---
title: "Gumbel analysis, 80-year and 1.4-year floods"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-80yr-620000-1-4yr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-80yr-620000-1-4yr
site: BCT Engineering
---

# Gumbel analysis, 80-year and 1.4-year floods

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2071 Shrawan, 6+4+4 marks
- Video: [Gumbel's method in Hydrology for flood estimation | Flood Routing | GATE 2022 |](https://www.youtube.com/watch?v=gtUiLfn_E8U) — TechnoCivil
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-80yr-620000-1-4yr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

An analysis of an annual flood series covering the period 1890 to 1966 on a certain river shows that the 80 year flood has a magnitude of 620000 units and 1.4 year flood has a magnitude of 215000 units. Assume the annual floods are Gumbel distributed.

i) What is the probability of having a flood as great as or greater than 440000 units?

ii) What is the magnitude of flood having a recurrence interval of 40 years?

iii) What is the probability of having 575000 units flood or a greater flood in the coming 25 years time?

## Short answer

Following Gumbel's linear relation in reduced variate, probability of flood≥440000 is 0.0926 (return period~10.8 years); 40-year flood magnitude≈558429 units; probability of at least one such flood in 25 years is 40.8%.

## Answer

Gumbel's equation is linear in the reduced variate: $x_T=\bar x+K\sigma=a+b\,y_T$, with $y_T=-\ln[\ln(T/(T-1))]$. The two given floods fix the line.

| T (yr) | $y_T$ | $x_T$ (units) |
|---|---|---|
| 1.4 | -0.2254 | 215 000 |
| 80 | 4.3757 | 620 000 |

$$
b=\frac{620000-215000}{4.3757-(-0.2254)}=88023 \qquad\Rightarrow\qquad x_T=215000+88023\,[y_T-(-0.2254)]
$$

#### (i) Probability of a flood ≥ 440 000
$$
y=-0.2254+\frac{440000-215000}{88023}=2.3308
$$

$$
P(X\ge440000)=1-e^{-e^{-y}}=1-e^{-e^{-2.3308}}=0.0926
$$

**Answer: probability ≈ 0.0926 (about 9.3%; return period ≈ 10.8 years).**

#### (ii) Flood with a 40-year recurrence interval
$$
y_{40}=-\ln\left[\ln\frac{40}{39}\right]=3.6762
$$

$$
x_{40}=215000+88023\,(3.6762-(-0.2254))=558429 \text{ units}
$$

**Answer: $x_{40}$ ≈ 558429 units.**

#### (iii) Probability of a flood ≥ 575 000 in the next 25 years
$$
y=-0.2254+\frac{575000-215000}{88023}=3.8645
$$

$$
p=1-e^{-e^{-3.8645}}=0.02075 \quad(T=48.2 \text{ years})
$$

The probability of at least one such flood in n = 25 years is

$$
1-(1-p)^{25}=1-(1-0.02075)^{25}=0.4080
$$

**Answer: probability ≈ 0.4080 (about 40.8%).**

## Check yourself

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

1. Why do two known floods fix the whole Gumbel line?
   A. Because Gumbel needs only the mean
   B. Because x_T is linear in the reduced variate y_T
   C. Because x_T is linear in T itself
   D. Because y_T is the same for all floods
   Answer: B. x_T = a + b y_T is a straight line in y_T, so two points give a and b. (Key point: Linearity of Gumbel)

2. How is the slope b calculated from the 80-year and 1.4-year floods?
   A. b = 620000/4.3757
   B. b = (4.3757 - 0.2254)/(620000 - 215000)
   C. b = (620000 - 215000)/(80 - 1.4)
   D. b = (620000 - 215000)/(4.3757 + 0.2254)
   Answer: D. b is change in x over change in y_T: 405000/4.6011 = 88023. (Key point: Slope of the line)

3. After finding the reduced variate y for 440000 units, how is P(X >= 440000) found?
   A. P = 1 - exp(-exp(-y))
   B. P = 1/y
   C. P = exp(-exp(-y))
   D. P = e^(-y)
   Answer: A. The Gumbel non-exceedance probability is exp(-exp(-y)), so exceedance is 1 minus that (0.0926). (Key point: Probability of exceedance)

4. A flood has annual exceedance probability p = 0.02075. What is the chance of at least one in 25 years?
   A. (1 - p)^25
   B. 25 x p
   C. 1 - (1 - p)^25
   D. p^25
   Answer: C. At least one occurrence in n years is 1 - (1 - p)^n, which is 0.408 here. 25p would ignore overlap. (Key point: Probability over several years)

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