298 Mathematics for Finance
10.3To achieve a return of 14%, we would have to sell the bond for 0.8700e14%∼=
1 .0007 dollars, which is impossible. (A zero-coupon bond can never fetch a
price higher than its face value.)
In general, we have to solve the equationB(0,12)ek=e−τy(6)to find
y(6), wherekis the prescribed logarithmic return. The left-hand side must be
smaller than 1.
10.4During the first six months the rate isy(n)∼= 8 .34%, forn=0,...,179,
and during the rest of the yeary(n)∼= 10 .34%, forn= 180,...,360. The
bond should be sold for 0.92e^4 .88%∼= 0 .9660 dollars or more. This cannot
be achieved during the first six months, since the highest price before the
rate changes isB(179,360)∼= 0 .9589 dollars. On the day of the rate change
B(180,360)∼= 0 .9496 dollars, and we have to wait until dayn= 240, on which
the bond price will exceed the required $0.9660 for the first time.
10.5The rate can be found by using a spreadsheet with goal seek facility to solve
the equation10. 896 ×(
10 + 10e−y(1)+ 10e−^2 y(1)+ 110e−^3 y(1))
=1,000ek.This givesy(1)∼= 12 .00% fork= 12% in case a),y(1)∼= 12 .81% fork= 10%
in case b) andy(1)∼= 11 .19% fork= 14% in case c).
10.6The numbers were found using an Excel spreadsheet with accuracy higher
than the displayed 2 decimal points.
10.7Scenario 1: $1, 427 .10; Scenario 2: $1, 439 .69.
10.8Formula (10.2) can be applied directly to findD∼= 1 .6846.
10.9The duration is equal to 4 if the face value is $73.97. The smallest possible
duration, which corresponds to face valueF= $0, is about 2.80 years. For
very high face valuesFthe duration is close to 5, approaching this number as
Fgoes to infinity.
WhenF= 100, the coupon valueC∼= 13 .52 gives duration of 4 years. If
the coupon value is zero, then the duration is 5 years. For very high coupon
valuesCtending to infinity the duration approaches about 2.80 years.
10.10Since the second derivative ofP(y) is positive,
d^2
dy^2 P(y)=(τn^1 )(^2) C 1 e−τn 1 y+(τn 2 ) (^2) C 2 e−τn 2 y+···+(τnN) (^2) (CN+F)e−τnNy
0 ,
Pis a convex function ofy.
10.11Solving the system 6 = 2wA+3. 4 wB,wA+wB= 1, we findwA∼=− 1. 8571
andwB∼= 2 .8571. As a result, we invest $14, 285 .71 to buy 14, 005 .60 bondsB,
raising the shortfall of $9, 285 .71 by issuing 9, 475 .22 bondsA.
10.12The yield on the coupon bondAis about 13.37%, so the price of the zero-
coupon bondBis $87.48. The coupon bond has duration 3.29 and we find
the weights to bewA∼= 0 .4366 andwB∼= 0. 5634 .This means that we invest
$436.59 to buy 4.2802 bondsAand $563.41 to buy 6.4403 bondsB.
10.13Directly from the definition (10.2) of duration we compute the durationDtat