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+// Example 32_33
+clc;funcprot(0);
+//Given data
+L_cap=100;// MW
+// n=0.4*L;(given)
+L=[20 80 30];// MW
+T=[0 8 16 24];// Time in hours
+CV=35;// MJ/kg
+C=2;// Coal cost in Rs./kg
+Sc=2.5;// Rs./kWh
+n_com=95/100;// Combustion efficiency
+
+// Calculation
+E=(L(1)*(T(2)-T(1)))+(L(2)*(T(3)-T(2)))+(L(3)*(T(4)-T(3)));// Total energy consumed a day in MWh
+L_a=E/24;// Average load of the plant in MW
+L_max=80;// MW
+LF=L_a/L_max;
+CF=L_a/L_cap;
+// Outputs
+O_1=(L(1)*(T(2)-T(1)));// MWh
+n_1=0.4*(L(1)/100);
+I_1=O_1/n_1;// MWh
+O_2=(L(2)*(T(3)-T(2)));// MWh
+n_2=0.4*(L(2)/100);
+I_2=O_2/n_2;// MWh
+O_3=(L(3)*(T(4)-T(3)));// MWh
+n_3=0.4*(L(3)/100);
+I_3=O_3/n_3;// MWh
+I=(I_1+I_2+I_3)*10^3;// Total input in MWh
+m_f=(I*3600)/(CV*10^3*n_com*24);//  kg/hr
+m_f=(m_f*24)/10^3;// tons/day
+Cf=m_f*10^3*C;// The cost of fuel in Rs./day
+Mg=E*10^3*Sc;// The money gained by selling the energy generated in rupees
+Pr=(Mg-Cf);// Profit gained during the day in rupees/day
+n_o=(E/(I/10^3))*100;// The overall efficiency of the plant in %
+printf('\n(a)The load factor of the plant=%0.2f \n   The capacity factor of the plant=%0.3f \n(b)The fuel consumed in tonnes per day=%0.1f tons/day \n(c)Profit gained by the plant=%0.0e rupees/day \n(d)The overall efficiency of the plant=%0.2f percentage',LF,CF,m_f,Pr,n_o);
+// The answer provided in the textbook is wrong