Chartered Accountant
2731 Points
Joined January 2008
Dear KARAN, its is as simple as that..............
COMPOUNDED ANNUAL GROWTH RATE.
We have to find the annual growth rate for a certain investment i.e. Rs.40 which become Rs.137 after 5 yrs....Here is how to calculate it........
Initial investment (1+r)^5 = total amount
i.e. Rs.40 (1+r)^5 = Rs.137
or , (1+r)^5 = 137 / 40
or , (1+r)^5 = 4.425 [ Here u can alternatively write (1 + r) = (4.425)^1/5 ]
or , Log (1+r)^5 = Log 4.425,
or, 5 Log (1+r) = .6459
or, Log (1+r) = .6459/5
or, Log(1+r) = .1291
or, Antilog (1+r) = antilog .1291
or, (1+r) = 1.346
or, r = 1.346 - 1
or, r = 34.6%
IT IS JUST THAT THE ABOVE FORMULA CONVERGES INTO IN ALL THESE EQUATIONS BUT IT ACTUALLY MADE IT WORSE. U CAN DO IT IN THIS WAY TO SIMPLIFY THINGS..
AS CONCERNED TO UR SECOND QUERY
What is TFP = S * e^r.t
It is another formula for compounding with the exception that when we need to find compunding at continuous rate (just like daily, monthly, quartly, yearly, it is continously without a greater difference of even a day), we use 'e' as a base and the rate of interst and time as its power to find the compounded amount of interest at continuous basis..........
FOR EG. We want to find the amount of Rs.100 compounded continuously at 10% for 2 yrs..
it would be like this..
100 * (e)^.2 (.1*2) = 100 * 1.2214 = 122.14/-
THERE COMES SPECIFIC TABLE FOR SEEING 'E' 's VALUES..
In the table, under 'x' column, see the compunded amt. of r*t i.e in our case above, its .1*2 = .2 and correspondingly see at 'e^x' column for the concerned value i.e. 1.2214 or if u want present value, it is at 'e^-x'.................
I THING ABOVE EXPLANATION WOULD SUFFICE..!!!!!!!!!!!!!!!!