Question

Table 1 below lists the annual peak discharge for the Ogeechee River, located on the coastal plain of Georgia. Total number of observations are 25. (1) (2) Annual Peak Discharge (cfs) (4) Recurrence Interval (RI) (5) Chance/Probability of occurrence in any given year (%| (3 Rank (M) Year 1974 9,000 1975 16,200 1976 7,720 1977 11,500 1978 17,300 1979 18,000 1980 27,900 1981 6,600 8,320 1982 1983 10,000 1984 6,560 1985 8,600 1986 8,200 1987 16,500 1988 2,700 5,030 1989 1990 5,620 1991 37,300 1992 6,544 1993 15,900 6,850 1994 1995 20,800 1996 8,550 1997 8,510 1998 28,200 Examine the data and enter the rank of each annual discharge values in column 3 (remember, the highest discharge will be ranked as 1 (already given on the table), the second highest will be ranked as 2...and so on and so forth). Now, determine the RI of each annual peak discharge event and enter them in the column 4 Now determine the annual exceedance probability of each peak discharge event using the RI that you calculated in column 4 and enter them it in the column 5 QUESTION 1 Q1: Using information from column 3 of Table 1, enter the rank of the following peak discharge events that occurred in 1977 1988 1994 1998 and 1980 . Enter only numbers. DO NOT ENTER text, comma, decimal, or symbol. QUESTION 2 PART-1: Q2 Using information from column 4 of Table 1, enter the RI of the following peak discharge events that occurred in 1978 1982 1987 . Do not round off. For example,12.591 should be entered as 12.59. and 0.7 should be 1993 and 1998 s0.70 entered QUESTION 3 PART-1: Q3 Using information from columns and 4 of Table 1, what is the probability ) peak discharge events with the following ranks: 1 ,10 15 and 25 any given year. Note: make sure you give your answers up to the second place after the decimal point only). Do not round off. For example, 12.591 should be entered as 12.59. and 0.7 should be entered as 0.70 QUESTION 4 The annual peak discharge data for the Des Moines River in Ohio is available for the last 124 vears. The largest flood on this river record occurred in 1993. The recurrence interval this flood this magnitude can occur in the year 2019. % chance that flood is. years. Additionally, there is a Note: Enter only numbers. Do not include text, comma, % or otther symbols with your responses. Do not round off. For example, 12.591 should be entered as 12.59. and 0.7 should be entered

Answer 1

Recurrence interval = (Number of years present on record+1)/(Rank)

Highest Annual peak discharge means to rank 1, next rank 2 and so on.

Chance/Probability in any given year (P)

P = [m/(n+1)]*100% = (1/RI)*100%

Where m is the rank of the annual peak discharge and n is the total number of events  

Year	Rank	Reccurence Interval	Chance/Probability in any given year (in %)
1974	12	(25+1)/12 = 2.167	(1/2.167)*100 = 46.15 %
1975	8	26/8 = 3.25	(1/3.25)*100 = 30.77 %
1976	18	26/18 = 1.44	(1/1.44)*100 = 69.44 %
1977	10	26/10 = 2.6	(1/2.6)*100 = 38.46 %
1978	6	26/6 = 4.33	(1/4.33)*100 = 23.09 %
1979	5	26/5 = 5.2	(1/5.2)*100 = 19.23 %
1980	3	26/3 = 8.67	(1/8.67)*100 = 11.53 %
1981	20	26/20 = 1.3	(1/1.3)*100 = 76.92 %
1982	16	26/16 = 1.625	(1/1.625)*100 = 61.54 %
1983	11	26/11 = 2.36	(1/2.36)*100 = 42.37 %
1984	21	26/21 = 1.2381	(1/1.2381)*100 = 80.77 %
1985	13	26/13 = 2	(1/2)*100 = 50 %
1986	17	26/17 = 1.5294	(1/1.5294)*100 = 65.39 %
1987	7	26/7 = 3.714	(1/3.714)*100 = 26.93 %
1988	25	26/25 = 1.04	(1/1.04)*100 = 96.15 %
1989	24	26/24 = 1.083	(1/1.083)*100 = 92.34 %
1990	23	26/23 = 1.1304	(1/1.1304)*100 = 88.46 %
1991	1	26/1 = 26	(1/26)*100 = 3.85%
1992	22	26/22 = 1.18	(1/1.18)*100 = 84.75 %
1993	9	26/9 = 2.89	(1/2.89)*100 = 34.60%
1994	19	26/19 = 1.3684	(1/1.3684)*100 = 73.08 %
1995	4	26/4 = 6.5	(1/6.5)*100 = 15.38 %
1996	14	26/14 = 1.857	(1/1.857)*100 = 53.85 %
1997	15	26/15 = 1.73	(1/1.73)*100 = 57.80 %
1998	2	26/2 = 13	(1/13)*100 = 7.69 %

Question 1

1977 - 10, 1988 - 25, 1994 - 19, 1998 - 2, 1980- 3

Question 2

1978 - 4.33, 1982 - 1.625, 1987 - 3.714, 1993 - 2.89, 1998- 13

Question 3

1 - 3.85% , 5 - 19.23% , 10 - 38.46% , 15 - 57.80% , 25 - 96.15%

Question 4

Reccurrence interval for this flood as per 1993 is (124+1)/ 1 = 125

Probability of occurrence in 2019 will be (1/125)*100% = 0.8%