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Weibull Distribution average calculations do not consider applied shift? #31

Open crsf29 opened 4 years ago

crsf29 commented 4 years ago

I have a variable on a Weibull distribution modelling queue time for machines waiting to be loaded in a value driver calculation for mining equipment. I'm applying a shape of 2.1, scale of 1.018, and a shift of 2. Essentially, no machine load time is under 2 minutes.

The average output I'm getting in the cell is .90, which is mathematically correct if we only consider the average is created by the shape and scale. The average value displayed does not appear to consider the shift applied to the distribution.

Changing the cell values to display random sampled values has never yielded a value under 2.

When I view the output of the distribution being used, I am seeing that there are no values under the shifted value being sampled, which is expected behavior. See blue arrow on distribution view. However, OFAT output also carries the same error, and does not respect the shifted distribution.

I am using the 64 bit version, and have no ability to use a 32 bit due to enterprise restrictions. I'm including the sheet, cell E23 is the one in question.

OFAT Queue Weibull plus 2 Weibull plus 2 shift Monte Carlo Driver Tree for Fleet Comparison.xlsx

blakeboswell commented 4 years ago

Hey Cory,

I can confirm what you are seeing. This appears to be a bug. To outline what i'm seeing:

This bug is unfortunate. A suggested work around is to define the distribution as Weibull without the shift and then define an rtaResult as "2 + the cell containing the Weibull". This will produce the same results but will be more explicit.

Even though it ignores the shift, the OAT analysis is still useful because it outlines the impact of the uncertain part of service time The width of the tornado bar and slant of the spider line are driven by the uncertain part of the service time and the shift is will only move them along the x and y axis respectively.

Thanks for bringing the issue to our attention!

crsf29 commented 4 years ago

Thanks for the timely response!!!

I wanted to make sure that I was using the SW correctly, as it was my first day of use. After about two hours of my own research I convinced myself I was right, and I'm glad those hours weren't wasted!

Have a good day.

-Cory

On Thu, Nov 7, 2019 at 4:38 PM boz notifications@github.com wrote:

Hey Cory,

I can confirm what you are seeing. This appears to be a bug. To outline what i'm seeing:

  • The value being displayed in the cell does not include the shift. I tested this for other distributions as well... it is not only a condition of weibull.
  • The simulated values in both RTA and Native Excel mode do consider the shift and create appropriate distributions (this is why the analysis wizard looks correct and op functions return expected results)
  • The OAT analysis chooses the evaluation percentiles based on the unshifted distribution.

This bug is unfortunate. A suggested work around is to define the distribution as Weibull without the shift and then define an rtaResult 2 + the cell containing the Weibull. This will produce the same results but will be more explicit.

Even though it ignores the shift, the OAT analysis is still useful because it outlines the impact of the uncertain part of service time The width of the tornado bar and slant of the spider line are driven by the uncertain part of the service time and will scale with the shift.... the shift is will only move them along the x and y axis respectively.

Thanks for bringing the issue to our attention!

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