Present value of cash flow

Question

You are given interest rates $i_0=i_1=0.25$ and $i_2=i_3=1$. You have entered into a business transaction where you will receive $2$ at time $0$, $5$ at time $3$, $10$ at time $4$, in return for a payment by you of $3$ at time $2$. In place of all these cash flows, you are offered a single payment made to you at time $1$. What is the smallest payment you would accept?

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$i_k=v(k+1,k)-1$

$\therefore i_0=v(1,0)-1 \rightarrow v(0,1)=0.8$

Similarly, $v(1,2)=0.8$

$v(2,3)=v(3,4)=0.5$

Therefore,

$v(0)=1, v(1)=0.8, v(2)=0.8^2, v(3)=(0.8)^2(0.5),v(4)=(0.8)^2(0.5)^2$

Present value of first cash flow

$=(2)(1)+(0)(.8)+(-3)(.8)^2+(5)(.8)^2(.5)+(10)(.8)^2(.5)^2=3.28$

For an equivalent cash flow at time $1$,

$3.28=(x)(0.8) \rightarrow x=4.1$

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But I am uncertain of my work. Please crosscheck the working.

Answer 1

$$ FV_1 = 2*(1+i_0)^2(1+i_2)^2 - 3(1+i_2)^2 + 5*(1+i_2)+10 = 20.5$$ $$FV_2 = C(1+i_1)(1+i_2)^2 = 5C$$

where $i_1 = i_0$ and $i_2 = i_3$ Equating these two

You get $C = \frac{20.5}{5} = 4.1$

Yes, You are correct on your answer.

Answer 2

You do not need to discount to time 0 and then yield interest $i_0$. But that aside, your solution is more than sufficient.

I like to illustrate my work progress, so that I am sure I get the right expression for the present value at time 1. I find that a good tool for many students who are new to this stuff. See my figure below: