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FRM Part I · FRM Exam Part I · Applying Duration, Convexity, and DV01

A pension fund has a liability with present value $120 million and modified duration 7. It will fund a hedge with two bonds with total market value of $120 million: Bond A with modified duration 4 and Bond B with modified duration 12. The hedge must match both the liability's present value and its duration. How much should be invested in Bond B?

Invest $45 million in Bond B and $75 million in Bond A. Solving 4(1 - w) + 12w = 7 gives a weight of 0.375 in Bond B, or $45 million of the $120 million. This matches the liability's value and its duration of 7.

  1. A$45 millionCorrect
  2. B$75 million
  3. C$70 million
  4. D$60 million

Explanation

Let the weight in B be w. Then 4(1 - w) + 12w = 7, so 8w = 3 and w = 0.375, which is $45m in B and $75m in A. Check: 75 x 4 + 45 x 12 = 840, and 840/120 = 7. The $75m answer swaps the two bonds, and $70m ignores Bond A's duration (7/12 x 120).

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