Q:

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment of 169 immediately and another payment of 169 at the end of two years. (ii) A payment of 225 at the end of two years and another payment of 225 at the end of four years. Calculate K.

Accepted Solution

A:
Answer:The present value of K is, [tex]K=251.35[/tex] Step-by-step explanation:HiFirst of all, we need to construct an equation system, so [tex](1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}[/tex][tex](2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}[/tex]Then we equalize both of them so we can find [tex]i[/tex][tex](3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}[/tex]To solve it we can multiply [tex](3)*(1+i)^{4}[/tex] to obtain [tex](1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})[/tex], then we have [tex]225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}[/tex].This leads to a third-grade polynomial [tex]169i^{3}+451i^{2}+395i-112=0[/tex], after computing this expression, we find only one real root [tex]i=0.2224[/tex].Finally, we replace it in (1) or (2), let's do it in (1) [tex]K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35[/tex]