This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Gumbel’s Method”.
1. The Gumbel method involves the use of which type of distribution function?
a) Generalized normal distribution
b) Generalized logarithmic distribution
c) Person distribution
d) Extreme value distribution
View Answer
Explanation: Gumbel’s distribution is a generalized extreme value distribution type-I. It is one on of the many probability distributions supported on the whole real number line. It is widely used in hydrologic studies for flood peak prediction, etc.
2. What is the definition of a flood according to Gumbel?
a) Largest flow of a particular month in a year
b) Largest of 365 daily flows in a year
c) 29th largest flow in a year
d) Sum of three largest daily flows in a year
View Answer
Explanation: According to Gumbel, a flood is defined as the largest of 365 daily flows in a year and the annual flood series is a group of these values for a number of successive years.
3. As per Gumbel’s distribution, what is the probability of occurrence of a flood X greater than or equal to a flood x? y is some dimensional variable.
a) ee-y
b) e-e-y
c) 1-ee-y
d) 1-e-e-y
View Answer
Explanation: As per Gumbel’s distribution, P(X<x)=e-e-y.
So, for a predicting a flood of magnitude greater than or equal to x, it is subtracted from 1.
∴P(X≥x)=1-P(X<x) = 1-e-e-y
4. For an infinite sample size, what is the correct expression for the dimensional variable y in the cumulative density function of Gumbel’s distribution? \(\overline{x}\) is the mean of the variate X and σ is the standard deviation of the variate X.
a) \(\frac{1.2825(x-\overline{x})}{σ}+0.577\)
b) \(\frac{1.2825(\overline{x}-x)}{σ}+0.577\)
c) \(\frac{1.2825(x+\overline{x})}{σ}-0.577\)
d) \(\frac{1.2825(\overline{x}-x)}{σ}-0.577\)
View Answer
Explanation: As per Gumbel’s distribution, P(X<x)=e-e-y
Where, y=α(x-a)⋯⋯⋯(1)
a=\(\overline{x}\)-0.45σ⋯⋯⋯(2) and α=1.2825/σ⋯⋯⋯(3), for infinite sample size.
Substitute (2) and (3) in (1),
y=\(\frac{1.2825}{σ}(x-(\overline{x}-0.45σ))=1.2825(\frac{x-\overline{x}}{σ}+\frac{0.45σ}{σ})\)
⇒y=1.2825\((\frac{x-\overline{x}}{σ}+0.45)=\frac{1.2825(x-\overline{x})}{σ}\)+(1.2825*0.45)
∴y=\(\frac{1.2825(x-\overline{x})}{σ}+0.577\).
5. The reduced variate in Gumbel method depends on the return period of flood and the available sample size.
a) True
b) False
View Answer
Explanation: The reduced variate in Gumbel method is a function of the probability of exceedance and hence only depends on the time period of the flood, irrespective of the sample size available.
6. What is the correct equation for the reduced variate y in Gumbel’s method? T is the time period.
a) -ln.ln\(\frac{T}{T-1}\)
b) -ln.ln\(\frac{T-1}{T}\)
c) -ln.ln\(\frac{1}{T-1}\)
d) -ln.ln\(\frac{1}{T}\)
View Answer
Explanation: In the hydrologic analysis, usually the value of flood is required for a given probability of occurrence. So, the cumulative distribution function can be written as,
y=-ln[-ln(1-P)], where P = 1/T is the probability of occurrence.
⇒y=\(-ln[-ln(1-\frac{1}{T})]=-ln[-ln(\frac{T-1}{T})]=-ln[ln(\frac{T-1}{T})^{-1}]\)
∴y=-ln.ln\(\frac{T}{T-1}\).
7. What is the correct expression of frequency factor (K) Gumbel’s flood equation for an infinite sample size? y is the reduced variate.
a) K=\(\frac{y+0.577}{1.2825}\)
b) K=\(\frac{y-0.577}{1.2825}\)
c) K=\(\frac{y+1.2825}{0.577}\)
d) K=\(\frac{y-1.2825}{0.577}\)
View Answer
Explanation: The reduced variate for an infinite sample size is given as,
y=\(\frac{1.2825(x-\overline{x})}{σ}+0.577 ⇒σ(y-0.577)=1.2825(x-\overline{x})\)
⇒\(\frac{σ(y-0.577)}{1.2825}=x-\overline{x} ⇒x=\overline{x}+\frac{(y-0.577)}{1.2825}.σ\)⋯⋯⋯(1)
Equation (1) can be compared to the general hydrologic frequency analysis equation x=\(\overline{x}\)+Kσ.
∴Frequncy factor,K=\(\frac{y-0.577}{1.2825}\).
8. Find the Gumbel’s reduced variate for a flood of return period 75 years?
a) -3.62
b) 3.62
c) -4.31
d) 4.31
View Answer
Explanation: The reduced variate is given as,
\(y=-ln.ln\frac{T}{T-1} =-ln.ln\frac{75}{75-1}=-ln.ln\frac{75}{74}\)=-ln(0.013423)=4.31
9. Find the Gumbel’s reduced variate of a flood with a probability of non-occurrence of 0.978?
a) -1.339
b) 1.339
c) -3.805
d) 3.805
View Answer
Explanation: Given q = 0.977. So, p = 1 – 0.978 = 0.022
⇒T=\(\frac{1}{p}=\frac{1}{0.022}\)=45.4545……years
∴Reduced variate, y=-ln.ln\(\frac{T}{T-1}=-ln.ln\frac{45.4545}{45.4545-1}\)=-ln(0.02225)=3.805.
10. What is the return period (in years) of a flood of magnitude 500 m3/s with a Gumbel’s reduced variate of 4?
a) 51
b) 54.5
c) 55.1
d) 56.6
View Answer
Explanation: Given y=\(-ln.ln\frac{T}{T-1}\)=4
⇒\(ln.ln\frac{T}{T-1}=-4 ⇒ln\frac{T}{T-1}=e^{-4}\)=0.01831564
⇒\(\frac{T}{T-1}=e^{0.01831564}\)=1.0184844
∴T=\(\frac{1.0184844}{1.0184844-1}=\frac{1.0184844}{0.0184844}\)=55.0997≅55.1 years
Sanfoundry Global Education & Learning Series – Engineering Hydrology.
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