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4-71 A set of cash flows begins at $20,000 the first year, with a decrease each year until n = 10 years. If the interest rate
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Answer #1

a)

Cash Flow at the end of year 1=$20000

Annual gradient=G=-$2000

Rate of interest=i=8%

Time period=n=10

Present value=20000*(P/A,0.08,10)-2000*(P/G,0.08,10)

Let us calculate the interest factors

(1+i) (P/A, i,n)=-

1- (1+0.08)106.710081 (P/A, 0.08, 10) = – 0.08

(P/G,i,n) = (1+i) -1 2(1+i) i(1+i)n

10 (P/G,0.08, 10) = (1 +0.08) 10 – 1 0.082(1 + 0.08)10 0.08(1 +0.08)10 09110 = 25.976831

So,

Present value=20000*6.710081-2000*25.976831=$82247.96

b)

In the case,

Cash Flow at the end of year 1=$20000

Growth rate=g=-10%

Rate of interest=i=8%

Time period=n=10

Present value, PVg is given by

PVg=\frac{R}{i-g}*\left [ 1-\frac{(1+g)^{n}}{(1+i)^{n})} \right ]

20000 0.08 +0.10 (1-0.1) (1+0.08)10)」 2

PV g = 111111.111111* 0.838494 = 93166.05

PVg=$93166.05

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