# LO 11.8: Calculate the value o f a constant maturity Treasury swap, given an

LO 11.8: Calculate the value o f a constant maturity Treasury swap, given an interest rate tree and the risk-neutral probabilities.
In addition to valuing options with binomial interest rate trees, we can also value other derivatives such as swaps. The following example calculates the price of a constant maturity Treasury (CMT) swap. A CM T swap is an agreement to swap a floating rate for a Treasury rate such as the 10-year rate.
Example: CM T swap
Assume that you want to value a constant maturity Treasury (CMT) swap. The swap pays the following every six months until maturity:
/
\$ 1,000,000
2
X (y
V C M T
Tc m t ls a semiannually compounded yield, of a predetermined maturity, at the time of payment (y( is equivalent to 6-month spot rates). Assume there is a 76% risk-neutral probability of an increase in the 6-month spot rate and a 60% risk-neutral probability of an increase in the 1-year spot rate.
Fill in the missing data in the binomial tree, and calculate the value of the swap.
Figure 5: Incomplete Binomial Tree for CM T Swap
Int. rate = 7.25% Swap price = ?
0.76
Int. rate = 6.75% Swap price = ?
Int. rate = 7.50% Swap price = ?
0.60
Int. rate = 7.00% Swap price = ?
Int. rate = 6.50% Swap price = ?
0.40
Today
Six Months
One Year
2018 Kaplan, Inc.
Page 139
Topic 11 Cross Reference to GARP Assigned Reading – Tuckman, Chapter 7
In six months, the top node and bottom node payoffs are, respectively:
payoff^ = ———— x (7.25% 7.00%) = \$1,250
\$ 1,000,000
payoff^ = ————-x (6.75% 7.00%) = \$1,250
\$ 1,000,000
rr
rr
2
2
^
^
Similarly in one year, the top, middle, and bottom payoffs are, respectively:
r r
\$ 1, 000,000
/ r r i r A f t /
payoff2ju = ————-x (7.50% 7.00%) = \$2,500
^
^ payoff2jM = ————-x (7.00% 7.00%) = \$0
/ r r / w w w
r
r
\$ 1,000,000
2
rr
\$ 1,000,000
payoff2 L = ————-x (6.50% 7.00%) = \$2,500
2
The possible prices in six months are given by the expected discounted value of the 1-year payoffs under the risk-neutral probabilities, plus the 6-month payoffs (\$1,250 and
\$1,250). Hence, the 6-month values for the top and bottom node are, respectively:
l,u Vi
V
1>L
(\$2,500×0.6)+ (\$0x0.4)
1+ 0.0725
+ \$1,250 = \$2,697.53
(\$ 0 x 0 .6 )+ (-\$2,500×0.4)
+ 0.0675 1
-\$1,250 = -\$2,217.35
Today the price is \$1,466.63, calculated as follows:
w ^ V Q — —————————ri'”r\~v———————- \$1,466.63
(\$2,697.53 x 0.76) + (\$2,217.35 x 0.24)
.
1+ 0.07
Page 140
2018 Kaplan, Inc.
Topic 11 Cross Reference to GARP Assigned Reading – Tuckman, Chapter 7
Figure 6 shows the binomial tree with all values included.
Figure 6: Completed Binomial Tree for CM T Swap
0.60
Int. rate = 7.50%
Swap price = \$2,500.00
Int. rate = 7.00%
Swap price = \$1,466.63
0.76
0.24
Int. rate = 7.25%
Swap price = \$2,697.53
Int. rate = 6.75%
Swap price = -\$2,217.35
Int. rate = 7.00%
Swap price =
\$0
Int. rate = 6.50%
Swap price = -\$2,500.00
0.40
Today
Six Months
One Year
O p t i o n -Ad j u s t e d S p r e a d