---
title: "Gumbel flood, confidence limits, hydrologic risk"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-100-500-n30-risk"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-100-500-n30-risk
site: BCT Engineering
---

# Gumbel flood, confidence limits, hydrologic risk

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2076 Chaitra, 4+4+3+3 marks; 2074 Asoj, 4+4+3+3 marks (asked 2 times)
- Video: [gumbel method numerical hydrology](https://www.youtube.com/watch?v=c5nTqXIWyxE) — CE (simplified engineering)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-100-500-n30-risk
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

A river, whose annual flood peak can be represented by Gumbel distribution, has 100-year and 500-year return period flood of magnitude 9900 m$^3$/s and 12100 m$^3$/s respectively. The sample size is n = 30 ($\bar{y}_n = 0.536$, $S_n = 1.1124$).

i) What is the magnitude of 200 year and 1000 year flood?

ii) What are 95% and 80% confidence limits for 200 year and 1000 year flood if $f(95\%) = 1.96$ and $f(80\%) = 1.28$.

iii) A hydraulic structure of 25 year life was designed for 12300 m$^3$/s peak flow. What is the hydrologic risk of the structure?

iv) What peak flow should be taken into consideration if you want the structure to be 99% reliable for a structure life of 25 years.

## Short answer

From Gumbel fit using given floods, mean=4358.4 m³/s, σ=1516.8 m³/s. 200-yr flood: 10849 m³/s, 1000-yr: 13046 m³/s. 95% confidence limits: 200-yr 8044–13653, 1000-yr 9424–16668 m³/s. Risk for 12300 m³/s over 25 yr: 4.2%. Design flow for 99% reliability over 25 yr: 14289 m³/s.

## Answer

Gumbel's method: $x_T = \bar x + K\sigma$, with $K=\dfrac{y_T-\bar y_n}{S_n}$ and $y_T = -\ln\!\left[\ln\dfrac{T}{T-1}\right]$. Here $\bar y_n=0.536$, $S_n=1.1124$ (n = 30).

#### (i) 200-year and 1000-year floods
First find $\bar x$ and $\sigma$ from the two given floods.

| T (yr) | $y_T$ | $K=(y_T-0.536)/1.1124$ |
|---|---|---|
| 100 | 4.600 | 3.653 |
| 500 | 6.214 | 5.104 |

$$
\sigma = \frac{12100-9900}{5.104-3.653} = 1516.8 \text{ m}^3/\text{s}, \qquad \bar x = 9900 - 3.653\times 1516.8 = 4358.4 \text{ m}^3/\text{s}
$$

| T (yr) | $y_T$ | $K$ | $x_T=\bar x+K\sigma$ (m³/s) |
|---|---|---|---|
| 200 | 5.296 | 4.279 | 10849 |
| 1000 | 6.907 | 5.727 | 13046 |

**Answer: $x_{200}$ ≈ 10849 m³/s and $x_{1000}$ ≈ 13046 m³/s.**

#### (ii) Confidence limits
$$
x_T' = x_T \pm f(c)\,S_e, \qquad S_e = \frac{b\,\sigma}{\sqrt{N}},\quad b=\sqrt{1+1.3K+1.1K^2}
$$

| T | K | b | $S_e$ (m³/s) | 95% limits (f = 1.96) | 80% limits (f = 1.28) |
|---|---|---|---|---|---|
| 200 | 4.279 | 5.167 | 1431 | 8044 to 13653 | 9017 to 12680 |
| 1000 | 5.727 | 6.673 | 1848 | 9424 to 16668 | 10680 to 15411 |

#### (iii) Hydrologic risk for 12300 m³/s, n = 25 years
$$
K=\frac{12300-4358.4}{1516.8}=5.236, \quad y_T=0.536+5.236\times 1.1124=6.360
$$

$$
T=\frac{1}{1-e^{-e^{-y_T}}} = 579 \text{ years}
$$

$$
\text{Risk}=1-\left(1-\frac1T\right)^{n}=1-\left(1-\frac{1}{579}\right)^{25} = 0.0423
$$

**Answer: T ≈ 579 years and the risk ≈ 4.2%.**

#### (iv) Flood for 99% reliability over 25 years
Reliability = 0.99 means risk = 0.01.

$$
1-\frac1T = (0.99)^{1/25} \Rightarrow T = 2488 \text{ years}
$$

$$
y_T=7.819, \quad K=6.547, \quad x_T = 4358.4 + 6.547\times1516.8 = 14289 \text{ m}^3/\text{s}
$$

**Answer: design peak flow ≈ 14289 m³/s (return period ≈ 2488 years).**

## Check yourself

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

1. How are the mean and standard deviation found from the two given floods?
   A. Compute K for T = 100 and 500, then sigma = difference in x divided by difference in K
   B. Use the sample size only
   C. Average the two floods
   D. Use x/T for each
   Answer: A. Two equations x = mean + K sigma give sigma = (12100-9900)/(5.104-3.653) and then the mean. (Key point: Method to find mean and sigma)

2. The 200-year flood is about:
   A. 9900 m3/s
   B. 10849 m3/s
   C. 12100 m3/s
   D. 13046 m3/s
   Answer: B. x = 4358.4 + 4.279 x 1516.8 = 10849. (Key point: 200-year flood)

3. The risk over n years for return period T is:
   A. n/T^2
   B. (1/T)^n
   C. 1 - (1 - 1/T)^n
   D. 1 - 1/(nT)
   Answer: C. Probability of at least one exceedance in n years is 1-(1-1/T)^n; here about 4.2% for T = 579, n = 25. (Key point: Hydrologic risk)

4. For 99% reliability over 25 years the return period is found from:
   A. 1/T = 0.99
   B. 1 - 1/T = 0.99 x 25
   C. T = 25/0.01
   D. 1 - 1/T = 0.99^(1/25)
   Answer: D. Risk = 0.01, so (1-1/T)^25 = 0.99, giving T of about 2488 years. (Key point: Reliability)

## Related in Engineering Hydrology

- [Gumbel 500-year flood and 80% limits, Kamalamai River](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-kamalamai-80cl) — Engineering Hydrology, chapter 6
- [Prove mean annual flood has 2.33-year return period](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-mean-flood-2-33yr) — Engineering Hydrology, chapter 6
- [Gumbel flood for 500-year return period](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-100-50-500) — Engineering Hydrology, chapter 6
- [Gumbel 500-year flood, 95% limits, 30 years](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-30yr-100-50-1200-1060-2) — Engineering Hydrology, chapter 6
- [Gumbel peak flood and confidence limits, 75 years](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-peak-0-4pct-75yr) — Engineering Hydrology, chapter 6
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