Final Exam Fall '12

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FinalExam-FINE 4140Answers

Show all your work to receive full creditfor the problems!

All calculations should be done with 4decimals!

Cristina DanciulescuTulane University

Date: December 10th, 2012Total points: 160

Problem 1. (12 points)A company enters into a short futures contract to sell5,000 bushels of wheat for 450 cents per bushel. The initial margin is $3,000 and themaintenance margin is$2,000.

a. What price change would lead to a margin call?b. Under what circumstances could$1,500 be withdrawn from the margin account?a. There is a margin call if $3,000-$2,000=$1,000 is lost on the contract. This will

happen if the price of wheat futures rises by 1,0005,000 = 20 cents from 450 cents to 470cents per bushel.

b. $1500 can be withdrawn if the futures price falls by 1,5005,000 = 30 cents to 420 centsper bushel.

Problem 2. (12 points) The two-month interest rates in Switzerland and theUnited States are 2% and 5% per annum, respectively, with continuous compound-ing. The spot price of the Swiss franc is $0.8000. The futures price for a contractdeliverable in two months is $0.8100. What arbitrage opportunities does this create?(For this problem you do not need to calculate the profits obtained from the arbitrageopportunities.)

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The theoretical futures price is $0.8000 e(0.050.02) 212 = 0.8040. The actual futuresprice is too high ($0.8100 > $0.8040). This suggests that an arbitrageur should buySwiss francs and short Swiss francs futures.

Problem 3. (12 points)It is January 30. You are managing a bond portfolioworth $6 million. The duration of the portfolio in six months will be 8.2 years. TheSeptember Treasury bond futures price is currently 108-15, and the cheapest-to-deliverbond will have a duration of 7.6 years in September. How should you hedge againstchanges in interest rates over the next six months?

The value of a contract is 108 1532 1, 000 = $108, 468.75 (i.e it is $108.46875 per$100 face value and the futures contract is for $100,000 face value). The number ofcontracts that should be shorted is 6,000,000108,468.75 8.27.6 = 59.7 Rounding to the nearest wholenumber, 60 contracts should be shorted. The position should be closed out at the endof July.

Problem 4. (26 points) Companies A and B face the following interest rates(adjusted for the differential impact of taxes):

US dollars (floating rate) Canadian dollars (fixed rate)Company A LIBOR+0.5% 5.0%Company B LIBOR+1.0% 6.5%

Assume that A wants to borrow U.S. dollars at a floating rate of interest and Bwants to borrow Canadian dollars at a fixed rate of interest. A financial institution isplanning to arrange a swap and requires a 50-basis-point spread.

Design a swap which is equally attractive to both A and B and nets a 50-basis-pointspread for the financial institution. What rates of interest will A and B end up paying?

Suppose that LIBOR=4%, then the opportunity cost table looks as follows

US dollars (floating rate) Canadian dollars (fixed rate)Company A 4.55.0 = 0.9

5.04.5

= 1.11

Company B 5.0

6.5 = 0.76 6.5

5.0 = 1.3

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Figure 1: Problem 4

The table shows that company A has a comparative advantage in the Canadian

dollar fixed-rate market and company B has a comparative advantage in the U.S.dollar floating-rate market. (This may be because of their tax positions.) However,company A wants to borrow in the U.S. dollar floating-rate market and company Bwants to borrow in the Canadian dollar fixed-rate market. This gives rise to the swapopportunity.

The differential between the U.S. dollar floating rates is 0.5% per annum, and thedifferential between the Canadian dollar fixed rates is 1.5% per annum. The differencebetween the differentials is 1% per annum. The total potential gain to all parties fromthe swap is therefore 1% per annum, or 100 basis points. If the financial intermediaryrequires 50 basis points, each of A and B can be made 25 basis points better off. Thus

a swap can be designed so that it provides A with U.S. dollars at LIBOR+0.25% perannum, and B with Canadian dollars at 6.25% per annum. The swap is shown inFigure 1.

Principal payments flow in the opposite direction to the arrows at the start of thelife of the swap and in the same direction as the arrows at the end of the life of theswap. The financial institution would be exposed to some foreign exchange risk whichcould be hedged using forward contracts.

Problem 5. (20 points)An European call option and put option on a stock bothhave a strike price of $20 and an expiration date in three months. Both sell for $3.The risk-free interest rate is 10% per annum, the current stock price is$19, and a$1dividend is expected in one month.

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a. Identify the arbitrage opportunity open to a trader.b. How much does the trader gain from the arbitrage opportunity?a. If the call is worth $3, put-call parity shows that the put should be worth

3 + 20 e0.10

3

12

+e0.1

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19 = 4.50. This is greater than $3. The put is thereforeundervalued relative to the call. The correct arbitrage strategy is to buy the put, buy

the stock, and short the call. This costs $19.b. If the stock price in three months is greater than $20, the call is exercised. If it

is less than $20, the put is exercised. In either case the arbitrageur sells the stock for$20 and collects the $1 dividend in one month. The present value of the gain to thearbitrageur is 3 1 9 + 3 + 2 0 e0.10 312+ e0.1 112 = $1.50 at t0 (today) (or $1.536 atT (in three months)).

Problem 6. (26 points)Three put options on a stock have the same expirationdate and strike prices of $55, $60, and $65. The market prices are $3, $5, and $8,

respectively.a. Explain how a butterfly spread can be created with the respective put options.b. Construct tables showing the payoff and profit from the butterfly strategy.c. For what range of stock prices would the butterfly spread lead to a loss?a. A butterfly spread is created by buying the $55 put, buying the $65 put and

selling two of the $60 puts. This costs 3 + 8 2 5 = $1 initially.b. The following tables shows the profit and payoff from the strategy.

Table 1: Profit from the butterfly spread.

Stock price Profit from Profit from Profit from Totalrange first long second long short puts profit

putl putST $55 ($55 ST) $3 ($65 ST) $8 2($60 ST) + $10 -$1

$55< ST $60 -$3 ($65 ST) $8 2($60 ST) + $10 ST $56$60< ST $65 -$3 ($65 ST) $8 $10 $64 ST

ST $65 -$3 -$8 $10 -$1

c. The butterfly spread leads to a loss when the final stock price is greater than $64 (or

$65) or less than $56 ( $55).

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Table 2: Payoff from the butterfly spread.

Stock price Payoff from Payoff from Payoff from Total

range first long second long short puts payoff put put

ST $55 $55 ST $65 ST 2($60 ST) $0$55< ST $60 $0 $65 ST 2($60 ST) ST $55$60< ST $65 $0 $65 ST $0 $65 ST

ST $65 $0 $0 $0 $0

Problem 7. (26 points) A stock price is currently $40. Over each of the next twothree-month periods it is expected to go up by 10% or down by 10%. The risk-free interest

rate is 12% per annum with continuous compounding.a. What is the value of a six-month European put option with a strike price of $42?

b. What is the value of a six-month American put option with a strike price of $42?

a. A tree describing the behavior of the stock price is shown in Figure 2. The upand down movements are u = 100100 +

10100 = 1.1 and d =

100100 10100 = 0.9. This implies

that S0u = 1.1 $40 = $44, S0d = 0.9 $40 = $36, S0uu = 1.1 1.1 $40 = $48.400,S0ud= 1.1 1.9 $40 = $39.600, S0dd= 0.9 0.9 $40 = $32.400.

The risk-neutral probability of an up move, p, is given by p = e0.12

3120.90

1.10.9 = 0.6523. Thisimplies that 1 p= 0.3477.

We then calculate fuu=max(42 48.400, 0) = $0, fud=max(42 39.600, 0) = $2.400,fdd= max(42

32.400, 0) = $9.600

The value of the European put option is given by f=e

2rt[p2fuu+ 2p(1p)fud+ (1p)2fdd] =e

20.12 312 [(0.6523)2 0 + 2 0.6523 0.3477 2.400 + (0.3477)2 9.600] = $2.118.

This can also be calculated by working back through the tree as shown in Figure 2. Thesecond number at each node is the value of the European option.

b. To obtain the value of the American option we need to calculate the following values:fu = e

rt[pfuu+ (1p)fud] = e0.12 3

12 [0.65230 + 0.34772.4] = $0.810 and fd =ert[pfud+ (1 p)fdd] =e0.12

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12 [0.6523 2.400 + 0.3477 9.600] = $4.759.Early exercise at node B implies that max(42 44, 0.810) = $0.810.Early exercise at node C implies that max(42 36, 4.759) = $6.000.In this casef=ert[pfu+(1p)fd] =e0.12

3

12 [0.65230.810+0.34776.000] = $2.537.Early exercise at node A implies max(42

40, 2.537) = $2.537. Therefore the value of

the American option today is $2.537. This is greater than the value of the European optionbecause it is optimal to exercise early at node C.

The value of the American option is shown as the third number at each node on the tree.

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Figure 2: Tree to evaluate European and American put options in Problem 7. At eachnode, upper number is the stock price, the next number is the European put price, andthe final number is the American put price.

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Problem 8. (26 points)Consider an American call option on a stock. The stock priceis$50, the time to maturity is 15 months, the risk-free rate of interest is 8% per annum, the

exercise price is $55, and the volatility is 25%. Dividends of $1.50 are expected in 4 months

and 10 months.

a. Show that it can never be optimal to exercise the option on either of the two dividend

dates.

b. Calculate the price of the option.

a. We have that D1 = D2 = 1.50, t1 = 412 = 0.3333, t2 =

1012 = 0.8333, T =

1512 = 1.25,

r= 0.08, and K= 55. It follows that

K[1 er(Tt2)] = 55(1 e0.080.4167) = 1.80. (1)

Hence,D2 < K[1 er(Tt2)]. (2)

Also,

K[1 er(t2t1)] = 55(1 e0.080.5) = 2.16. (3)Hence,

D1 < K[1 er(t2t1)]. (4)It follows that the option should never be exercised early.

b. The present value of dividends is

1.5e0.33330.08 + 1.5e0.83330.08 = 2.864. (5)

The option can be valued using the European pricing formula with: S0= 502.864 = 47.136,K= 55, = 0.25, r= 0.08, T = 1.25,

d1= ln(47.13655 ) + (0.08 +

0.2522 )1.25

0.25

1.25= 0.0545, (6)

d2= d1 0.25

1.25 = 0.3340, (7)N(d1) = 0.4783, N(d2) = 0.3692 and the call price is

47.136 0.4783 55 e0.081.25 0.3692 = 4.17 (8)

or $4.17.

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Final Exam Fall '12