108156.pdf

268 Mathematics for Finance

2.9The initial principalPsatisfies the equation

(1 + 0.12)^2 P=1, 000.

It follows thatP∼= 797 .19 dollars.

2.10a) Under daily compounding the present value is

100 , 000

(

1+^0365.^05

)− 100 × 365

∼= 674. 03

dollars.

b) If annual compounding applies, then the present value is

100 ,000 (1 + 0.05)−^100 ∼= 760. 45

dollars.

2.11The return will be

K(0,1) =

(

1+^012.^1

) 12

− 1 ∼= 0. 1047 ,

that is, about 10.47%.

2.12Using the binomial formula to expand themth power, we obtain

K(0,1) = (1 +mr)m− 1

=1+r+

(1−m^1 )

2! r

(^2) +···+(1−mm−^1 )
m! r
m− 1 >r
ifmis an integer greater than 1.
2.13Denote the interest rate byr, the amount borrowed byPand the amount of
each instalment byC,
C=PA(Pr,5)= 1 −(1 +Prr)− 5 ,
see Example 2.4. Letn= 1, 2, 3, 4 or 5. The present value of the outstanding
balance aftern−1 instalments are paid is equal to the amount borrowed
reduced by the present value of the firstn−1 instalments:
P−1+Cr−···−(1 +Cr)n− 1 =P(1 +r)
6 −n− 1
(1 +r)^5 − 1
.
The actual outstanding balance remaining aftern−1 instalments are paid can
be found by dividing the above by the discount factor (1 +r)−(n−1), which
gives
P(1 +r)
(^5) −(1 +r)n− 1
(1 +r)^5 − 1

. (S.1)

The interest included in thenth instalment is, therefore,

P(1 +r)

(^5) −(1 +r)n− 1
(1 +r)^5 − 1
r. (S.2)