Extra Practice 2
Problem 1: Choice C
1
Butterfly Spread
X1 < X2 < X3
•
•
•
•
A Butterfly combines a long bull spread with
another short bull spread
The Long Spread = C(S,X1,t) – C(S,X2,t)
The Short Spread = C(S,X3,t) – C(S,X2,t)
Resulting Butterfly Spread
C(S,X1,t) – 2C(S,X2,t) + C(S,X3,t)
C(S,45,t) – 2C(S,50,t) + C(S,55,t)
S
C
P
X
C
P
R
C
P
t
C
P
σ
C
P
2: B
4: A
5: C
Call
Problems 2, 4, 5
Keys for using OPT as an analytical tool
C(S,X,t) = S – B(X,t) + P(S,X,t)
B(X,t)
Stock
2
0
-C
S
X
X+C
Long Put
$
X-P
0
X
-P
$
0
-(P+C)
S
Short Call
$
Short Put
Long Call
Problem 3
Long Straddle (Choice C)
$
C
0
X+C
S
X
$
P
0
X
S
X-P
X-P-C
X
S
X+P+C
Problem 6
•  ΔC= ΔS * $0.10
•  ΔC = 0.7 * $0.10 = increase 7¢
3
Problems 7 & 8
•  C = (0.7* $50) – (0.6 * $43.75 )
•  C = $8.75
•  Call + Bond = Stock + Put
•  Put = Call + Bond – Stock
•  P = $8.75 + $43.75 – $50 = $2.50
Problem 9
•  F = ($5.8875+0.45)*e0.0135*180/365
•  F = $6.3798
If using 360-day year, F = $6.3804
4
Problem 10
•  PV = $85*e–.0125*220/365
•  PV = $84.3620
Equities as commodities (Problem 11)
$ 2108
Eur 1960.44
FRA
today
Eur 0.93
=$1
R = 2%
Eur 1979.87
FRA
later
Eur 1979.87
NY
today
Spot 2108
Dividend 1%
Eur 0.90
=$1
NY
later
$ 2199.8566
Future = 2178.7766
5
Problem 12
C(S,X,t) + B(X,t) = S + P(S,X,t)
$10.40 + $39.80
$48 + $2.40
$50.40
$50.20
Profit = $0.20
Problem 13
Initially:
$8900+ $4150 – $12000 = $1050
$1050
$1050
$71.05
70
$78.95
75
80
$3950
6
Problem 14
C(S,X,t) + B(X,t) = S + P(S,X,t)
$28 + $98.54
$100 + $25.00
$126.54
$125.00
$30 + $88.68
$100 + $21.00
$121.00
$118.68
Build a Box!
Problem 14
So, what comes from building the box?
Initially: $28 + $21+ $98.54 – $30 – $25 – $88.68
= $3.86
No matter what, buy stock for $90 and sell for $100, receive $90
from one bond and pay $100 for the other, so
At expiry, pay $10 on the bonds
receive $10 from the options
net flow will be zero
7
Problem 15
•  Start with moneyness: S/Xe–rt
•  Then take the log of it: ln(S/Xe–rt)
•  Then build
S
d1 =
( ) + .5σ
ln
Xe − rt
σ t
t
• At-the-money, when S/Xe–rt = 1
Since the log of 1 is zero, d1 reduces to
€
.5σ t
€
Problem 15
•  The first hedge ratio is N(d1)
•  d1 = 0.5 * .6 = .3
•  So we find the probability of that number
occurring in a normal distribution
•  .6179
8
Problem 15
•  Now let’s build
d2 =
( ) − .5σ
ln
S
Xe − rt
σ t
t
• At-the-money, when S/Xe–rt = 1
d2 reduces to
€
−.5σ t
€
Problem 15
•  The second hedge ratio is N(d2)
•  d2 = – 0.5 * .6 = – 0.3
•  So again we find the probability of that
number occurring in a normal distribution
•  .3821
9
Problems 15 & 16
•  C = (0.6179* $79.20) – (0.3821 * $79.20)
•  C = 18.6754
•  Call + Bond = Stock + Put
•  Put = Call + Bond – Stock
•  P = $18.6754 + $79.20 – $79.20 = $18.6754
10