Question

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?

Answer 1

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`