Hi
I've just got an assignment in school, where I've to determine the design factor 'z' to a given failure probability.
This baisically means that I have to find the lowest value of beta (vector containing of u values) that in combination with design parameter z gives an failure function which is equal to zero.
Want to make a loop which can check for all values of z in the interval of 1 and 7. Is that possible?
You can turn the solve block into a function u(z).
Then you can calculate the vector u for any value of z.
A second solve block with "find" or the "root" function (less space consuming and less ugly looking) then could be used to find a solution z of the equation u(z)-beta=0. No need for the for-loop you suggested - let Primes algorithms do the iteration work.
For future questions you should always attach your worksheet and state which version of Prime you are using (especially if you are not using the latest (at the time being thats Prime 9).
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