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Today is 1/1/2009. For 20 years, I receive $50,000 on the first day of each year and $25,000 on the first of July each year. If I discount cash flows at 10% annually, what is the present value (rounded to the nearest thousand dollars) of these payments?

User JohnColvin
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1 Answer

4 votes

Solution:

Cash Flow(F )= $ 50,000

Rate = 10 %

Discounted Cash Flow of amount $50,000 (D)=
F * ((1)/(1+(10)/(100)))^1+ F * ((1)/(1+(10)/(100)))^2+F * ((1)/(1+(10)/(100)))^3+F * ((1)/(1+(10)/(100)))^4+F * ((1)/(1+(10)/(100)))^5+..........\\\\= F * (10)/(11) + F * [(10)/(11)]^2+ F [* (10)/(11)]^3+..................................+ F [* (10)/(11)]^(20)

As, this is a geometric Progression.

Formula for Sum of n terms of geometric Progression having common ratio r,


S_(n)=( a * (1- r^n))/(1-r)

For, r < 1 and

for , r>1


S_(n)=( a * (r^n-1))/(r-1)


S_(20)=50,000 *(1 * ( [(11)/(10)]^(20)-1))/((11)/(10)-1)\\\\ S_(20)=50,000 * (6.7274-1)/(1.1-1)\\\\ S_(20)=50,000 * 57.27=2863500

If cash flow(K) = $ 25,000

Then at the rate of 10 % ,

Value after 20 years is given by:

Discounted Cash Flow of amount $25,000 (D)=
K * ((1)/(1+(10)/(100)))^1+ K * ((1)/(1+(10)/(100)))^2+K * ((1)/(1+(10)/(100)))^3+K * ((1)/(1+(10)/(100)))^4+K * ((1)/(1+(10)/(100)))^5+..........\\\\ =K * (10)/(11) + K * [(10)/(11)]^2+ K [* (10)/(11)]^3+..................................+ K [* (10)/(11)]^(20)

for , r>1, formula for sum of n terms of geometric series


S_(n)=( a * (r^n-1))/(r-1)


K_(20)=25,000 *(1 * ( [(11)/(10)]^(20)-1))/((11)/(10)-1)\\\\ K_(20)=25,000 * (6.7274-1)/(1.1-1)\\\\ K_(20)=25,000 * 57.27=1431750

User Roger Stewart
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