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+
+
+//example6.2
+//calculate required useful storage
+clc;funcprot(0);
+//given
+in=[8.6 2.2 1.8 0 0 13.5 280.6 510.2 136 52.5 20.6 12.3];      //inflow(ha-m)
+pan=[2.2 2.3 3.1 8.6 12.8 15.6 12.3 10.6 10 8.2 5.8 3];       //pan evaporation
+p=[0.8 1.2 0 0 0 4.8 12.2 18.6 8.6 1.5 0 0]                   //precipitation
+D=[14.5 15.8 16.2 16.8 17.5 18 18 17 16.5 16 15.8 15];        //Demand
+s=0;
+for i=1:12
+    if in(i)<10 then
+        r(i)=in(i);                                       //D/S requirement
+    else
+        r(i)=10;
+    end
+    E(i)=3.6*pan(i);                                     //Evaporation over reservior area
+    P(i)=3.5*p(i);                                       //Precipitation
+    I(i)=in(i)-r(i)-E(i)+P(i);                           //Adjusted inflow
+    S(i)=D(i)-I(i);                                      //Water required from storage
+    if S(i)<0 then
+        S(i)=0;
+    end
+    s=s+S(i);
+end
+mprintf("required useful storage=%f ha-m.",s);