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+// Example 34_3

+clc;funcprot(0);

+//Given data

+C=40000;//Capital cost in Rupees

+V_1=4000;// Salvage value in Rupees

+n=20;// Useful life in years

+r=6/100;//Interest rate

+

+//Calculation

+//(a)

+TD=C-V_1;//Total depreciation in rupees

+D=TD/2;//Depreciation in 10 years in rupees

+V_2a=C-D;//Plant value at the end of 10 years in rupees

+//(b)

+// Assume  p_d=(1-p)

+p_d=(V_1/C)^(1/n);

+n=10;// Plant life in years

+V_2b=C*(p_d)^(n);// Plant value at the end of 10 years in rupees

+//(c)

+n=20;// Plant life in years

+Q=C-V_1;// Replacement cost  in rupees

+q=(Q*r)/(((1+r)^n)-1);//The amonut set aside per year

+n=10;// Plant life in years

+Q=(q*(((1+r)^n)-1))/(r);// The amount deposited in sinking fund in rupees

+V_2c=C-Q;// Plant value at the end of 10 years in rupees

+printf('\n(a)Plant value at the end of 10 year,V_2=Rs.%0.0f\n(b)Plant value at the end of 10 year,V_2=Rs.%0.0f\n(c)Plant value at the end of 10 year,V_2=Rs.%0.0f',V_2a,V_2b,V_2c)

+// The answer vary due to round off error