1. A. Call buyeri. CT = max(O,ST - X) = max(0,55 - 50) = 5n=CT-Co=5-4=1

ii. CT = max(O,ST - X) = max(0,51 - 50) = 1 n=c -c -1-4=-3 ... T 0-

HI. CT = max(O,ST - X) = max(0,48 - 50) = 0 n=c -c -o-4=-4 T 0-

B. Call selleri. Value = -CT = -max(O,ST - X) = -max(0,49 - 50) = 0n=-CT+Co=-~+4=4ii. Value = -CT = -max(O,ST - X) = -max(0,52 - 50) = -2n=-CT+Co=-2+4=2iii. Value = -CT = -max(O,ST - X) = -max(0,55 - 50) = -5

n = -c,+co= -5 +4= -1C. Maximum and minimumi. Maximum profit to buyer (loss to seller) =ii. Maximum loss to buyer (profit to seller) = co = 4

D. ST* = X + co = 50 + 4 = 542. A. i. CT = max(O,ST - X) = max(0,99 - 105) = 0

.. n=cT-C0=O-7=-7 11. c, = max(O,ST - X) = max(0,104 - 105) = 0n=cT-co=o-7= -7iii. CT = max(O,ST - X) = max(0,105 - 105) = 0n=cT-c0=0-7=-7iv. CT = max(O,ST - X) = max(0,109 - 105) = 4JJ=c -c -4-7=-3 T 0-

v. CT = max(O,ST - X) = max(O,ll2 - 105) = 7n=c~-Co=7-7=0vi. CT = max(O,ST - X) = max(O,ll5 - 105) = 10n=cT-co=10-7=3

B. ST* =X +CO= 105 + 7= 112Clearly, this result is consistent with our solution above, where the profit is exactlyzero in Part A(v), in which the price at expiration is 112.3. A. Put buyeri. pT = max(0,X - ST) = max(0,2100 - 2125) = 0

n = PT - Po = o - 106.25 = - 106.25ii. p, = max(0,X - ST) = max(0,2100 - 2050) = 50

n = p, - Po = 50 - 106.25 = -56.25iii. pT = max(0,X - ST) = max(0,2100 - 1950) = 150II = p~ - po = 150 - 106.25 = 43.75B. Put selleri. Value = -pT = -max(O,X - ST) = -max(0,2100 - 1975) = - 125

.. n = -p, + Po = -125 + 106.25 = -18.75 11. Value = -pT = -max(O,X - ST) = -max(0,2100 - 2150) = 0

n = -PT + Po = -0 + 106.25 = 106.25C. Maximum and minimumi. Maximum profit to buyer (loss to seller) = X - pa = 2100 - 106.25 =1993.75Solutions 493ii. Maximum loss to buyer (profit to seller) = po = 106.25D. ST* = X - PO = 2100 - 106.25 = 1993.75

4. A. i. p, = max(0,X - ST) = max(0,95 - 100) = 0n=p,-po=o-5=-5ii. pT = max(0,X - ST) = max(0,95 - 95) = 0rI=p,-po=o-5=-5iii. p, = max(0,X - ST) = max(0,95 - 93) = 2n=pT-p0=2-5=-3iv. pT = max(0,X - ST) = max(0,95 - 90) = 5n=pT-p0=5-5=ov. pT = max(0,X - ST) = max(0,95 - 85) = 10n=pT-po= 10-5=5B. ST* = X - PO = 95 - 5 = 90Clearly, this result is consistent with our solution above, where the profit is exactlyzero in Part A(iv), in which the price at expiration is 90.C. i. Maximum profit (to put buyer) = X - po = 95 - 5 = 90.ii. This profit would be realized in the unlikely scenario of the price of the underlyingfalling all the way down to zero.5. A. This position is commonly called a covered call.B. i. VT = ST - max(O,ST - X) = 70 - max(0,70 - 80) = 70 - 0 = 70~=vT-vo=70~(so-~o)=70-(77-6)=70-71=-1ii. VT = ST - max(O,ST - X) = 75 - max(0,75 - 80) = 75 - 0 = 75

n = VT - Vo = 75 - (So - co) = 75 - (77 - 6) = 4iii. VT = ST - max(O,ST - X) = 80 - max(0,80 - 80) = 80 - 0 = 80II = VT - Vo = 80 - (So - c0) = 80 - (77 - 6) = 9iv. VT = ST - max(O,ST - X) = 85 - max(0,85 - 80) = 85 - 5 = 80IT = VT - Vo = 80 - (SO - c0) = 80 - (77 - 6) = 9

C. i. Maximum profit = X - So + co = 80 - 77 + 6 = 9ii. Maximum loss = So - co = 77 - 6 = 71iii. The maximum profit would be realized if the expiration price of the underlyingis at or above the exercise price of $80.iv. The maximum loss would be incurred if the underlying price drops to zero. 'D. ST* = SO- co = 77 - 6 = 716. A. This position is commonly called a protective put.

B. i. VT = ST + max(0,X - ST) = 70 + max(0,75 - 70) = 70 + 5 = 75

n = vT - vo = 75 - (so + po) = 75 - (77 + 3) = 75 - 80 = -5

ii. VT = ST + max(0,X - ST) = 75 + max(0,75 - 75) = 75 + 0 = 75

II = VT - ... Vo = 75 - (So + po) = 75 - (77 + 3) = 75 - 80 = -5 111. VT = ST + max(0,X - ST) = 80 + max(0,75 - 80) = 80 + 0 = 80

n = vT - vo = 80 - (so + = 80 - (77 + 3) = 80 - 80 = oiv. VT = ST + max(0,X - ST) = 85 + max(0,75 - 85) = 85 + 0 = 85

n = vT - vo = 8s - (so + P,) = 85 - (77 + 3) = 85 - 80 = 5

v. VT = ST + max(0,X - ST) = 90 + max(0,75 - 90) = 90 + 0 = 90

=VT - Vo = 90 - (So + po) = 90 - (77 + 3) = 90 - 80 = 10

C. ..i . Maximum profit = rn 11. Maximum loss = - (X - So - po) = - (75 - 77 - 3) = 5iii. The maximum loss would be incurred if the expiration price of the underlyingwere at or below the exercise price of $75.

D. ST* = SO + PO = 77 + 3 = 80494 Chapter 7 Risk Management Applications of Option Strategies7. A. This position is commonly called a bull spread.B. Let XI be the lower of the two strike prices and X2 be the higher of the two strikeprices.i. VT = max(O,ST - X1) - max(O,ST - X2) = max(0,89 - 75) - max(0,89 -85) = 14 - 4 = 10II = VT - Vo = VT - (cl - c2) = 10 - (10 - 2) = 2ii. V, = max(O,S, - X1) - max(O,ST - X2) = max(0,78 - 75) - max(0,70 -85)=3-0=3IT =VT -VO =VT - (cl - c2) = 3 - (10 - 2) = -5iii. V, = max(O,ST - X1) - max(O,ST - X2) = max(0,70 - 75) - max(0,70 -85)=0-O=O

n =vT -v, = v, - (c, - c2) = o - (10 - 2) = -8C. i. Maximum profit = X2 - XI - (c1 - c2) = 85 - 75 - (10 - 2) = 2ii. Maximum loss = cl - c2 = 10 - 2 = 8

D. ST* = XI + (c, - c2) = 75 + (10 - 2) = 83E. VT = max(O,ST - XI) - max(O,ST - X2) = max(0,83 - 75) - max(0,83 -85)=8-0=8IT =VT -Vo=VT - (c] - c2) = 8 - (10 - 2) = 0Therefore, the profit or loss if the price of the underlying increases to 83 at expirationis indeed zero.8. A. This position is commonly called a bear spread.B. Let X1 be the lower of the two strike prices and X2 be the higher of the two strikeprices.i. VT = max(0,X2 - ST) - max(O,X1 - ST) = max(0,0.85 - 0.87) -max(0,0.70 - 0.87) = 0 - 0 = 0

n = v, - v, = v, - (p2 - pl) = o - (0.15 - 0.03) = -0.12ii. VT = max(0,X2 - ST) - max(O,X1 - ST) = max(0,0.85 - 0.78) -max(0,0.70 - 0.78) = 0.07 - 0 = 0.07

n = v, - v, = v, - (p2 - pl) = 0.07 - (0.15 - 0.03) = -0.05iii. VT = max(0,X2 - ST) - max(O,X1 - ST) = max(0,0.85 - 0.68) -max(0,0.70 - 0.68) = 0.17 - 0.02 = 0.15

n =vT-v,=v,- (p2- p,) =OM - (0.15 -0.03) = 0.03C. i. Maximum profit = X2 - X1 - (p2 - p,) = 0.85 - 0.70 - (0.15 - 0.03) =0.03ii. Maximum loss = p2 - pl= 0.15 - 0.03 = 0.12D. Breakeven point = X2 - (p2 - pI) = 0.85 - (0.15 - 0.03) = 0.73E. VT = max(0,X2 - ST) - max(O,X1 - ST) = max(0,0.85 - 0.73) -max(0,0.70 - 0.73) = 0.12 - 0 = 0.12II =VT -VO=VT - (p2 - PI) = 0.12 - (0.15 - 0.03) = 0Therefore, the profit or loss if the price of the currency decreases to $0.73 at expirationof the puts is indeed zero.9. A. Let XI be 110, X2 be 115, and X3 be 120.

Vo = cl - 2c2 + ~3 = 8 - 2(5) + 3 = 1

i. VT = max(O,S, - X1) - 2max(0,ST - X2) + max(O,ST - X3)

VT = max(0,106 - 1 10) - 2max(0,106 - 1 15) + max(0,106 - 120) = 0n=v,-v,=o-I=-1

ii. VT = max(O,S, - X1) - 2max(0,ST - X2) + max(O,ST - X3)

V, = max(0,llO - 110) - 2max(O,llO - 115) + max(0,llO - 120) = 0II=VT-VO=O- 1=-1A-- - - - - - - - -- .- -

Solutions 495

iii. VT = max(O,S, - XI) - 2max(0,ST - Xz) + max(O,ST - X3)

V, = max(O,ll5 - 110) - 2max(0,115 - 115) + max(O,ll5 - 120) = 5II=VT-Vo=5-1=4

iv. VT = max(O,ST - XI) - 2max(0,ST - X2) + max(O,ST - X3)

VT = max(O,l20 - 110) - 2max(0,120 - 115) + max(0,120 - 120) = 10-10+0=0n=v,-v,=o-I=-1

v. VT = max(O,S, - XI) - 2max(0,ST - X2) + max(O,ST - X3)

VT = max(0,123 - 1 10) - 2max(0,123 - 115) + max(0,123 - 120) = 13 -16+3=0n=vT-v,=o-I=-1

B. i. Maximumprofit = X2 - XI - (c, - 2c2 + c3) = 115 - 110 - 1 = 4

ii. Maximum loss = cl - 2c2 + c3 = 1iii. The maximum profit would be realized if the price of the stock at expirationof the options is at the exercise price of $1 15.iv. The maximum loss would be incurred if the price of the stock is at or belowthe exercise price of $1 10, or if the price of the stock is at or above the exerciseprice of $120.

C. Breakeven: ST* = XI + (c, - 2c2 + c3) and ST* = 2X2 - XI - (cl - 2c2 +c3) SO, ST* = 110 + 1 = 11 1 and ST* = 2(115) - 110 - 1 = 119.10. A. LetX, be 110,X2be 115, andX3 be 120.

Vo = p1 - 2p2 + p3 = 3.50 - 2(6) + 9 = 0.50

i. VT = max(0, XI - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

VT = max(0,llO - 106) - 2max(0,115 - 106) + max(0,120 - 106) =

4 - 2(9) + 14 = 0

n = v, - v, = o - 0.50 = -0.50

ii. VT = max(0, XI - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

V, = max(0,llO - 110) - 2max(0,115 - 110) + max(O,l20 - 110) =

0 - 2(5) + 10 = 0

... n = vT - v, = o - 0.50 = -0.50 111. VT = max(0, XI - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

VT = max(0,llO - 115) - 2max(0,115 - 115) + max(0,120 - 115) =

0 - 2(0) + 5 = 511 = VT - Vo = 5 - 0.50 = 4.50

iv. V, = max(0, XI - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

VT = max(0,llO - 120) - 2max(0,115 - 120) + max(0,120 - 120) = 0

n = VT - Vo = 0 - 0.50 = -0.50

v. VT = max(0, XI - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

VT = max(0,llO - 123) - 2max(0,115 - 123) + max(0,120 - 123) = 0

n = v, - v, = o - 0.50 = -0.50

B. i. Maximum profit = X2 - XI - (pl - 2p2 + p3) = 115 - 110 - 0.50 = 4.50

ii. Maximum loss = p, - 2p2 + p3 = 0.50

iii. The maximum profit would be realized if the expiration price of the stock isat the exercise price of $1 15.iv. The maximum loss would be incurred if the expiration price of the stock is ator below the exercise price of $1 10, or if the expiration price of the stock is ator above the exercise price of $120.

C. Breakeven: ST* = XI + (pl - 2p2 + p3) and ST* = 2X2 - XI - (pl - 2p2 + p3)

So, ST* = 110 + 0.50 = 110.50 and ST* = 2(115) - 110 - 0.50 = 119.50496 Chapter 7 Risk Management Applications of Option StrategiesD. For ST = 110.50:

VT = max(0, X1 - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)VT = max(0,llO - 110.50) - 2max(0,115 - 110.50)+ max(0,120 -

110.50) = -2(4.50) + 9.50 = 0.50IT = VT - Vo = 0.50 - 0.50 = 0For ST = 1 19.50:

VT = max(0, X1 - ST) - 2max(0,X2 - ST) + max(0,X3 - ST)

VT = max(0,llO - 119.50) - 2max(0,115 - 119.50) + max(0,120 -119.50) = 0.50II = VT - Vo = 0.50 - 0.50 = 0Therefore, we see that the profit or loss at the breakeven points computed in PartD above is indeed zero.

11. A. i. VT = ST + max(O,X, - ST) - max(O,ST - X2) =

92 + max(0,75 - 92) - max(0,92 - 90) = 92 + 0 - 2 = 90II = VT - So = 90 - 80 = 10

ii. VT = ST + max(O,X1 - ST) - max(O,ST - XZ) =

90 + max(0,75 - 90) - max(0,90 - 90) = 90 + 0 - 0 = 90

IT = VT - So = 90 - 80 = 10

iii. VT = ST + max(O,X1 - ST) - max(O,ST - X2) =

82 + max(0,75 - 82) - max(0,82 - 90) = 82 + 0 - 0 = 82II=VT-So=82-80=2

iv. VT = ST + max(O,Xl - ST) - max(O,ST - X2) =

75 + max(0,75 - 75) - max(0,75 - 90) = 75 + 0 - 0 = 75

n = v, - so = 75 - 80 = -5

v. VT = ST + max(O,X, - ST) - max(O,ST - X2) =

70 + max(0,75 - 70) - max(0,70 - 90) = 70 + 5 - 0 = 75Il= VT - So = 75 - 80 = -5B. i. Maximum profit = X2 - So = 90 - 80 = 10ii. Maximum loss = - (X, - So) = - (75 - 80) = 5

iii. The maximum profit would be realized if the price of the stock at the expirationof options is at or above the exercise price of $90.iv. The maximum loss would be incurred if the price of the stock at the expirationof options were at or below the exercise price of $75.C. Breakeven: ST* = SO = 8012. A. This position is commonly called a straddle.

B. i. VT = max(O,ST - X) + max(0,X - ST) = max(0,35 - 25) + max(0,25 -

35) = 10 + 0 = 10

n =vT - (c, + pol = 10 - (4 + 1) = 5

ii. VT = max(O,ST - X) + max(0,X - ST) = max(0,29 - 25) + max(0,25 - 29) =

4 + 0 = 4

II =VT- (c0 + po) =4 - (4 + 1) = -1

iii. VT = max(O,ST - X) + max(0,X - ST) = max(0,25 - 25) + max(0,25 - 25) =

0 + 0 = 0

n =vT - (c, + Po) =O - (4 + 1) = -5

iv. VT = max(O,ST - X) + max(0,X - ST) = max(0,20 - 25) + max(0,25 - 20) =0+5=5

n = vT - (c, + Po) = 5 - (4 + 1) = oV. VT = max(O,ST - X) + max(0,X - ST) = max(0,15 - 25) + max(0,25 - 15) =

0 + 10 = 10

n =vT - (c, + = 10 - (4 + 1) = 5Solutions 497C. i. Maximum profit = m

ii. Maximum loss = co + po = 4 + 1 = 5

D. ST* = X * (c0 + po) = 25 5 (4 + 1) = 30,2013. Regardless of the stock price at the expiration of the options, the box spread alwayshas a value at expiration of X2 - X1 = 30 - 25 = 5. So,Because the transaction is risk free, the present value of the payoff, discounted usingthe risk-free rate, should equal the initial outlay if the options are correctly priced.Hence, we should have

Because X2 - X1 = 5, T = 6/12 = 0.50, and initial outlay = cl - c2 + p2 - pl =4.85, we haveSolving for r, we have r = 0.0628 or 6.28 percent.14. The payoff from the call at expiration on 15 July is given below and will be received90 days later on 13 October:

$lO,OOO,OOO max(0, LIBOR -o : 0.: 055) (- 1The premium on the call is paid on 18 May. So, we need to compute its value compoundedfrom 18 May to 15 July. Because there are 58 days between 18 May and 15 ,July, the compounded value of the call premium of $25,000, based on the currentLIBOR plus 200 basis points isSo the call costs $25,000 on 18 May, which is equivalent to $25,322 on 15 July. Theeffective loan proceeds are

The loan interest is

$10,000,000 (LIBOR on 15 July + 200 bas io s : po: ints) (- 1The call payoff was given above. The loan interest minus the call payoff is the effectiveinterest. The effective rate on the loan is498 Chapter 7 Risk Management Applications of Option Strategies$10,000,000 plus effective interest$9,974,678The results are shown below for a range of LIBORs on 15 July.LlBOR on Loan Interest Effective Effective15 July Loan Rate Paid 13 October Call Payoff lnterest Loan Rate15. The payoff from the call at expiration on 1 May is given below and will be received180 days later on 28 October:The premium on the call is paid on 15 March. So, we need to compute its value compoundedfrom 15 March to 1 May. Because there are 47 days between 15 March and1 May, the compounded value of the call premium of $8,000, based on the currentLIBOR plus 3 percent isSo the call costs $8,000 on 15 March, which is equivalent to $8,089 on 1 May. Theeffective loan proceeds areThe loan interest is

$3,000,000(LIBOR on 1 May + 0.03)k -::)The call payoff was given above. The loan interest minus the call payoff is the effectiveinterest. The effective rate on the loan is$3,000,000 plus effective interest$2,99 1,9 11The effective interest rate for the company will continue to increase until LIBORreaches the strike rate. Beyond this point, the effective interest rate will remain thesame regardless of the increase in LIBOR. So, to get the effective ceiling on the rate,we could compute the effective rate when LIBOR on 1 May equals the strike rate, or5.25 percent. Then the loan interest isSolutions 499The call payoff is zero. So, the effective interest is $123,750.The effective rate on the loan isThis is the effective ceiling on the rate.16. The payoff at expiration is given below and will be received 180 days later.

$100,000,000 max(0,0.0625 - LIBOR) G-:)The option premium compounded from 16 April to 1 June isSo the put costs $120,000 on 16 April, which is equivalent to $121,380 on 1 June. Theeffective amount lent isThe loan interest isLIBOR on 1 June plus 250 basis pointsThe put payoff was given above. The loan interest plus the put payoff is the effectiveinterest on the loan. The effective rate on the loan is$100,000,000 plus effective interest$100,121,380The results are shown below for a range of LIBORs on 1 June.LlBORon 1 June0.030

0.0450.0600.0750.0900.105Loan Rate0.0550.0700.0850.1000.1150.130Loan InterestPaid on 28 Nov Put Payoff$1,625,000875,000125,000000EffectiveInterestEffectiveLoan Rate17. The number of days in each period is as follows:500 Chapter 7 Risk Management Applications of Option Strategies1 March - 1 June: 92 days1 June - 1 September: 92 days1 September - 1 December: 9 1 days1 December - 1 March: 90 daysThe interest due for each period is computed as

(~a~s $10,0~~,000(LIBOoRn previous reset date + 200 basis points) ,,,For example, the first interest payment is calculated asEach caplet payoff is computed as:(Days inpod$10,000,000 max (0,LIBOR on previous reset date - 0.0725)where the "previous reset date" is the caplet expiration, and 0.0725 corresponds to thestrike rate. Payment is deferred until the date on which the interest is paid at the givenLIBOR. For example, the caplet expiring on 1 September is worth $10,000,000 max(0,0.0750 - 0.0725)(91/360) = $6,319, which is paid on 1 December. The effectiveinterest is the actual interest minus the caplet payoff. The payments are shown in thetable below:Days in Caplet EffectiveDate LIBOR Loan Rate Period Interest Due Payoff Interest1 March 0.0700 0.09001 June 0.0725 0.0925 92 $230,000 $230,0001 September 0.0750 0.0950 92 236,389 $0 236,3891 December 0.0800 0.1000 91 240,139 6,319 233,819

1 March 90 250,000 18,750 23 1,25018. The number of days in each period is as follows:16 July - 16 October: 92 days16 October - 16 January: 92 days16 January - 16 April: 90 days16April - 16 July: 91 daysThe interest due for each period is computed as

$20,000,000(LIBOR on previous reset date + 0.0250) (Days inopenod)Solutions 501For example, the first interest payment isA. Each floorlet payoff is computed as(Days inn$20,000,000 max(0,0.0575 - LIBOR on previous reset date)where the "previous reset date" is the floorlet expiration. Payment is deferred untilthe date on which the interest is paid at the given LIBOR. For example, the floorletexpiring on 16 October is worth $20,000,000 max(0,0.0575 - 0.0525)(92/360) =$25,556, which is paid on 16 January. Similarly, the floorlet expiring on 16 Januaryis worth $20,000,000 max(0,0.0575 - 0.055)(90/360) = 12,500, which is paid on16 April. The effective interest is the actual interest plus the floorlet payoff. Thepayments are shown in the table below:Days in Floorlet EffectiveDate LlBOR Loan Rate Period Interest Due Payoff Interest16 July 0.0600 0.085016 October 0.0525 0.0775 92 $434,444 $434,44416 January 0.0550 0.0800 92 396,111 $25,556 370,55616 April 0.0575 0.0825 90 400,OOO 12,500 387,50016 July 91 417,083 0 417,083B. Each floorlet payoff is computed as$20,000,000 max(0,0.0600 - LIBOR on previous reset date)(Days inmhd)where the "previous reset date" is the floorlet expiration. The payments are shownin the table below. Note that there is no change in the first four columns. Becauseof the higher strike price of the call, the floorlet payoff is higher than in Part Aabove and the effective interest is lower.Days in Floorlet EffectiveDate LlBOR Loan Rate Period Interest Due Payoff Interest- --

16 July 0.0600 0.085016 October 0.0525 0.0775 92 $434,444 $434,44416 January 0.0550 0.0800 92 396,111 $38,333 357,77816 April 0.0575 0.0825 90 400,000 25,000 375,00016 July 9 1 417,083 12,639 404,444C. Each floorlet payoff is computed as(Day s $20,000,000 max(0,0.0550 inmbd - LIBOR on previous reset date)502 Chapter 7 Risk Management Applications of Option Strategieswhere the "previous reset date" is the floorlet expiration. The payments are shownin the table below. Note that there is no change in the first four columns. Becauseof the lower strike price of the call, the floorlet payoff is lower than in Part A aboveand the effective interest is higher.Days in Floorlet EffectiveDate LlBOR Loan Rate Period Interest Due Payoff Interest16 July 0.0600 0.085016 October 0.0525 0.0775 92 $434,444 $434,44416 January 0.0550 0.0800 92 396,111 $12,778 383,33316 April 0.0575 0.0825 90 400,000 0 400,000

16 July 9 1 417,083 0 417,08319. The loan interest is computed asDays in settlement period

$20,000,000(LIBOR on previous reset date + 0.015)360The caplet payoff isDays in settlement period$20,000,000max(O,LIBOR on previous reset date - 0.0625)360The floorlet payoff isDays in settlement periodLIBOR on previous reset date)360The effective interest is the interest due plus the caplet payoff plus the floorlet payoff(note that the floorlet payoff is negative).The following table shows the payments on the loan and collar:Loan Days in lnterest Caplet Floorlet EffectiveDate LlBOR Rate Period Due Payoffs Payoffs Interest1 March 0.0600 0.0751 June 0.0650 0.080 92 $383,333 $383,3331 September 0.0550 0.070 92 408,889 $12,778 $0 396,1111 December 0.0500 0.065 91 353,889 0 0 353,8891 March 90 325,000 0 -12,500 337,50020. A. If the stock moves up (down) by $1, 1,000 calls should change by 1,000(0.64) =$640. Thus, 640 shares are needed.B. Shares worth 640($46) = $29,440Options worth - 1 ,OOO($5) = - $5,000Bonds worth - $3,000Total = 29,440 - 5,000 - 3,000 = $21,440C. The new required number of shares is 1,000(0.5420) = 542. So we need to have542 shares instead of 640 shares. That means we would sell 640 - 542 = 98Solutions 503shares, generating 98($46) = $4,508. We would invest this amount in risk-freebonds. We had a bond balance of -$3,000, so this amount pays off all of this debt,leaving a balance of $4,508 -$3,000 = $1,508 going into the next day.The composition of the portfolio will then be as follows:Shares worth 542($46) = $24,932Options worth - 1 ,OOO($5) = -$5,000Bonds worth $1,508

Total = 24,932 - 5,000 + 1,508 = $21,440D. Shares worth 542($45.50) = $24,661Options worth - 1,000($4.7 1) = - $4,7 10Bonds worth 1,508 exp(0.0451365) = $1,508

Total = 24,661 - 4,710 + 1,508 = $21,459The benchmark is $21,440 exp(0.0451365) = $21,442, which is about $17 lessthan the market value computed above