Independent solution

How to solve this Redington Convexity Range question

Setup

Setup

Use the supplied present values as weights for liability duration and convexity.

DL=24,000(7.5)+1,000(20)25,000=8D_L=\frac{24{,}000(7.5)+1{,}000(20)}{25{,}000}=8
CL=24,000(7.5)2+1,000(20)225,000=70C_L=\frac{24{,}000(7.5)^2+1{,}000(20)^2}{25{,}000}=70

Model

Model

Let f be the present-value fraction in the n-year zero. Duration matching determines f and restricts it to a valid allocation.

8=fn+(1−f)2n8=fn+(1-f)2n
f=2−8nf=2-\frac8n
4≤n≤84\le n\le8

Compute

Compute

Substitute that weight into asset convexity and require it to exceed 70.

CA=fn2+(1−f)4n2=−2n2+24nC_A=fn^2+(1-f)4n^2=-2n^2+24n
−2n2+24n>70-2n^2+24n>70
(n−5)(n−7)<0(n-5)(n-7)<0

Answer

Answer

The permissible maturities satisfy 5 < n < 7, choice B.

5<n<7(B)\boxed{5<n<7\quad\text{(B)}}