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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART I : GENERATION
+// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION
+
+// EXAMPLE : 7.12 :
+// Page number 76-77
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_installed = 500.0           // Installed capacity of the station(MW)
+CF = 0.45                       // Capacity factor
+LF = 0.6                        // Annual laod factor
+cost_fueloil = 10.0*10**7       // Annual cost of fuel,oil etc(Rs)
+capital_cost = 10**9            // Capital cost(Rs)
+interest = 0.15                 // Interest and depreciation
+
+// Calculations
+// Case(i)
+MD = cap_installed*CF/LF                                    // Maximum demand(MW)
+cap_reserve = cap_installed-MD                              // Reserve capacity(MW)
+// Case(ii)
+hours_year = 365.0*24                                       // Total hours in a year
+units_gen = MD*10**3*LF*hours_year                          // Units generated per annum(kWh)
+fixed_charge = interest*capital_cost                        // Annual fixed charges(Rs)
+running_charge = cost_fueloil                               // Annual running charges(Rs)
+annual_charge = fixed_charge+running_charge                 // Total annual charges(Rs)
+cost_unit = annual_charge*100/units_gen                     // Cost per kWh generated(Paise)
+
+// Results
+disp("PART I - EXAMPLE : 7.12 : SOLUTION :-")
+printf("\nCase(i) : Minimum reserve capacity of station = %.f MW", cap_reserve)
+printf("\nCase(ii): Cost per kWh generated = %.f paise", cost_unit)