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Interest Rate Options

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Title: Interest Rate Options


1
Interest Rate Options
  • Interest rate options provide the right to
    receive one interest rate and pay another.
  • An interest rate call pays off if the interest
    rate ends up above the strike rate. The holder
    pays the strike rate and receives the market
    rate, usually LIBOR.
  • An interest rate put pays off if the interest
    rate ends up below the strike rate. The holder
    pays the market rate LIBOR - and receives the
    exercise rate.

2
  • Interest rate options usually are written by
    dealers and are tailored to the needs of a
    specific clientele.
  • The options are typically European, i.e., they
    can be exercised only at expiration. The
    exercise price, or as it is referred to the
    strike rate usually is set at the current level
    of the spot rate for example, the current six
    -month LIBOR. In this case, the options pays off
    on the basis of the difference between the
    six-month LIBOR at expiration and the exercise or
    strike rate.

3
  • The payoff of an interest rate call
  •  
  • (principal)MaxO,LIBOR-Xn/360,
  • Where
  • X the exercise rate and
  • n the number of days from the options
    expiration to the actual payment
    day.
  • when exercised, the payment by the writer is
    made not at the options expiration but at a
    future date that corresponds to the maturity of
    the underlying spot instrument.

4
  • Example 1
  • A call option written on the 90-day LIBOR with
    an exercise rate of 11 which is the current
    LIBOR, based on a principal amount of 25M and
    with an expiration date 30 days hence.
  • Upon expiration, the call holder exercises the
    call if the market 90-day LIBOR exceeds 11. In
    this case, the payment from the call call holder
    will be made 90 days from the exercise date. The
    call holders profit will be
  • 25MLIBOR 1190/360.
  • If, the 90 day LIBOR at expiration were 13.45,
    for instance, the Call holders profit would be
  • 25M.1345-.1190/360 153,125.

5
Hedging a Planned Loan with an Interest Rate Call
  • Example 2
  • A firm plans to borrow 20 million in 30 days at
    the 90-day LIBOR100 bps. The loan is taken out
    in 30 days and will mature 90 days later. The
    loan is paid back in one lump sum. The firm
    faces the risk of increasing LIBOR during the
    next 30 days and would like to establish a
    maximum rate it will pay on the loan. Thus, the
    firm buys an interest rate call based on 90-day
    LIBOR with an exercise rate at the current
    90-days LIBOR X 10. The payoff is based on
    90 days and a 360-day year. The firm pays now the
    call premium 50,000.
  • The table below describes different possible
    results of the loan rate hedged by the call.

6
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7
  • The results in the table are obtained as follows
  • First, we put the call premium and the loan on
    the same time footings.
  • The premium of 50,000 today,
  • will be valued
  • 50,0001.11(30/360)
  • 50,458
  • in 30 days. Thus, in 30 days, the firm
    effectively receives
  • 20,000,000-50,458
  • 19,949,542.

8
  • Next, consider two outcomes, one of
  • which leaves the call out-of-the-money
  • and one of which leaves the call in-the-
  • money.
  • The Call is out-of-the-money.
  • LIBOR at Expiration is 6 . Interest on
  • the loan
  • 20,000,000.07(90/360) 350.000.
  • Total effective interest 350,000
  • Amount borrowed 19,949,542
  • Amount paid back 20,350,000
  • Effective annual rate

9
  • The call is in-the-money
  • LIBOR at Expiration is 14.
  •  Interest on the loan
  •   20,000,000.15(90/360) 750,000.
  • Call is in-the-money, exercised, and pays
  • 20,000,000(.14-.10)(90/360)
  • 200.000.
  • Total effective interest
  • 750,000-200,000550,000.
  • Amount borrowed 19,949,542
  • Amount paid back 20,550.000
  • Effective annual rate

10
  • The next figure illustrates the cost of
  • the planned fixed-rate loan with and
  • without the interest rate call.
  • The fixed-rate loan plus the call
  • creates a maximum cost of 12.78,
  • which is reached if LIBOR ends up at
  • 10 or above.
  • Note that this payoff graph looks
  • similar to a covered call or short put.
  • The difference is that previously, the
  • loss/profit profile was based on the
  • underlying asset price. Here, the
  • payoff is based on an interest rate.

11
The Cost of Planned Loan with and without the
Interest Rate Call
Annualized Cost of Loan (Percent)
Loan
Loan plus the Interest Rate Call
LIBOR at Expiration (Percent)
12
  • AN INTEREST RATE PUT
  • Next we illustrates an interest rate put. A
  • very common use of an interest rate put
  • is by a bank that lends at LIBOR plus
  • possibly a spread. It thus, faces the risk
  • of a decline in LIBOR before the loan is
  • given out.
  • In general the payoff from an interest
  • rate put is
  • (principal)MaxO,X-LIBORn/360,
  • Where
  • X the exercise rate and
  • n the number of days from the
    options expiration to payment day.

13
  • Example 3
  • A bank will lend 10 million in 90 days at
    180-day LIBOR 150 bps. The loan will mature
    180 days from the day the loan is given out and
    is paid back in one lump sum.
  • The bank buys an interest rate put for 26,500
    with a strike rate of 9, thereby putting a
    floor to the rate it will receive. The put will
    pay off to the bank in 90 days from now, the
    180-day LIBOR is below the strike rate of 9.
    The payoff is based on 180 days and a 360-day
    year. The exercise price X is the current LIBOR
    is 9.
  • The following table describe some possible
    results of the loan with the protective interest
    rate put.

14
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15
  • Here, the bank plans to lend in 90 days at
    LIBOR150 bps. The amount of the loan is 10
    million, and the loan will be for 180 days and be
    paid back in one lump sum. The payoff is based on
    180 days and a 360-day year. The put premium is
    26,500. Its payoff is
  • (10M)Max(0,.09-LIBOR)180/360
  • First, we compound the premium forward for 90
    days at today's LIBOR 150bps
  •  26,500l .105(90/360)
  • 27,196.
  • The effective proceeds paid by the bank when the
    loan is taken out are
  • 10,000,000 27,196
  • 10,027,196.

16
  • The put is out-of-the-money
  • LIBOR at Expiration is 12
  • Interest on the loan 
  • 10,000,000.135(180/360)
  • 675,000.
  • Again, the put premium value in 90 days
  • 26,5001 .105(90/360)
  • 27,196
  • Total effective interest 675,000
  • Amount paid out on loan 10,027,196
  • Amount repaid on loan 10,675,000
  • Effective annual rate
  •  

17
  • The put is in-the-money
  • LIBOR at Expiration is 7
  • Interest on the loan
  • 10,000,000.085(180/360)
  • 425,000.
  • Put is exercised, and the bank receives
  •   10,000,000(.09 - .07)(180/360)
  • 100,000.
  • Total effective interest received
  • 425,000 100,000 525,000.
  • Amount paid out on loan 10,027,196
  • Amount repaid on loan 10,525,000
  • Effective annual rate

18
  • The next figure illustrates the return on the
    loan with and without the interest rate put.
  • Notice that the payoff looks like a call or
    protective put.
  • In previous examples, these were
  • bullish strategies that paid off if an
  • asset price rose. Here they pay off if
  • an LIBOR increases. The PUT
  • provides a minimum return of 10.32,
  • which is reached if LIBOR is at 9 percent
  • or below.

19
Return on Planned Loan with and without interest
Rate Put
Annualized Cost of Loan (Percent)
Loan
Loan Interest Rate Put
LIBOR at Expiration (Percent)
20
  • Pricing Interest Rate Options
  • The Black model requires the forward price of the
    underlying asset, the exercise price, the
    risk-free rate, the time to expiration, and the
    volatility of the forward price. Here, we use
    the forward rate for the forward price, the
    strike rate for the exercise price, and the
    volatility of the forward rate for the volatility
    of the forward price. The risk-free rate and the
    time to expiration are the same variables as
    before. Because of the delay between the option
    expiration and the day the payoff is actually
    made, the computed price must be discounted using
    the forward rate.

21
  • Consider example 2. The interest rate call we
    examined expires in 30 days and has a strike rate
    of 10.
  • Suppose that the 30-day continuously compounded
    risk-free rate is 8 and the volatility - the
    standard deviation - of the forward LIBOR- is s
    0.2. Let the 30-day forward LIBOR for 90-day
    deposits be 11.01 percent. The time to
    expiration is 30/365, or T .0822. Let c be the
    premium obtained from the Black model, then the
    total cost of the option is
  • c(principal)(n/360)
  • n the number of days in the underlying LIBOR
    instrument

22
Calculating the Black Price for an Interest Rate
Call
  • Lf0 .1101X.10r.08s.2T.0822
  • d2 1.71 -.2 ?.0822 1.65.
  • N(1.71) .9564.
  • N(1.65) .9505.
  • c e(.08)(.0822).1101(.9564) -
    .10(.9505)e -.1101 (90/365)
  • .0099.
  • (20,000,000)(90/360)(.0099)
  • 49,500.
  • Notice that this amount is very close to the
    50,000 charged by the dealer.

23
  • A similar approach is used for an interest rate
    put.
  • By employing the put-call parity for options on
    futures,
  • P C - (F0 - X)e rT
  • one easily obtains the put price once the call
    price is obtained.
  • Alternatively, a direct computation of the
    Black's model for put options is
  • P e rTXN(-d2) -Lf0N(-dl).

24
Interest Rate Caps, Floors, and Collars
  • An interest rate cap is a series of European
    interest rate calls that pay off at dates
    corresponding to the interest payment dates on a
    loan. Each option is a separate interest rate
    call. These individual component options are
    called caplets.
  • At each interest payment date of a cap, the
    holder of the cap decides whether to exercise the
    call based on whether the market LIBOR has risen
    above the exercise rate.
  • A price is paid up front for the cap. The price
    corresponds to the sum of the prices of the
    series of calls that make up the cap.

25
  • A cap example On January 2, a Firm borrows 25
    million over one year. It will make payments on
    April 2, July 2, October 2, and next January 2.
    On each date, starting with January 2, LIBOR in
    effect on that day will be the interest- rate
    paid over the next three months. The current
    LIBOR is 10. The firm wishes to fix its loan
    rate at or below 10, so it buys a cap for an
    up-front cap premium of 70,000. The payoffs are
    based on the exact number of days and a 360-day
    year. At each interest payment date, the cap
    will be worth
  • 25M(n/360)Max0, LIBOR - .10.
  • In the formula, LIBOR is the rate that was in
    effect at the beginning of each quarter.

26
  • For the first quarter, the firm will pay
  • LIBOR of 10 percent in effect on
  • January 2. Thus, on April 2 it will owe
  • 25,000,000(.10)(91/360) 631,944,
  • based on 91 days from January 2 to
  • April 2 . Then, on April 2, LIBOR is
  • 10.68. This is greater than 10 thus,
  • the cap will pay off at the next interest
  • payment date and the holder of the cap
  • will receive a payment of
  • 25M(91/360)(.1068 - .10)
  • 42,972.

27
  • This will help offset the interest of 674,917,
    based on a rate of 10.68 for 91 days from April
    2 to July 2. On July 2, LIBOR is 12.31, so the
    cap will pay off on October 2. The net effect of
    these cash flows is seen in the table below. On
    January 2, the firm received 25 million from the
    lender but paid out 70,000 for the cap for a net
    cash inflow of 24,930,000. It made periodic
    payments as shown and on the next January 2, made
    the final interest payment less the cap payoff
    and repaid the principal. Notice that because of
    the cap, the interest payments differ only
    because of the different number of days in each
    quarter and not because of the rate. The
    interest rate is capped at LIBOR of 10 percent.

28
  • If we wish to know what annualized rate the firm
    actually paid, we essentially must solve for the
    internal rate of return, which requires a
    computer or financial calculator. We are solving
    for the cash flow that equates the present value
    of the four payouts to the initial receipt

The solution is y .026. Annualizing y gives a
rate of (1.026) 4 1 .108 or, 10.8.
29
  • Solving for the internal rate of return for the
    cash flows of the un capped loan gives an annual
    rate of .117 or 11.7. Thus, the cap saved the
    firm 90 basis points, because during the life of
    the loan, interest rates generally were higher
    than they were at the time the loan was
    initiated. Of course, if rates had fallen, the
    firm would have benefited less because the
    caplets would have been out-of-the-money and
    would not have been exercised, but the premium
    was paid up front.
  • The next two tables summarize an hypothetical
    outcome, assuming different LIBOR rates for the
    duration of the loan and the cap.

30
Summary of the interest Rate Cap.
  • Scenario On January 2, a firm takes out a
    25M, one-year loan with interest paid quarterly
    at LIBOR. The firm buys an interest rate cap with
    a strike of 10 percent for a premium of 70,000.
    The payoffs are based on the exact number of days
    and a 360-day year.
  • n the number of days in the quarter.
  • I Interest due.

31
  • The next table summarizes The cash flow with and
    without the cap
  • Effective annual rate paid on the loan
  • With cap 10.8 Without cap 11.7.
  • Note This is but one of infinite number of
    possible outcomes to the cap. It is used only in
    order to illustrate how the payments are
    determined and not their likely amounts.

32
Interest Rate Floors
  • The lender in a floating-rate loan may want
    protection against falling rates. This type of
    protection can be purchased with an
  • interest rate floor,
  • which is a series of interest rate put options
    expiring at the interest payment dates. Each
    component put is called a floorlet.

33
  • At each interest payment date, the
  • payoff of an interest rate floor tied to
  • LIBOR with an exercise rate of, say,
  • X, payoffs based on the exact
  • number of days, n and a 360-day
  • Year and a notional principal N will be
  •  
  • (N)Max(0,X - LIBOR)(n/360).
  • As previously, LIBOR is determined at
  • the beginning of the interest payment
  • period.

34
  • An Interest Rate Floor
  • An Example
  • Suppose that on December 16 a bank makes a
    one-year, 15 million loan with payments made at
    3-month LIBOR on March 16, June 16, September 15,
    and next December 16. Currently, it is December
    16, and the 3-month LIBOR is 7.93. Thus, on
    March 16 the bank will receive
  • 15M.0793(90/360)) 297,375
  • in interest.
  • Assume that the new rate on March 16 is 7.50
    percent. Thus, the floor is in-the-money and
    will pay off 
  • 15M(.08 - .075)(92/360)
  • 19,167
  • on the next interest payment day.

35
  • This pay off from the floor will add to
  • the interest payment of 287,500,
  • which is lower because of the fall in
  • interest rates.
  • The complete results for the one-year
  • loan are shown in the table below.
  • The floor is in-the-money and thus is
  • exercised on each of the last three
  • interest payment dates. This is
  • because in this example, the interest
  • rates were lower than 8 during the
  • entire year.

36
  • The lender paid out 15,000,000 up front to the
    borrower and another 30,000 for the floor.
  • Following the same procedure as in the cap, we
    can solve for the periodic rate that equates the
    present value of the inflows to the outflow.
    This rate turns out to be about 1.97 percent.
    Annualizing this gives a rate of (1.0197) 4 - 1
    .081 or 8.1. The cash flows without the floor
    yield an annualizes return associated with these
    cash flows is 7.4. Thus, the floor boosted the
    banks return by 70 basis points. Of course, in
    a period of rising rates, the bank will gain less
    from the increase in interest rates.

37
Summary of the interest Rate floor.
  • Scenario On December 16, a bank makes a 25M,
    one-year loan with interest paid quarterly at
    LIBOR. The bank buys an interest rate floor with
    a strike of 8 for a premium of 30,000. The
    payoffs are based on the exact number of days and
    a 360-day year.
  • n the number of days in the quarter.
  • I Interest due.

38
  • The next table summarizes The cash flow with and
    without the floor
  • Effective annual rate received on the loan
    With floor 8.1 Without floor 7.4.
  • Note This is but one of infinite number of
    possible outcomes to the floor. It is used only
    in order to illustrate how the payments are
    determined and not their likely amounts.

39
Interest Rate Collars
  • Consider a firm planning to borrow money that
    decides to purchase an interest rate cap. In so
    doing, the firm is trying to place a ceiling on
    the rate it will pay on its loan if rates
    increase. If rates fall, the firm can gain by
    paying lower rates. In some cases, however, a
    firm will find it more advantageous to give up
    the right to gain from falling rates in order to
    lower the cost of the cap. One way to do this is
    to sell a floor. That is, the firm sells the
    floor in order to finance the cap.
  • The combination of a long interest rate cap and
    short interest rate floor is called an interest
    rate collar. The premium received from selling
    the floor helps finance part or all of the
    purchased cap.

40
  • It is even possible to structure the exercise
    rates on the collar so that the premium received
    from the sale of the floor exactly equals the
    premium paid for the purchase of the cap. This
    is called a zero cost collar. If interest rates
    fall, the options that comprise the floor will be
    exercised.
  • The net effect of the collar is that the strategy
    will establish both a floor and a ceiling on the
    interest cost.
  • The existence of both limited gains and losses
    should remind you of a money spread with options.

41
Interest Rate Zero Cost Collars
  • Example a zero cost collar.
  • Consider a firm borrowing 50M over
  • Two years buys a cap for 250,000
  • with an exercise price of 10 and sells
  • a floor for 250,000 with an exercise
  • price of 8.5.
  • The loan begins on March 15 and will
  • require payments at approximately 91-
  • day intervals at LIBOR.
  • Remember that the principal amount
  • of 50M is paid on the final date
  • March 14, two years hence.

42
Summary of the interest Rate Collar.
  • n the number of days in the quarter.
  • I Interest due.

43
Summary of the interest Rate Collar.
  • n the number of days in the quarter.
  • I Interest due.

44
  • By now, you should be able to verify the numbers
    in the tables. The interest paid on June 15 is
    based on LIBOR on March 15 of 10.5 and 92 days
    during the period. The cap pays off on September
    14 and December 14 because those are the ends of
    the periods in which LIBOR at the beginning of
    the period turned out to be greater than 10.
    The floor pays off on September 14 and December
    15 of the next year and on March 14 of the
    following year, the due date on the loan. Note
    that when the floor pays off, the firm makes,
    rather than receives, the payment.

45
  • Following the procedure previously described to
    solve for the internal rate of return we can
    solve for the borrowing rate
  • With the collar 9.82
  • With the cap only 9.91
  • Without unhedged 10.08.
  • It is seen that the cap by itself would have
    helped lower the firm's cost of borrowing from
    10.08 to 9.91.
  • By selling the floor and thus creating a collar,
    the cost of the loan was lowered even more to
    9.82.
  • This completed the collar example.

46
Interest Rate Options and Swaps
  • Now that we have introduced caps and
  • floors as options on interest rates, it is
  • helpful to see how they are related to
  • swaps.
  • Let us consider a simple,
  • One - Payment swap,
  • which is basically an FRA.
  • Suppose you pay, the floating rate,
  • LIBOR, and receive the swap rate. The
  • payoff will be
  • Swap Rate - LIBOR.
  • Now suppose in addition you buy a
  • cap and sell a floor, both with a
  • strike rate of X.

47
  • The cap payoff is
  • 0 for LIBOR ? X.
  • LIBOR - X for LIBOR gt X.
  • The floor payoff is
  •  
  • - (X - LIBOR) for LIBOR ? X.
  • 0 for LIBOR gt X.
  •  
  • Thus, the combined
  • swap, cap and floor payoff is
  •  
  • Swap Rate - X for LIBOR ? X.
  • Swap Rate - X for LIBOR gt X. 

48
  • The swap rate is set in the market. It usually is
    determined by the term structure of interest
    rates. The strike rate X, however, is chosen by
    the investor. If one chooses X to be equal to the
    market-determined swap rate, the transaction
    becomes not only risk-free but also guarantees a
    payoff of zero. In that case, the transaction
    must have zero cash flow up front. Since the
    swap has no initial cash flow, then the long cap
    and short floor must have no initial cash flow.
    Thus, in this case, the floor premium must equal
    the cap premium.
  • The implication of this result is that
  • a swap is equivalent to a combination of a long
    cap and short floor where the strikes on the cap
    and floor are equivalent and equal to the swap
    rate.

49
  • The cap-floor-swap parity.
  • A pay-fixed and receive-floating swap
  • a long cap and short floor.
  • or
  • A pay-floating, receive-fixed swap
  • a short cap and long floor.

50
  • A final caveat
  • Remember that the above are for
  • single payment swaps and caps and
  • floors that have just one caplet and
  • floorlet. When the swap has multiple
  • payments, as it usually does, the cap
  • and floor must have corresponding
  • multiple caplets and floorlets. Each
  • caplet and floorlet must have a strike
  • equal to the swap rate. While the sum
  • of the overall premiums for the cap
  • and floor must be equal, it is not the
  • case that the premium for an
  • individual caplet and its corresponding
  • floorlet will be equal.
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