chapter 5: interest rate derivatives: forwards and swaps

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Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

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Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS. 5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS. 5.1.1 Forward Rates by No Arbitrage 5.1.2 The Forward Curve 5.1.3 Extracting the Spot Rate Curve from Forward Rates. 5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS. - PowerPoint PPT Presentation

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Page 1: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

Page 2: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS•5.1.1 Forward Rates by No Arbitrage

•5.1.2 The Forward Curve

•5.1.3 Extracting the Spot Rate Curve from Forward Rates

Page 3: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS• Suppose that a firm would like to commit to a rate for a loan

to occur in the future, can a bank offer such a rate?• Yes, this is the forward rate• If such a rate exists it should also be related to a discount

factor, just as it occurs with normal interest rates• The forward discount factor at time t defines the time value

of money between two future dates, T1 and T2 > T1. Given the discount factors Z(t, T1) and Z(t, T2), the forward discount factor is given by

.

21 2

1

,, ,

,

Z t TF t T T

Z t T

Page 4: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS (cont.)• The forward discount factor has the following properties:▫ 1. F(t, T1, T2) = 1 for T2 = T1

▫ 2. F(t, T1, T2) is decreasing in T2

• The forward rate at time t for a risk free investment from T1 to T2, and with compounding frequency n, is the interest rate determined by the forward discount factor

• The continuously compounded forward rate is obtained for n that grows to infinity:

.

2 1

1 2 1

1 2

1, , 1

, ,n

n T T

f t T T nF t T T

1 21 2

2 1

ln , ,, ,n

F t T Tf t T T

T T

Page 5: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS (cont.)• An example:

▫ Let today be March 1, 2001, suppose a firm sold a piece of equipment to a client for $100 million

▫ The client will pay in six months, i.e. on T1 = September 4, 2001 ▫ Suppose the firm does not need that cash immediately, but it will need it six month

later, at T2 = March 1, 2002,to fund some capital investment ▫ Today, the firm would like to fix the interest rate to be applied on the receivable of

$100 million for the six month period from T1 to T2 ▫ The firm calls up its bank to ask for a quote, and the bank quotes the (semi-annually

compounded) annualized rate of f2 = 4.21%. ▫ That is, the bank is ready to commit today to receive in six months (at T1) $100

million from the firm, and return six months later (at T2) the amount $102.105 = $100 × (1 + f2/2) million

▫ The rate, f2, which the bank commits to today is called forward rate▫ How does the bank determine the forward rate f2? By no arbitrage▫ On March 1, 2001 (today), the value of 6-months Treasury bills is $97.728 and the

value of 1-year Treasury bills is $95.713

Page 6: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1 FORWARD RATES AND FORWARD DISCOUNT FACTORS (cont.)• An example (cont’d):

▫ In order to guarantee the rate f2 to the client, the bank can perform the following strategy: Today the bank:

Borrows T-bills with maturity T1 = 6 months and sells them for $97.728 million This amount of cash is then invested in 1-year T-bills expiring in T2 Given the price of the latter of $95.713 the bank can now purchase M = 1.02105 = $97.728/$95.713

million of 1-year T-bills (for $100 of face value)The net cash flow remaining to the bank is zero, as all the cash obtained from the sale of the T1 T-bill has been used to purchase T2 T-bills

At time T1 the bank: Must pay back $100 million to the counterparty it borrowed the T1 T-bills from However, this is also the time in which the firm will give the bank $100 million

So, also at T1 the net cash flow is zero, as the bank receives $100 million from the firm and uses it to close the short position

At time T2: The M = 1.02105 T-bills mature The bank receives M × $100 = $102.105 million

This is the cash flow promised to the firm in return on the investment of $100 million at T1

Indeed, the return for the firm from T1 to T2 is 2.105% = ($102.105−$100)/$100, which implies the annualized interest rate of f2 = 2.105%×2 = 4.21%.

Page 7: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS
Page 8: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS
Page 9: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1.1 Forward Rates by No Arbitrage

•Consider the following investment strategies:▫Invest $W in T2 zero coupons; payoff at T2 is:

▫Invest $W in T1 zero coupons, and at T1 invest in forward rate f2 to be applied for an investment from T1 to T2 (agreed on today); payoff at T2 is:

2

$$100

,z

Wpayoff

P t T

2 12

21

$$100 1 / 2

,T T

z

Wpayoff f

P t T

Page 10: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1.1 Forward Rates by No Arbitrage (cont.)

▫Both investments should give the same payoff

▫Which leads us to:

2 12

21 2

$ $$100 1 / 2 $100

, ,T T

z z

W Wf

P t T P t T

2 12 12

2

,1 / 2

,T T Z t T

fZ t T

2 1

2 1 2 1

21 2

1, , 2 1

, , T T

f t T TF t T T

Page 11: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1.2 The Forward Curve

•Consider the following continuously compounded rate:

• The forward curve gives the relation between the forward rate f (0,T,T+∆) and the time of the investment T

• Forward curve is related, to some extent, to the market expectations of future nominal interest rates, as well as market participants degree of aversion to risky investments

ln 0, ,0, ,

F T Tf T T

Page 12: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS
Page 13: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1.3 Extracting the Spot Rate Curve from Forward Rates• In previous sections we computed discount factors

from available bond data, yet this might not always be available

•From the previous formulas it can be seen that missing data to compute the spot rate can be completed when the forward rates are known

Page 14: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.1.3 Extracting the Spot Rate Curve from Forward Rates• An example:▫ Let today be May 5, 2008 ▫ The third column in Table 5.2 contains continuously compounded

forward rates at the quarterly frequency (Δ = 0.25), note that the first forward rate corresponds to the spot rate f(0,0,0.25) = r(0,0.25)

▫ The first entry in the third column determines the first discount: Z(0,0.25) = e-f(0,0,0.25)×0.25 = e-2.7605%×0.25 = 0.993123

▫ Next, we compute the discount for the maturity T2 = 0.5:Z(0,0.5) = Z(0,0.25) × e-2.9460%×0.25 = 0.986521

▫ We can proceed for every i = 2, ..., 19 and use the formula:Z(0,Ti) = Z(0,Ti-1) × e-f(0,Ti-1,Ti)×0.25

▫ Column 4 contains the resulting discount factors for every maturity▫ Column 5 uses these discount factors to obtain the continuously

compounded spot rate:r(0,Ti) = - ln(Z(0,Ti)) / Ti

Page 15: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS
Page 16: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS

•5.2.1 The Value of a Forward Rate Agreement

Page 17: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS• A Forward Rate Agreement (FRA) is a contract between

two counterparties, according to which one counterparty agrees to pay the forward rate fn(0,T1,T2) on a given notional amount N during a given future period of time from T1 toT2 = T1 +Δ, while the other counterparty agrees to pay according to future market floating rate rn(T1,T2). The net payment between the two counterparties at the maturity T2 of the contract is then given by:

Net Payment at T2 = N × Δ × [fn(0,T1,T2) – rn(T1,T2)]Above, Δ = T2 − T1, typically a quarter or six months, while the subscript n = 1/Δ denotes the corresponding compounding frequency, e.g., n = 4 or n = 2 for quarterly or semi-annual compounding

Page 18: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS• An example:

▫ Recall from a previous example that today (t = 0) is March 1, 2001, that a firm has a receivable of $100 million in six months (T1 = 0.5), and the firm wishes to invest this amount for an additional six months (until T2 = 1)

▫ An alternative strategy is for the firm to enter into a six-month FRA with a bank for the period T1 to T2, and notional N = $100 million

▫ That is, today the bank agrees to pay in one year (T2 = 1) the amount N × f2(0, 0.5, 1), where f2(0, 0.5, 1) is the current semi-annually compounded forward rate for the period T1 to T2, while the firm agrees to pay on the same day the amount N × r2(0.5, 1), where r2(0.5, 1) is the semi-annually compounded spot interest rate at time T1 = 0.5

▫ That is, they exchange the payment at T2 = 1Net payment at T2 = N × [f2(0,0.5,1) – r2(0.5,1)] / 2

where we must divide by two because the interest is only over six months▫ Recall f2(0, 0.5, 1) = 4.21%

Page 19: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS• An example (cont’d)▫At T1 = 0.5 when the firm receives its $100 million receivable, the

firm can simply invest this amount at the market interest rate r2(0.5, 1)

▫How much money will the firm have at time T2 = 1? At this time, the firm receives the payoff from the investment, plus the net payment:

▫ The firm is in exactly the same position as in the previous example

million 105.102$

2

1,5.0,01million 100$

paymentFRA

1,5.01,5.0,02

Investmenton Return

2

1,5.01million 100$Tat amount Total

2

222

2

f

rfNr

Page 20: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS• An example (cont’d)▫ The bank is now exposed to interest rate risk, as the FRA yields a

negative payoff if f2(0, 0.5, 1) > r2(0.5, 1); however, a little modification to the strategy ensures the bank is hedged as well: At time 0, the bank strategy is the same:

Shorts T1 T-bills Purchases M = 1.02105 T2 T-bills

At time T1: The bank must come up with $100 million to pay the short position The bank can borrow this amount of money, at the current rate r2(0.5,1)

At time T2, the bank total cash flows are as follows:

A perfect hedge

0

paymentFRA

1,5.0,01,5.02

m 100$

mature billsT

m 100$02105.1

loanback Pay

2

1,5.01m 100$Tat Total 22

2

22

fr

r

Page 21: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2 FORWARD RATE AGREEMENTS

Page 22: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2.1 The Value of a Forward Rate Agreement•The value of a forward agreement is zero at inception•The value at time t of a Forward Rate Agreement

(FRA) is given by:VFRA(t) = N × [M × Z(t,T2) – Z(t,T1)]

where: M = 1 + fn(0,T1,T2) Δ = Z(0,T1) / Z(0,T2)•Equivalently, the value of a FRA can be expressed as

follows:VFRA(t)= N × Z(t,T2) × Δ × [fn(0,T1,T2) – fn(t,T1,T2)]

recall that given the definition of a forward rate we have: M = 1 + fn(t,T1,T2) Δ = Z(t,T1) / Z(t,T2)

Page 23: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2.1 The Value of a Forward Rate Agreement• An example:

▫ Consider the previous example and suppose that three months after the initiation of the contract (June 1, 2001) the firm decides to close its FRA with the bank

▫ So far, the two counterparties have not exchanged any money, so it may appear that the firm could simply call the bank and ask to close the contract, but as interest rates changed between March 1 and June 1, so did the value of the FRA

▫ We note that at initiation (today = 0) the bank sold one T-bill maturing at T1 and bought M = 1.02105 T-bills maturing at T2; this portfolio (long T2 T-bills and short T1 T-bills) exactly hedges the bank commitment to the FRA

▫ Since it produces exactly the cash flow that the firm will receive, the value of this portfolio must reflect the value of the FRA for the firm

▫ Thus, for every t ≤ T1, we have:VFRA(t) = M × Pbill(t,T2) – Pbill(t,T1)

recall M = Pbill(0,T1) / Pbill(0,T2)▫ Thus at initiation we have:

VFRA(0) = 0showing that at initiation there is no exchange of money

Page 24: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2.1 The Value of a Forward Rate Agreement•An example (cont’d):▫On June 1, 2001 (= t), the T1 T-bill price was Pbill(t,T1) =

$99.10 and the T2 T-bill price was Pbill(t,T2) = $97.37; therefore:

VFRA(t) = 1.02105 × Pbill(t,T2) – Pbill(t,T1) = 1.02105 × $97.37 – $99.1 = $0.319638

▫On June 1, 2001, the value of the forward rate agreement initiated 3 months earlier was worth $319,638 to the firm

▫Simply calling up the bank and exiting from the FRA would be a mistake, as the firm would lose this amount of money

▫Of course, knowing this, the bank would pay this sum of money to the firm in case it wanted to exit from the contract, minus some transaction costs

Page 25: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.2.1 The Value of a Forward Rate Agreement• Another example:

▫ Instead of calling up the counterparty and asking to close the contract, the firm can achieve the same result by entering into a new FRA at time t with the reversed payoff:

Payoff (reverse) FRA at T2 = N × [r2(T1,T2) – f2(t,T1,T2)] / 2▫ Clearly, the payoff of the FRA at time t depends on the current forward rate f2(t,T1,T2)

instead of the old one f(0,T1,T2) , so:

Total payoff of both (old and new) FRA at T2 = N × [f2(0,T1,T2) – r2(T1,T2)] / 2 + N × [r2(T1,T2) – f2(t,T1,T2)] / 2

= N × [f2(0,T1,T2) – f2(t,T1,T2)] / 2▫ That is, at the time of the decision to close the original FRA contract, the firm will end up

with a positive payoff if the current forward rate declined since inception, and vice versa ▫ Since Pbill(t,T1) = $99.10 and Pbill(t,T2) = $97.37, the current forward is:

f2(t,T1,T2) = 2×(.991/.9737− 1) = 3.55% < f2(0,T1,T2) = 4.21%▫ The time T2 payoff is then N × [f2(0,T1,T2) − f2(t,T1,T2)] / 2= $328, 272, which is known at

the earlier time t ▫ Therefore, the present value is:

VFRA = Z(t,T2) × [f2(0,T1,T2) − f2(t,T1,T2)] / 2= 0.9737 × $328,272 = $319,638

Page 26: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3 FORWARD CONTRACTS

•5.3.1 A No Arbitrage Argument

•5.3.2 Forward Contracts on Treasury Bonds

•5.3.3 The Value of Forward Contracts

Page 27: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3 FORWARD CONTRACTS

•A forward contract is a contract between two counterparties in which one counterparty agrees to purchase and the other agrees to sell a given security at a given future time, and at a given price, called the forward price. Denoting by Pfwd(0,T,T*) the forward price at time 0 for delivery at time T of a security expiring at T*, whose value at T we denote by P(T,T*), the payoff at T of being long the forward contract is

Payoff = P(T,T*) – Pfwd(0,T,T*) The payoff from being short the forward contract is the negative of the equation

Page 28: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3 FORWARD CONTRACTS

• An example:▫ On March 1, 2001 (today = time 0), the firm may enter into a forward contract with a bank to

purchase six months later (T1 = 0.5) $100 million-worth of 6-months Treasury bills for a price Pfwd, for $100 par value, specified today

▫ What purchase price would the bank quote to the firm for the 6-month T-bills? ▫ Pfwd = 100 × F(0,0.5,1) = $97.938▫ Recall that from the (short) sale of T1 T-bills, the bank purchases M = 1.02105 million of T2 T-

bills ▫ Recall also that the net cash flow to the bank at time 0 is zero▫ At time T1 the bank has to cover the short position, and uses the $100 million from the firm▫ Note that at this point the bank holds an amount M = 1.02105 million of T2 T-bills, which now

have maturity T2 − T1 = six months ▫ Therefore, the bank can use these M 6-month T-bills to honor the terms of the forward contract▫ That is, at T1 the bank simply delivers its own holdings M of 6-month T-bills to the firm ▫ This number M of T2 T-bills is exactly the number of 6-month T-bills that are needed to ensure

the firm gets $100 million-worth of 6-month T-bills, as the firm requested; in fact, given Pfwd, we have:

$100 million / Pfwd = 1.02105 million = M

Page 29: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3 FORWARD CONTRACTS

• Another example:▫ Let Pfwd(0,0.5,1) = 100 × F(0,0.5,1) = $97.938 be the forward price quoted

at time 0 for the investment between T1 = 0.5 to T2 = 1, as determined in the previous example

▫ Let the firm enter into a forward contract to purchase M = 1.02105 million of 6-month T-bills on T1 = 0.5 (with $100 principal)

▫ The payoff from the forward contract at time T1 = 0.5 is then given byPayoff forward contract at T1 = M × (Pbill(T1,T2) – $97.938)

▫ Ex post, the fear of the firm that the interest rate would decline is in fact realized, and the price of the 6-month T-bill at T1 turns out to be Pbill(T1,T2) = $98.89 > $97.938

▫ Thus, the payoff from the forward contract is:

▫ The firm at T1 can then invest this additional amount, $972,043.54, together with the receivable $100, 000, 000 into the new T-bill

136.054,021,189.98$

54.043,972$000,000,100$at bills-Tin Investment 1

T

Page 30: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3 FORWARD CONTRACTS

• Another example (cont’d):▫ In particular, it will be purchasing an amount of T-bills equal to

▫ This is exactly equal to the forward rate determined in a previous example and is in fact independent of the realized Treasury bill T1 = 0.5, Pbill(T1,T2)

▫ The higher the value of the T-bill at maturity, the higher is the payoff from the forward contract ; but more it becomes expensive to purchase T-bills for an investment between T1 and T2

▫ These two effects exactly cancel each other out, and the firm is guaranteed the forward rate

%21.41000,000,100

136.054,021,12

1at Investment

at Payoff1return of rate Annualized

1

2

12

T

T

TT

Page 31: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3.1 A No Arbitrage Argument•Consider a forward contract in which one counterparty

agrees to buy at a future date T a zero coupon bond Pz(T,T*). The forward price is given by the forward discount factor (multiplied by the notional)

Pzfwd(0,T,T*) = F(0,T,T*) × 100

• It is useful to look at the arbitrage argument that makes this true

•Assume Pzfwd(0,T,T*) > F(0,T,T*) × 100

•Recalling that:.

0, * 0, *0, , *

0, 0,z

z

Z T P TF T T

Z T P T

Page 32: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3.1 A No Arbitrage Argument (cont.)• At time 0 an arbitrageur can (with net cash flow = 0)

1. Sell the forward zero coupon Pz(T,T*) at forward price Pz

fwd(0,T,T*)2. Short exactly F(0,T,T*) zero coupon bonds with maturity T for

the amount F(0,T,T*) × Pz(0,T) = Pz(0,T*), the price of a zero coupon with maturity T*

3. Use the proceeds from 2 to purchase one zero coupon with maturity T*

• At time T the arbitrageur:1. Delivers the zero coupon Pz(T,T*), which he or she owns, and

receives Pzfwd(0,T,T*), thereby closing the forward contract

2. Covers the short position paying F(0,T,T*) × 100• The net cash position is positive, with no uncertainty or risk in

this strategy, thus it is an arbitrage opportunity.

Page 33: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3.2 Forward Contracts on Treasury Bonds•Consider a forward contract in which one counter-party

agrees to purchase at a future date T a Treasury bond with coupon rate c and with maturity T * > T. Let T1, T2,..., Tn = T* be the coupon dates after T, the maturity of the forward contract. Then, the forward price is given by

.

1 2

1

100 1000, , * 0, , 0, ,

2 2100

... 100 0, ,2

0, , 0, ,2

fwdc

n

nfwd fwd

z i z ni

c cP T T F T T F T T

cF T T

cP T T P T T

Page 34: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.3.3 The Value of Forward Contracts

•Consider a forward contract in which one counterparty agrees to purchase at a future date T a Treasury bond with coupon rate c and with maturity T * > T. Let T1, T2,..., Tn = T* be the coupon dates after T. The value of the forward contract for every t < T is

Vfwd(t) = Z(t,T) × [Pcfwd(t,T,T*) – K]

where K = Pcfwd(0, T, T*) is the delivery price agreed

upon at initiation

Page 35: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

•5.4.1 The Value of a Swap•5.4.2 The Swap Rate•5.4.3 The Swap Curve•5.4.4 The LIBOR Yield Curve and the Swap

Spread•5.4.5 The Forward Swap Contract and the

Forward Swap Rate•5.4.6 Payment Frequency and Day Count

Conventions

Page 36: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

•A plain vanilla fixed-for-floating interest rate swap contract is an agreement between two counterparties in which one counterparty agrees to make n fixed payments per year at an (annualized) rate c on a notional N up to a maturity date T, while at the same time the other counterparty commits to make payments linked to a floating rate index rn(t). Denote by T1, T2,..., Tn = T the payment dates, with Ti = Ti−1 + Δ and Δ = 1/n, the net payment between the two counterparties at each of these dates is

Net payment at Ti = N × ∆ × [rn(Ti-1) – c]• The constant c is called swap rate

Page 37: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

• An example:▫ A firm and bank decide to enter into a fixed for floating, semi-annual, 5-year swap with

swap rate c = 5.46% and notional amount N = $200 million; the reference floating rate is the 6-month LIBOR

▫ In this swap contract, the firm agrees to pay to the bank every six months (Ti = 0.5,1,1.5,...,5):

Cash flow from firm to bank at Ti = $200 m × 0.5 × 5.46% = $4.56 million▫ In exchange, the bank pays the firm at every Ti an amount that depends on the 6-month

LIBOR r2(Ti−1) ▫ Note that the reference rate for the payment at time Ti is not the LIBOR at Ti, but the one

determined six months before, at Ti−1

Cash flow from bank to firm at Ti = $200 m × 0.5 × r2(Ti-1)%▫ Table 5.4 illustrates the cash flows from the bank to the firm and vice versa ▫ The noteworthy point is that the cash flows from the bank to the firm in Column 3 vary

over time, and in particular they have a six months lag from the time the LIBOR, in Column 2, is realized

▫ In this particular instance, the firm would receive a negative net cash flow, as the reference floating rate declined from 4.951% at initiation to a much lower number

Page 38: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

Page 39: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

• Another example:▫ Today is t = 0 = March 1, 2001, consider a firm that sold a piece of equipment

to a highly rated corporation, and it is then due to receive payments in 10 equal installments of $5.5 million each over 5 years

▫ The firm would like to use these $5.5 million semi-annual cash flows to hedge against the coupon payments the firm must make to service a $200 million floating rate bond that it issued in the past, and also expiring in 5 years

▫ Suppose that the floating rate on the corporate bond is tied to the LIBOR, at LIBOR + 4 bps

▫ The 6-month LIBOR on March 1, 2001 is currently at 4.95% and so the next interest rate payment the firm must make is (4.95 + 0.04)% / 2×200 million = $4.9 million; thus, the next floating rate coupon payment is covered

▫ However, if the LIBOR were to increase by more than 0.51% in the next 5 years, the cash flows from the installments would not be sufficient to service the debt

▫ A solution is to enter into a fixed-for-floating swap with an investment bank, in which the firm pays the fixed semi-annual swap rate c, over a notional of $200 million, and the bank pays the 6-month LIBOR to the firm

Page 40: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4 INTEREST RATE SWAPS

• Another example (cont’d):▫On March 1, 2001, the swap rate for a 5-year fixed-for-floating

swap was quoted at c = 5.46% ▫ So, in this case, the net cash flow to the firm from the swap

contract isNet cash flow at Ti = $200 million × [r2(Ti-1) – 5.46%] / 2

where r2(t) is the six month LIBOR at time t▫Why does this swap resolve the problem? Consider the net

position of the firm: at every Ti the firm: receives 5.5 million pays (r2(Ti − 0.5) + 4bps) / 2 × 200 million on its outstanding floating

rate debt receives r2(Ti − 0.5)/2 × 200 million from the bank as part of the swap pays 5.46%× 0.5 × 200 million to the bank as part of the swap

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5.4 INTEREST RATE SWAPS

• Another example (cont’d):▫ Summing up, the firm’s net cash flow position from the

receivable, debt, and swap isTotal cash flow at Ti = 5.5 million (Receivable)

– (r2(Ti – 0.5) + 4 bps) / 2 × 200 million (Debt)+ 0.5 × [r2(Ti – 0.5) – 5.46%] × 200 million (Swap)= 5.5 – 0.04% × 100 – 5.46% × 100= 0

▫ That is, the firm is perfectly hedged: The risk in the fluctuations of the LIBOR stemming from its liabilities has been eliminated by the swap (the firm receives the LIBOR from the bank, and pays the LIBOR + 0.04% to bond holders)

▫ The remaining fixed components sum up to zero

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5.4.1 The Value of a Swap

•How do we value a swap?• The sequence of net cash flows is the same as the one of a

portfolio that is long a floating rate bond, and short a fixed rate bond with coupon c

Vswap(t;c,T) = PFR(t,T) – Pc(t,T)where Vswap(t;c,T) is the value of the swap at time t, with swap rate c and maturity TM; PFR(t,T) is the value of the floating rate bond and Pc(t,T) is the value of the fixed coupon bond with coupon rate c

•At payment dates Ti, we have PFR(t,T) = 100; so:

.

1

, , 100 100 , 100 ,2

Mswap

i M i j i Mj i

cV T c T Z T T Z T T

Page 44: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.1 The Value of a Swap

•An example:▫ Recall the previous example: the discount factors Z(0,T) on

March 1, 2001 are in the second column of Table 5.5, so we can obtain:

▫ The value of the swap in the previous example is (almost) zero, reflecting the fact that it does not cost anything to enter a swap contract at initiation

.

M

jMj

swap TZTZn

cTcV

1

100,0,0100100,;0

Page 45: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.1 The Value of a Swap

Page 46: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.2 The Swap Rate

•The swap rate c is given by that number that makes Vswap(0;c,T) equal to zero. Rewriting the equation generically for any payment frequency n and payment dates T1,...,TM, we have

•Solving this equation for c we find

1

0, , 100 100 0, 100 0,M

swapM j M

j

cV c T Z T Z T

n

1

1 0,

0,

MM

jj

Z Tc n

Z T

Page 47: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.2 The Swap Rate

•An example:▫In the previous example, given the discount factors in

the second column of Table 5.5, the swap rate 5.46% makes the swap value equal to zero

▫Thus, c = 5.46% is exactly the proper swap rate, as it would be quoted by the bank

Page 48: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.3 The Swap Curve

• Over the years, the swap market grew so much that market forces determine the swap rate for every possible future maturity

• The swap curve at time t is the set of swap rates (at time t) for all maturities T1, T2,...,TM. We denote the swap curve at time t by c(t,Ti) for i = 1,...,M

• Given the swap curve we can determine the implicit discount functions through the bootstrap methodology

• For i = 1

• while for i = 2,…,M

.

1,

1

,

1 ,,

1

i

i

ic t Tjn j

i c t Tn

Z t TZ t T

1 ,

1,

1 ic t Tn

Z t T

Page 49: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.3 The Swap Curve

• An example:▫ The last column of Table 5.5 contains the swap curve data as of March 1,

2001 ▫ What is the zero curve that is implicit in the swap curve? ▫ For i = 1:

▫ For i = 2:

▫ For i = 3:

▫ and so on… The result is the set of discount factors Z(0, T) in the second column of Table 5.5

9758.01

15.0,0

204951.0

Z

9527.002455.01

9758.002455.01

1

)5.0,0(11,0

20491.0

20491.0

Z

Z

9289.0

0249.01

9527.09758.00249.01

1

)1.0,0()5.0,0(15.1,0

20498.0

20498.0

ZZ

Z

Page 50: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.4.4 The LIBOR Yield Curve and the Swap Spread

•What is the relationship between the LIBOR Yield curve (computed in the previous slide) and the one obtained from Treasuries

•The difference between two zero coupon yields with same maturities, from different yield curves is called the swap spread, and it is generally assumed that▫It is unlikely to become negative (probability of default

by swap dealers is higher than the government’s)▫It is unlikely to move to infinity

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5.4.5 The Forward Swap Contract and the Forward Swap Rate•The forward swap contract is a contract in which

two counterparties agree to enter into a swap contract at a predetermined future date and for a predeter-mined swap rate f s, called the forward swap rate

•The forward swap rate of a forward swap contract to enter into a swap at time T, with maturity T*, payment frequency n, and payment dates T1,T2,...,Tm (with Tm = T*) is given by

1

1 0, , *0, , *

0, ,

sn m

jj

F T Tf T T n

F T T

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5.4.5 The Forward Swap Contract and the Forward Swap Rate• An example:

▫ Consider again the previous example, but assume now that on March 1, 2001, the firm signed a contract to deliver one year later (on March 1, 2002) a large piece of equipment

▫ The payment will be made in 8 equal installments of $5.5 million each over the next 4 years, starting on September 1, 2003

▫ Assume the firm plans to use these cash inflows to meet the payments of a floating rate bond issued some time in the past

▫ As explained in the previous example, the firm can enter into a fixed-for-floating swap in which it pays fixed and receives floating

▫ The problem is that the firm will start receiving payments much further in the future, and therefore it will need to enter into such a fixed-for-floating swap one year from now, on March 1, 2002

▫ The firm is worried however that the 4-year swap rate may increase between now and March 1, 2002, an event that may unduly increase its cash outflows from the hedging program

▫ Therefore, the firm decides to enter into a forward contract with a bank, in which the bank and the firm agree today that the 4-year swap rate in the future will be fs

2 = 5.616%, to be paid semi-annually in exchange of the 6-month LIBOR

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5.4.5 The Forward Swap Contract and the Forward Swap Rate•An example (cont’d):▫How did the bank obtain fs

2?▫We can use the LIBOR discount data in Table 5.5, reported

also in the Column 2 of Table 5.6, to compute the forward swap rate to enter on March 1, 2002 (one year from now) into a 4-year swap

▫For every Tj = 1.5, 2, ...,5, we can compute the forward discount factor

F(0,1,Tj) = Z(0,1) / Z(0,Tj)▫This calculation is in Column 3 of Table 5.6 ▫Applying the formula for the forward swap rate, we obtain:

fs2(0,1,5) = 5.616%

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5.4.5 The Forward Swap Contract and the Forward Swap Rate

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5.4.6 Payment Frequency and Day Count Conventions

•Many swap contracts have payment date at different frequency:▫For example a common fixed-for-floating swap:

Is tied to the three month LIBOR and pays at quarterly frequency and uses the actual / 360 day count convention

While the fixed rate counterparty pays at semiannual frequency and uses the 360 / 360 day count convention

•These usually don’t pose a problem for valuation since each leg is priced separately

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5.5 INTEREST RATE RISK MANAGEMENT USING DERIVATIVE SECURITIES•A fixed-for-floating swap doesn’t cost anything to

enter, but can change the duration of a portfolio•The dollar duration of a swap can be considered as

D$swap = D$

floating – D$fixed

•Thus by choosing the adequate amount of notional we can make sure that parallel shifts in term structure won’t have an impact on equity value

D$E = D$

A + N × D$swap – D$

L

Page 59: Chapter 5: INTEREST RATE DERIVATIVES: FORWARDS AND SWAPS

5.5 INTEREST RATE RISK MANAGEMENT USING DERIVATIVE SECURITIES• An example:

▫ Consider a financial institution with a total asset size of around $2.4 billion and assets’ dollar duration of $19.74 billion, and total liabilities at $1.8 billion with a liabilities’ dollar duration of only $5 billion; this leads to a market value of equity is $600 million, but with a dollar duration is $14.740 billion

▫ The implications of this mismatch between the assets and liabilities dollar durations is that a parallel upward shift in interest rates of 1% generates a decline in assets far greater than in liabilities, implying an equity decline of $147.4 million, which in percentage, corresponds to a 24% decline in market value of equity

▫ How can derivatives, and in particular swaps, help stabilize the value of equity?▫ Assume that the term structure of interest rates is flat at 4% (semi-annually compounded)

and that in this case, the current swap rate is also 4% ▫ The dollar duration of a fixed-for-floating swap (receiving fixed) can be found to be $784

for $100 of notional ▫ Clearly, the financial institution has a duration of assets that is higher than the one of its

liabilities, and therefore would like to enter into a swap in which it pays fixed and receives floating, which has the opposite duration of −$784 (per $100 notional)

▫ We can choose the notional N to make the dollar duration of equity D$E = 0

▫ The notional that solves the equation is $1.889 billion.