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Can't get the formula to work, pass/fail based on tolerance

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    Can't get the formula to work, pass/fail based on tolerance

    Hi everyone.

    I really need some help here. I have tried many different formulas, and have trailed the internet to no avail..

    I am going to measure something that is supposed to have a nominal value of 487. The real "read" value is allowed to deviate with 10%. I have calculated the allowed upper and lower tolerances (C3 and C4). I want the status field in G3 to show a pass if the value is between the lower and upper tolerance, and a fail if it is outside the tolerance.
    I have tried the following formulas:
    =IF(ABS(E3-B3)<=D3,"Pass","Fail") *
    =IF(MEDIAN(438,535,E5)=E5,"Pass","Fail")
    =IF(AND(E5>=438,E5<=535),"Pass","Fail")
    =IF(OR(E5<438,E5>535),"Fail","Pass")

    **I can get this one to work for the positive side of the tolerance, but not the negative.

    I get the exact same error message on most of them (the regular one about "we found a problem about this formula"). No matter what I try, the result stays the same and I am nearly going out of my mind. I have never had this kind of problem before, I have always managed to find the correct solution. But now, I am stumped (the solution is probably dead simple and I am just temporarily blind or stupid).

    So i really need some help here.
    Anyone know a solution?
    Attached Files Attached Files

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    Re: Can't get the formula to work, pass/fail based on tolerance

    Perhaps
    =IF(ABS(F3)>=B3*E6%,"Fail","Pass")


    you need to compare the absolute difference with the tolerance (10% of 487)
    If its larger fail, lower pass

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    Re: Can't get the formula to work, pass/fail based on tolerance

    Hi davsth.

    Thanks for your suggestion, it worked. But I had to tweak it a little. I had to exchange the , with ; to get it to work (it kept giving me the error message until i exchanged them). So now it looks like this: =IF(ABS(F3)>=B3*E6%;"Fail";"Pass")

    And now it works. Thank you, thank you, thank you I really appreciate your help. Have a nice day, a lovely christmas and a happy new year!

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    Re: Can't get the formula to work, pass/fail based on tolerance

    That's because you are using a European locale: the delimiter on the continent is ; because , is used as a decimal separator. In English-speaking locales we use , as a delimiter and . as a decimal separator.

    If that takes care of your original question, please select Thread Tools from the menu link above and mark this thread as SOLVED.
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    Re: Can't get the formula to work, pass/fail based on tolerance

    Aha, that explains a lot. Thanks for that crucial bit of information, it will make future problem solving easier.

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