---
title: "Gumbel 200-year flood from 50 and 100-year floods"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-35yr-200yr-flood"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-35yr-200yr-flood
site: BCT Engineering
---

# Gumbel 200-year flood from 50 and 100-year floods

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2081 Baisakh, 8 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-35yr-200yr-flood
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

From the analysis of available data on annual flood peaks of a small stream for a period of 35 years, 50 years and 100 years flood have been estimated to be 660 m$^3$/s and 740 m$^3$/s. Using Gumbel's method, estimate the 200 years flood for the stream. (Take $Y_n = 0.54034$ and $S_n = 1.12843$ for n = 35 years)

## Short answer

Using yT: b=(740-660)/(4.6001-3.9019)=114.58 m³/s. 200-yr flood = 660+114.58×(5.2958-3.9019)=820 m³/s.

## Answer

Gumbel's equation is linear in the reduced variate: $x_T=a+b\,y_T$, with $y_T=-\ln[\ln(T/(T-1))]$. The two given floods fix the line, so $Y_n$ and $S_n$ are not needed for $x_{200}$ (they are needed only to separate the line into $\bar x$ and $\sigma$).

| T (yr) | $y_T$ | $x_T$ (m³/s) |
|---|---|---|
| 50 | 3.9019 | 660 |
| 100 | 4.6001 | 740 |
| 200 | 5.2958 | ? |

$$
b=\frac{\sigma}{S_n}=\frac{740-660}{4.6001-3.9019}=114.58 \text{ m}^3/\text{s}
$$

$$
x_{200}=660+114.58\,(5.2958-3.9019)=820 \text{ m}^3/\text{s}
$$

(Check with the statistics: $\sigma=b\,S_n=114.58\times1.12843=129.3$ m³/s and $\bar x=660-b(y_{50}-Y_n)=274.8$ m³/s.)

**Answer: 200-year flood ≈ 820 m³/s.**

## Check yourself

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

1. Why are Y_n and S_n not needed to find x200?
   A. They are always 0
   B. Two known floods fix the line x = a + b y_T directly
   C. x200 equals x100
   D. Gumbel does not use them
   Answer: B. Y_n and S_n only split the line into mean and sigma. (Key point: Linearity)

2. What is the slope?
   A. (740 - 660)/(4.6001 - 3.9019) = 114.58
   B. (740 - 660)/(100 - 50)
   C. 740/4.6001
   D. (740 + 660)/2
   Answer: A. Slope is change in flood over change in y_T. (Key point: Slope)

3. x200 =
   A. 740 + 114.58 x 200
   B. 660 + 114.58 x 5.2958
   C. 740 x 2
   D. 660 + 114.58 x (5.2958 - 3.9019) = 820 m^3/s
   Answer: D. Step along the line from the 50-year point. (Key point: Extrapolation)

4. What does the slope represent?
   A. sigma x S_n
   B. sigma/S_n
   C. mean/sigma
   D. Y_n/S_n
   Answer: B. x = mean + sigma (y - Y_n)/S_n, so the slope is sigma/S_n. (Key point: Slope meaning)

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