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+// Example 32_34
+clc;funcprot(0);
+//Given data
+L_cap=1500;// kW
+// n=0.43*(L)^0.48;(given)
+T=[0 4 8 12 16 20 24];// Time in hours
+L_a=[200 600 1000 400 200 100];// Load in kW
+L_b=[800 400 200 200 600 400];// Load in kW
+L_t=[1000 1000 1200 600 800 500];// Load in kW
+CV=45*10^3;// MJ/kg
+Dc=30;// The cost of diesel in Rs./liter
+SG=0.85;// Specific gravity
+pr=15/100;// The profit required
+oc=30/100;// The other costs
+n_com=92/100;// Combustion efficiency
+
+// Calculation
+t=[0 4 4 8 8 12 12 16 16 20 20 24 24];// Time in hrs for load curve
+L_A=[200 200 600 600 1000 1000 400 400 200 200 100 100 1500];// Load A in kW for load curve
+subplot(3,1,1);
+xlabel('Time in hours');
+ylabel('Load in kW');
+xtitle('Load of consumer-A');
+plot(t,L_A,'b');
+legend('Load curve for (A)');
+t=[0 0 4 4 8 8 12 12 16 16 20 20 24 24];// Time in hrs for load curve
+L_B=[0 800 800 400 400 200 200 200 200 600 600 400 400 1500];// Load B in kW for load curve
+subplot(3,1,2);
+xlabel('Time in hours');
+ylabel('Load in kW');
+xtitle('Load of consumer-B');
+plot(t,L_B,'b');
+legend('Load curve for (B)');
+t=[0 0 4 4 8 8 12 12 16 16 20 20 24 24];// Time in hrs for load curve
+L_AB=[0 1000 1000 1000 1000 1200 1200 600 600 800 800 500 500 1500];// Load A+B in kW for load curve
+subplot(3,1,3);
+xlabel('Time in hours');
+ylabel('Load in kW');
+xtitle('Load of on plant for consumerA and B');
+plot(t,L_AB,'b');
+legend('Load curve for (A+B)');
+//(i)
+E=(L_a(1)*(T(2)-T(1)))+(L_a(2)*(T(3)-T(2)))+(L_a(3)*(T(4)-T(3)))+(L_a(4)*(T(5)-T(4)))+(L_a(5)*(T(6)-T(5)))+(L_a(6)*(T(7)-T(6)));// Total energy consumed a day in kWh
+L_a1=E/24;// kW
+L_max1=1000;// kW
+LF_A=L_a1/L_max1;
+E=(L_b(1)*(T(2)-T(1)))+(L_b(2)*(T(3)-T(2)))+(L_b(3)*(T(4)-T(3)))+(L_b(4)*(T(5)-T(4)))+(L_b(5)*(T(6)-T(5)))+(L_b(6)*(T(7)-T(6)));// Total energy consumed a day in kWh
+L_b1=E/24;// kW
+L_max2=800;// kW
+LF_B=L_b1/L_max2;
+E=(L_t(1)*(T(2)-T(1)))+(L_t(2)*(T(3)-T(2)))+(L_t(3)*(T(4)-T(3)))+(L_t(4)*(T(5)-T(4)))+(L_t(5)*(T(6)-T(5)))+(L_t(6)*(T(7)-T(6)));// Total energy consumed a day in kWh
+L_ab=E/24;// kW
+L_max=1200;
+LF_AB=L_ab/L_max;
+PL=((LF_AB-LF_A)/LF_A)*100;// Maximum percentage increase in load factor
+DF=(L_max1+L_max2)/L_max;// Diversity factor
+O_1=(L_t(1)*(T(3)-T(1)));// kWh
+n_1=0.43*(L_t(1)/L_cap)^0.48;
+I_1=O_1/n_1;// kWh
+O_2=(L_t(3)*(T(4)-T(3)));// kWh
+n_2=0.43*(L_t(3)/L_cap)^0.48;
+I_2=O_2/n_2;// kWh
+O_3=(L_t(4)*(T(5)-T(4)));// kWh
+n_3=0.43*(L_t(4)/L_cap)^0.48;
+I_3=O_3/n_3;// kWh
+O_4=(L_t(5)*(T(6)-T(5)));// kWh
+n_4=0.43*(L_t(5)/L_cap)^0.48;
+I_4=O_4/n_4;// kWh
+O_5=(L_t(6)*(T(7)-T(6)));// kWh
+n_5=0.43*(L_t(6)/L_cap)^0.48;
+I_5=O_5/n_5;// kWh
+I_t=(I_1+I_2+I_3+I_4+I_5)*3600;// Total input in kJ
+m_f=I_t/(CV*n_com*24);// kg/hr
+V_f=m_f/0.85;// liters/hr
+V_f=V_f*24;// liters
+C_f=V_f*Dc;// Cost of fuel in Rs./day
+Oc=C_f*oc;// The other cost running the plants in Rs./day
+Tc=C_f+Oc;// The total cost running the plants in Rs./day
+Pr=Tc*pr;// The profit required in Rs./day
+Tcs=Tc+Pr;// Total cost of saling the energy generated/day in rupees
+O_t=O_1+O_2+O_3+O_4+O_5;// Total energy generated in kWh
+Cs=Tcs/O_t;// The cost of sailing the energy in Rs./kWh
+printf('\n(i)The individual load factor of consumer A=%0.3f \n   The individual load factor of consumer B=%0.3f \n(ii)Load factor of the system=%0.3f \n    Diversity factor of the system=%0.1f \n(iii)The cost of selling the power=Rs.%0.2f/kWh',LF_A,LF_B,LF_AB,DF,Cs);
+// The answer vary due to round off error
+