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Oxymoron
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I have this homework question that I need checking. If anyone has some expertise in financial mathematics, could you please have a look at my solution and tell me if/where I have made any mistakes.
Cheers.
Question:
If money is worth 4% effective, what equal payments [tex] X [/tex] at the end of 1 year, 3 years, and 5 years, will equitably replace the following obligations:
$1200.00 due in 2 years with interest (from today) at 2% compounded quarterly
$2000.00 due in 3 years without interest
$5600.00 due in 6 years with interest (from today) at 4.5% compounded semi-annually.
Answer:
Use Maple (or equivalent maths package) to solve equations.
Choose as focal time, present (k=0).
The present value of the debt at time k=0 is
[tex]
V = 1200\left(1+\frac{0.02}{4}\right)^{-8} + 2000(1+0) + 5600\left(1+\frac{0.045}{2}\right)^{-12}
= 7440.80
[/tex]
Therefore the borrower owes $7440.80 today.
Now instead of repaying the loan this way, he wants to make 3 equal payments in 1 year, 3 years, and 5 years, at 4% p.a.
So we have to solve
[tex]
X\left(1+0.04\right)^{-1} + X\left(1+0.04\right)^{-3} + X\left(1+0.04\right)^{-5} = 7440.80
[/tex]
Since we know the value of the loan at k=0 to be $7440.80, we can find the [tex] X [/tex]'s this way using the Equation of Value as written above.
Therefore
[tex]
X = 2784.25
[/tex]
In other words, If he pays $2784.25 in 1 year time, $2784.25 in 3 years time, and $2784.25 in 5 years time, at 4% p.a. he would have paid the loan off.
Cheers.
Question:
If money is worth 4% effective, what equal payments [tex] X [/tex] at the end of 1 year, 3 years, and 5 years, will equitably replace the following obligations:
$1200.00 due in 2 years with interest (from today) at 2% compounded quarterly
$2000.00 due in 3 years without interest
$5600.00 due in 6 years with interest (from today) at 4.5% compounded semi-annually.
Answer:
Use Maple (or equivalent maths package) to solve equations.
Choose as focal time, present (k=0).
The present value of the debt at time k=0 is
[tex]
V = 1200\left(1+\frac{0.02}{4}\right)^{-8} + 2000(1+0) + 5600\left(1+\frac{0.045}{2}\right)^{-12}
= 7440.80
[/tex]
Therefore the borrower owes $7440.80 today.
Now instead of repaying the loan this way, he wants to make 3 equal payments in 1 year, 3 years, and 5 years, at 4% p.a.
So we have to solve
[tex]
X\left(1+0.04\right)^{-1} + X\left(1+0.04\right)^{-3} + X\left(1+0.04\right)^{-5} = 7440.80
[/tex]
Since we know the value of the loan at k=0 to be $7440.80, we can find the [tex] X [/tex]'s this way using the Equation of Value as written above.
Therefore
[tex]
X = 2784.25
[/tex]
In other words, If he pays $2784.25 in 1 year time, $2784.25 in 3 years time, and $2784.25 in 5 years time, at 4% p.a. he would have paid the loan off.