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    IF Formula with Multiple Criteria

    ✭✭✭✭✭
    edited 09/01/22 in Formulas and Functions

    Hi all,

    Working on an IF formula with 3 conditions.

    IF Value = 0, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use."

    IF Value, 2-36, "Low probability of Risk: No Computer Validation testing required."

    IF Value, =>36, "Medium probability of Risk: Computer Validation testing required."

    I've used the following formulas, but I'm not having luck with the text appearing if the values are 2-36 or =>36.

    #INCORRECT ARGUEMENT SET Error appears.

    =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.", "Low probability of Risk: No Computer Validation testing required."))

    =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row, <36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

    Assuming I can also using an IFERROR formula if the value is 0 also?

    Thanks,

    Jack

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    Best Answer

    • Employee
      Answer ✓

      It looks like you have an extra comma in your logical_expression of a few IF statements. For example:

      IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

      Should look like this:

      IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))


      Here are your functions without the stray commas in the logical_expressions

      =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.", "Low probability of Risk: No Computer Validation testing required."))


      =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row < 36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))


      Hope this helps!

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    Answers

    • Employee
      Answer ✓

      It looks like you have an extra comma in your logical_expression of a few IF statements. For example:

      IF([RPN (Severity x Probability x Detection)]@row, >=36, "Medium probability of Risk: Computer Validation testing required.")))

      Should look like this:

      IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))


      Here are your functions without the stray commas in the logical_expressions

      =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.", "Low probability of Risk: No Computer Validation testing required."))


      =IF([RPN (Severity x Probability x Detection)]@row = 1, "No probability of Risk: Functionality requires evaluation and/or validation when/ if the functionality is intended for production use.", IF([RPN (Severity x Probability x Detection)]@row < 36, "Low probability of Risk: No Computer Validation testing required.", IF([RPN (Severity x Probability x Detection)]@row >=36, "Medium probability of Risk: Computer Validation testing required.")))


      Hope this helps!

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    • ✭✭✭✭✭

      @SolutionSal This resolve the issue. Thank you!

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