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+// Example 35_2
+clc;funcprot(0);
+//Given data
+// C_1=2500 kW+0.550 kWh
+// C_2=2400 kW+0.6 kWh
+A_1=2500;
+A_2=2400;
+B_1=0.55;
+B_2=0.6;
+MD=100;// Maximum demand in MW
+md=10;// Minimum demand in MW
+t=8760;// hours
+
+// Calculation
+L=(A_1-A_2)/(B_2-B_1);// Time in hours
+P_b=MD-((L/t)*(MD-md));// MW
+P_p=100-P_b;// MW
+// (a)
+Ic_b=P_b;// MW
+Ic_p=P_p*1.2;// MW
+// (b)
+// For base load plant
+E_g1=((1/2)*(L+t)*(P_b-md))+(md*t);// MW-hrs
+LF_b=(E_g1/(P_b*t))*100;// Load factor
+CF_b=LF_b;// Capacity factor
+UF_b=CF_b/LF_b;// Use factor
+// For peak load plant
+E_g2=((1/2)*(L)*(MD-P_b));// MW-hrs
+LF_p=(E_g2/(P_p*t))*100;// Load factor
+CF_p=(E_g2/(24*t))*100;// Capacity factor
+UF_p=(CF_p/LF_p)*100;// Use factor
+// (c)For I-plant(base)
+E_g1=(((1/2)*(L+t)*(P_b-md))+(md*t))*1000;// kW-hrs
+C_1=(P_b*10^3*A_1)+(E_g1*B_1);// rupees
+// (c)For I-plant(peak)
+E_g2=((1/2)*(L)*(P_b-md))*1000;// kWh
+C_2=(P_p*10^3*A_2)+(E_g2*B_2);// rupees
+E_t=E_g1+E_g2;// kWh
+C=C_1+C_2;// Total generating cost in rupees
+C_a=C/E_t;// Average cost in Rs./kWh
+printf('\nAverage cost=Rs.%0.3f/kWh \nLoad shared by peak load plant=%0.0f MW \nLoad factor(base load plant)=%0.0f percentage \nLoad factor(peak load plant)=%0.1f percentage',C_a,P_p,LF_b,LF_p);
+// The answer vary due to round off error