Question

1. Perpetuities in arithmetic progression. If a perpetuity has first payment P and each payment increases...

1. Perpetuities in arithmetic progression. If a perpetuity has first payment P and each payment increases by Q, then its present value, one period before the first payment, is P/i + Q/i^2 Using this formula, find the present value of a perpetuity-immediate which has annual payments with first payment \$360 and each subsequent payment increasing by \$40, at annual interest rate 1.3%.

2. Filip buys a perpetuity-immediate with varying annual payments. During the first 5 years, the payment is constant and equal to 10. Beginning in year 6, the payments start to increase. For year 6 and all future years, the current year’s payment is K% larger than the previous year’s payment. At an annual effective interest rate of 9.2%, the perpetuity has a present value of 167.50. Calculate K, given that K < 9.2.

1.

Present Value = 360/(0.013) + 40/(0.013)2

Present Value = \$264,378.70

2.

For First 5 years,

Present Value,

Using TVM calculation,

PV = [FV = 0, PMT = 10, T = 5, I = 9.20%]

PV = \$38.7

so,

Present Value of Growing perpetual payment = 167.50 - 38.7 = \$128.80

At the end of 5th year,

Value of perpetual payment = 128.80(1.092)5

Value of perpetual payment = \$200

Present Value of Growing Perpetuity = 10(1 + k)/(0.092 - k)

200(0.092 - k) = 10 + 10k

18.4 - 200k = 10 + 10k

k = 4.00016%

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