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+//the examples are continuously numbered throughout the textbook
+//Example 6_25 page no:345
+clc;
+//given
+beg_cost = 240000;//in rupees
+salvage_val = 24000;//in rupees
+t = 20;//in years
+t1 = 10;//in years
+tot_dep = beg_cost - salvage_val;
+tot_dep_af10 = beg_cost - 108000;
+val = beg_cost * (0.891)^10;
+tot_sink_fund = 216000;//in rupees
+annual_deposit = (0.08 * tot_sink_fund)/((1.08)^20 -1);
+annual_deposit_af10 = (annual_deposit *( 1.08^10-1))/0.08;
+val_plant = beg_cost - annual_deposit_af10;
+disp(tot_dep_af10,"the value calculated in straight line depreciation at the end of 10 years will be ( in rupees)");
+disp(val,"the value calculated in reducing balance depreciation at the end of 10 years will be ( in rupees)");
+disp(val_plant,"the value calculated in sinking fund depreciation at the end of 10 years will be ( in rupees)");
+//the result vary slightly hence values are rounded off in textbook
+