---
title: "Gumbel return period and confidence limit"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-return-period-300"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-return-period-300
site: BCT Engineering
---

# Gumbel return period and confidence limit

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2079 Baisakh, 5+3 marks
- Video: [Flood discharge at various return periods using Gumbel's extreme value distribution | Hydrology](https://www.youtube.com/watch?v=RE7Q4NDVPQQ) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-return-period-300
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

A hydraulic structure is designed for a discharge of 300 m$^3$/s. If the available flood data is for N years (reduced mean = 0.5224, reduced standard deviation = 1.1124) and the mean and standard deviation of the annual flood series are 140 m$^3$/s and 50 m$^3$/s respectively, calculate the return period for the design flood using Gumbel's method. Also estimate the 90% confidence limit, if $f(90\%) = 1.645$.

## Short answer

Using Gumbel method with mean=140 m³/s, σ=50 m³/s, reduced mean=0.5224, Sn=1.1124 (N=30), K=(300-140)/50=3.20, yT=4.082, return period ≈60 years. 90% confidence limits: 239–361 m³/s.

## Answer

The number of years N is not given; N = 30 is assumed because $S_n=1.1124$ is the Gumbel value for N = 30.

#### Return period
$$
K=\frac{x_T-\bar x}{\sigma}=\frac{300-140}{50}=3.20
$$

$$
y_T = \bar y_n + K S_n = 0.5224 + 3.20\times 1.1124 = 4.082
$$

$$
T=\frac{1}{1-e^{-e^{-y_T}}} = \frac{1}{1-e^{-e^{-4.082}}} = 60 \text{ years}
$$

**Answer: return period of the design flood ≈ 60 years** (probability of exceedance in any year = 0.0167).

#### 90% confidence limit
$$
b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(3.20)+1.1(3.20)^2}=4.053
$$

$$
S_e=\frac{b\,\sigma}{\sqrt N}=\frac{4.053\times 50}{\sqrt{30}}=37.00 \text{ m}^3/\text{s}
$$

$$
x_T' = x_T \pm f(c)S_e = 300 \pm 1.645\times37.00 = 300 \pm 60.9
$$

**Answer: 90% confidence limits are 239 m³/s to 361 m³/s** (for N = 30).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The frequency factor K for the 300 m3/s design flood (mean 140, sigma 50) is:
   A. 3.20
   B. 2.80
   C. 6.0
   D. 0.32
   Answer: A. K = (300-140)/50 = 3.2. (Key point: Frequency factor)

2. Which relation gives y_T from K?
   A. y_T = K / S_n
   B. y_T = y_n + K S_n
   C. y_T = y_n - K S_n
   D. y_T = K x y_n
   Answer: B. From K = (y_T - y_n)/S_n, so y_T = 0.5224 + 3.2 x 1.1124 = 4.082. (Key point: Reduced variate)

3. The return period of the design flood is about:
   A. 6 years
   B. 600 years
   C. 60 years
   D. 30 years
   Answer: C. T = 1/(1-exp(-exp(-4.082))) is about 60 years. (Key point: Return period)

4. The confidence limits are found from:
   A. x_T +/- sigma
   B. x_T x f(c)
   C. x_T +/- S_e / f(c)
   D. x_T +/- f(c) S_e, with S_e = b sigma / sqrt(N)
   Answer: D. S_e = 37 m3/s so limits are 300 +/- 1.645 x 37. (Key point: Confidence limits)

## Close variants set in other exams

- [Return period of 350 m3/s design flood by Gumbel](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-return-period-350-25yr) — Engineering Hydrology, chapter 6, asked 2×

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