Question

You are evaluating an investment that will pay ​$80 in 1​ year, and it will continue to make payments at annual intervals​ thereafter, but the payments will grow by 5% forever.

a. What is the present value of the first ​$80 payment if the discount rate is 12%?

b. How much cash will this investment pay 100 years from​ now? What is the present value of the 100th​ payment? Again use a 12​% discount rate.

c. What is the present value of the entire growing stream of perpetual cash​ flows?

d. Explain why the answers to parts a and b help to explain why an infinite stream of growing cash flows has a finite present value.

Answer 1

Present value of a cash flow received after n years at discount rate r is given as:

$\small PV = \frac{FV}{(1+r)^n}$

a)

Payment after 1 year is $80 at discount rate 12%, therefore

$\small PV = \frac{FV}{(1+r)^n} = \frac{80}{(1+0.12)^1} = \frac{80}{1.12} = 71.43$

b)

Value of investment after 100 years = 80 + 80*(1+0.05) + 80*(1+0.05)² + .......... + 80*(1+0.05)⁹⁹

   = 80 + 80*1.05 + 80*(1.05)² + .............+ 80*(1.05)⁹⁹

$\small \frac{80((1.05)^{100} - 1)}{1.05 - 1} = 208802.01$

Sum of a finite GP with n terms and first term a, common ration r = a(rⁿ - 1)/(r-1) when r > 1

Present value of the 100th payment:

$\small PV = \frac{80((1.05)^{99})}{(1+0.12)^{100}} = \frac{80*125.24}{83522.27} = 0.12$

c) Future value of the growing stream = 80 + 80*(1+0.05) + 80*(1+0.05)² + .......... + 80*(1+0.05)⁹⁹

Present discounted value

$\small = \frac{80}{(1+0.12)} + \frac{80(1+0.05)}{(1+0.12)^2} + \frac{80(1+0.05)^2}{(1+0.12)^3} +..............+ \frac{80(1+0.05)^{99}}{(1+0.12)^{100}}$

This is a finite GP of n terms with first term = 80/1.12 = 71.43 and common ratio r = 1.05/1.12 = 0.94

Sum of GP = Present discounted value of growing stream

$\small = \frac{71.43(1-(0.94)^{99})}{1-0.94} = 1187.9$

d)

From part (a) and (b), we know that present value of the payment after nth period keeps on decreasing as n increases. So even if payment stream is for infinite years, its present value keeps on decreasing and there is very less contribution from later terms. Hence this stream has a finite present value.