**Rmazing**, and kindly contributed to R-bloggers)

I added an `interval`

function to my ‘propagate’ package (now on CRAN) that conducts error propagation based on interval arithmetics. It calculates the uncertainty of a model by using interval arithmetics based on (what I call) a “combinatorial sequence grid evaluation” approach, thereby avoiding the classical *dependency problem* that often inflates the result interval.

This is how it works:

For two variables with intervals and , the four basic arithmetic operations are

So for a function with variables, we have to create all combinations , evaluate their function values and select .

The so-called *dependency problem* is a major obstacle to the application of interval arithmetic and arises when the same variable exists in several terms of a complicated and often nonlinear function. In these cases, over-estimation can cover a range that is significantly larger, i.e. . For an example, see here under “Dependency problem”. A partial solution to this problem is to refine by dividing into smaller subranges to obtain sequence . Again, all combinations are evaluated as described above, resulting in a larger number of in which and may be closer to and , respectively. This is the “combinatorial sequence grid evaluation” approach which works quite well in scenarios where monotonicity changes direction, obviating the need to create multivariate derivatives (Hessians) or use some multivariate minimization algorithm.

If the interval is of type

## Example 2: A complicated nonlinear model.

## Reduce sequence length to 2 => original interval

## for quicker evaluation.

EXPR2 <- expression(C * sqrt((520 * H * P)/(M *(t + 460))))

H <- c(64, 65)

M <- c(16, 16.2)

P <- c(361, 365)

t <- c(165, 170)

C <- c(38.4, 38.5)

DAT2 <- makeDat(EXPR2)

interval(DAT2, EXPR2, seq = 2)

[1317.494, 1352.277]

## Example 5: Overestimation from dependency problem.

# Original interval with seq = 2 => [1, 7]

EXPR5 <- expression(x^2 - x + 1)

x <- c(-2, 1)

DAT5 <- makeDat(EXPR5)

interval(DAT5, EXPR5, seq = 2)

[1, 7]

# Refine with large sequence => [0.75, 7]

interval(DAT5, EXPR5, seq = 100)

[0.7502296, 7]

# Tallies with curve function.

curve(x^2 - x + 1, -2, 1)

Have fun!

Cheers,

-ans

Filed under: General Tagged: dependency problem, error propagation, interval arithmetic, uncertainty

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