---
title: "Gumbel design flood, 40-year life, 85% reliability"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-21yr-life40-rel85"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-21yr-life40-rel85
site: BCT Engineering
---

# Gumbel design flood, 40-year life, 85% reliability

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2075 Chaitra, 4+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-21yr-life40-rel85
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Analysis of the annual flood peak of a river of 21 years yielded a mean of 8520 m$^3$/s and standard deviation of 3900 m$^3$/s. A proposed water control project on this river is to have an expected life of 40 years. The acceptable reliability by the design policy is 85%.

i) Using Gumbel's method recommend the flood discharge for this project. Take $y_n = 0.5252$ and $S_n = 1.0696$ for 21 years.

ii) What would the 80% confidence limit of the above flood be if $f(c) = 1.282$ at 80% confidence level.

## Short answer

Return period from 40-yr life and 85% reliability: T = 1/[1 - (0.85)^(1/40)] = 247 yr. Frequency factor K = (5.506-0.5252)/1.0696 =4.657; design flood =8520+4.657×3900 =26681 m³/s. 80% confidence limits: 26681 ± 1.282×4731 → 20615 and 32746 m³/s.

## Answer

#### (i) Design flood
Reliability = 0.85, so the risk $R=1-0.85=0.15$ in the life of n = 40 years.

$$
R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/40}\ \Rightarrow\ T=247 \text{ years}
$$

$$
y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.506, \qquad K=\frac{y_T-\bar y_n}{S_n}=\frac{5.506-0.5252}{1.0696}=4.657
$$

$$
x_T=\bar x+K\sigma=8520+4.657\times3900=26681 \text{ m}^3/\text{s}
$$

**Answer: recommended design flood ≈ 26681 m³/s (return period ≈ 247 years).**

#### (ii) 80% confidence limits
$$
b=\sqrt{1+1.3K+1.1K^2}=5.559, \qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.559\times3900}{\sqrt{21}}=4731 \text{ m}^3/\text{s}
$$

$$
x_T'=26681\pm1.282\times4731=26681\pm6065
$$

**Answer: 80% confidence limits are 20615 m³/s and 32746 m³/s.**

## Check yourself

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

1. For 85% reliability, the risk is:
   A. 0.15
   B. 0.85
   C. 0.015
   D. 0.35
   Answer: A. Risk = 1 - reliability. (Key point: Risk)

2. The design return period comes from:
   A. 1/T = 0.15/40
   B. 1/T = 1 - (0.85)^(1/40)
   C. T = 40 x 0.15
   D. 1/T = 1 - 0.85 x 40
   Answer: B. Since reliability is (1-1/T)^n, T is about 247 years. (Key point: Return period)

3. The recommended flood is about:
   A. 8520 m3/s
   B. 20615 m3/s
   C. 26681 m3/s
   D. 32746 m3/s
   Answer: C. 8520 + 4.657 x 3900. (Key point: Design flood)

4. The 80% limits are:
   A. 26681 +/- 4731
   B. 26681 +/- 1.282
   C. 26681 +/- 3900
   D. 26681 +/- 6065 m3/s
   Answer: D. 1.282 x S_e (4731) = 6065. (Key point: Confidence limits)

## 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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