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All combination factors must be unique

  • 9 August 2021
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T
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Good Morning,

In walking through an older exercise for CTO Setup of a manufactured chair, getting error "All combination factors must be unique" when attempting to use the same factor type (Characteristic Value) and its associated factor value twice, in order to create a combination table using the same for ranges and associated return values. The help area explicitly states this should be possible, but doesn't seem to be the case, the way I'm doing it anyhow. Screenshots:

Desired end game:

Thanks

jt

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Best answer by matt.watters 20 August 2021, 13:59

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J
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Hi @thejtluv,

Below is much layman, but it is the only way which I can think of.
Let me know your ideas.

 


 

Thanks,

Jagath

T
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Appreciate what you’re doing there, but it’s cumbersome and wouldn’t be feasible for larger ranges. I forgot to include screenshot above, but thought it strange the Factor Value help actually states you should be able to use the same factor more than once:

 

matt.watters
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Not sure in which release this restriction was introduced, but for this exercise where a characteristic value can vary, this would be a possible setup. Sequence value is important. The documentation reads: “This sequence is used to break a tie if two combinations both evaluate to true.  If more than one row is true, the last combination in the sequence is returned.”

 

Regards - Matt
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