(*Take 𝑡 = 0 be the current time in this problem)
We consider the following three sets of cashflows:
Set A: It pays 200, 300 and 500 ( dollars) at time 1, time 2 and time 3 respectively.
Set B: It pays 300 and 700 at time 1 and time 3 respectively.
Set C: It pays 2𝑋 and 3𝑋 (dollars ) at time 2 and time 3 respectively.
It is given that the money grows at an annual effective interest rate 𝑖 (where 𝑖 > 0).
Suppose that the present values of these 3 sets of cashflows at time 1 are equal, calculate the present value of set 𝐶 at time
If the present value of all cash flow is equal at t=1 and $a(n)=(1+i)^n $
$ 200 + \frac{300×a(1)}{a(2)} + \frac{500×a(1)}{a(3)} = 300 + \frac{700×a(1)}{a(3)} $
$\frac{200}{a(2)}-\frac{300}{a(1)}+100=0$
$100(1+i)^2-300(1+i)+200=0$
$i^2-i=0$
$i=1$
To find $X$,
At $t=1$ , present value of all cash flow is same
$300 + \frac{700×a(1)}{a(3)}=2X\frac{a(1)}{a(2)}+3X\frac{a(1)}{a(3)}$
$300 +\frac{700}{(1+i)^2}=2X\frac{(1+i)}{(1+i)^2}+3X\frac{(1+i)}{(1+i)^3}$
Put i=1
$X=345.45454545$
To get the present value of set C cash flow at t=0
$$\frac{2×X}{a(2)} +\frac {3×X}{a(3)}= 302.272727$$
Is this method correct as im not sure about calculating present value at time t=1 instead of t=0 ( in order to find X and i)
Best Answer
The value of $i=1$ is right. And your equation to evaluate $X$ is right as well. But the result is $X=\frac{1900}{7}\approx271.43$, see here.
Then your next equation is right as well. With the right result from above the term for the present value of set C cash flow at $t=0$
$$2\cdot \frac{1900}{7}\cdot \frac{1}{2^2}+3\cdot \frac{1900}{7}\cdot \frac{1}{2^3}=237.5$$