 Hello everyone, Myself Dr. Sunil Lathayan Kulkarni, Assistant Professor, Department of Mechanical Engineering, Valchan Institute of Technology, SolarPur. Today I am going to take the video session on methods of determining depreciation too. In the last video session, we have discussed the two methods that is straight line method and diminishing value method. In this video session, we are going to discuss the next method of determining depreciation. At the end of this session, the student will be able to explain the sinking fund method of calculating depreciation and solve numerical problems on methods of depreciation. The content of this video session are sinking fund method, its numerical treatment or mathematical treatment and then numerical problems on methods of depreciation. Now let us see what is sinking fund method. So in the sinking fund method, as we have seen earlier in straight line method as well as diminishing value method, we are saving certain amount. But in those methods, there was no interest accumulated on the amount which is set aside. Whereas in sinking fund method, in this method, some amount is set aside every year for n years and invested to earn compound interest. Now let us think for a while that what is the difference exactly in the previous methods and this method. In previous method, that is straight line and diminished value method, we were not accounting or considering any interest share. Whereas in case of sinking fund method, whatever amount is set aside or accumulated every year, it is invested to earn compound interest. Now you know the simple and compound interest are different. In compound interest, we get the and we are entitled to get the interest on the amount principle amount plus interest amount. Now let us assume that A is the annual deposit that is the amount which is set aside at the end of every year for n years and R is the rate of interest that is compounded annually. So amount set at the end of first year will be equal to capital A. So let us assume that this amount is the fixed amount which we are keeping aside every year. Now at the end of first year, we have kept set aside the amount A. So amount at the end of second year will be equal to A plus interest on A that is A plus A into R that is A into 1 plus R. Similarly, if we calculate the amount at the end of third year will be equal to whatever amount was there at the end of second year that is A into 1 plus R plus the interest on A into 1 plus R. So it will be equal to A into 1 plus R plus A into 1 plus R into R where R is the rate of interest that is compounded interest annually. Therefore, if we take A common what we get it as A into bracket 1 plus R into again 1 plus R by taking A common. So it becomes A into 1 plus R square. So as we have seen here the amount at the end of first year that is second year it is equal to A into 1 plus R means it is equal to A into 1 plus R raise to 1. Similarly, amount at the end of third year is equal to A into 1 plus R square. Similarly, we can write by analogy the amount at the end of end year will be equal to A into 1 plus R bracket raise to n minus 1 because it is at the end of third year it is 3 minus 1 that is 2. So on the same analogy we can write amount at the end of end year will be equal to A into 1 plus R raise to n minus 1. So total amount accumulated in n years let us call this amount as x. So x will be equal to sum of the amounts accumulated in n years. So it will be equal to amount set aside at the end of first year A plus amount accumulated in the second year, third year and so on and at the end of end year. So if we add them so we get A plus A into 1 plus R plus A into 1 plus R square plus A into 1 plus R raise to n minus 1. Now let us call that equation by taking A common we get x is equal to A into bracket 1 plus 1 plus R plus 1 plus R bracket square plus dy dy dash plus 1 plus R raise to n minus 1. Let us call it as equation number 1. Now if we multiply the above equation by 1 plus R on both the sides we will get x into 1 plus R is equal to A into bracket taking the 1 plus R inside the bracket we will get 1 plus R plus 1 plus R square plus 1 plus R cube and so on plus 1 plus R raise to n because here it is 1 n minus 1 into 1 plus R 1 plus R into bracket raise to n minus 1 into 1 plus R A become 1 plus R raise to n. Now if we subtract the equation 1 from 2 that is equation 2 minus 1 will give us now if you compare the equation 1 and 2 what you are finding is that okay x into 1 plus R minus x so we get x into R. Now here all the terms most of the terms which are common okay those we will get cancelled and what we will get it as A into 1 plus this these terms are getting cancelled only we will get A into 1 plus R bracket raise to n minus A so that is equal to x into R and therefore x into R will be equal to if I take A common again x into R is equal to A into bracket 1 plus R raise to n minus 1 or x will be equal to A into bracket 1 plus R bracket raise to n minus 1 divided by R where x is the actually total amount accumulated okay now if A is the depreciation charge per year C is the initial cost S is the salvage value then x is equal to C minus S because this is the amount which we need to collect okay initial cost minus salvage value that is the depreciation cost that we need to collect x is equal to C minus S therefore we can write C minus S is equal to putting instead of x we write it as C minus S is equal to A into bracket 1 plus R raise to n minus 1 upon R where C is the initial cost S is the salvage value and A is the depreciation charge per year and R is the rate of interest compounded annually and N is the useful life of the equipment or if I want to calculate the depreciation charge per year rearranging this equation I can write it as A is equal to C minus S into bracket R upon 1 plus R raise to 1 bracket raise to n so A is equal to C minus S into bracket R divided by into bracket 1 plus R raise to n minus 1 now let us see how to find out the depreciation charge by looking at the various numericals now in problem one let us see the problem one a transformer is costing 90,000 rupees as a useful life of 20 years determine the annual depreciation charge by using straight line method assume the salvage value of the equipment or transformer to be 10,000 rupees so it is a very simple problem let us see the solution so let initial cost P is equal to 90,000 useful life is 20 years and salvage value is 10,000 rupees S therefore using straight line method we can calculate the depreciation charge by using the equation P minus S upon N that is annual depreciation charge will be equal to 90,000 rupees minus 10,000 divided by 20 which comes out to be rupees 4000 rupees now let us see another numerical in which the equipment in a power station cost rupees 15,60,000 and has a salvage value of 60,000 at the end of 25 years so this is the initial cost and this is the salvage value at the end of 25 years now we want to find out the depreciated value of the equipment at the end of 20 years for the following methods so using straight line method using diminishing value method and using sinking fund method with 5% rate of interest compounded annually now here how to find out this so let initial cost of the equipment is 15,60,000 salvage value is 60,000 and useful life is 25 so by straight line method we can calculate the annual depreciation charge as P minus S upon S that is 15,60,000 minus 60,000 upon 25, 60,000 therefore this is the depreciation which will occur in 25 years and therefore this is the depreciation which will occur per year 60,000 rupees therefore the value of the equipment after 20 years will be equal to initial cost minus annual depreciation 60,000 multiplied by in 20 years what will be the depreciation so when we calculate this it comes to be the cost of equipment is 3,60,000 rupees now if we consider the diminishing value method in this method we first of all calculate the annual unit depreciation X which we have already derived X is equal to 1 minus S by P raised to 1 by N where X is equal to 1 minus S is the salvage value 60,000 P is initial cost 15,60,000 raised to 1 by 25 useful life of the year so this comes out to be 0.122 therefore the value of the equipment directly we can calculate after 20 years P into 1 minus X raised to 20 so which comes out to be rupees 115,615 and using sinking fund method if rate of interest is 5% okay that is 0.05 then annual deposit will be Q will be equal to P minus S into R upon 1 plus R raised to N minus 1 so when we put the initial cost salvage value and value of R and N we get the annual deposit in sinking fund is 31,433 so sinking fund which will be available at the end of 20 years will be equal to this value Q into 1 plus R raised to N minus 1 upon R so this comes out to be putting Q as 31,433 into 1 plus 0.05 raised to 20 minus 1 upon 0.05 it comes to be 10,39,362 therefore the value of plant after 20 years will be rupees initial cost 15,60,000 minus 10,39,362 which comes to be 5,20,638 these are the references thank you