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+//Chapter 7
+//Example 7.4
+//page 263
+//To find required generation for each plant and losses incurred
+clear;clc;
+
+///Let us use the program given in the Appendix G in the textbook which includes penalty factor also to write
+//a function that returns the value of lamda,Loading of each generator and losses
+//when the total load on the plant is sent to the function
+
+function [lamda,Pg,PL]=optimum2(Pd)
+n=2; //no of generators
+Alpha=[0.02 0.04];
+Beta=[16 20];
+lamda=20; //initial value of lamda
+lamdaprev=lamda;
+eps=1; //tolerance
+deltalamda=0.1;
+Pgmax=[200 200];
+Pgmin=[0 0];
+B=[0.001 0;0 0];
+Pg=zeros(n,1);
+noofiter=0;
+PL=0;
+Pg=zeros(n,1);
+while abs(sum(Pg)-Pd-PL)>eps
+    for i=1:n
+        sigma=B(i,:)*Pg-B(i,i)*Pg(i);
+        Pg(i)=(1-(Beta(i)/lamda)-(2*sigma))/(Alpha(i)/lamda+2*B(i,i));
+        PL=Pg.'*B*Pg;
+        if Pg(i)>Pgmax(i) then
+            Pg(i)=Pgmax(i);
+        end
+        if Pg(i)<Pgmin(i) then
+            Pg(i)=Pgmin(i);
+        end
+    end
+    PL=Pg.'*B*Pg;
+    if(sum(Pg)-Pd-PL)<0 then
+        lamdaprev=lamda;
+        lamda=lamda+deltalamda;
+    else
+        lamdaprev=lamda;
+        lamda=lamda-deltalamda;
+    end
+    noofiter=noofiter+1;
+    Pg;
+end
+endfunction
+
+//In this example let us take the answer .i.e load(Pd)=237.04MW and calculate 
+//lamda so that we can use the algorithm used in the textbook
+Pd=237.04
+[lamda_test,Pg_test,PL_test]=optimum2(Pd);
+printf('\nLagrange''s multiplier (lamda) is\n Lamda =%0.1f',lamda_test);
+printf('\n\nRequired generation for optimum loading are \n Pg1=%0.2f MW \n Pg2=%d MW\n',Pg_test(1),Pg_test(2));
+printf('\nThe transmission power loss is\n PL=%0.2f MW',PL_test);
+printf('\n\nThe load is \n Pd=%0.2f MW',Pd);
+