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+//Chapter 7
+//Example 7.3
+//page 249
+//To find the economical operation
+clear;clc;
+
+//from the table we got as the output in the example_7_1
+//for optimum operation of load 220MW,unit 1&2 must be loaded 100MW and 120MW respwctively
+//and for a load of 76MW,unit 1&2 must be loaded 20MW and 56MW respwctively
+start_up=400;
+//case(i)
+printf('\nCase(i)');
+//total fuel cost for the load of 220MW during 6AM to 6PM 
+Pg1=100;
+Pg2=120;
+C1=0.1*Pg1^2+40*Pg1+120;
+C2=0.125*Pg2^2+30*Pg2+100;
+total1=(C1+C2)*12;
+printf('\nTotal fuel cost for the load of 220MW during 6AM to 6PM = Rs. %d',total1);
+
+//total fuel cost for the load of 76MW during 6PM to 6AM 
+Pg1=20;
+Pg2=56;
+C1=0.1*Pg1^2+40*Pg1+120;
+C2=0.125*Pg2^2+30*Pg2+100;
+total2=(C1+C2)*12;
+printf('\nTotal fuel cost for the load of 76MW during 6PM to 6AM if both the units run = Rs. %d',total2);
+
+total=total1+total2; //total fuel cost for 24hrs
+
+printf('\nTotal fuel cost for the load during 24hrs if both the units run = Rs. %d',total);
+
+//case(ii)
+printf('\n\nCase(ii)');
+//If during light load condition unit2 is On and Unit1 is Off then
+Pg2=76;
+C2=0.125*Pg2^2+30*Pg2+100;
+total2=C2*12;
+total_case2=total1+total2+start_up;
+
+printf('\nTotal fuel cost for the 24hrs laod if only unit 2 run during light loads is = Rs. %d',total_case2);
+\ No newline at end of file