Hi Tore

>If I understand correctly, what about something like yourresult=yourresult*(0.95 + (0.1*rand(-1))). This will give you a result which is max ±5% from the correct answer.

Thanks a lot. What if I only want a 2.5% deviation? If I understand correctly yourresult=yourresult*(0.975 + (0.1*rand(-1)))

Please advise.