108156.pdf

Solutions 271

2.24The ratersatisfies

er=

(

1+^012.^12

) 12

.

The solution isr∼= 0 .1194, about 11.94%.

2.25The frequencymsatisfies

(

1+^0 m.^2

)m

=1+0. 21.

Whencem=2.0.

2.26If monthly compounding at a raterapplies, then

(

1+ 12 r

) 12

=1+reand the

present value of the annuity is

V(0) =1+Cr

12

+( C

1+ 12 r

) 2 +···+

C

(

1+ 12 r

) 12 n

= C

(1 +re)^1 /^12

+ C

(1 +re)^2 /^12

+···+(1 +Cr

e)n

=C^1 −(1 +re)

−n

(1 +re)^1 /^12 − 1

.

2.27If bimonthly compounding at a raterapplies, then

(

1+r 6

) 6

=1+reand the

present value of the perpetuity is

V(0) =1+Cr

6

+( C

1+r 6

) 2 +

C

(

1+r 6

) 3 +···

= C

(1 +re)^1 /^6

+ C

(1 +re)^2 /^6

+ C

(1 +re)^3 /^6

+···

=C^1

(1 +re)^1 /^6 − 1

.

2.28We solve the equation

100 = 95 (1 +r)^12

forrto find the implied effective rate to be about 10.80%. If this rate remains

constant, then the bond price will reach $99 at a timetsuch that

100 = 99 (1 +r)(

(^12) −t)
.
The solution ist∼= 0 .402 years, that is, about 0. 402 × 365 ∼= 146 .73 days. The
bond price will reach $99 on day 147.
2.29The interest rate for annual compounding implied by the bond can be found
by solving the equation
(1 +r)−(1−^0 .5)=0. 9455
forr. The solution is about 11.86%. By solving the equation
(
1+r 2
)−2(1− 0 .5)
=0. 9455 ,
we obtain the semi-annual rate of about 11.53%, and solving
e−r(1−^0 .5)=0. 9455 ,
we find the continuous rate to be about 11.21%.