Independent solution

How to solve this Cash-Flow Modified Duration question

Setup

Setup

Write Investment A's price at yield i and use modified duration as negative price derivative divided by price.

PA(i)=1,000(1+i)−2+3,000(1+i)−3+X(1+i)−5P_A(i)=1{,}000(1+i)^{-2}+3{,}000(1+i)^{-3}+X(1+i)^{-5}
Dmod,A=−PA′(i)PA(i)D_{\mathrm{mod},A}=-\frac{P_A'(i)}{P_A(i)}

Model

Model

At i = 0.10, impose the stated duration of 4 and solve for X.

4=7,649.75+2.82237X3,089.39+0.62092X4=\frac{7{,}649.75+2.82237X}{3{,}089.39+0.62092X}
X=13,793.98X=13{,}793.98

Compute

Compute

Insert X into Investment B's price and derivative.

PB=9,878.85P_B=9{,}878.85
−PB′=54,545.05-P_B'=54{,}545.05
Y=54,545.059,878.85=5.5214Y=\frac{54{,}545.05}{9{,}878.85}=5.5214

Answer

Answer

Investment B's modified duration is 5.52, choice B.

Y≈5.52(B)\boxed{Y\approx5.52\quad\text{(B)}}