---
title: "Gumbel peak flood and confidence limits, 75 years"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-peak-0-4pct-75yr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-peak-0-4pct-75yr
site: BCT Engineering
---

# Gumbel peak flood and confidence limits, 75 years

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2076 Asoj, 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-peak-0-4pct-75yr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

In the time series data of annual peak flood for 75 years, the mean and standard deviation are found to be equal to 5561 m$^3$/s and 1718 m$^3$/s respectively. Using $y_n = 0.556$ and $S_n = 1.189$ (for 75 yrs):

i) Determine the peak flood for 0.4% probability of exceedance by Gumbel's method.

ii) Compute 90% confidence limits for the above floods, using $f(c) = 1.6$ for 90% confidence level respectively.

## Short answer

For P =0.4 %, T =250 yr; frequency factor K = (y_T - y_n)/S_n =4.174; peak flood =5561 +4.174×1718 =12733 m³/s. 90% confidence limits: 12733 ± 1.6×1004 → 11127 and 14339 m³/s.

## Answer

#### (i) Peak flood for 0.4% probability of exceedance
$$
T=\frac1P=\frac{1}{0.004}=250 \text{ years}
$$

$$
y_T=-\ln\left[\ln\frac{T}{T-1}\right]=-\ln\left[\ln\frac{250}{249}\right]=5.519
$$

$$
K=\frac{y_T-\bar y_n}{S_n}=\frac{5.519-0.556}{1.189}=4.174
$$

$$
x_T=\bar x+K\sigma=5561+4.174\times1718=12733 \text{ m}^3/\text{s}
$$

**Answer: $x_{250}$ ≈ 12733 m³/s.**

#### (ii) 90% confidence limits
$$
b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(4.174)+1.1(4.174)^2}=5.059
$$

$$
S_e=\frac{b\sigma}{\sqrt N}=\frac{5.059\times1718}{\sqrt{75}}=1004 \text{ m}^3/\text{s}
$$

$$
x_T'=x_T\pm f(c)S_e=12733\pm1.6\times1004=12733\pm1606
$$

**Answer: the 90% confidence limits are 11127 m³/s and 14339 m³/s.**

## Check yourself

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

1. 0.4% probability of exceedance corresponds to a return period of:
   A. 25 years
   B. 250 years
   C. 400 years
   D. 40 years
   Answer: B. T = 1/0.004 = 250. (Key point: Return period)

2. The flood x_250 is about:
   A. 5561 m3/s
   B. 10000 m3/s
   C. 12733 m3/s
   D. 14339 m3/s
   Answer: C. K = 4.174, so x = 5561 + 4.174 x 1718. (Key point: Flood magnitude)

3. The 90% confidence limits are about:
   A. 12733 +/- 1004
   B. 12733 +/- 6000
   C. 12733 x 1.6
   D. 12733 +/- 1606 m3/s
   Answer: D. S_e = 1004, limits +/- 1.6 x 1004. (Key point: Confidence limits)

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