Question

please show all work.

Recent flooding on the Big Bad River at Goodtown has motivated the community to reconsider building a flood levee. A nearby stream gage on the river has a 60-year record. The ranked sample of annual maximum peak discharges are: Rank Peak Discharge (cfs) 1 5850 2 4120 4020 3 58 59 60 410 390 200 The 3rd largest flood (shown above) occurred in the past year. What is the estimated return period for the 3rd largest flow based on the plotting position? a) Assume a 10-year return period flood (or greater) produces flood damage at Goodtown. A damaging flood has occurred in each of the past two years. What is the probability of this occurrence? Before the two floods, there was an extended period of 20-year with no floods of this size or greater. What is the probability of this occurrence? b) The town decides to build a flood levee for a design life of 30 years. The risk of failure that they are willing to accept during the design life is 1 chance in 5 (or 0.2). For what return period should the levees be designed? 1

Answer 1

• The return period for 3 rd largest flood(by california formula ) T=N/m,where

N= number of ranks and m= rank of the desired flood

this way ,for 3rd largest flood,return period(T) =60/3=20 years

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(a)

since return period of the flood is 10 years,T=10

for a binomial event,as in this case flood if occuring twice in 10 years of return period

hence probability (p) of the flood is 1/10 =0.1 and probability of not occuring (q) = 1-0.1 =0.9

and probability(P)is given as  , where n= return period of flood,r= number occurance in the return period,p= probability of occuring an event and q= probability of not occuring(p-1)

7C*p' *q"

P = =45 * (0.1)² * (0.9)⁸ =45*0.01*0.43 =0.1937 =19.37 %

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• for a 20 year no flood occurance before this ,the probability would be calculated as calculated above,

just n=20+2 =22 and r=2 and p=1/22 =0.045 and q= 0.954

hence ,P= =231 * (0.045)² * (0.954)²⁰ =231*0.002*0.389

22-2 2C *p'* qr

P=0.1801 =18.01 %

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(b)

For a designe life of 30 years thus n=30 ,risk (R) acceptable is 1/5 =0.2 and

ris is given as ,R =1-(1-p)ⁿ ,from here we can calculate p as,

p=1-(1-R)^1/n,hence p=1-(1-0.2)³⁰ =1-0.00123 =0.99877 ,hence the probability of the flood to occure during designe life of 30 years of the leaves is 99.87%