Thanks.Step 1.
Step 4.Determine the FTY for each operation
Current State VerficationFTY = 1((number of scrapped pcs + number of ocs reworked)/(Total number of pcs produced)
First Time Yield at Operation 1:Operation 1 FTY:
1210.8
First Time Yield at Operation 2:Operation 2 FTY:
1090.9
First Time Yield at Operation 3:Operation 3 FTY:
1000.92
Step 2.
Step 5.RTY = FTY1xFTY2xFTY3
Reverse the problem and check how the line works0.6624
RTY = 0.6624, implies that the 3 operations must have the following FTY:Step 3.
0.872In order to 100 good parts from a stream with 3 operations and an RTY = 0.6624,what is the total number of parts (gross) that need to be produced at the first operation?
Future State with Average FTY = 0.872151
First Time Yield at Operation 1:
132
First Time Yield at Operation 2:
115
First Time Yield at Operation 3:
1000November 5, 2004 at 3:44 pm #110338Guys, Guys . You are confounding the ideal of a RTY with the need to fill an order and getting yourselves lost in the numbers. The purpose of RTY discussions center on exposing the hidden factory in which rework gets added back into the mix inflating the typical reported yield of a multistep process. Its based on the idea of 1st pass quality. Stans point about not having the data to answer this question is right on the money. In this discussion, past the first step you lose your ability to determine what the step yield is from the original parts and therefore have to make assumption on which parts were defective in the step the original part or a reworked part. In a RTY calculation you dont add more (reworked?) parts to the denominator, calculate the step yield, and apply it to the process RTY. If you do you just continue the process of hiding the factory.
0November 5, 2004 at 7:52 pm #110346Ted,
You seem to be more convincing.Can you give a detailed calculationcumexplanation based on your concept you just said.The example will further support your stance.
So far Arvind’s approach is exciting me…the purest RTY.
SSMBB0November 6, 2004 at 2:32 am #110351
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.Hey Jones, you know better than that. In fact, it was you who taught me the difference between FPY and RTY and how to do the calculations correctly. Or has Ivan and old age addled your brain? Pretty soon you and Rubberdude will be back in AR, rockin on the porch, suckin on toads and talkin bout the good ole days. Btw, possible CD in CLT 11/18 will keep you posted. Pass the word.
0November 6, 2004 at 4:42 am #110354Yes I do know better. Problem is, “FPY” in the classic sense assumes rework is not possible to get a theoretical result (presuming rework is possible in the application of the discussion). I was not clear making my point: That he started out calling it FPY but everyone else is talking RTY… So I was awkwardly confirming that we were talking about RTY, not FPY…
Drop a line w/ details to pass around.0November 8, 2004 at 7:03 am #110382the answer is that RTY = FPY_step1*FPY_step2…FPY_stepn
FPY represents the probability of a something passing through that particular step
RTY represents the probability of something making it through ALL steps0November 8, 2004 at 2:30 pm #110407That’s easy – RTY = 0
0November 8, 2004 at 3:08 pm #110410
GaneshpsParticipant@Ganeshps Include @Ganeshps in your post and this person will
be notified via email.Hi
Iam a proof reader…
how can i implement six sigma in proof reading….
how to calculate the number of oppurtunities…
Pls enlighten me..0November 8, 2004 at 3:09 pm #110411SSMBB.
For starters, I hope you are not really a MBB. To get excited about a wrong answer and call it pure is scarey.
Let’s take the original example.
Step 1 – 100 units, 80 good, 10 are reworked and passed on to step 2, and 10 are scrapped. Yield is 80%. Easy, should be no disagreements. it is definitely not the 81.9% as suggested by Fer.
Step 2 – 100 units, 80 good from step 1, 10 reworked from step 1, and 10 “added” units which we are to assume had no defectives when they too went through step 1 (odds are there were two). We have 10 defectives, 5 of which are scrapped, and 5 are reworked. Arvind’s approach wants us to assume all of the defectives come from either the 80 good from step 1 or the “added” units. The flaw in this logic is that the reworked units don’t fail, but the odds are that one of the defectives come from the reworked units and one comes from the “added” units. Yield is 90%. It is not the 91.3 as suggested by Fer or the .89 as suggested by Darth. What the original post told us is that 100 units went through step 2 and 10 failed. There was nothing to tell us that the reworked units or the “added” units had some mystical power that allowed them not to fail. Yield 90%.
Step 3 – What we know is that 95 units come forward from Step 2. 5 have been reworked. We don’t know if any of the reworked units were from the rework from step 1, the “added” from step 2, or from the original 100 units. It could be from any of those, but the most likely scenario is that 4 are from the original units and 1 is from either the reworked from step 1 or the “added” units. We also have 5 more “added” units which we have no information on their performance is step 1 or step 2, but it is likely that they had at least 1 defective as they passed through step 1 and 2. 8 units fail and are reworked. Yield 92%, not the 90.6% as suggested by Darth or the 93.3% suggested by Fer. For Darth calculation to work, we are assuming that reworked and “added” units cannot fail. Odds are at least 2 of the defectives at this step are reworks from previous steps or “added” units.
This is easy – we take the data we are given and use it. We know from a limited sample of 100 units at each of three process steps that step 1 yields .8, step 2 yields .9 and step 3 yields .92 for a RTY of 66.24%.
The issue with Fer’s approach is that it assumes defects and we are given defectives. We don’t have a clue on how many defects make up the defectives (although some logical assumptions can be made and we ought to be able to test the assumptions).
To make assumptions about whether defectives at subsequent step are from pristine, reworked, or “added” units is crazy. If you believe this, you should also believe Praveen Gupta’s conjured sigma level concept.0November 8, 2004 at 3:45 pm #110419
Habib SdidiquiParticipant@HabibSdidiqui Include @HabibSdidiqui in your post and this person will
be notified via email.I agree with Arvind Pathak that the RTY for the system described is 64.4% (based on the assumptions made about rework).
0November 8, 2004 at 3:48 pm #110421You and Arvind are wrong.
0November 8, 2004 at 5:10 pm #110427
arvind pathakParticipant@arvindpathak Include @arvindpathak in your post and this person will
be notified via email.Stan,
A great reply.Now let’s look back at definition of First Pass Yield and Rolled Throughput Yield and discuss the same.
Are you not adding 10 reworked units in step 2 for calculating FPY? So what 100 units are to be there in step 2 to fulfill order requirement;We must deduct no. of reworked units from Denominator & Numerator at evry step for assessing how many good parts would go to the customer first time.It’s not necessary to keep 100 as Denominator always.Here the emphasis is on not to consider no. of reworked units, not on the logic”the reworked units don’t fail”.If we consider reworked units in calculation, it’s no more FTY, you have already incurred cost in making the “bad” “good” !
PS:Please see my post( the second one) again.0November 8, 2004 at 5:32 pm #110430I see different opinions on the same (apparently) simple calculation. In a real case what could be the final unassailable proof of the real process capability?
0November 8, 2004 at 6:44 pm #110435
Mike CarnellParticipant@MikeCarnell Include @MikeCarnell in your post and this person will
be notified via email.Arvind,
There are a couple serious issues with your logic:
1 You seem to assume that the rework process is not part of the process. It is.
2. You assume that rework makes bad units good. That certainly is not the case. If you have ever dealt with electronics you know that the probability of failure increases the minute you remove it from the production line and begin to handle it.
Stan has the most valid model in this situation.
Just my opinion.
Good luck.0November 8, 2004 at 7:38 pm #110446
Habib SdidiquiParticipant@HabibSdidiqui Include @HabibSdidiqui in your post and this person will
be notified via email.After reading Stan’s remark about Arvind and I being wrong, I took a look at Stan’s calculation, with the assumptions he has made. With the assumptions he makes, Stan is right in his calculation of RTY.
RTY is a cumulative multiplier of FPY from each step in a multistep process. So, the calculation boils down to what are the individual FPYs.
The original problem statement from Michael, however, because of a lack of enough descriptive element within it (say, e.g., how the reworked materials and newer additions were made), could open up the gate for reinterpretation of the statement. And this is what happened. This led to Arvind and I initially look at it from a different angle than what Stan has suggested. Sorry for any misunderstanding on this.0November 8, 2004 at 9:19 pm #110460
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.The correct answer for all of this lies in which assumptions you make. Since we all have made different assumptions, the answers will likely conflict. My assumptions were simple. Each item going back into the process for rework will pass. The FPY for each subsequent step will include in the denominator the previous steps good+rework although the yield for the previous step only includes those good first time through. Sure, we could assume some probability of failure for the reworked items but we don’t have that data. I assumed that “added” was the bad going back for rework not any additional pieces brought in from the “outside”. Realistic, who cares, all of this is hypothetical anyway but it is the assumption behind my calculations. Sort of like Harry’s 1.5 shift is valid given his assumptions, realistic or not.
0November 8, 2004 at 9:26 pm #110463
FTY_LearnerParticipant@FTY_Learner Include @FTY_Learner in your post and this person will
be notified via email.A printing press makes invitation cards(Unit).Defects are well defined by customers.There are three steps
Step1: Prepare Card by person A
Step2: Review the Card by the novice reviewer B
Step3: Review the same card by the expert reviewer C
100 cards are prepared by A. The reviewer B reviews and finds 5 defective cards with total 10 defects. These defective cards are sent back to A for correction.Now all 100 cards(inclusive of reworked 5 cards) are reviewed again by the expert reviewer C. He finds 2 defective but 12 defects.C sends these defective cards to A for correction. These corrected cards are first reviewed by the novice B and then by the expert C again and found 100% OK. But customers returns 2 cards with 4 defects.What would be the the First Time Yield of every step both unit and defects wise.
The reviewer B sometimes may not catch an error and sometimes may state an error which actually is not an error as certified by the expert reviewer C. C’s review is final and assumed as the authentic one.
Which approach is better to assess the true RTY and why?
Approach 1:
Unit Based FTY: step 1 : 95/100
step 2 : 98/100
step 3 : 98/100
Defect based FTY : Step 1 : exp(10/100)
Step 2 : exp(12/100)
Step 3: exp(4/100)
Approach 1:
Unit Based FTY: step 1 : 95/100
step 2 : 93/95
step 3 : 91/93
Defect based FTY : Step 1 : exp(10/100)
Step 2 : exp(12/100)
Step 3: exp( 4/100)
Approach 3: Any Other?0November 8, 2004 at 10:43 pm #110465You need to go back and reread the original question. The denominator for each step is 100.
0November 9, 2004 at 1:47 am #110474
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.I have reread the original post and replicated your answer. I bow to your brilliance.
0November 9, 2004 at 1:56 am #110476Hi All,
Actually my question has been simplified. But I think originally it will be better for you to give me an answer.
Since the confusion was raised, let me give you more detail. But it may confuse you more.
Revised information:
Suppose we have an order to complete 100 of product A. Product A includes 12 assembly components. Among them, there are 3 significant components which will be impossible to be dismantled and reworked after assembly. There will be 3 steps to complete it. 4 components (includes 1 significant components) will be assembled in each step.
The output is as below:
1st step: 10 scrap, 10 need rework. We rework them and add the new to ensure there are 100pcs passed to step 2. Step 1 wont tell step 2 whether there are any reworked parts passed over.
2nd step: 5 scrap, 5 need rework. We rework them and add the new to ensure there are 100pcs passed to step 3. The owner of step 2 wont care whether the 100pcs from step 1 contains reworked parts.
3rd step: 8 need rework. We rework them successfully.
Lets assume the added components and the reworked parts can be successfully done at first time. But frankly speaking, to fulfill the order is our ultimate goal. In other words, the reworked and added components may not be successfully done at the first time.
Obviously, we are talking about defective not defects.0November 9, 2004 at 12:32 pm #110492RTY – The likelihood, in a percentage, that any given unit can go through an entire process or procedure without receiving any defects. i.e. Defect free processes.
You don’t have defect calculation, you have defective products, so you output only 62 good products from 100. You start with 100 and follow them up to the end of steps.( Don’t take these 15 products in your calculation – mark these). You must remember for what you need this metric and what you want to know. You can look an example in isixsigma/dictionary/RTY
Kalju0November 11, 2004 at 2:09 pm #110591Don’t do that – you’ll take all of the fun out of this.
0October 18, 2007 at 1:49 pm #163342
casperlohParticipant@casperloh Include @casperloh in your post and this person will
be notified via email.Stan,
I agree with your analysis, but I still have some questions on your explanation.
1)If the reworked parts are taken into consideration and brought to the following process, why dont we consider these successfully reworked parts in its original process?
2) As you state, the output of every steps should be kept as 100units as required by the problem.Why the total no. of parts that you assume is 92 instead of 100?
In your calculation, the successfully reworked parts in step 1 are taken into account in step 2 but not in step 1; the successfully reworked parts in step 2 are considered in step 3 but not in step 2; the successfully reworked parts in step 3 are not considered.
The most important problem is the successfully reworked parts in steps three. Shouldn’t they be considered somewhere?
My opinion on the calculation is as follow:
FPY step 1: (80 good parts + 10 successfully reworked parts) / (100 input parts) = 90/100 = 0.9
FPY step 2: (90 good parts + 5 successfully reworked parts) / (80 good parts + 10 successfully reworked parts from step 1 + 10 additional parts) = 95/100 = 0.95
FPY step 3: (92 good parts + 8 successfully reworked parts) / (90 good parts + 5 successfully reworked parts + 5 addtional parts) = 100/100 = 1
RTY = 0.9 x 0.95 x 1=0.855
Thank you for your guidance!0 
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