CFA 2018 level 3 schweser practice exam CFA 2018 level 3 question bank CFA 2018 CFA 2018 r29 risk management applications of option strategies IFT notes

Risk Management Applications of Option Strategies IFT Notes Risk Management Applications of Option Strategies Introduction Option Strategies for Equity Portfolios 2.1 Standard Long and Short Positions 2.2 Risk Management Strategies with Options and the Underlying 2.3 Money Spreads 2.4 Combinations of Calls and Puts 11 Interest Rate Option Strategies 14 3.1 Using Interest Rate Calls with Borrowing 14 3.2 Using Interest Rate Puts with Lending 17 3.3 Using an Interest Rate Cap with a Floating-Rate Loan 20 3.4 Using an Interest Rate Floor with a Floating-Rate Loan 21 3.5 Using an Interest Rate Collar with a Floating-Rate Loan 23 Option Portfolio Risk Management Strategies 25 4.1 Delta Hedging an Option over Time 26 4.2 Gamma and the Risk of Delta 28 4.3 Vega and Volatility Risk 28 Final Comments 28 Summary 29 Examples from the Curriculum 34 Example 34 Example 35 Example 37 Example 37 Example 38 Example 39 Example 40 Example 42 Example 43 Example 10 43 Example 11 44 Example 12 45 IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes Example 13 47 Example 14 48 Example 15 48 Example 16 49 This document should be read in conjunction with the corresponding reading in the 2018 Level III CFA® Program curriculum Some of the graphs, charts, tables, examples, and figures are copyright 2017, CFA Institute Reproduced and republished with permission from CFA Institute All rights reserved Required disclaimer: CFA Institute does not endorse, promote, or warrant the accuracy or quality of the products or services offered by IFT CFA Institute, CFA®, and Chartered Financial Analyst® are trademarks owned by CFA Institute IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes Introduction Some basic notations used in options are: Time 0: It is the time at which the strategy is initiated Time T: It is the time the option expires c0, cT = price of the call option at time and time T p0, pT = price of the put option at time and time T X = exercise price S0, ST = price of the underlying at time and time T V0, VT = value of the position at time and time T Π = profit from the transaction: VT – V0 r = risk-free rate Option Strategies for Equity Portfolios 2.1 Standard Long and Short Positions Note: Section 2.1 is optional and is a review of concepts learned at earlier levels Buy Call (Long Call) Exhibit shows the profit diagram for a buyer of a call option IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes The important points are:  cT = max(0, ST – X)  Value at expiration = cT  Profit: Π = cT – c0  Maximum profit = ∞  Maximum loss = c0  Breakeven: ST* = X + c0 Sell Call (Short Call) Exhibit shows the profit diagram for a seller of a call option Note that it is the mirror image of the long call The important points are:  cT = max(0, ST – X)  Value at expiration = –cT  Profit: Π = –cT + c0  Maximum profit = c0  Maximum loss = ∞  Breakeven: ST* = X + c0 IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes Refer to Example from the curriculum Buy Put Exhibit shows the profit diagram for a buyer of a put option The important points are:  pT = max(0, X – ST)  Value at expiration = pT  Profit: Π = pT – p0  Maximum profit = X – p0  Maximum loss = p0  Breakeven: ST* = X – p0 Sell Put Exhibit shows the profit diagram for a seller of a put option IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes The important points to note are:  pT = max(0, X – ST)  Value at expiration = –pT  Profit: Π = –pT + p0  Maximum profit = p0  Maximum loss = X – p0  Breakeven: ST* = X – p0 Refer to Example from the curriculum 2.2 Risk Management Strategies with Options and the Underlying LO.a: Compare the use of covered calls and protective puts to manage risk exposure to individual securities Covered Call In this strategy we take a long position in the underlying and sell a call option A covered call provides some protection against a fall in the price of the underlying It also generates cash up front, but the covered call writer could miss out on the upside in a strong bull market Exhibit shows the profit diagram for a covered call strategy IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes The important points to note are:  Value at expiration: VT = ST – max(0, ST – X)  Profit: Π = VT – S0 + c0  Maximum profit = X – S0 + c0  Maximum loss = S0 – c0  Breakeven: ST* = S0 – c0 Refer to Example from the curriculum Protective Put In this strategy we are long on the underlying and buy a put It provides downside protection while retaining the upside potential However, we need to pay cash upfront to buy the put option Exhibit shows the profit diagram for a protective put IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes The important points to note are:  Value at expiration: VT = ST + max(0,X – ST)  Profit: Π = VT – S0 – p0  Maximum profit = ∞  Maximum loss = S0 + p0 – X  Breakeven: ST* = S0 + p0 Refer to Example from the curriculum LO.b: Calculate and interpret the value at expiration, profit, maximum profit, maximum loss, breakeven underlying price at expiration, and general shape of the graph for the following option strategies: bull spread, bear spread, butterfly spread, collar, straddle, box spread This LO is covered in sections 2.3 and 2.4 2.3 Money Spreads A spread is a strategy in which we buy one option and sell another option which is identical except for exercise price or time to expiration If the expiration time is different, then the spread is called a time spread (not covered in this reading) If the exercise price is different, then the spread is called a money spread This strategy is called a spread because the payoff is based on the difference, or spread, between option exercise prices We will cover the following money spread strategies:  Bull spread IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies   IFT Notes Bear spread Butterfly spread Bull Spread In this strategy we combine a long position in a call with exercise price X1 and a short position in a call with a higher exercise price X2 This strategy is designed to make money when the market goes up Exhibit shows the profit diagram for a bull spread The important points to note are:  Value at expiration: VT = max(0, ST – X1) – max(0, ST – X2)  Profit: Π = VT – c1 + c2  Maximum profit = X2 – X1 – c1 + c2  Maximum loss = c1 – c2  Breakeven: ST* = X1 + c1 – c2 Refer to Example from the curriculum Bear Spread This strategy is exactly opposite to a bull spread Here we sell a call with a lower exercise price X1 and buy a call with a higher exercise price X2 Alternatively we can also execute this strategy by buying puts, we would buy a put with a higher exercise price X2 and sell a put with a lower exercise price X1 This strategy is designed to make money when the market goes down IFT Notes for the Level III Exam www.ift.world Page Risk Management Applications of Option Strategies IFT Notes Exhibit shows the profit diagram for a bear spread Note that it is a mirror image of the bull spread The important points to note are:  Value at expiration: VT = max(0, X2 – ST) – max(0, X1 – ST)  Profit: Π = VT – p2 + p1  Maximum profit = X2 – X1 – p2 + p1  Maximum loss = p2 – p1  Breakeven: ST* = X2 – p2 + p1 Refer to Example from the curriculum Butterfly spreads This strategy combines a bull spread and a bear spread It is designed to make money when the market has low volatility and stays within a range Consider three exercise prices X1, X2, and X3 As shown above, we can construct a bull spread by buying the call with exercise price of X1 and selling the call with exercise price of X2 Similarly we can construct a bear spread by buying the call with exercise price X3 and selling the call with exercise price X2 The end result is that we own the calls with exercise price X1 and X3 and have sold two calls with exercise price X2 Exhibit 10 shows the profit diagram for a butterfly spread IFT Notes for the Level III Exam www.ift.world Page 10 Risk Management Applications of Option Strategies IFT Notes D ST* = X – p0 = 60 – = 56 Back to Notes Example Consider a bond selling for $98 per $100 face value A call option selling for $8 has an exercise price of $105 Answer the following questions about a covered call A Determine the value of the position at expiration and the profit under the following outcomes: i The price of the bond at expiration is $110 ii The price of the bond at expiration is $88 B Determine the following: i The maximum profit ii The maximum loss C Determine the breakeven bond price at expiration Solutions: A i 𝑉𝑇 = 𝑆𝑇 − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋) = 110 − 𝑚𝑎𝑥(0,110 − 105) = 110 − 110 + 105 = 105 𝛱 = 𝑉𝑇 − 𝑉0 = 105 − (𝑆0 − 𝑐0 ) = 105 − (98 − 8) = 15 ii 𝑉𝑇 = 𝑆𝑇 − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋) = 88 − 𝑚𝑎𝑥(0,88 − 105) = 88 − = 88 𝛱 = 𝑉𝑇 − 𝑉0 = 88 − (𝑆0 − 𝑐0 ) = 88 − (98 − 8) = −2 B i Maximum profit = X – S0 + c0 = 105 – 98 + = 15 ii Maximum loss = S0 – c0 = 98 – = 90 C ST* = S0 – c0 = 98 – = 90 Back to Notes Example Consider a currency selling for $0.875 A put option selling for $0.075 has an exercise price of $0.90 Answer the following questions about a protective put A Determine the value at expiration and the profit under the following outcomes: IFT Notes for the Level III Exam www.ift.world Page 37 Risk Management Applications of Option Strategies i The price of the currency at expiration is $0.96 ii The price of the currency at expiration is $0.75 IFT Notes B Determine the following: i the maximum profit ii the maximum loss C Determine the breakeven price of the currency at expiration Solutions: A i 𝑉𝑇 = 𝑆𝑇 + 𝑚𝑎𝑥(0, 𝑋𝑇 − 𝑆𝑇 ) = 0.96 + 𝑚𝑎𝑥(0,0.90 − 0.96) = 0.96 𝛱 = 𝑉𝑇 − 𝑉0 = 0.96 − (𝑆0 + 𝑝0 ) = 0.96 − (0.875 + 0.075) = 0.01 ii 𝑉𝑇 = 𝑆𝑇 + 𝑚𝑎𝑥(0, 𝑋𝑇 − 𝑆𝑇 ) = 0.75 + 𝑚𝑎𝑥(0,0.90 − 0.75) = 0.90 𝛱 = 𝑉𝑇 − 𝑉0 = 0.90 − (𝑆0 + 𝑝0 ) = 0.90 − (0.875 + 0.075) = −0.05 B i Maximum profit = ∞ ii Maximum loss = S0 + p0 – X = 0.875 + 0.075 – 0.90 = 0.05 C ST* = S0 + p0 = 0.875 + 0.075 = 0.95 Back to Notes Example Consider two call options on a stock selling for $72 One call has an exercise price of $65 and is selling for $9 The other call has an exercise price of $75 and is selling for $4 Both calls expire at the same time Answer the following questions about a bull spread: A Determine the value at expiration and the profit under the following outcomes: i The price of the stock at expiration is $78 ii The price of the stock at expiration is $69 iii The price of the stock at expiration is $62 B Determine the following: i the maximum profit IFT Notes for the Level III Exam www.ift.world Page 38 Risk Management Applications of Option Strategies ii IFT Notes the maximum loss C Determine the breakeven stock price at expiration Solutions: A i 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋1 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 𝑚𝑎𝑥(0,78 − 65) − 𝑚𝑎𝑥(0,78 − 75) = 13 − = 10 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑐1 − 𝑐2 ) = 10 − (9 − 4) = ii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋1 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 𝑚𝑎𝑥(0,69 − 65) − 𝑚𝑎𝑥(0,69 − 75) = − = 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑐1 − 𝑐2 ) = − (9 − 4) = −1 iii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋1 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 𝑚𝑎𝑥(0,62 − 65) − 𝑚𝑎𝑥(0,62 − 75) = − = 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑐1 − 𝑐2 ) = − (9 − 4) = −5 B i Maximum profit = X2 – X1 – (c1 – c2) = 75 – 65 – (9 – 4) = ii Maximum loss = c1 – c2 = – = C ST* = X1 + c1 – c2 = 65 + – = 70 Back to Notes Example Consider two put options on a bond selling for $92 per $100 par One put has an exercise price of $85 and is selling for $3 The other put has an exercise price of $95 and is selling for $11 Both puts expire at the same time Answer the following questions about a bear spread: A Determine the value at expiration and the profit under the following outcomes: i The price of the bond at expiration is $98 ii The price of the bond at expiration is $91 iii The price of the bond at expiration is $82 B Determine the following: i the maximum profit ii the maximum loss IFT Notes for the Level III Exam www.ift.world Page 39 Risk Management Applications of Option Strategies IFT Notes C Determine the breakeven bond price at expiration Solutions: A i 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,95 − 98) − 𝑚𝑎𝑥(0,85 − 98) = − = 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝2 − 𝑝1 ) = − (11 − 3) = −8 ii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,95 − 91) − 𝑚𝑎𝑥(0,85 − 91) = − = 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝2 − 𝑝1 ) = − (11 − 3) = −4 iii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,95 − 82) − 𝑚𝑎𝑥(0,85 − 82) = 13 − = 10 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝2 − 𝑝1 ) = 10 − (11 − 3) = B i Maximum profit = X2 – X1 – (p2 – p1) = 95 – 85 – (11 – 3) = ii Maximum loss = p2 – p1 = 11 – = C ST* = X2 – p2 + p1 = 95 – 11 + = 87 Back to Notes Example Consider three put options on a currency that is currently selling for $1.45 The exercise prices are $1.30, $1.40, and $1.50 The put prices are $0.08, $0.125, and $0.18, respectively The puts all expire at the same time Answer the following questions about a butterfly spread A Determine the value at expiration and the profit under the following outcomes: i The price of the currency at expiration is $1.26 ii The price of the currency at expiration is $1.35 iii The price of the currency at expiration is $1.47 iv The price of the currency at expiration is $1.59 B Determine the following: i the maximum profit ii the maximum loss IFT Notes for the Level III Exam www.ift.world Page 40 Risk Management Applications of Option Strategies IFT Notes C Determine the breakeven currency price at expiration Solutions: A i 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 2𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) + 𝑚𝑎𝑥(0, 𝑋3 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,1.30 − 1.26) − 2𝑚𝑎𝑥(0,1.40 − 1.26) + 𝑚𝑎𝑥(0,1.50 − 1.26) = 0.04 − 2(0.14) + 0.24 = 0.0 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝1 − 2𝑝2 + 𝑝3 ) = 0.0 − [0.08 − 2(0.125) + 0.18] = −0.01 ii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 2𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) + 𝑚𝑎𝑥(0, 𝑋3 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,1.30 − 1.35) − 2𝑚𝑎𝑥(0,1.40 − 1.35) + 𝑚𝑎𝑥(0,1.50 − 1.35) = 0.0 − 2(0.05) + 0.15 = 0.05 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝1 − 2𝑝2 + 𝑝3 ) = 0.05 − [0.08 − 2(0.125) + 0.18] = 0.04 iii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 2𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) + 𝑚𝑎𝑥(0, 𝑋3 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,1.30 − 1.47) − 2𝑚𝑎𝑥(0,1.40 − 1.47) + 𝑚𝑎𝑥(0,1.50 − 1.47) = 0.0 − 2(0) + 0.03 = 0.03 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝1 − 2𝑝2 + 𝑝3 ) = 0.03 − [0.08 − 2(0.125) + 0.18] = 0.02 iv 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 2𝑚𝑎𝑥(0, 𝑋2 − 𝑆𝑇 ) + 𝑚𝑎𝑥(0, 𝑋3 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,1.30 − 1.59) − 2𝑚𝑎𝑥(0,1.40 − 1.59) + 𝑚𝑎 �(0,1.50 − 1.59) = 0.0 − 2(0) + 0.0 = 0.0 𝛱 = 𝑉𝑇 − 𝑉0 = 𝑉𝑇 − (𝑝1 − 2𝑝2 + 𝑝3 ) = 0.0 − [0.08 − 2(0.125) + 0.18] = −0.01 B i 𝑀𝑎𝑥𝑖𝑚𝑢𝑚𝑝𝑟𝑜𝑓𝑖𝑡 = 𝑋2 − 𝑋1 − (𝑝1 − 2𝑝2 + 𝑝3 ) = 1.40 − 1.30 − [0.08 − 2(0.125) + 0.18] = 0.09 ii 𝑀𝑎𝑥𝑖𝑚𝑢𝑚𝑙𝑜𝑠𝑠 = 𝑝1 − 2𝑝2 + 𝑝3 = 0.08 − 2(0.125) + 0.18 = 0.01 IFT Notes for the Level III Exam www.ift.world Page 41 Risk Management Applications of Option Strategies IFT Notes C 𝑆𝑇∗ = 𝑋1 + 𝑝1 − 2𝑝2 + 𝑝3 = 1.30 + 0.08 − 2(0.125) + 0.18 = 1.31 𝑆𝑇∗ = 2𝑋2 − 𝑋1 − 𝑝1 + 2𝑝2 − 𝑝3 = 2(1.40) − 1.30 − 0.08 + 2(0.125) − 0.18 = 1.49 Back to Notes Example The holder of a stock worth $42 is considering placing a collar on it A put with an exercise price of $40 costs $5.32 A call with the same premium would require an exercise price of $50.59 A Determine the value at expiration and the profit under the following outcomes: i The price of the stock at expiration is $55 ii The price of the stock at expiration is $48 iii The price of the stock at expiration is $35 B Determine the following: i the maximum profit ii the maximum loss C Determine the breakeven stock price at expiration Solutions: A i 𝑉𝑇 = 𝑆𝑇 + 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 55 + 𝑚𝑎𝑥(0,40 − 55) − 𝑚𝑎𝑥(0,55 − 50.59) = 55 + − (55 − 50.59) = 50.59 𝛱 = 𝑉𝑇 − 𝑆0 = 50.59 − 42 = 8.59 ii 𝑉𝑇 = 𝑆𝑇 + 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 48 + 𝑚𝑎𝑥(0,40 − 48) − 𝑚𝑎𝑥(0,48 − 50.59) = 48 + − = 48 𝛱 = 𝑉𝑇 − 𝑆0 = 48 − 42 = iii 𝑉𝑇 = 𝑆𝑇 + 𝑚𝑎𝑥(0, 𝑋1 − 𝑆𝑇 ) − 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋2 ) = 35 + 𝑚𝑎𝑥(0,40 − 35) − 𝑚𝑎𝑥(0,35 − 50.59) = 35 + − = 40 𝛱 = 𝑉𝑇 − 𝑆0 = 40 − 42 = −2 B IFT Notes for the Level III Exam www.ift.world Page 42 Risk Management Applications of Option Strategies i Maximum profit = X2 – S0 = 50.59 – 42 = 8.59 ii Maximum loss = S0 – X1 = 42 – 40 = IFT Notes C ST* = S0 = 42 Back to Notes Example Consider a stock worth $49 A call with an exercise price of $50 costs $6.25 and a put with an exercise price of $50 costs $5.875 An investor buys a straddle A Determine the value at expiration and the profit under the following outcomes: i The price of the stock at expiration is $61 ii The price of the stock at expiration is $37 B Determine the following: i the maximum profit ii the maximum loss C Determine the breakeven stock price at expiration Solutions: A i 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋) + 𝑚𝑎𝑥(0, 𝑋 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,61 − 50) + 𝑚𝑎𝑥(0,50 − 61) = 11 − = 11 𝛱 = 𝑉𝑇 − (𝑐0 + 𝑝0 ) = 11 − (6.25 + 5.875) = −1.125 ii 𝑉𝑇 = 𝑚𝑎𝑥(0, 𝑆𝑇 − 𝑋) + 𝑚𝑎𝑥(0, 𝑋 − 𝑆𝑇 ) = 𝑚𝑎𝑥(0,37 − 50) + 𝑚𝑎𝑥(0,50 − 37) = + 13 = 13 𝛱 = 𝑉𝑇 − 𝑆0 = 13 − (6.25 + 5.875) = 0.875 B i Maximum profit = ∞ ii Maximum loss = c0 + p0 = 6.25 + 5.875 = 12.125 C ST* = X ± (c0 + p0) = 50 ± (6.25 + 5.875) = 62.125, 37.875 Back to Notes Example 10 Consider a box spread consisting of options with exercise prices of 75 and 85 The call prices are 16.02 IFT Notes for the Level III Exam www.ift.world Page 43 Risk Management Applications of Option Strategies IFT Notes and 12.28 for exercise prices of 75 and 85, respectively The put prices are 9.72 and 15.18 for exercise prices of 75 and 85, respectively The options expire in six months and the discrete risk-free rate is 5.13 percent A Determine the value of the box spread and the profit for any value of the underlying at expiration B Show that this box spread is priced such that an attractive opportunity is available Solutions: A The box spread always has a value at expiration of X2 – X1 = 85 – 75 = 10 𝛱 = 𝑉𝑇 − (𝑐1 − 𝑐2 + 𝑝2 − 𝑝1 ) = 10 − (16.02 − 12.28 + 15.18 − 9.72) = 0.80 B The box spread should be worth (X2 – X1)/(1 + r)T, or (85−75)/(1.0513)0.5=9.75 The cost of the box spread is 16.02 – 12.28 + 15.18 – 9.72 = 9.20 The box spread is thus underpriced At least one of the long options is priced too low or at least one of the short options is priced too high; we cannot tell which Nonetheless, we can execute this box spread, buying the call with exercise price X1 = 75 and put with exercise price X2 = 85 and selling the call with exercise price X2 = 85 and put with exercise price X1 = 75 This would cost 9.20 The present value of the payoff is 9.75 Therefore, the box spread would generate an immediate increase in value of 0.55 Back to Notes Example 11 On 10 January, ResTex Ltd determines that it will need to borrow $5 million on 15 February at 90-day Libor plus 300 basis points The loan will be an add-on interest loan in which ResTex will receive $5 million and pay it back plus interest on 16 May To manage the risk associated with the interest rate on 15 February, ResTex buys an interest rate call that expires on 15 February and pays off on 16 May The exercise rate is percent, and the option premium is $10,000 The current 90-day Libor is 5.25 percent Assume that this rate, plus 300 basis points, is the rate it would borrow at for any period of up to 90 days if the loan were taken out today Interest is computed on the exact number of days divided by 360 Determine the effective annual rate on the loan for each of the following outcomes: 90-day Libor on 15 February is percent 90-day Libor on 15 February is percent Solutions: First we need to compound the premium from 10 January to 15 February, which is 36 days This calculation tells us the effective cost of the call as of the time the loan is taken out: IFT Notes for the Level III Exam www.ift.world Page 44 Risk Management Applications of Option Strategies IFT Notes 36 $10,000[1 + (0.0525 + 0.03)( )] = $10,083 360 The loan proceeds will therefore be $5,000,000 – $10,083 = $4,989,917 Solution to 1: Libor is percent The loan rate will be percent The interest on the loan will be $5,000,000(0.06 + 0.03) (90/360) = $112,500 The option payoff will be $5,000,000 max(0,0.06 – 0.05) (90/360) = $12,500 Therefore, the effective interest will be $112,500 – $12,500 = $100,000 The effective rate on the loan will be 365/90 $5,000,000 + $100,000 ( ) $4,989,917 − = 0.0925 Of course, a little more than 300 basis points of this amount is the spread Solution to 2: Libor is percent The loan rate will be percent The interest on the loan will be $5,000,000(0.04 + 0.03) (90/360) = $87,500 The option payoff will be $5,000,000 max(0,0.04 – 0.05) (90/360) = $0.00 The effective interest will, therefore, be $87,500 The effective rate on the loan will be $5,000,000 + $87,500 ( ) $4,989,917 365/90 − = 0.0817 Of course, a little more than 300 basis points of this amount is the spread Back to Notes Example 12 State Bank and Trust (SBT) is a lender in the floating-rate instrument market, but it has been hurt by recent interest rate decreases SBT often makes loan commitments for its customers and then accepts the rate in effect on the day the loan is taken out SBT has avoided floating-rate financing in the past It takes out a certain amount of fixed-rate financing in advance to cover its loan commitments One particularly large upcoming loan has it worried This is a $100 million loan to be made in 65 days at 180day Libor plus 100 basis points The loan will be paid back 182 days after being taken out, and interest will be based on an exact day count and 360 days in a year Current Libor is 7.125 percent, which is the rate it could borrow at now for any period less than 180 days SBT considers the purchase of an interest rate put to protect it against an interest rate decrease over the next 65 days The put will have an IFT Notes for the Level III Exam www.ift.world Page 45 Risk Management Applications of Option Strategies IFT Notes exercise price of percent and a premium of $475,000 Determine the effective annual rate on the loan for the following outcomes: 180-day Libor at the option expiration is percent 180-day Libor at the option expiration is percent Solutions: First we need to compound the premium for 65 days This calculation tells us the effective cost of the put as of the time the loan is made: 65 $475,000[1 + (0.07125 + 0.01)( )] = $481,968 360 The outlay will effectively be $100,000,000 + $481,968 = $100,481,968 Solution to 1: Libor is percent The loan rate will be 10 percent The interest on the loan will be $100,000,000 (0.09 + 0.01)(182/360) = $5,055,556 The option payoff will be $100,000,000 max (0,0.07 – 0.09)(182/360) = $0.0 Because there is no option payoff, the effective interest will be $5,055,556 The effective rate on the loan will be 365/182 ( $100,000,000 + $5,055,556 ) $100,481,968 − = 0.0934 Of course, a little more than 100 basis points of this amount is the spread Solution to 2: Libor is percent The loan will be percent The interest on the loan will be $100,000,000 (0.05 + 0.01) (182/360) = $3,033,333 The option payoff will be $100,000,000 max(0,0.07 – 0.05) (182/360) = $1,011,111 The effective interest will, therefore, be $3,033,333 + $1,011,111 = $4,044,444 The effective rate on the loan will be 365/182 $100,000,000 + $4,044,444 ( ) $100,481,968 − = 0.0724 Of course, a little more than 100 basis points of this amount is the spread Back to Notes IFT Notes for the Level III Exam www.ift.world Page 46 Risk Management Applications of Option Strategies IFT Notes Example 13 Healthy Biosystems (HBIO) is a typical floating-rate borrower, taking out loans at Libor plus a spread On 15 January 2002, it takes out a loan of $25 million for one year with quarterly payments on 12 April, 14 July, 16 October, and the following 14 January The underlying rate is 90-day Libor, and HBIO will pay a spread of 250 basis points Interest is based on the exact number of days in the period Current 90-day Libor is 6.5 percent HBIO purchases an interest rate cap for $20,000 that has an exercise rate of percent and has caplets expiring on the rate reset dates Determine the effective interest payments if Libor on the following dates is as given: 12 April 7.250 percent 14 July 6.875 percent 16 October 7.125 percent Solution: The interest due for each period is computed as $25,000,000(Libor on previous reset date + 0.0250)(Days in period/360) For example, the first interest payment is calculated as $25,000,000(0.065 + 0.025)(87/360) = $543,750, based on the fact that there are 87 days between 15 January and 12 April Each caplet payoff is computed as $25,000,000 max(0,Libor on previous reset date – 0.07)(Days in period/360), where the “previous reset date” is the caplet expiration Payment is deferred until the date on which the interest is paid at the given Libor For example, the caplet expiring on 12 April is worth $25,000,000 max(0,0.0725 – 0.07)(93/360) = $16,145, which is paid on 14 July and is based on the fact that there are 93 days between 12 April and 14 July The effective interest is the actual interest minus the caplet payoff The payments are shown in the table below: Date Libor Loan Rate Days in Period Interest Due 15 January 0.065 0.09 12 April 0.0725 14 July 16 October 0.0975 87 $543,750 0.06875 0.09375 93 629,688 $16,146 613,542 0.07125 0.09625 94 611,979 611,979 90 601,563 7,813 593,750 14 January Caplet Payoff Effective Interest $543,750 Back to Notes IFT Notes for the Level III Exam www.ift.world Page 47 Risk Management Applications of Option Strategies IFT Notes Example 14 Capitalized Bank (CAPBANK) is a lender in the floating-rate loan market It uses fixed-rate financing on its floating-rate loans and buys floors to hedge the rate On May 2002, it makes a loan of $40 million at 180-day Libor plus 150 basis points Interest will be paid on November, the following May, the following November, and the following May, at which time the principal will be repaid The exercise rate is 4.5 percent, the floorlets expire on the rate reset dates, and the premium will be $120,000 Interest will be calculated based on the actual number of days in the period over 360 The current 180day Libor is percent Determine the effective interest payments CAPBANK will receive if Libor on the following dates is as given: November 4.875 percent May 4.25 percent November 5.125 percent Solution: The interest due for each period is computed as $40,000,000(Libor on previous reset date + 0.0150)(Days in period/360) For example, the first interest payment is $40,000,000(0.05 + 0.0150)(184/360) = $1,328,889, based on the fact that there are 184 days between May and November Each floorlet payoff is computed as $40,000,000 max(0,0.045 – Libor on previous reset date)(Days in period/360), where the “previous reset date” is the floorlet expiration Payment is deferred until the date on which the interest is paid at the given Libor For example, the floorlet expiring on May is worth $40,000,000 max(0,0.045 – 0.0425)(180/360) = $50,000, which is paid on November and is based on the fact that there are 180 days between May and November The effective interest is the actual interest plus the floorlet payoff The payments are shown in the table below: Date Libor Loan Rate Days in Period Interest Due May 0.05 0.065 November 0.04875 May November 0.06375 184 $1,328,889 0.0425 0.0575 185 1,310,417 $0 1,310,417 0.05125 0.06625 180 1,150,000 50,000 1,200,000 182 1,339,722 1,339,722 May Floorlet Payoff Effective Interest $1,328,889 Back to Notes Example 15 Exegesis Systems (EXSYS) is a floating-rate borrower that manages its interest rate risk with collars, IFT Notes for the Level III Exam www.ift.world Page 48 Risk Management Applications of Option Strategies IFT Notes purchasing a cap and selling a floor in which the cost of the cap and floor are equivalent EXSYS takes out a $35 million one-year loan at 90-day Libor plus 200 basis points It establishes a collar with a cap exercise rate of percent and a floor exercise rate of percent Current 90-day Libor is 6.5 percent The interest payments will be based on the exact day count over 360 The caplets and floorlets expire on the rate reset dates The rates will be set on the current date (5 March), June, September, and December, and the loan will be paid off on the following March Determine the effective interest payments if Libor on the following dates is as given: June 7.25 percent September 6.5 percent December 5.875 percent Solution: The interest due for each period is computed as $35,000,000(Libor on previous reset date + 0.02)(Days in period/360) For example, the first interest payment is $35,000,000(0.065 + 0.02)(91/360) = $752,014, based on the fact that there are 91 days between March and June Each caplet payoff is computed as $35,000,000 max(0,Libor on previous reset date – 0.07)(Days in period/360), where the “previous reset date” is the caplet expiration Payment is deferred until the date on which the interest is paid at the given Libor For example, the caplet expiring on June is worth $35,000,000 max(0,0.0725 – 0.07)(93/360) = $22,604, which is paid on September and is based on the fact that there are 93 days between June and September Each floorlet payoff is computed as $35,000,000 max(0,0.06 – Libor on previous reset date)(Days in period/360) For example, the floorlet expiring on December is worth $35,000,000 max(0,0.06 – 0.05875) (90/360) = $10,938, based on the fact that there are 90 days between December and March The effective interest is the actual interest minus the caplet payoff minus the floorlet payoff The payments are shown in the table below: Date Libor Loan Rate March 0.065 0.085 June 0.0725 0.0925 91 $752,014 September 0.065 0.085 93 836,354 $22,604 $0 813,750 December 0.05875 0.07875 89 735,486 0 735,486 90 689,063 –10,938 700,001 March Days in Period Interest Due Caplet Payoff Floorlet Payoff Effective Interest $752,014 Back to Notes Example 16 DynaTrade is an options trading company that makes markets in a variety of derivative instruments IFT Notes for the Level III Exam www.ift.world Page 49 Risk Management Applications of Option Strategies IFT Notes DynaTrade has just sold 500 call options on a stock currently priced at $125.75 Suppose the trade date is 18 November The call has an exercise price of $125, 60 days until expiration, a price of $10.89, and a delta of 0.5649 DynaTrade will delta-hedge this transaction by purchasing an appropriate number of shares Any additional transactions required to adjust the delta hedge will be executed by borrowing or lending at the continuously compounded risk-free rate of percent DynaTrade has begun delta hedging the option Two days later, 20 November, the following information applies: Stock price $122.75 Option price $9.09 Delta 0.5176 Number of options 500 Number of shares 328 Bond balance –$6,072 Market value $29,645 A At the end of 19 November, the delta was 0.6564 Based on this number, show how 328 shares of stock is used to delta hedge 500 call options B Show the allocation of the $29,645 market value of DynaTrade’s total position among stock, options, and bonds on 20 November C Show what transactions must be done to adjust the portfolio to be delta hedged for the following day (21 November) D On 21 November, the stock is worth $120.50 and the call is worth $7.88 Calculate the market value of the delta-hedged portfolio and compare it with a benchmark, based on the market value on 20 November Solution to A: If the stock moves up (down) $1, the 328 shares should change by $328 The 500 calls should change by 500(0.6564) = $328.20, rounded off to $328 The calls are short, so any change in the value of the stock position is an opposite change in the value of the options Solution to B:  Stock worth 328($122.75) = $40,262  Options worth –500($9.09) = –$4,545  Bonds worth –$6,072 IFT Notes for the Level III Exam www.ift.world Page 50 Risk Management Applications of Option Strategies o IFT Notes Total of $29,645 Solution to C: The new required number of shares is 500(0.5176) = 258.80 Round this number to 259 So we need to have 259 shares instead of 328 shares and must sell 69 shares, generating 69($122.75) = $8,470 We invest this amount in risk-free bonds We had a bond balance of –$6,072, so the proceeds from the sale will pay off all of this debt, leaving a balance of $8,470 –$6,072 = $2,398 going into the next day The composition of the portfolio would then be as follows:  Shares worth 259($122.75) = $31,792  Options worth –500($9.09) = –$4,545  Bonds worth $2,398 o Total of $29,645 Solution to D: The benchmark is $29,645 exp(0.04/365) = $29,648 Also, the value of the bond one day later will be $2,398 exp(0.04/365) = $2,398 (This is less than a half-dollar’s interest, so it essentially leaves the balance unchanged.) Now we have  Shares worth 259($120.50) = $31,210  Options worth –500($7.88) = –$3,940  Bonds worth $2,398 o Total of $29,668 This is about $20 more than the benchmark Back to Notes IFT Notes for the Level III Exam www.ift.world Page 51 ... 0.6104 1 230 77. 53 78.08 0.55 0. 637 4 1240 83. 35 84.59 1.24 0.6 635 IFT Notes for the Level III Exam www .ift. world New Delta Page 26 Risk Management Applications of Option Strategies a b IFT Notes. .. interest rate call option IFT Notes for the Level III Exam www .ift. world Page 16 Risk Management Applications of Option Strategies IFT Notes Refer to Example 11 from the curriculum 3. 2 Using Interest... Put Exhibit shows the profit diagram for a seller of a put option IFT Notes for the Level III Exam www .ift. world Page Risk Management Applications of Option Strategies IFT Notes The important points