## multipliation rule for independent variables

## multipliation rule for independent variables

(OP)

We have an issue with our customer concerning the stack-up of tolerances. Its similar to the common multipliation rule for independent variables for reliability calculations but this concerns the stack up of tolerances.

I am looking for a MIL-standard paragaph which calls out the proper way to stack tolerances.

For example, if a part has 4 independent tolerances on an initial value of X:

a 25%, b 40%, c 5%, d 35%

The worst-case value if all of these tolerances stack up in the negative direction would be:

X*(1-.25)*(1-.4)*(1-.05)*(1-.35) = 0.278 by the multipliation rule.

Our customer is insisting that we have to add these, not subtract. In the case above, he would calculate:

X* (1-25%-40%-5%-35%) = -0.05X He would conclude that an intial mass of X with this tolerance stack up could end up with a mass of -0.05X (which is preposterous).

The Question: Do know which paragraph in which MIL-handbook calls out to use the multipliation rule for tolerance stack up?

thanks

I am looking for a MIL-standard paragaph which calls out the proper way to stack tolerances.

For example, if a part has 4 independent tolerances on an initial value of X:

a 25%, b 40%, c 5%, d 35%

The worst-case value if all of these tolerances stack up in the negative direction would be:

X*(1-.25)*(1-.4)*(1-.05)*(1-.35) = 0.278 by the multipliation rule.

Our customer is insisting that we have to add these, not subtract. In the case above, he would calculate:

X* (1-25%-40%-5%-35%) = -0.05X He would conclude that an intial mass of X with this tolerance stack up could end up with a mass of -0.05X (which is preposterous).

The Question: Do know which paragraph in which MIL-handbook calls out to use the multipliation rule for tolerance stack up?

thanks

## RE: multipliation rule for independent variables

For instance, if your article has a 0.0001 failure rate, and it can vary by 25% due to influence A, changing the failure rate to 0.000125 at worst or 0.000075 at best. The other influence factors will have the same effect, leading to upper and lower bounds on the reliability with all factors considered. The result you get would be STRONGLY dependent on the base failure rate of the component before the influence factors are applied.

(PS I think there is a 40-50% chance that I've misunderstood your question, haha).

I have found that the DOD 3235.1H primer on reliability to most helpful with my calculations. Its full title is "Test and Evaluation of System Reliability, Availability, and Maintainability"

printed in 1982.

STF

## RE: multipliation rule for independent variables

Are you are refering to system reliability (X) with four different failure modes (a,b,c,d) and their probability values?

You may be in the wrong forum if your are asking about mass/weight and tolerances rather than Statistics..

## RE: multipliation rule for independent variables

As you observe, adding tolerances results in not only absurd answers, but also gross over design. If all the tolerances are truly independent, then the probability of each tolerance hitting its 3 sigma limit simultaneously is that probability raised to the number of components involved, so if the individual probability is 0.1%, and there are ten tolerances, the probability of all of them occurring simultaneously is 10^-30

TTFN

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