# Best/Worst Case Uncertainty Propagation Problems Best/Worst Case Uncertainty Propagation Problems

Here are a few example problems using the best/worst case method to find the uncertainty in a calculated quantity. In all the problems below, assume that x = 5˚ ± 1˚ and y = 10 ± 1 m.

1. z = y2 + 2y – 6

zbest = 114 m2 (Note: Since y2 obviously has units of m2, the “2” in the second term must carry units of m, and the “6” in the last term must carry units of m2.)

To find zmax, we need to make the first two terms as big as we can, so we put in y + y in each term, and we find that zmax = 137 m2.

Then z = zmax – zbest = 137 – 114 = 23 m2.

So we have to round everything to the tens place, since that’s the first digit where the uncertainty shows up. We then have

z = 110 ± 20 m2.

2. (Don’t worry about the units in this one. You have to pull the same tricks as in #1 above and have the “1” take on units in each term.)

z = 0.09 (Notice that I’m dropping the “best” subscript and just calling it z.)

To find zmax, we want the first term to be as big as it can be, which means we use y–y, and we want the second term to be as small as possible (since it’s being subtracted), so we choose y+y in that term. Plug the numbers in and you get zmax = 0.1028

Then z = 0.0128. Round everything to the hundredths digit, and you get

z = 0.09 ± 0.01

1. z = y sin(x)

Plug in the numbers and get z = 0.8716

To get zmax, we want y as big as possible, and we want [sin(x)] to be as big as possible. For the y factor, we just use y + y, but for sin(x), we need to check both possibilities.

sin(4) = 0.0698

sin(6) = 0.1045

So we choose sin(x+x) = sin(6) = 0.1045

Then zmax = 11 sin(6) = 1.15 and z = 0.278

z = 0.9 ± 0.3 m (the units are easy on this one; only the y carries units. Trig functions never carry units.)

4. Here notice that z will have units of 1/m or m–1.

Plug in the numbers and get z = 0.09962

To get zmax, we want cos(x) to be as big as possible, and y to be as small as possible. Check out the cosine possibilities first:

cos(4) = 0.99756

cos(6) = 0.99452

So we need to choose cos(x–x). If we also choose y–y in the denominator, we find that zmax = 0.1108, which gives z = 0.011

z = 0.10 ± 0.01 m–1

5. z = y[1+tan(x)]Here we’ll assume the 1, like the tan(x), carries no units, so z has units of m.

Plug in the numbers and get z = 10.87

To find zmax, we need y+y and we need tan(x) to be as big as possible. Try the numbers yourself and convince yourself that zmax = 12.156, giving z = 1.286. Then z = 11 ±1 m.

6. Let’s not worry about the units on this one, since there’s a subtlety here we haven’t covered.

Z = 0.003038

To find zmax, we want y+y, the biggest sin(x) value, and x–x in the denominator. Compare the sine values and convince yourself that we want sin(6). Thus zmax = 0.0075, which gives z = 0.0075.

Rounding to the thousandths place, we get z = 0.003 ± 0.008. Notice that the uncertainty is bigger than the value itself!