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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 II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 17: ELECTRIC POWER SUPPLY SYSTEMS
+
+// EXAMPLE : 17.9 :
+// Page number 429
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 1000.0                 // Maximum demand(kW)
+energy_cons = 5.0*10**6     // Annual energy consumption(kWh)
+PF = 0.85                   // Power factor
+capital_cost = 80000.0      // Capital cost of cable(Rs/km)
+cost_energy = 5.0/100       // Energy cost(Rs/kWh)
+interest_per = 10.0/100     // Rate of interest and depreciation
+r_specific = 1.72*10**-6    // Specific resistance of copper(ohm/cubic.cm)
+V = 11.0                    // Voltage(kV)
+
+// Calculations
+I = MD/(3**0.5*V*PF)                                             // Line current corresponding to maximum demand(A)
+hours_year = 365.0*24                                            // Total hours in a year
+LF = energy_cons/(MD*hours_year)                                 // Load factor
+loss_LF = 0.25*LF+0.75*LF**2                                     // Loss load factor
+P_2 = capital_cost*interest_per                                  // Cost in terms of L(Rs)
+P_3 = 3.0*I**2*r_specific*10**4*hours_year*loss_LF*cost_energy   // Cost in terms of I^2 & L(Rs)
+a = (P_3/P_2)**0.5                                               // Most economical cross-section of conductor(sq.cm)
+
+// Results
+disp("PART II - EXAMPLE : 17.9 : SOLUTION :-")
+printf("\nMost economical cross-section of the conductor, a = %.2f cm^2 \n", a)
+printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")