How to Use Performance Rating Calculating BasicTime?

In the article Calculation of SAM using Time Study I have used operator performance rating to calculate basic time. Many readers asked me a question on basic time calculation using performance rating. I understand that basic time calculation method has not been clearly explained in the earlier post. I thought to clarify this question to OCS readers in this post.

Two recent questions are published here (name of the question sender is changed).

Question#1: In your article, you indicate that if an operator has an 80% rating that you multiply the cycle time by the rating (which reduces the cycle time for the operation). Wouldn't you actually divide the cycle time by the rating to INCREASE the cycle time for the operation before you add the other allowances? ... asked by Rick

Question#2: Actually I am confused why basic time getting lower when the performance rating is lower in your articles whereas it should be another way, which is basic time should be more when the rating resulting in lower. So instead of multiplying you need to divide the basic time from rating. Can you please explain this as I don't understand this? ... asked by Kumar

I replied to Rick and I had a discussion with him to clarify the confusion over whether to multiply or divide the rating% to cycle time to calculate basic time. Read the following discussion with Rick, if you have the same question.

Later, I had just sent this discussion to Kumar. He understood this completely. I hope you will also understand once you read the following conversation.

OCS: Hi Rick, Rating factor is used to normalize the cycle time. 100% rated the operator's time considered as normal time (basic time). So, if an operator is working at 80% she is actually doing work at a slower pace compared to a 100% standard operator. That’s why she is taking a longer time than the 100% performer. Secondly, the normal time to do the job is less than the time taken by her (80% rate). That is why the rating factor is multiplied to the cycle time instead of dividing. I hope you understand my point.

Rick: Thank you for responding. Yes, I believe I understand what you are saying, however, I still believe that you need to divide the standard cycle time by the operator rating. 
Let's take an example... If the cycle time of an operation is 10 seconds and the operator is rated as 80% then if as you describe above he/she is working slower than the normal cycle time, so his/her overall time must be LONGER than the normal cycle time, correct? 10 seconds divided by .80 equals 12.5 seconds (longer). If we were to multiply the cycle time by the rating we would get a shorter cycle time (10 seconds times .80 equals 8 seconds). I'm very new to all of this so I want to be sure that I'm getting this correct. Thanks in advance for the help.

OCS: Rick, you are not following the purpose of application of Rating. The purpose is to normalize the cycle time or making equivalent to 100% rated operator. Here cycle time is observed time. The purpose to not increasing or lowering cycle time.
Just ask this question - in your example at 80% operator is completing a task in 10 seconds. How much time she would take if worked at 100%. Result must be lower than 10 seconds. That is why you need to multiply rating to cycle time to calculate normal time.

I hope I am able to clarify this to you too.

Check more about the performance rating in Revolvy and in GSD website.

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