---
title: "Gumbel analysis from 100 and 50-year floods"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-30yr-100-50-1200-1060"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-30yr-100-50-1200-1060
site: BCT Engineering
---

# Gumbel analysis from 100 and 50-year floods

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2067 Shrawan (old course), 16 marks
- Video: [Gumbel's method in Hydrology for flood estimation | Flood Routing | GATE 2022 |](https://www.youtube.com/watch?v=gtUiLfn_E8U) — TechnoCivil
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-30yr-100-50-1200-1060
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Using 30 years data and Gumbel's method the flood magnitudes, for return periods of 100 and 50 years for a river are found to be 1200 and 1060 m$^3$/sec respectively. ($\bar{y}_n = 0.5362$ and $S_n = 1.1124$).

a) Determine the mean and standard deviation of the data used.
b) Estimate the magnitude of a flood with a return period of 500 years.
c) What are the 95% confidence limits for this estimate if $f(95\%) = 1.96$.
d) What is the probability of the flood equal to or greater than a 500-year flood occurring three times in the next 10 years?

## Short answer

a) Mean = 385.1 m³/s, σ = 223.1 m³/s. b) T=500 flood = 1524 m³/s. c) 95% confidence limits ≈ 1043–2004 m³/s. d) Probability of three ≥500-yr floods in 10 years ≈ 9.5×10⁻⁷.

## Answer

#### a) Mean and standard deviation
$\bar y_n=0.5362$, $S_n=1.1124$, $y_T=-\ln[\ln(T/(T-1))]$, $K=(y_T-\bar y_n)/S_n$, $x_T=\bar x+K\sigma$.

| T (yr) | $y_T$ | $K$ | $x_T$ (m³/s) |
|---|---|---|---|
| 100 | 4.600 | 3.653 | 1200 |
| 50 | 3.902 | 3.026 | 1060 |

Subtract the two equations:

$$
\sigma=\frac{1200-1060}{3.653-3.026}=223.1 \text{ m}^3/\text{s}
$$

$$
\bar x=1200-3.653\times223.1=385.1 \text{ m}^3/\text{s}
$$

**Answer: mean = 385.1 m³/s, standard deviation = 223.1 m³/s.**

#### b) 500-year flood
$$
y_{500}=-\ln[\ln(500/499)]=6.214,\qquad K=\frac{6.214-0.5362}{1.1124}=5.104
$$

$$
x_{500}=385.1+5.104\times223.1=1524 \text{ m}^3/\text{s}
$$

**Answer: $x_{500}$ ≈ 1524 m³/s.**

#### c) 95% confidence limits of the 500-year flood
$$
b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(5.104)+1.1(5.104)^2}=6.024
$$

$$
S_e=\frac{b\sigma}{\sqrt N}=\frac{6.024\times223.1}{\sqrt{30}}=245.3 \text{ m}^3/\text{s}
$$

$$
x_T'=x_T\pm f(c)S_e=1524\pm1.96\times245.3=1524\pm481
$$

**Answer: 95% confidence limits are about 1043 m³/s and 2004 m³/s.**

#### d) Probability of three floods of 500-year size (or larger) in the next 10 years
The annual probability is $p=1/500=0.002$. Floods in different years are independent, so use the binomial distribution with n = 10, r = 3:

$$
P=\binom{10}{3}p^3(1-p)^7=120\times(0.002)^3\times(0.998)^7 \approx 9.47\times10^{-7}
$$

**Answer: probability ≈ $9.47\times10^{-7}$** (about 1 in a million), practically negligible.

## Check yourself

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

1. How are the mean and sigma found from the 100- and 50-year floods?
   A. Subtract the two equations x_T = mean + K sigma, giving sigma = 140/(3.653 - 3.026)
   B. sigma = 1200 - 1060
   C. Add the equations and divide by 2
   D. sigma = 1200/3.653
   Answer: A. Subtracting eliminates the mean: sigma = 140/0.627 = 223.1, then mean = 1200 - 3.653 x 223.1 = 385.1. (Key point: Finding sigma)

2. With mean 385.1, sigma 223.1 and K = 5.104 for T = 500, what is x500?
   A. 223.1 + 5.104 x 385.1
   B. 385.1 x 5.104 = 1966
   C. 1200 x 5 = 6000
   D. 385.1 + 5.104 x 223.1 = 1524 m^3/s
   Answer: D. x_T = mean + K sigma = 385.1 + 1138.7 = 1524 m^3/s. (Key point: 500-year flood)

3. How are 95% limits computed?
   A. x_T ± 1.96 S_e, S_e = b sigma/sqrt(N), b = sqrt(1 + 1.3K + 1.1K^2)
   B. x_T ± 1.96 sigma
   C. x_T ± sigma/N
   D. x_T ± 0.95 S_e
   Answer: A. The standard error uses b and sqrt(N); limits are x_T ± f(c) S_e = 1524 ± 481. (Key point: Confidence limits)

4. Why is the probability of three 500-year floods in 10 years computed with the binomial formula?
   A. Floods in different years are independent with p = 0.002 each year
   B. Because the floods are normally distributed
   C. It is always 3/10
   D. Floods occur exactly every 500 years
   Answer: A. P = C(10,3) p^3 (1-p)^7 for independent annual trials, about 9.47e-07. (Key point: Binomial probability)

## Related in Engineering Hydrology

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