Welcome To A Session On Annuity ©. S. B. Bhattacharjee

Business Mathematics (Eleventh Edition) Your Textbook Business Mathematics (Eleventh Edition) By D.C. Sancheti  V.K. Kapoor Chapter 8 ©. S. B. Bhattacharjee

What is meant by Annuity? Suppose, we want to make payment of Tk.5000 each year for 10 years to an account which earns 6 percent interest, compounded annually, and we want to calculate the total amount which will be in the account after 10 payments. This series of periodic payments is called an annuity. An ordinary annuity is a series of equal payments, each payment of which is made at the end of a period. Different annuities may be stated as under: ©. S. B. Bhattacharjee

i = Interest rate per period M = Sum of the annuity A = Payment per period i = Interest rate per period n = Number of periods Continued….. ©. S. B. Bhattacharjee

What are various types of annuity? Continued…… ©. S. B. Bhattacharjee

Different types of annuities Annuity Annuity Contingent (Payments are to be made till the happening of some contingent event such as the death of a person, marriage of a girl ete.) Annuity Certain (Payments are to be made for a certain or fixed number of years) Annuity due (Where the first payment falls due at the beginning of the ist interval, and so on) Annuity immediate (Where the first payment falls due at the end of the first interval, and so on) ©. S. B. Bhattacharjee

Calculation of different types of annuities: 1. Formula for the present value of an Annuity due Continued….. ©. S. B. Bhattacharjee

2. Formula for the present value of an Immediate Annuity 3. Formula for the amount of an Immediate Annuity 4. Formula for the amount of Annuity due 5. Present Value of an Annuity The present value of an Annuity is the sum of the present values of its installments. ©. S. B. Bhattacharjee

Example: 1 How much must be deposited each year at 5 percent compounded annually to accumulate Tk.1000 after 10 years from now? ©. S. B. Bhattacharjee Continued…

Log (1.05)10 =10 log 1.05 = 10× .021189299 = 0.21189299 Antilog 0.21189299 =1.628894 = 1.628894 ©. S. B. Bhattacharjee

A = Annuity (annual payment) i = Rate of interest n = Number of years Example: 2 A man borrows Tk. 6000 at 6% and promises to pay off the loan in 20 annual payments beginning at the end of the first year. What is the annual payment necessary? , where V = Present value A = Annuity (annual payment) i = Rate of interest n = Number of years Continued.. ©. S. B. Bhattacharjee

Continued…… ©. S. B. Bhattacharjee

Log (1.06)20 = 20 log 1.06 =20×0.0253 =0.506 Antilog 0.506 =3.206 ©. S. B. Bhattacharjee

The sum will increase by 25% Example: 3 How long will it take for a sum of money to increase by 25 percent at compound interest of 5 percent per year? [ Mathematics- Bowen ] Answer : The sum will increase by 25% This means if the principal amount is Tk. 100, the increased compounded amount will be Tk (100+25) =Tk.125 A = Tk.125 P = Tk.100 Continued… ©. S. B. Bhattacharjee

©. S. B. Bhattacharjee

Example: 4 How long will it take for sum of money to double itself at 6 percent, compounded annually? [Bowen] Answer : The compound amount will be double. Hence, if the principal amount is x, the compounded amount will be 2x. We know, A= P(1+r)n, where A= Compounded amount P = Principal amount r = rate of interest n = period Here, A = 2x Continued… ©. S. B. Bhattacharjee

©. S. B. Bhattacharjee

b) What will be the simple interest in the above case? Example: 5 Find the compound interest on Tk. 1,00,000 for 4 years at 5% per annum. b) What will be the simple interest in the above case? Answer : We know, A= P (1+r)n, where A=compounded amount P = Principal amount r= Rate of interest n = years Continued… ©. S. B. Bhattacharjee

©. S. B. Bhattacharjee

Now compound Interest = A - P =Tk.1,21,500 -1,00,000 = Tk.21,500/- ©. S. B. Bhattacharjee Continued…

What will be the simple interest in the above case? Simple Interest = P n r = 1,00,000×4×0.05 = Tk. 20,000/- ©. S. B. Bhattacharjee

Find the compound interest on 25,800 for 5 years, if the Example:6 Find the compound interest on 25,800 for 5 years, if the rate of interest be 2% in the 1st year, interest in the second year,3% in the 3rd year and thereafter at 4% p.a. We know, A = P( 1+r)n, where A = Compounded amount r = Rate of interest per annum n = Period ©. S. B. Bhattacharjee Continued…

©. S. B. Bhattacharjee Continued…

= 2.5% Continued… Rate of interest in the 2nd year = Rate of interest in the 3rd year=3% ©. S. B. Bhattacharjee Continued…

0.04 log x = log (1.04)2 =2 log 1.04 =2×0.0170 = 0.034 Let x = (1.04)2 Rate of interest after 3rd year 0.04 log x = log (1.04)2 =2 log 1.04 =2×0.0170 = 0.034 Let x = (1.04)2 x = Antilog 0.034 =1.0816 (1.04)2=1.0816 ©. S. B. Bhattacharjee

Example: 7 Find the compound interest on Tk. 6,950 for 3 years, if interest is payable half yearly, rate for the first two years being 6%, and for the third year 9% p.a. P = 6950 n= 2, .06 ©. S. B. Bhattacharjee Continued…

©. S. B. Bhattacharjee

Example: 8 Depreciation ©. S. B. Bhattacharjee Continued…

We know, SV = PV (1-d)n, where SV = Scrap value PV = Purchase value d = depreciation n = period Here, SV = Tk. 2250 PV = Tk. 5810 ©. S. B. Bhattacharjee Continued…

©. S. B. Bhattacharjee

Example: 9 A company buys a machine for TK 1,00,000. Its estimated life is 12 years and scrap value is TK. 5000. What amount is to be retained every year from the profit and allowed to accumulate at 5 % C.I. for buying a new machine at the same price after 12 years. Or The cost of a machine is TK. 1,00,000 and its effective life is 12 years. If the scrap realises only TK. 5,000, what amount should be retained out of profits at the end of each year to accumulate at 5 % C.I. p.a. Continued……… ©. S. B. Bhattacharjee

Amount of the annuity = Cost – Scrap value = TK 1,00,000 – TK 5,000 Answer. Cost of the machine = TK. 1,00,000 Its scrap value = TK 5,000 Amount of the annuity = Cost – Scrap value = TK 1,00,000 – TK 5,000 = TK.95,000 n = 12, i = 5%= = 0.05 Annuity , A = ? Continued…… ©. S. B. Bhattacharjee

Let X = ( 1.05)12 log X = log (1.05)12 = 12 log 1.05 = 12×0.0212 = 0.2544 X = Antilog 0.2544 = 1.79639 (1.05)12 = 1.79639 Continued…. ©. S. B. Bhattacharjee

©. S. B. Bhattacharjee

Example: 10 A machine is depreciated in a such way that the value of the machine at the end of any year is 90 % of the value at the beginning of the year. The cost of the machine was TK. 12,000 and it was sold eventually as waste material for Tk 200. Find out the number of years during which the machine was in use. We know SV = PV (1- d )n, where SV= Scrap value PV= Purchase value d = depreciation n = period Continued… ©. S. B. Bhattacharjee

Selling price i.e. Scrap value = TK. 200 PV = TK. 12,000 d = Here, Selling price i.e. Scrap value = TK. 200 PV = TK. 12,000 d = Continued…. ©. S. B. Bhattacharjee

the number of years during which the machine was in use = 39 years. ©. S. B. Bhattacharjee

Thank you For Attending The Session ©. S. B. Bhattacharjee