An investment offers to pay you $10,000 a year for five years. If it costs $33,5

Question

An investment offers to pay you $10,000 a year for five years. If it costs $33,520, what will be your rate of return on the investment?

Explanation / Answer

the equation or to say the formula for present value of an ordinary annuity is as follows PVA = PMT [(1+r)^n - 1]/r As you have noticed from the equation, it is not possible to solve this equation for the variable r thus we either resort to using Newton Raphson method, the secant method to find the rate of annuity The other venue at disposal is to use linear interpolation but this will only get you an approximate value for the rate and not the actual rate You could also restate the problem to lay out the cash flows as series of discounted cash flows and find the rate of return yet this too entails using the methods listed earlier to find the rate of return -33520 + 10000(1+r)^-1 + 10000(1+r)^-2 + 10000(1+r)^-3 + 10000(1+r)^-4 + 10000(1+r)^-5 = 0 You can make use of these online tools to find the rate This first calculator http://thinkanddone.com/finance/present-… calculates one of the four variables I listed in the first formula above for present value of annuity where PVA is 33520 the payment R is 10000 and length of annuity terms is 5. this calculator reports a rate of return of 15% The second calculator http://thinkanddone.com/finance/online-i… finds IRR or internal rate of return when we supply the cash flows as -33520 10000 10000 10000 10000 10000 It too finds the rate as 15% If you wish to see how it calculated the rate with Newton Raphson method then this tool http://thinkanddone.com/finance/online-i… presents the calculations required that are listed below f(x) = -33520(1+i)^0 +10000(1+i)^-1 +10000(1+i)^-2 +10000(1+i)^-3 +10000(1+i)^-4 +10000(1+i)^-5 f'(x) = -10000(1+i)^-2 -20000(1+i)^-3 -30000(1+i)^-4 -40000(1+i)^-5 -50000(1+i)^-6 x0 = 0.1 f(x0) = 4387.8677 f'(x0) = -96841.6776 x1 = 0.1 - 4387.8677/-96841.6776 = 0.145309703467 Error Bound = 0.145309703467 - 0.1 = 0.04531 > 0.000001 x1 = 0.145309703467 f(x1) = 377.2062 f'(x1) = -80821.9405 x2 = 0.145309703467 - 377.2062/-80821.9405 = 0.149976829953 Error Bound = 0.149976829953 - 0.145309703467 = 0.004667 > 0.000001 x2 = 0.149976829953 f(x2) = 3.39 f'(x2) = -79374.8841 x3 = 0.149976829953 - 3.39/-79374.8841 = 0.150019538906 Error Bound = 0.150019538906 - 0.149976829953 = 4.3E-5 > 0.000001 x3 = 0.150019538906 f(x3) = 0.0003 f'(x3) = -79361.7995 x4 = 0.150019538906 - 0.0003/-79361.7995 = 0.150019542415 Error Bound = 0.150019542415 - 0.150019538906 = 0 < 0.000001 IRR = x4 = 0.150019542415 or 15%

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