Frequency Analysis
by Gumbel Method:
Principle and Steps
Principle and Steps
• As a simple method, frequencies (or probabilities), P(X ≥ x), of the
observed flood peaks could be calculated.
• The Gumbel method of frequency analysis is based on extreme value
distribution and uses frequency factors developed for theoretical
distribution.
x = ẍ + ∆x
Where x is magnitude of flood of some given
probability (P) or return period
x is mean of floods on record
∆x is departure of variate from the mean
• ∆x depends on dispersion characteristics, recurrence interval (T) and
other statistical parameters. It can be expressed as
• ∆x = S K
• where S is standard deviation of the sample and K is frequency factor
Thus, equation (i) above can be expressed as
x = ẍ + KS
Steps Involved in Frequency
Analysis:
• List and arrange annual floods (x) in descending order of magnitude
• Assign rank ‘m’, m = 1 for highest value and so on
• Calculate return period (T) and/or probability of exceedence (P) by
equations n + 1/m and m/n +1 respectively.
• Using tabular form calculate x2 and ∑x and Ex2.
• Now calculate mean x; squared mean x2; mean of squares x2 and
standard deviation S.
• From the table find the frequency factors value for desired return
periods
• Using relation x = x + KS calculate flood values for various return periods
• The annual flood series for a river is available for 21 years. The
observed flood peaks are as given below. Calculate the 100 year flood
using Gumbel’s method.
Solution
Now, using equation x = x + KS and adopting values of x and S from above
and different K and T values from Table flood flows of various return periods
can be calculated
The mean annual flood of a river is 600m3/s and the standard deviation of the annual flood time
series is 150m3/s. Determine the return period of a flood of magnitude 1000m3/s occurring in the
river. Use Gumbel’s method and assume sample size to be very large
• XT= ͞x +KS,
Given that x mean=600m3 /s and s=150 m3 /s and XT=1000 m3 /s, then substituting in the above
equation,
= 1000=600+k 150
400/150=k
K=2.667
K=Yt-Yn /Sn
Given for N large, yn and Sn are 0.577 and 1.2825 respectively
2.667= (Yt-0.577)/1.2825
Yt = 3.9970