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+//example 2.5

+//calculate 

+//size of cut-back stream.

+//time required for putting 37.5 mm depth of water

+//average depth of water applied

+

+clc;

+//given

+d=37.5//crop water requirement

+W=1//furrow spacing

+L=120//length of furrow

+n=-0.49;

+k=38;

+Ttotal=143;//Total time of irrigation

+A=[0 23 52 88 127]//given values of time of advance

+

+for i=1:5//loop to find respective values of time of ponding

+    B(i)=143-A(i);

+end

+

+

+for j=1:5//loop to find respective furrow infiltration

+    C(j)=B(j)^(n)*k;

+end

+

+

+for K=1:4//loop to find respective average infiltration

+   

+   D(K)=(C(K)+C(K+1))/2;

+end

+

+E(1)=D(1);

+for l=2:4//loop to determine cumulative infiltration

+    E(l)=D(l)+E(l-1);

+end

+I=E(4);

+

+T=(30*d*W*(n+1)/k)^(1/(n+1));

+dav=((24.5*Ttotal)+(I*(T-Ttotal)))/L;

+q=((120*37.5)-(24.5*143))/62;

+T=round(T);

+dav=round(dav*10)/10;

+q=round(q*100)/100;

+I=round(I*100)/100;

+mprintf("Maximum size of cut-back stream=%f lpm.",I);

+mprintf("\nMinimum size of cut-back stream=%f lpm.",q);

+mprintf("\nTime required for putting 37.5mm depth of water=%f minutes.",T);

+mprintf("\nAverage depth of water required=%f mm.",dav);