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RAIN FALL ANALYSIS - Standardized Precipitation Index (SPI)

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Station Name: Angoda Mental Hospital

Type: Precipitatiom

Data Interval: Daily

Unit: Millimeters

Period 1960/01/01 to 2016/12/31

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4 appVersion 0.99.7109.29964

Climate\0020\Data\0020\Guide\0020\SPI\002E\pdf SPEI.pdf

1. Select period

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Assume 30 day months input.1 1 Month 30 2 Months 60 3 Months 90 4 Months 120 5 Months 150 6 Months 180 7 Months 210 8 Months 240 9 Months 270 10 Months 300 11 Months 330 12 Months 360122 mat :

"1 Month"
"2 Months"
"3 Months"
"4 Months"
"5 Months"
"6 Months"
"7 Months"
"8 Months"
"9 Months"
"10 Months"
"11 Months"
"12 Months"

Select how many Months

n_day

Starting time t.0 0 time :

n_day 30

2. Import rainfall data from EXCEL file

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NOTE : SMATH treats blanks in EXCEL as zeros. Hence, in case of

missing data, first use a suitable method to fill in such data.

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MyDirectory DocumentDirectory : E:\All Smath Files in ASUS\0001. Training ID\0001. IRESH Irrig. Dept\

MyFile MyDirectory Rainfal_Data.xlsx concat :

Read daily rainfall data file. Define correct cell range (in this case B7:B20826

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RF_DATA MyFile Sheet1 B7 B20826 importData.XLSX :

RF_DATA rows 20820

RF_DATA cols 1

n.DATA RF_DATA length : 20820

Find 'p' day Total of rainfall series

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Program 1: To find 'p' day totals of vector 'A' A p n_day_total Find 'p' day total of raw data r 1 A rows p / trunc range C r el A 1 r 1 - (p * + (r p * (range el sum : for C 31 line :

period.tot n.DATA 360 / yr * :

period.tot yr 58

ΔT 0 time t.0 - : 1 s *

Apply Program 1 with appropriate variables X RF_DATA n_day n_day_total : 30.4 278.7 63.9 387.1 330 187.3 495.9 18 163.3 128.8 795.3 111 mat

X max 983.6239

X min 0

n X rows : 694

RF_DATA rows n_day / 694

n.bal RF_DATA rows n_day / n - (n_day * : 0

n n_day * n.bal + 20820

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3. Dealing with zero values

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ΔT 0 time t.0 - : 6 s *

Program 2: To deal 
with ln(0) n.zero 0 : j 1 X rows range X j el 0 ≤ LNX j el 0 : n.zero n.zero 1 + : pos n.zero el j : 31 line LNX j el X j el ln : if for n.zero 0 ≡ pos Undefined : pos pos : if 31 line

ln(x) values LNX 3.41 5.63 4.16 5.96 5.8 5.23 6.21 2.89 5.1 4.86 6.68 4.14 3.8 3.88 141 mat

Index of the LNX 
array where LNX = 0 pos 99148172196197221245269300301320328331131 mat

LNX 28 el 5.0907

j 694

Clear j from memory j Clear 1

X.mean X sum n / : 202.7543

LNX.mean LNX sum n / : 4.768

ΔT 0 time t.0 - : 6.2 s *

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4. Gamma PDF and CDF

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Number of 
zero values n.zero 29

Prob of zero values Pr.zero n.zero n / : 0.0418

A X.mean ln LNX.mean - : 0.544

α 14 A * / 114 A 3 / * + sqrt + (* : 1.0633

β X.mean α / : 190.6882

Eqn for Gamma function PDF x g 1 β α^α Gamma * / x α 1 -^* e x β / -^* :

α Gamma 0.9672

5 Gamma 24

Eqn. 1

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12 / Gamma 1.7725

π sqrt 1.7725

Eqn for Gamma function CDF x_val G 1 β α^α Gamma * / x α 1 -^e x β / -^* x 0 x_val int * :

0 G 0

Eqn. 2

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0 g 0

ΔT 0 time t.0 - : 6.2 s *

PDF of Gamma function. See 
how vectorize function works. g1 X g vectorize : 0.0041 0.0013 0.0036 0.0007 0.001 0.002 0.0004 0.0042 0.0023 0.0027 0.0001 111 mat

CDF of Gamma function,
using vectorize function G1 X G vectorize : 0.1271 0.7456 0.2567 0.8525 0.8035 0.597 0.914 0.0752 0.5455 0.4601 0.9775 111 mat

X 4 el G 0.8525

X 4 el g 0.0007

X 4 el 387.1

ΔT 0 time t.0 - : 25.9 s *

Method to deal with 
zero probabilites G1 Pr.zero 1 Pr.zero - (G1 * + :

where Pr.zero 0.04179

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Returns Inv. Cumulative 
Normal Distribution values F.inv.normal G1 ICDF.Normal :

G1 0.16358 0.75623 0.28774 0.85864 0.81174 0.61389 0.9176 0.11388 0.56452 0.4827 0.97846 0.28288 0.2193 0.2323 0.59786 0.85857 0.9152 171 mat

F.inv.normal 0.16358 0.97983 - 0.75623 0.69423 0.28774 0.56 - 0.85864 1.07424 0.81174 0.88434 0.61389 0.28947 0.9176 1.3891 0.11388 1.20613 - 0.56452 0.16243 0.4827 0.04337 - 0.97846 2.02285 0.28288 0.5743 - 0.2193 0.77457 - 0.2323 0.73129 - 0.59786 0.24781 0.85857 1.07393 0.9152 1.37349 172 mat

G1 rows 694

G1 Max 0.987886

G1 Min 0.04179

ΔT 0 time t.0 - : 28.1 s *

5. Calculation of SPI values and Probabilities

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SPI Values in Column 2 of F.inv.normal Z F.inv.normal 2 col : 0.97983 - 0.69423 0.56 - 1.07424 0.88434 0.28947 1.3891 1.20613 - 0.16243 0.04337 - 2.02285 0.5743 - 121 mat

This gives a guide to determine upper and lower limits for condition statement

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Z Max 2.2535

Z Min 1.7303 -

To extract values satisfying above condition statements, it is necessary to round off Z values to 2 decimals.

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Z Z 2 round vectorize :

n.tot Z rows : 694

Z Max 2.250

Z Min 1.730 -

ΔT 0 time t.0 - : 28.1 s *

Condition statements x cond1 x 2.0 ≥ : x cond2 x 1.5 1.99 lele : x cond3 x 1.0 1.49 lele : x cond4 x 0.99 - 0.99 lele : x cond5 x 1.49 - 1.0 - lele : x cond6 x 1.99 - 1.5 - lele : x cond7 x 2.0 - ≤ : 71 line

Example: Vector 
satisfying "cond4" Z cond4 vectorize 1110110081 mat

Z 0.980 - 0.690 0.560 - 1.070 0.880 0.290 1.390 1.210 - 0.160 0.040 - 2.020 111 mat

Very handy & powerful. Using 
a vector of All Conditions cc All_Conds cc cond1 cc cond2 cc cond3 cc cond4 cc cond5 cc cond6 cc cond7 71 mat :

Not necessary to vectorize 
each element cc All_Conds cc cond1 vectorize cc cond2 vectorize cc cond3 vectorize cc cond4 vectorize cc cond5 vectorize cc cond6 vectorize cc cond7 vectorize 71 mat :

SPI: using vectorize function SPI Z All_Conds vectorize sum : 926805073240071 mat

Probability vector (%) Prr SPI SPI sum / 100 * : 1.3 3.75 11.53 73.05 4.61 5.76 071 mat

Alternatively. This looks better Z All_Conds vectorize Mean 100 * 1.3 3.75 11.53 73.05 4.61 5.76 071 mat

cond(4) total Z cond4 vectorize sum 507

Z cond7 vectorize sum 0

Z cond4 vectorize Mean 100 * 73.05

Z cond7 vectorize rows 694

Results Res SPI Prr augment : 9 1.3 26 3.75 80 11.53 507 73.05 32 4.61 40 5.76 0072 mat

507694 / 100 * 73.05

Sum of SPI SPI sum 694

n 694

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Sum of probabilities Prr sum 100

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ΔT1 0 time t.0 - : 28.5 s *

6. Data for Table format for Results

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Define sp n_day 30 / var2str Month SPI concat : 1 Month SPI

Define Header 
of the TABLE my_header sp Category Number of Times Probability % 14 mat :

tp Tot. Period = period.tot yr / 0 var2str yr. concat : Tot. Period = 58 yr.

Footer of 
the Table my_footer tp --- SPI sum Prr sum 14 mat : Tot. Period = 58 yr. --- 69410014 mat

Define 
Category Cat 2.0+ Extremely Wet 1.5 to 1.99 Very Wet 1.0 to 1.49 Moderately Wet -0.99 to +0.99 Near Normal -1.0 to -1.49 Moderately Dry -1.5 to -1.99 Severely Dry ≤ -2.0 Extremely Dry 72 mat :

To find 90th percentile P.90 0.90 768 ICDF.Normal : 86.25

Reverse P.90 768 CDF.Normal 0.9

Body of 
the Table Res Cat Res augment : 2.0+ Extremely Wet 9 1.3 1.5 to 1.99 Very Wet 26 3.75 1.0 to 1.49 Moderately Wet 80 11.53 -0.99 to +0.99 Near Normal 507 73.05 -1.0 to -1.49 Moderately Dry 32 4.61 -1.5 to -1.99 Severely Dry 40 5.76 ≤ -2.0 Extremely Dry 0074 mat

50th percentile is Mean P.50 0.50 768 ICDF.Normal : 76

Reverse P.50 768 CDF.Normal 0.5

Mean

7. Results

]]> Cat my_footer my_header ΔT1 Res Prr SPI All_Conds cond7 cond6 cond5 cond4 cond3 cond2 cond1 G g β α A Pr.zero LNX.mean X.mean ΔT period.tot n_day_total t.0 my_footer my_header ΔT1 Res Prr SPI All_Conds cond7 cond6 cond5 cond4 cond3 cond2 cond1 Z G1 g1 G g β α A Pr.zero LNX.mean X.mean pos LNX X ΔT period.tot n_day_total RF_DATA t.0 ΔT1 Prr SPI All_Conds cond7 cond6 cond5 cond4 cond3 cond2 cond1 G g β α A Pr.zero LNX.mean X.mean ΔT period.tot t.0

Res
2.0+	Extremely Wet	9	1
1.5 to 1.99	Very Wet	26	4
1.0 to 1.49	Moderately Wet	80	12
-0.99 to +0.99	Near Normal	507	73
-1.0 to -1.49	Moderately Dry	32	5
-1.5 to -1.99	Severely Dry	40	6
≤ -2.0	Extremely Dry	0	0
#
1 Month SPI	Category	Number of Times	Probability %
#
#
Tot. Period = 58 yr.	---	694	100

ΔT2 1 time t.0 - : 28.6 s *