---
title: "Gumbel design flood, safety factor, confidence limits"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-20yr-life35-rel87"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-20yr-life35-rel87
site: BCT Engineering
---

# Gumbel design flood, safety factor, confidence limits

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2075 Asoj, 12 marks
- Video: [5.6.1 Risk, Reliability and Safety Factor | CE-603 (A) |](https://www.youtube.com/watch?v=EiH8QNzoI-g) — RGPV Official
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-20yr-life35-rel87
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Annual flood peak flood of a river for 20 years yielded a mean value of 5460 m$^3$/s and the standard deviation of 2950 m$^3$/s. The proposed hydraulic project on this river has an expected life of 35 years and reliability of project is 87%.

(i) Using Gumbel's method predict the flood discharge for the project if the value of $\bar{y}_n = 0.5402$ and $S_n = 1.1285$.

(ii) What discharge is to be adopted if the safety factor for flood magnitude is taken as 1.5 and also determine safety margin on this basis.

(iii) Calculate the confidence limits at 95% confidence probability, $f(c) = 1.96$.

## Short answer

T =252 yr from 35-yr life and 87% reliability; Gumbel K =4.419, predicted flood =18495 m³/s. With safety factr 1.5, adopt 27743 m³/s (margin 9248 m³/s). 95% confidence limits: 18495 ± 1.96×3504 → 11627 and 25364 m³/s.

## Answer

#### (i) Design flood
Reliability = 0.87, so risk $R=0.13$ over n = 35 years.

$$
\frac1T=1-(0.87)^{1/35}\ \Rightarrow\ T=252 \text{ years}
$$

$$
y_T=5.527,\qquad K=\frac{5.527-0.5402}{1.1285}=4.419
$$

$$
x_T=5460+4.419\times2950=18495 \text{ m}^3/\text{s}
$$

**Answer: predicted flood for the project ≈ 18495 m³/s (T ≈ 252 years).**

#### (ii) Safety factor 1.5
$$
Q_{adopted}=1.5\times18495=27743 \text{ m}^3/\text{s}
$$

Safety margin = adopted − predicted flood = 27743 − 18495 = 9248 m³/s (that is 50% of the predicted flood).

**Answer: adopt 27743 m³/s; safety margin = 9248 m³/s.**

#### (iii) 95% confidence limits
$$
b=\sqrt{1+1.3K+1.1K^2}=5.312,\qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.312\times2950}{\sqrt{20}}=3504 \text{ m}^3/\text{s}
$$

$$
x_T'=18495\pm1.96\times3504=18495\pm6868
$$

**Answer: 95% confidence limits are 11627 m³/s and 25364 m³/s.**

## Check yourself

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

1. For 87% reliability over 35 years the return period is about:
   A. 252 years
   B. 35 years
   C. 87 years
   D. 13 years
   Answer: A. 1/T = 1 - 0.87^(1/35). (Key point: Return period)

2. The predicted flood is:
   A. 5460 m3/s
   B. 5460 + 4.419 x 2950 = 18495 m3/s
   C. 27743 m3/s
   D. 11627 m3/s
   Answer: B. x = mean + K sigma. (Key point: Design flood)

3. With a safety factor 1.5, the adopted discharge and margin are:
   A. 18495 m3/s and 0
   B. 27743 m3/s and 27743
   C. 27743 m3/s and 9248 m3/s
   D. 12330 m3/s and 6165
   Answer: C. 1.5 x 18495 = 27743; margin = 27743 - 18495 = 9248. (Key point: Safety factor)

4. The 95% limits are:
   A. 18495 +/- 3504
   B. 18495 +/- 1.96
   C. 18495 +/- 9248
   D. 18495 +/- 6868 m3/s
   Answer: D. 1.96 x 3504 = 6868. (Key point: Confidence limits)

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