Exceedance Probability


Sometimes a hydrologist may need to know what the chances are over a given time period that a flood will reach or exceed a specific magnitude. This is called the probability of occurrence or the exceedance probability.

Let’s say the value “p” is the exceedance probability, in any given year. The exceedance probability may be formulated simply as the inverse of the return period. For example, for a two-year return period the exceedance probability in any given year is one divided by two = 0.5, or 50 percent.

But we want to know how to calculate the exceedance probability for a period of years, not just one given year. To do this, we use the formula

Exceedance probability = 1 – (1 – p)n

In this formula we consider all possible flows over the period of interest “n” and we can represent the whole set of flows with “1.” Then (1–p) is the chance of the flow not occurring, or the non–exceedance probability, for any given year.

(1–p)n is all the flows that are less than our flood of interest for the whole time period.

Finally, “1,” all possible flows, minus (1–p)n, all flows during the time period than are lower than our flood of interest, leaves us with 1 – (1–p)n, the probability of those flows of interest occurring within the stated time period.

This table shows the relationship between the return period, the annual exceedance probability and the annual non–exceedance probability for any single given year.

So, if we want to calculate the chances for a 100–year flood (a table value of p = 0.01) over a 30–year time period (in other words, n = 30), we can then use these values in the formula for the exceedance probability.

We can also use these same values of p and n to calculate the probability of the event not occurring in a 30–year period, or the non–exceedance probability.

Example of exceedance probability

Let’s say you want to know what the probability is for a 50–year flood over a 50–year period. It’s not 100 percent!

Calculation for Probability of 50–Year Flood Over 50–Year Period
1 – (1 – p)n
n = 50
p = 0.02

We know that n = 50 since we are looking at a 50–year period of time and using the probability of occurrence table we see that p=0.02 for a 50–year return period.

1 – (1 – 0.02)50
= 1 – (0.98)50

So, applying these values in the equation, the (1–p) value is (1–0.02), or 0.98.

= 1 – 0.36
= 0.64 or 64%

(1–p) to the n is 0.98 raised to the 50th power. That comes out to 0.36.

Now we have (1–0.36), which is 0.64.

There is a 64 percent chance of a 50–year flood in a 50–year period. That means there is a 36 percent chance we won’t see a 50–year flood in the 50–year period.

Now let’s determine the probability of a 100–year flood occurring over a 30–year period of a home mortgage where the home is within the 100–year floodplain of a river.

Calculation for Probability of 100–Year Flood Over 30–Year Period
1 – (1 – p)n
n = 30
p = 0.01

n=30 and we see from the table, p=0.01 .

1 – (1 – 0.01) 30
= 1 – (0.99) 30
= 1 – 0.74
(probability of non–occurrence = 0.74)
= 0.26 or 26% probability of occurrence

The 1–p is 0.99, and .9930 is 0.74.

There is a 0.74 or 74 percent chance of the 100–year flood not occurring in the next 30 years.

But 1–0.74 is 0.26, which shows there is a 26 percent chance of the 100–year flood in that time.

Source: http://stream2.cma.gov.cn/pub/comet/HydrologyFlooding/FloodFrequencyAnalysisInternationalEdition/comet/hydro/basic_int/flood_frequency/print.htm

Leave a Reply

Your email address will not be published. Required fields are marked *