# Effeciency calculation, need help

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• #36024

Can anyone in this forum help me out here. I had a long discussion with my friend regarding calculating effeciency, based on the available hours in month versus consumed. Here are the details what he is saying
” If we take data for the month of december, here we have total available time is 1261680 minutes and the total shipment time (consumed) was 666636 minutes give us effeciency of 52.84%. If we suppose 50000 minutes workers did rework job, then wherever we adjust this our effeciency will improve!!. For instance, if we add this 50000 in shipment time then the total worked time will be 716636 and effeciecny will be 56.8%. If we decrease our total available time by reworking time, then shipment time will remain the same as 666636 and available time will decrease to 1211680 minutes providing us 55.02% efficiency”

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#102721 grover
Participant

Effeciency will be reduced due to rework. It can not be increased. Your friend’s concepts does not seems right. Just think this scenerio as:
I have given five parts to be made in one hour. Now, lets suppose that i made only four parts in one hour and one part is made wrong which has to be reworked. In that case i really consumed my one hour. I will be paid for one hour, irrespective i made it right or wrong. Now that rejected part should be reworked and suppose i take five more minutes to make that part correct. In that case i will charge money for those hours too. So, i will be paid more amount than should be. Meaning effeciency of the whole work is decreased, not increased!!!
Hope this helps

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#102738 jediblackbelt
Participant

I calculate efficiency based on earned hours – which is usually a standard way to calculate it in the IE costing world.  A part has an amount of time that it requires to make it.  This is the standard time and includes PFD allowances (Personal/Fatigue/Delay).  Every good part you make then earns you that amount of time.  You then take those hours and divide it against the time you used.  This is your efficiency.
Example.  A widget takes 2 minutes to make from start to finish (with all PFD added).  You make 160 good pieces in an 8 hour shift.
(160*2) = 320 earned minutes;  8 hour shift is 480 minutes.  SO….
320/480= 67% efficiency.
It is possible to go over 100% efficiency, but this leads to questions about the standard being accurate, did they run through breaks and lunches, etc…

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#102741 V. Laxmanan
Member

Dear Naveen:
You have given a very nice example about the meaning of efficiency.  I just want a clarification about 4 parts versus 5 parts.
In the example, you were supposed to make 5 parts per hour.  You made 4 parts in that hour, or only 80% of what you expected to do.  Still you got paid for the whole hour. Who said bosses are not kind!
Now, about the part that spent reworking for 5 minutes.  Was that also made within the hour?  Or, are you talking about reworking one of the four parts ONLY that you made in one hour.
I am sure it is just an oversight and I also commend you for the example that you have given to illustrate the idea of efficiency – I was confused by the example given that started this thread.
Now that you have given us a good example, please just also clear up the point about the part that was being reworked and how many were made in the one hour for which the person got paid – don’t mind the extra five minutes that were (I hope not) overtime wages.  Just having some fun!  That ‘s very important too. Regards.
Laxman

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#102751 V. Laxmanan
Member

Dear JediBlackBelt:
Very nice example as well. If you don’t mind, let’s continue this a bit.
160 good ones made in 8 hours. At 2 minutes per widget, that’s 320 PFD “points” or minutes.  The 8 hours equals 480 minutes. So efficiency is 320/480 = 3/3 = 66.67%.  What will happen when they person comes back for another 8 hour shift?  I contend that efficiency as calculated is not a good indicator of the operator’s performance, or efficiency.
It cannot be calculated unless you have also taken this second 8 hour shift, and then a third 8 hour shift, then a fourth and so on.  Very soon you are calculating averages, standard deviations etc.
But, there is a simpler way to approach this.  Just throw some number at me for the second eight hour shift and I will describe my view of the efficiency of this operator. Cheers.  Regards.
Laxman
P. S.  We are not going to discuss what was happening for 160 minutes (320 minutes spent on the good ones, out of 480 minutes) and if any bad ones were made during the 160 minutes.  That’s a whole new question.
Just throw a number at me for good ones on the second eight hour shift – for the same operator and the same widget made with the same machine, under identical conditions as in the first 8 hour shift.  I am sure you are smart enough to figure this out for yourself but let’s post this anyway.

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#102753 Ken Feldman
Participant

Hey V, did you mean 3/3=67%?  Given your last thread I wondered if this was part of that work function thing.

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#102754 V. Laxmanan
Member

Dear Darth:
So efficiency is 320/480 = 3/3 = 66.67%.  What will happen when they person comes back for another 8 hour shift?
I thought you were smarter than that, Darth. I counted 91 characters and one was mistyped. (May be I miscounted too, see PFD allowance, where F stands for Fatigue)
Hey, I am new to this Six Sigma thing! Not at 3.4 DPMO yet. Trying …..Cheers.
Laxman

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#102757 jediblackbelt
Participant

Alright, lets say the next shift comes in and produces 200 parts.  Now for two shifts I have 360 parts with 2 minutes per part totals 720 earned minutes and I have used 960 minutes of time.  So I have 720/960 or 75% Efficiency.
What am I missing?  Always up for some teaching especially if there is some improvements to make.

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#102763 V. Laxmanan
Member

Dear JediBlackbelt:
First of all, thanks for the response. You said,
the next shift comes in and produces 200 parts.
Are we talking a second operator who comes to work on the same machine, in the second shift?
Or, should we be thinking about the same operator coming back the next day, to work on the same machine, on the same shift as on the first day.
As for as I am concerned, I just needed a number for the second “day” , or a second shift. It doesn’t matter. But, it may help set the stage for a nice discussion.
I am also glad you did what you did by adding up the good ones produced from two shifts (or days) and recalculating the efficiency.  It is now up to 75%.  Wonderful. I will get back on this soon. Regards.
Laxman

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#102765 Ken Feldman
Participant

Wasn’t being critical.  Given the gist of the last few day’s worth of discussion, I wanted to make sure it was merely a typo and not some mathematical wizardry.

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#102767 V. Laxmanan
Member

Dear Darth:
Thanks.  Looks like we are in this together.
But wait, the mathematical wizardry is just beginning! I have to prepare a couple of graphs for my next post on efficiency. Regards.
Laxman

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#102774 grover
Participant

Dear Laxmanan,
I am sorry i was not very much clear in that example. Let me explain it again.
Effeciency will be reduced due to rework. It can not be increased. Your friend’s concepts does not seems right. Just think this scenerio as:
“I have given five parts to be made in one hour. Now, lets suppose that i made four ACCEPTABLE parts AND ONE DEFECTIVE part IN THAT GIVEN ONE HOUR. THAT DEFECTIVE PART has to be reworked. In that case i HAVE ALREADY consumed my one hour ON THOSE FIVE PARTS. I will be paid for one hour, irrespective i made FIVE PARTS right or wrong. Now that rejected part should be reworked and suppose i take five more minutes, IN ADDITION TO ONE HOUR, to make that part correct. In that case i will charge money for  ONE HOUR AND FIVE MINUTES. So, i will be paid more amount than should be. Meaning effeciency of the whole work is decreased, not increased!!!”
Hope this clarify the scenerio.
Regards,

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#102775 grover
Participant

What is the definition of “earned minutes”. Lets see the formula for effeciecncy again.
Effeciency = earned minutes / Actual minutes consumed
Earned minutes = estimated time * progress
Progress = Good parts produced / Total # of given parts
Now, lets see example.
Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
progress =  900/1000= 0.9
Earned minutes = 100 * 0.9 = 90 min
Actual time utilized in processing 1000 parts = 100 min
Effeciency = 90/100 = 0.9 = 90%!!
Now please see my first thread where my friend’s logic is explained and then tell me whats wrong in it. Regards,
Naveen

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#102777 V. Laxmanan
Member

Dear Naveen:
Thanks for the clarification and it makes perfect sense to me.
I wanted to be sure if 5 parts (4 good 1 bad) were made in hour or if you were talking about 4 parts made in one hour of which one was reworked.
Now, this is purely philosophically speaking, see how nice the world is. We get paid even to produce defective parts!  Of coures, I think there is a reason for this.  Nobody really knows why defective parts are being made in a process and we don’t necessarily want to blame the operator (the human being) alone.  He or she gets paid, the machine continues to be used on a overtime basis
But, wait, again we are missing something! We did not really include all the costs associated with reworking (the machine overtime) and so our idea of efficiency is at best a vague one – Welcome to work function – this is what I was getting at but I wanted us to be very clear.  Regards.
Laxman
P. S. I must admit I understand the logic behind the other post where you go through a series of ratios to arrive at 90% efficiency. If you plug the other defined ratios back into the basic definition of efficiency, it appears that estimated time = actual time! Efficiency becomes simply number of good one divided by total made. Why not use this simpler ratio instead of the more complicated calculation?  Estimated time cannot be equal to actual time in all cases.  How do you tell the difference?  Can you clarify that also.  Thanks.

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#102778 V. Laxmanan
Member

Dear Naveen:
Just another thought. From your example, it is clear that on “average” time to make parts is 0.2 hours (5 per hour) or 12 minutes each.
However, in this example there are 4 goods ones and one bad part.
The bad one took five more minutes of reworking.  Let’s say the last one made was the bad one.  Then the total time is given by
4x + y = 65
where x is the “average” time to make good ones is given and y is the observed time to make at the one bad one. The sum equals 65. What is the average time to make the parts?  What is the efficiency? we really do not know the time taken to make a good part.  There are two unknowns here and we are missing one equation.  Is x = 12?
What do you think? Are we missing one equation? What will an actual process show if there is a continuous inspection of good and bad parts.  Again, this is just my opinion. Regards.
Laxman

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#102783 grover
Participant

You said:
“I must admit I understand the logic behind the other post where you go through a series of ratios to arrive at 90% efficiency. If you plug the other defined ratios back into the basic definition of efficiency, it appears that estimated time = actual time! Efficiency becomes simply number of good one divided by total made. Why not use this simpler ratio instead of the more complicated calculation?  Estimated time cannot be equal to actual time in all cases.  How do you tell the difference?”
Well,  I took the very simple case where i ASSUMED that the estimated time is equal to actual time. Also, i am emphaising over there that estimated time is not really playing key role to define effeciency. It is the EARNED TIME which matters. For instance, someone says, base on his/her time & motion study, that the estimated time for the said job is 100 min. Now the point is ” whether we EARNED that 100 minutes or not, and hence we have the philosophy of PROGRESS. product of progress and estimated time is really your output. Hence we have to consider Earned time instead of estimated time in our effeciency calcualtion.
As far as considering machine time with labor time in order to get total rework time, Yes i am totally agree with you that we have to consider this also in our calcuation. Lets carry on the same example (its a learning experience for me also Lax. and i am very much enjoying this),
CASE 2:

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 1000
total rejected parts out of 1000 parts = 0
progress =  1000/1000= 1.0
Earned minutes = 100 * 1.0 = 100 min
Actual time utilized in processing 1000 parts = 110 min
Effeciency = 100/110 = 0.91 = 91%!!

CASE 3

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
progress =  900/1000= 0.9
Earned minutes = 100 * 0.9 = 90 min
Actual time utilized in processing 1000 parts = 110 min
Effeciency = 90/110 = 0.82 = 82%!!

Now, lets see effeciency with reworking

CASE 4

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
total parts get cured after reworking= 100 (simple case, one can also have scrap too!)
Total time consumed in reworking of 100 parts = 30 min (including man+machine)
progress =  1000/1000= 1.0
Earned minutes = 100 * 1.0 = 100 min
Actual time utilized in processing 1000 parts = 130 min
Effeciency = 100/130 = 0.77 = 77%!!

So we can conclude here that by reworking, effeciency is decreased, not increased.

I need your comments now, where i am wrong ….

Regards

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#102784 tundrafrost
Member

Try the OEE method (Overall equipment effectiveness) It takes into account Availability, preformance and Quality. Go to Google and search OEE.,There are some sample work sheets on some of the sights

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#102786 tundrafrost
Member
#102797 V. Laxmanan
Member

Dear Naveen:
Thanks for more examples. I have to study them.
I did not say anything about being wrong.  Actually, every one is right.  No one does anything wrong intentionally.  We use the skills we have and apply them.  As, we learn, we improve.  That is Six Sigma in a nut shell. I have to study the new examples. Regards.
Laxman

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#102807 V. Laxmanan
Member

Dear Naveen:
Thanks again for the new cases provided in this post, in addition to case 1 from your earlier post (49204, on June 30, 2004).
Interestingly, you have provided examples where the total number of parts is held constant at 1000 and the estimate time is also constant at 100 minutes.
What would happen if we observe the process as 1000 parts are being made one by one?  Let’s assume that it is possible to inspect them and decide immediately whether they are good or bad. We are also able to determine the time taken to make each part in this fashion.  These are the missing pieces of information, not your fault really, but something we tend to overlook today in the way manufacturing processes are being controlled.
The efficiency E,  defined as earned time/actual time is actually the product of two ratios.  Let me call them E1 and E2.
E1  = Estimated time/Actual time
E2 = Good Parts/Total Parts
E = E1 * E2
In case 4, the efficiency is very low.  This is because actual time is very high (130 minutes) although the ratio E1 = 1.  But, the actual time is high because in case 4, the parts were being reworked and this time is being added to the actual time.  In other cases (case 1 and 3) where parts were rejected, the actual time did not include the time for reworking of the parts.  This introduces contradictions although we miss it since we are using a mathematical formula.
If you can develop actual data for a continuous process, you will soom find out the law relating x and y is
y = hx + c = hx – W = h [x – (W/h)] never really y = hx since W > 0.
Here x is estimated time and y the actual time.  Or, x is total parts and y is good parts. The two ratios interact in a complex way because each has its own c that we must properly account for.  Indeed, we don’t even bother to find the relation between x and y.
I realize that you have just provided some hypothetical examples. This is a good way to start thinking about “complex” real world processes, where there is a lot of “chaos”.  This “chaos” is called entropy in quantum theory of Planck and Einstein and can be handled via the work function W = – c.
If we take your example and convert into a real world scenario by developing the x and y value, we will begin to see how the work function affects the very idea of efficiency of a process.  For what Einstein did, we can define efficiency as follow.
K = hf – W = h[f – (W/h)]
Efficiency = K/hf = 1 – (W/h)
The efficiency is the ratio of the actual energy of the electron divided by the maximum possible energy.  This is like actual time (K) and estimated time (hf).  The estimated time is proportional to money (or cost) for producing the parts. Energy is just like money.
The discussion in my posts thus relates directly to the Six Sigma efforts of many in this forum.  Again, thanks for these nice examples. I hope this has been useful to you personally and that it can soon be integrated quickly into your Six Sigma activities. Regards.
Laxman
P. S. In brief, Observe the process. Develop x, y data for estimated versus actual times (this means many different types of parts) and good parts versus total parts (holding estimated time constant) to see how the function y = hx – W evolves in a more complex situation.  I feel we may have to pay attention to only one of the two ratios here to gain an understanding of a more complex situation.

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#102825

Dear Laxmanan & Naveen,
Your discussion is very knowledgeable to me but i am not getting my answer which i posted in my first thread.

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#102827 V. Laxmanan
Member

Dear Kanan:
I had some difficulty understanding the info you provided, which is pasted below. I was hoping you would return to see the thread. Thanks for doing so.
” If we take data for the month of december, here we have total available time is 1261680 minutes and the total shipment time (consumed) was 666636 minutes give us effeciency of 52.84%. If we suppose 50000 minutes workers did rework job, then wherever we adjust this our effeciency will improve!!. For instance, if we add this 50000 in shipment time then the total worked time will be 716636 and effeciecny will be 56.8%. If we decrease our total available time by reworking time, then shipment time will remain the same as 666636 and available time will decrease to 1211680 minutes providing us 55.02% efficiency”
In very simple terms what you are doing, or observing in your efficiency calculation, is easily understood, as follows.  Think of what will happen to the ratio y/x where x and y are any two numbers. The ratio will go up if y increases, with x remaining constant.  The ratio will also go up is x decreases, with y remaining constant.  This is exactly what you are doing.
You are starting with y/x = 666636/1261660 = 0.5284 or 52.84%. Then you add 50,000 to the numerator. The ratio will no doubt go up.  Then you subtract 50,000 from the denominator.  Again, no doubt, the ratio will go up.
This is elementary math.  To understand the dilemma you seem to be facing with the efficiency calculation, you have to clarify, for me, some of the following points.
How did you arrive at available time of 1261680 minutes for the month of December?
You are using the term shipment (consumed) time.  Is the same as the terminology being used by Naveen?  There has to be a rigorous and consistent use of terms to carry on such a discussion. Can you please clarify, after reviewing Naveen’s post as well.  It would certainly help me a lot.
The same applies for the other times you have given.  At least I am not able to grasp the scenario you are describing.  I found it easier to follow what Naveen and Jediblackbelt were for examples and so I tried to add my small bit to the discussion.
Again, this is entirely my limitation, not yours.  I look forward to your clarifications.  Regards.
Laxman

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#102828

Kanan,
Naveen got it right in the first reply.  Efficiency does not include rework time in the numerator…
Best Regards,
Bob J

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#102829 V. Laxmanan
Member

Dear Kannan:
I mistyped 1261660 instead of 1261680 and it was also used in my Excel program for the calculations.  But this number is so huge that the error of 20 does not make any difference to the percentages calculated.  Just wanted to clarify what I did. Regards.
Laxman

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#102830 V. Laxmanan
Member

Dear Kanan:
This is what I think you have to do. In the meanwhile Bob J has also replied to you.
Efficiency = y/x = 666636/1261680 = 0.5284 = 52.84%
With rework time the denominator must increase, not decrease, by 50000 minutes.  Imagine a customer is at your door waiting to pick up something.  You thought you will be done. But you are busy reworing and the customer is waiting. Your efficiency has decreased.
The same applies now to the bigger problem. If there is reworking time, you will be late for shipment compared to when you are not. Hence, the total shipment time, or consumed time, is higher.
Efficiency = y/x = 666636/1311680 = 0.5082 or 50.82%
It is now clear that efficiency has reduced due to rework time. This is what Naveen was saying and Bob J is also pointing out. But, I am still curious about 1261680 minutes for the month of December. Regards.
Laxman

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#102838

Kanan,If you include the extra rework time, thenm you are becoming more efficient at USING UP MINUTES, not at delivering product. For example, you could improve “efficiency” by going around breaking products! Break just enough products that your workers are busy all the time and your efficiency (at keeping the workers busy) goes up to 100%.From a production standpoint, efficiency would be a ratio of products delivered to minutes worked. From a managment standpoint, efficiency would be a ratio of minutes worked to minutes paid for.
Tim F

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#102840 V. Laxmanan
Member

Dear Tim F:
From a managment standpoint, efficiency would be a ratio of minutes worked to minutes paid for.
Right on. You hit the nail right on the head. Happy July 4th.
Laxman

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#102846 jediblackbelt
Participant

This is where I would calculate two ratios.  The one is what I posted earlier about the efficiency and then you can calculate utilization.  Take the two and multiply them to get productivity.  Now you are finding out if you are spending money making parts and you are finding out how good to standard you are at making parts.  You don’t pay for rework or scrap and you base it against a standard metric.  By multiplying the two you are now seeing how productive you are for the company.
Efficiency = earned hours/time making parts
earned hours = (part standard, time/piece* good pieces)
Utilization = time making parts / time paid
Productivity = Efficiency * Utilization
Any problems with this calculation?  I believe this has been the IE standard when I was taught and has been around for years.  Not saying it is right (mass versus lean production), but would love some more discussion on thoughts for this subject.

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#102861

Dear Everyone,
Thanks for everyone for giving me such a wonderful threads on this issue. This forum is owesome!!!
It is confirmed now that by reworking effeciency will decrease. One more confusion.
How do we include reworking time into our effeciency calculation. Naveen added reworking time directly into effeciency calculation. I dont understand this point… Should not we do the effeciency calculation for the reworked parts separately than our original calculations.
Regards,

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#102900 grover
Participant

Dear Naveen,
Sorry for delayed answer.
With reference to your query, here is my opinion. Lets take the same case which i took in the previous threads.

I said:

CASE 3

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
progress =  900/1000= 0.9
Earned minutes = 100 * 0.9 = 90 min
Actual time utilized in processing 1000 parts = 110 min
Effeciency = 90/110 = 0.82 = 82%!!

Now, lets see effeciency with reworking

CASE 4

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
total parts get cured after reworking= 100 (simple case, one can also have scrap too!)
Total time consumed in reworking of 100 parts = 30 min (including man+machine)
progress =  1000/1000= 1.0
Earned minutes = 100 * 1.0 = 100 min
Actual time utilized in processing 1000 parts = 130 min
Effeciency = 100/130 = 0.77 = 77%!!

The reason i am saying that progress is 100% is because initially we had 100 rejected parts (to be reworked). In that case progress is 90%. Now in this case we reworked all the parts (assume no scrap). This will complete our count of 1000 parts out of 1000 initially given. So now the progress is 100%. But look, our time has increased from the given time (100 min. vs 130 min) which really decrease our effeciency. Customer has to wait 30 minutes more than the original time given to him (100 min)!!!
Well, what you can also do is to calculate the effeciency of the rework  separately if you really want to see that in doing the rework activity how much scrap we got at the end of the work.
Hope this helps

Regards,

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#103099 Participant

Naveen, You mentioned in your last post:
CASE 3

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
progress =  900/1000= 0.9
Earned minutes = 100 * 0.9 = 90 min
Actual time utilized in processing 1000 parts = 110 min
Effeciency = 90/110 = 0.82 = 82%!!

Now, lets see effeciency with reworking

CASE 4

Total part = 1000
estimated time to complete 1000 parts = 100 min
total accepted parts out of 1000 parts after processing = 900
total rejected parts out of 1000 parts = 100
total parts get cured after reworking= 100 (simple case, one can also have scrap too!)
Total time consumed in reworking of 100 parts = 30 min (including man+machine)
progress =  1000/1000= 1.0
Earned minutes = 100 * 1.0 = 100 min
Actual time utilized in processing 1000 parts = 130 min
Effeciency = 100/130 = 0.77 = 77%!!

Now here is my opinion in this regard:

The actual time which is mentioned in your CASE4, should be 140 minutes, not 130. It is due to the fact that now we are looking how much time it took to clear the whole lot of 1000 parts. The time is really 110+ 30 = 140. Because company is actually paying for 140 minutes for the whole lot instead of estimated 100 minutes.
Therefore, the TRUE efficiency of the whole job is 71.42%
This is just my opinion, lets see if someone in this forum comes up with different idea.

Regards,

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#106877 Giorgio Parodi Horsky
Participant

May OEE will not be the best tool here …
OEE is related to equipment Effectiveness and it is different from the Efficiency.
An equipment (or process) can be Effective (doing what the equipment was designed to do ), while not been Efficient … (using more resources and “cost” to perform a task than it should)

Best Regards,

Giorgio Horsky

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#106878 Giorgio Parodi Horsky
Participant

Hello ,
With the numbers provided by you , and assuming the following :
Your company works 3 shifts a day (24 hours) , five days a week, you should be talkig about 38 FTE working to obtain the “available time”
In reality the ” rework” time must be considered a “waste of resource” and therefore it reduces the results.
The correct calculation should be (total worked time – rework time) / total available time,
You will see that the result of it will give you  48.9% , ou better talking … if you paid 1 for every minute available (salaries) the real “productive time” was only the one that produced results “first time right”
the 55% efficiency that you calculate, only would be correct if you do reduce from the “cost of time available” the “cost of rework”, or basically talking … you will not pay the workers for the rework time … what somehow I think will be difficult to happen.
Best Regards,
Giorgio Parodi Horsky

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#106890

Giorgio,
I would say its the over way around, I have reported OEE at 25% and for the same process and time period measured 80% efficiency.
Basically because performance in OEE was measured against absolute speed of process, and not target units. Using OEE as a measure of opportunity, bit like a project hopper.

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#106908

Go to the fundamentals…
Efficiency = output energy/input energy (theoretically 100% efficiency is impossible)
Now what is enegry in a business, or in a factory \$, Rs…Efficiency = value output units/cost of producing them
A quick formula for efficiency = (1-s)/(1+r), wheres and r are the proportions of scrap and rework respectively..
Can somebody derive the formula for effectiveness?

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#106910 Giorgio Parodi Horsky
Participant

As you mentioned

Efficiency = output Energy / Input Energy

But in case of time available here, and assuming the numbers presented are related are FTE, then the input energy is equal to the total available time (or FTE)

Rework is always to be considered waste of resources

So if you input in your system 1261680 minutes of time available, and obtain 666636 minutes of worked time , where 50000 minutes are rework .

Efficiency = (666636  50000) / 1261680 =  48.9%

If you search, you will find two types of efficiency

Technical and economic effciency

On the technical it is said that the idea is to obtain the same or more output with less input, irrespective the price.

The economic efficiency measures the ability of an organisation to produce and distribute its product at the lowest possible cost.

A company can have a high technical efficiency but a low economic efficiency, because their prices are too high to meet competition.

When talking specifically about labour the efficiency ratio normally used is std hours divided by the actual hours used to produce a good / service.

If you want to track effectiviness you need to think on the process as an equipment:

Time = Time running / Time available (not the same as FTE availability)

Rate  =  Number of orders ou packs shipped  per running hour / BDP (Best Demonstrated practice)

That is assuming a shipping company

Quality  = Amount of orders first time right / total orders processed or shipped

Overall Effectiveness = Time x Rate x Quality

But of course as said before , you can be highly effective , but you can have low economic efficiency.

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