---
title: "Gumbel design flood, 43 years data, 85% reliability"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-43yr-330-187-life50"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-43yr-330-187-life50
site: BCT Engineering
---

# Gumbel design flood, 43 years data, 85% reliability

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2070 Chaitra, 9 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-43yr-330-187-life50
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Analysis of the annual flood peak of a river for 43 years yielded a mean of 330 m$^3$/s and a standard deviation of 187 m$^3$/s. A proposed water control project on this river is to have an expected life of 50 years. Policy decision of the project allows an acceptable reliability of 85%. Using Gumbel's method, recommend the flood discharge for this project.

A table for reduced mean ($\bar{y}_n$) and reduced standard deviation ($S_n$) is given below:

| N | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|
| $\bar{y}_n$ | 0.5436 | 0.5442 | 0.5448 | 0.5453 | 0.5458 | 0.5463 |
| $S_n$ | 1.1413 | 1.1436 | 1.1458 | 1.1480 | 1.1499 | 1.1519 |

## Short answer

With 85% reliability over 50-year life, design return period is 308 years. Using Gumbel with sample size 43 (y_bar_n=0.5453, S_n=1.1480), frequency factor K=4.515 and design flood=1174 m³/s.

## Answer

For N = 43 years the table gives $\bar y_n=0.5453$ and $S_n=1.1480$.

#### Step 1: Design return period
Reliability 85% means risk $R=1-0.85=0.15$ during the life n = 50 years.

$$
R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/50}\ \Rightarrow\ T=308 \text{ years}
$$

#### Step 2: Reduced variate and frequency factor
$$
y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.729
$$

$$
K=\frac{y_T-\bar y_n}{S_n}=\frac{5.729-0.5453}{1.1480}=4.515
$$

#### Step 3: Design flood
$$
x_T=\bar x+K\sigma=330+4.515\times187=1174 \text{ m}^3/\text{s}
$$

**Answer: recommended design flood ≈ 1174 m³/s (return period ≈ 308 years).**

## Check yourself

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

1. A reliability of 85% over a 50-year life means what risk R?
   A. R = 0.85
   B. R = 1/50
   C. R = 0.15
   D. R = 0.015
   Answer: C. Risk is 1 - reliability = 0.15. (Key point: Risk from reliability)

2. How is the design return period T found from R and n?
   A. T = n x R
   B. T = n/R
   C. From R = 1 - (1 - 1/T)^n, so 1/T = 1 - (1 - R)^(1/n)
   D. T = 1/R
   Answer: C. Solving the risk equation for T gives 1/T = 1 - 0.85^(1/50), so T is about 308 years. (Key point: Design return period)

3. Which expression gives K for Gumbel's method with a finite sample?
   A. K = (y_T - y_n) x S_n
   B. K = (y_T - y_n)/S_n
   C. K = y_T/S_n
   D. K = (y_T + y_n)/S_n
   Answer: B. K = (y_T - y_n-bar)/S_n uses the reduced mean and reduced standard deviation for N = 43. (Key point: Frequency factor)

4. With K = 4.515, mean 330 and sigma 187, what is the design flood?
   A. 4.515 x 187 = 844 m^3/s
   B. 330 x 4.515 = 1490 m^3/s
   C. 187 + 4.515 x 330 = 1677 m^3/s
   D. 330 + 4.515 x 187 = 1174 m^3/s
   Answer: D. x_T = mean + K x sigma = 330 + 844 = 1174 m^3/s. (Key point: Design flood)

## Close variants set in other exams

- [Gumbel design flood, 21 years data, 40-year life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-21yr-life40-rel85-b) — Engineering Hydrology, chapter 6, asked 3×

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