Problem Set 7
Portfolio Insurance
Problem 1
How many “shares” will you have in the
revised portfolio, and how many puts?
• NS + NP = 21,000(1224.36)
• N(1224.36) + N(32.70) = 21000(1224.36)
• N(1224.36+ 32.70) = 21000(1224.36)
• N(1257.06) = 25,711,560
• N = 20,453.7254
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Problem 1
What is the value of the revised stock
portfolio combined with this insurance?
• NS + NP = N(1224.36) + N(32.70) =
1257.06 * N
• = $25,711,560
Problem 2
• Value of the insured portfolio = (1201.25 *
N) + [(1210 – 1201.25) * N]
• Value of the insured portfolio = 1210 * N =
$24,749,008 (rounded to nearest dollar).
• This is the minimum value of the insured
portfolio, no matter how far the value of
equity might drop.
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Problem 2
• Without portfolio insurance, the value of the
uninsured portfolio would have been
$1201.25 * 21000 = $25,226,250. Only if
the equity fell below the breakeven would
the insured portfolio outperform the
uninsured portfolio.
• The breakeven point would be
24,749,008/21000 = 1178.52
Problem 3
• Puts would expire worthless, so value of the
insured portfolio = 1234.36 * N
=$25,247,260 (rounded to nearest dollar)
• Without portfolio insurance, the value of the
uninsured portfolio would have
been$1234.36 * 21000 = $25,921,560
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Problem 3
• The cost of insurance is the difference
between the outcomes; in this case,
$25,921,560 – $25,247,260 = $674,300 (or
about 2.6% of the original portfolio value.
• Upside capture is 100% – 2.6% = 97.4%
Problem 4
• Nnew = Nold S / ∂1 S + (1– ∂2)B(X,t)
• Nnew = (21000 * 1224.36)/ [ (.5933 *
1224.36) + (1– .5592)*1204.05]
• Nnew= 20,453.71
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Problem 4
• Number of equity shares = ∂1 * Nnew
• Number of equity shares = .5933 *
20453.71
• Number of equity shares = 12,134.58
Problem 4
• Number of bonds = (1–∂2) * Nnew
• Number of bonds = (1 – .5592) * 20453.71
• Number of bonds = 9014.98
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Problem 4
• Value of insured portfolio = (12134.58 *
$1224.36) + (9014.98 * $(1204.05)
• Value of insured portfolio = $25,711,560
(again, recall that the result given by the
options calculator is more precise than you
would get by just running the rounded
intermediate values through a handheld
calculator).
Problem 5
• ∂C/ ∂S S/(S+P) = .5933 *
1224.36/(1224.36+32.70) = .5778
• (again, recall that the result given by the
options calculator is more precise than
you would get by just running the
rounded intermediate values through a
handheld calculator).
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Problem 6
Nf = Nold {[∂C/ ∂S S/(S+P)]–1} e–(r–d)t
Nf = 21000 * [ (0.5778–1) e–(.05–.03)(90/365) ]
Nf = –8822.58
Number of contracts = –8822.58 /250
= –35.29
• The futures price in the contract would
be $1224.36 e(.05–.03)t =$1230.41
•
•
•
•
Problem 7
• For the portfolio in problem 1, the value of
the put would become $32.30 and the new
value of the insured portfolio would be
($32.30 + $1225.36) * N = $25,723,732
(rounded to nearest dollar). This is a gain of
about $12,172.
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Problem 7
• For the portfolio in problem 6, the change in
futures price would be
($1225.36 –$1224.36) e.05–.03)t = $1.0049 per
share of the futures contract.
• So, you would lose $8866.20 on the futures
contracts and make $21,000 on the stock,
netting a gain of $12,133.80
Problem 7
• Thus the outcomes from the two strategies
are close, assuming you can buy fractional
futures contracts. Without fractional
contracts, you might have chosen to sell 35
contracts (with the multiplier of 250, this
represents 8750 shares).
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Problem 7
• Then the futures contracts would lose $8793
on the futures contracts and make $21,000
on the stock, netting a gain of $12,207. So,
the lack of fractional shares makes a
relatively small difference in this case (the
difference between fractional contracts and
whole numbers is about 0.6% in this case).
Problem 8
• Forward price = $43.75 e0.0375*(270/365)
= $44.98
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Problem 9
• The puts and calls are each worth $2.70 per
share, so at the risk-free rate the face value
of the “loan” would be
• $2.70 e0.0375*(270/365) = $2.7759 per share
(call it $2.78)
Problem 10
• Payoff per share = $46.50 – $44.98 – $2.78
= –$1.26
• If the stock price at expiration is $42.50, the
option expires out of the money. Then the
payoff per share = – $2.78
• Breakeven point is $44.98 + $2.78 = $47.76
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Problem 11
• Oaon = S0 N(d1) = .5564 * 65 = $36.17
• If the stock price is above $65 at
expiration, the holder of the option would
receive the underlying stock
• If the stock price is $65 or lower at
expiration, the option-holder would
receive nothing
Problem 12
• Ocon = e –rt N(d2) = (64.47/65) * .5242 =
52¢
• If the stock price is above $65 at
expiration, the holder of the option would
receive one dollar
• If the stock price is $65 or lower at
expiration, the option-holder would
receive nothing
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Problem 13
• C(S,X,t) = Oaon – XOcon
= $36.17 – ($65 * .52) = $2.37
Problem 14
• Chooser = Call(S,65,60/365)
+ P(S,65e –.05(30/365) , 30/365)
= $2.37 + $1.23 = $3.60
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Problem 15
• Ccp = $2.37/0.52 = $4.56
Problems 16-22
• For class discussion
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Put-Call Parity: Another Look
C(S,X,t) + B(X,t) = S + P(S,X,t)
C(S,X,t) – P(S,X,t) = S – B(X,t)
C(S,X,t) – P(S,X,t) = F(X,t)
F(X,t) = 0 if X = Se(R-d)t
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S
C
P
X
C
P
R
C
P
t
C
P

C
P
Call
Keys for using OPT as an analytical tool
C(S,X,t) = S - B(X,t) + P(S,X,t)
B(X,t)
Stock
Building Black-Scholes
• C(S,X.t) = S*N(d1) – e-Rt X*N(d2)
 Start with the “moneyness ratio”
S
Xe Rt
“At the money” = 1
“In the money” > 1
“Out of the money” < 1
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Building Black-Scholes
• C(S,X.t) = S*N(d1) – e-Rt X*N(d2)
 Consider the log of the “moneyness ratio”
 S 
ln  Rt 
 Xe 
“At the money” = 0
“In the money” > 0
“Out of the money” < 0
Building Black-Scholes
• C(S,X.t) = S*N(d1) – e-Rt X*N(d2)
 S 
ln   Rt 
 Xe  1
d1 
  t
2
 t
 S 
ln   Rt 
 Xe  1
d2 
  t
2
 t
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Building Black-Scholes
• C(S,X.t) = S*N(d1) – e-(R-d)t X*N(d2)
When the option is “at the money” and the
“moneyness ratio” is 1, the calculation is
greatly simplified, because ln (1) is zero. This
is the “golden point” where the put and call
have the same value. Then
d1  1  t
2
d2   1  t
2
Digital Options Break a Call in
Two
• Digital options, sometimes called binary
options, are of two types:
– Asset-or-nothing options
• pay the holder the asset if the option expires in the
money and nothing otherwise.
– Cash-or-nothing options
• pay the holder a fixed amount of cash (usually $1) if
the option expires in the money and nothing
otherwise
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Decomposition of a European
Call
• A balanced portfolio of long asset-ornothing options and short cash-or-nothing
options is equivalent to a European Call
– Oaon = S0 N(d1)
– Ocon = e–Rt N(d2)
• So, C(S,X,t) = Oaon – XOcon
Synthetic Options
• Portfolio Insurance
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