Under logarithmic interpolation, if he $t _ { i }$ -tear yield is $y \left( t _ { i } \right)$ , $i = 1,2$ , then the interpolated yield for $t$ lying between $t _ { 1 }$ and $t _ { 2 }$ is given by $y ( t ) = y \left( t _ { 1 } \right) \left[ 1 + \ln \left( 1 + x \cdot \left( t - t _ { 1 } \right) \right) \right]$ where $x$ is a parameter. Suppose the yield at one year is 4% and the yield at two years is 5%. Then, the closest yield at one and a half years, using logarithmic interpolation in time $t$ , is
A) 4.47%
B) 4.50%
C) 4.53%
D) 4.55%

Question

Quizplus · Accepted Answer

The given formula for logarithmic interpolation is $y(t) = y(t_1)[1 + \ln(1 + x \cdot (t - t_1))]$. To find the yield at one and a half years ($t = 1.5$), we first need to determine the parameter $x$. We know $y(t_1) = 4\%$, $y(t_2) = 5\%$, $t_1 = 1$, and $t_2 = 2$. The yield increase from $t_1$ to $t_2$ is $1\%$, so we solve for $x$ using the formula at $t_2$: $5\% = 4\%[1 + \ln(1 + x \cdot (2 - 1))]$. Solving for $x$ gives us the value to use in the interpolation formula for $t = 1.5$. After finding $x$, we substitute $t = 1.5$ into the interpolation formula to calculate the yield at one and a half years. The calculation yields a result closest to 4.53%.

Under Logarithmic Interpolation, If He tit _ { i }ti​ -Tear Yield Is

Under Logarithmic Interpolation, If He $t _ { i }$ -Tear Yield Is