CHAPTER 22
Real Options
Answers to Practice Questions
1.
a.
A fiveyear American call option on oil.
The initial exercise price is $32 a
barrel, but the exercise price rises by 5 percent per year.
b.
An American put option to abandon the restaurant at an exercise price of
$5 million.
The restaurant’s current value is ($700,000/r).
The annual
standard deviation of the changes in the value of the restaurant as a going
concern is 15 percent.
c.
A put option, as in (b), except that the exercise price should be interpreted
as $5 million in real estate value plus the present value of the future fixed
costs avoided by closing down the restaurant.
Thus, the exercise price is:
$5,000,000 + ($300,000/0.10) = $8,000,000.
Note:
The underlying asset
is now PV(revenue – variable cost), with annual standard deviation of 10.5
percent.
d.
A complex option that allows the company to abandon temporarily (an
American put) and (if the put is exercised) to subsequently restart (an
American call).
e.
An inthemoney American option to choose between two assets; that is,
the developer can defer exercise and then determine whether it is more
profitable to build a hotel or an apartment building.
By waiting, however,
the developer loses the cash flows from immediate development.
f.
A call option that allows Air France to fix the delivery date and price.
2.
A
commitment
to invest in the Mark II would have a negative NPV.
The
option
to
invest has a positive NPV.
The value of the option more than offsets the
negative NPV of the Mark I.
3. a. P = 467
EX = 800
σ
= 0.35
t = 3.0
r
f
= 0.10
.7194
)
3.0
(0.35
.1132
t
σ
d
d
.1132
/2
)
3.0
(0.35
)
3.0
)]/(0.35
00/1.10
log[467/(8
/2
t
σ
t
X)]/σ
log[P/PV(E
d
1
2
3
1
0
0
0

=
×


=

=

=
×
+
×
=
+
=
N(d
1
) = N(0.1132) = 0.4549
N(d
2
) = N(0.7194) = 0.2359
Call value = [N(d
1
)
×
P] – [N(d
2
)
×
PV(EX)]
203
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document= [0.4549
×
467] – [0.2359
×
(800/1.10
3
)] = $70.65
b.
P = 500
EX = 900
σ
= 0.35
t = 3.0
r
f
= 0.10
0.8010
)
3.0
(0.35
0.1948
t
σ
d
d
0.1948
/2
)
3.0
(0.35
)
3.0
)]/(0.35
00/1.10
log[500/(9
/2
t
σ
t
X)]/σ
log[P/PV(E
d
1
2
3
1

=
×


=

=

=
×
+
×
=
+
=
N(d
1
) = N(0.1948) = 0.4228
N(d
2
) = N(0.8010) = 0.2116
Call value = [N(d
1
)
×
P] – [N(d
2
)
×
PV(EX)]
= [0.4228
×
500] – [0.2116
×
(900/1.10
3
)] = $68.32
c.
P = 467
EX = 900
σ
= 0.20
t = 3.0
r
f
= 0.10
1.2417
)
3.0
(0.20
0.8953
t
σ
d
d
0.8953
/2
)
3.0
(0.20
)
3.0
)]/(0.20
00/1.10
log[467/(9
/2
t
σ
t
X)]/σ
log[P/PV(E
d
1
2
3
1

=
×


=

=

=
×
+
×
=
+
=
N(d
1
) = N(0.8953) = 0.1853
N(d
2
) = N(1.2417) = 0.1072
Call value = [N(d
1
)
×
P] – [N(d
2
)
×
PV(EX)]
= [0.1853
×
467] – [0.1072
×
(900/1.10
3
)] = $14.05
4. P = 1.7
EX = 2
σ
= 0.15
t = 1.0
r
f
= 0.12
.4029
)
1.0
(0.15
0.2529
t
σ
d
d
0.2529
/2
)
1.0
(0.15
)
1.0
)]/(0.15
/1.12
log[1.7/(2
/2
t
σ
t
X)]/σ
log[P/PV(E
d
1
2
1
1
0

=
×


=

=

=
×
+
×
=
+
=
N(d
1
) = N(0.2529) = 0.4002
N(d
2
) = N(0.4029) = 0.3435
Call value = [N(d
1
)
×
P] – [N(d
2
)
×
PV(EX)]
= [0.4002
×
1.7] – [0.3435
This is the end of the preview.
Sign up
to
access the rest of the document.