Business And Financial Mathematics-annuity

Annuity Sub Topics I Definition of annuity II Annuity Formula III Application of annuity IV. Problems

I

Definition of annuity An Annuity is a sequence of payments made at equal intervals of time and usually equal in amount.

II Formula Annuity Formula Amount of Annuity n

Sn =A

(1 + i ) – 1 i

Where: Sn = Amount of annuity A = periodic payment (usally every month) i = interest rate (usually per month) n = number of periods The formula above can be written as follows: Sn = A Sni Where: n (1 + I ) – 1 Sni = i The value of Sni can be seen in a table of annuity. Example. If a man puts Rp 20,000 in the bank every month at interest rate 1% per month, find the total amount of annuity for 4 years. Answer. Given A=20,000, i=0.01, n=48 Answer: 48 (1 + 0.01) – 1 Sn =20000 0.01 =20000 x 61.2226078

23

=1,224,452.156

Formula Present Value of Annuity PVn =A

1- (1 + i ) i

-n

Where: PVn = Present value of annuity The formula above can be written as follows: PVn =A PVni Where: -n 1- (1 + i ) PVni = i The value of PVni can be seen in a table of annuity. Example. A debt is to be discharged by making equal payments Rp 287.680 at the end of each month for five years. If the interest rate charged is 2% per month , find the size of the debt ? Answer. Given A=287.680, i=0,02, n=60 1- (1 + 0,02 ) PVn=287680 0,02 =287680 x 34,76089 =10.000.000

-60

III Application of annuity

A father deposit in a bank Rp 100.000 every month at interest rate 1% per month for 8 years. Then, he wants his son to withdraw at equal at the end of each month during his 5 years university course. How much can he withdraw every month? Answer: For 8 years can be seen as an amount of annuity. Given A=100,000, i=0.01, n=96 Answer: 96 (1 + 0.01) – 1 Sn =100000 0.01

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=100000 x 159.9273 =15,992,730 For the next 5 years is a process of present value of annuity. Given PVn= 15,992,730 , i=0.01, n=60, A= …..? PVn =A

1- (1 + i ) I

-n

1- (1 + 0.01 ) 0.01 15,992,730 = A x 44.95504 A = 15,992,730 / 44.95504 = 355,749. 4

-60

15,992,730=A

IV Problems Problem 1 A man deposit Rp 200,000 at the end of each month. At the end of 6 years the amount of his deposit is 21,786,930. Find the interest rate per month? Problem 2 A debt Rp 1,000,000 is to be discharged by making equal payments Rp 339,479.9 every month at interest rate 2.5% per month. Find how long time will it take to pay off the debt?

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Interpolation Method Related to the formula Sn or PV, suppose we want to find the value of variable i if A, Sn and PV are given, we can use interpolation method : For the formula Sn

(Sni –Sni1) (Sni2 –Sni1)

I = i1 +

(i2-i1)

i1 = the smaller i i2 = the greater i Sni1 = can be found in the table or computed using the formula. Example : A man deposit at the bank Rp 250,000 at the end of each month. In 5 years the amount of his deposit will be Rp 21,784,318. Find the interest rate per month? Answer: Given : Sn

=

21,784,318

A n

= =

250,000 60

Steps in evaluating i using interpolation method. Step 1 (Evaluate Sni) Sni = Sn/A = 21,784,318 /250,000 = 87.137272 Step 2 (To Find position of Sni in the tabel ) Look at the table of Sn, for n=60,we can see i between i=1,125% and 1,25% So: I1 I2 Sni1 Sni2

= = = =

1.12500% 1.25000% 85.03512704 88.57450776

Steps 3 (Substitute to the interpolation formula )

i = 0.01125 +

(87.137272 – 85.03512704 ) (88.57450776 –85.03512704)

= 0.01199 =1.199%

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(0.0125 – 0.01125 )

For formula PV

(PVni –PVni1) (PVni2 –PVni1)

i = i1 +

(i2-i1)

i1 = the smaller i i2 = the greater i PVni1 = can be found in the table or computed using the formula. Example : A man has a debt amounts Rp 8 millions at a bank with annuity system for 3 years. A sequence of payments should be done Rp 347,000 at the end of each month. Find the interest rate charged? Answer: Given : PV A N

= = =

8000000 347,000 36

Steps Steps 1 (Evaluate PVni) PVni = PV/A = 8 million/347,000 = 23.054755 Step 2 (To find position of PVni in the table ) (Look at table ani, for n=36, we can see i between i=2,5% dan 2,75%. Jadi: i1 i2 Pvni1 PVni2

= = = =

2.5% 2.75% 23.5562511 22.6699175

Steps 3 (Substitute to the interpolation formula )

i = 0.025 +

(23.054755 - 23.5562511) (22.6699175–23.5562511)

= 0.0264145 =2.64145%

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(0.0275 – 0.025

)