無代寫-CHAPTER 7:

時間：2021-07-12

PRACTICE PROBLEMS

CHAPTER 7: BOND MATHEMATICS

A word about BEY (i.e., Bond Equivalent Yield).

T-bills use semi-annual compounding. So, for n <

365

2

, we use simple interest.

Principle + interest is given by P(1 + y?

n

365

) = 100 at maturity.

Solving for y gives

y ? BEY =

100? P

P

?

365

n

.

For n =

365

2

, we have Principal + Interest = P(1 +

y

2

)

Thereafter, i.e., for n >

365

2

, we use simple interest again:

P(1 +

y

2

)(1 + y?

n ?

365

2

365

) = 100,

which is equivalent to the text’s equation on p. 132.

We can use the quadratic equation to solve for y:

BEY =

? ? ?

?

? ? ? ?

?

2

365 365

2 2

365

100

2

365

2 1 1

1

n n n

P

n

( ) ( )( )

.

Exercise 7.1: An investor buys a face amount $1 million of a six-month (182 days)

U.S. Treasury bill at a discount yield of 9.25%. What is the cost of purchasing

these bills? Calculate the bond equivalent yield. Indicate clearly the formula

you used and show all the steps in your calculations. Recalculate the bond

equivalent yield if the T-bill has a maturity of 275 days.

Exercise 7.2: On November 18, 2014, a 7 78 % U.S. T-bond maturing on May 15,

2017, was quoted for settlement on November 20, 2014. The last coupon was

paid on November 15, 2014. The number of days between coupon payments is

182. The bond’s YTM is 7.91%.

(a) What is the invoice price (cash price) of the T-bond?

(b) What is the accrued interest on the T-bond?

(c) What is the quoted price of the T-bond?

(d) What is the exact duration of the T-bond?

(e) What is the approximate convexity of the T-bond?

(f) Re-do part (a) using the next coupon date; show that the prices are the same.

Exercise 7.3: What is the price of a ten-year zero-coupon bond priced to yield 10%

under each of the following assumptions?

(a) Annual yield. (b) Semi-annual yield.

(c) Monthly yield. (d) Daily yield.

What is the continuous limit?

2

Exercise 7.4: In the table below, fill in the indicated blanks. Show all the steps in

your calculations. Duration is denoted by D, coupon by C, and yield by y

(compounded semi-annually). The settlement date is February 15, 2014. Now

suppose you have a liability of $100 million that will be paid as a lump sum in

5.5 years. Using the 9% coupon bond and the 10% bond how can you hedge

this liability? When calculating the number of bonds you need to purchase,

assume that the coupon bonds have a face value of $100,000.

C

y

Maturity

Date

Cash

Price

Exact

D

DV01

(per $100)

9% 9.00% 5/15/xx 102.26 4.8 xxxx

0% 9.00% 2/15/2020 xxx.xx x.x xxxx

10% 9.00% 2/15/2024 106.50 6.66 xxxx

Exercise 7.5: Supply the missing information in the following table. The settlement

date is February 15, 2014. Using this information, how would you use these

two bonds to hedge a liability with a present value of $100 million and a

Macaulay duration of 5.0? (Zero coupon bonds have a face value of

$1,000,000, and coupon bonds have a face value of $100,000.)

C

y

Maturity

Date

Cash

Price

D

0% 9.12% 2/15/2018 70.00 x.x

9% 7.33% 2/15/2022 110.00 6.0

Exercise 7.6: On November 15, 2014, you bought $10 million (face amount) of a

7.50% T-bond maturing on November 15, 2044, at a yield of 7.60%. All

coupons were reinvested at 5% yield (with simple interest for this yield only).

The bond was sold on June 20, 2015, (36 days after the last coupon payment,

and there are 184 days between coupon payments at the time that the bond is

sold) at a yield of 7.50%. Calculate the annualized rate of return from this

transaction using simple interest. (Note: there are 181 days from 15/11/2014 to

the first coupon payment on 15/05/2015.)

Exercise 7.7: Assuming that the settlement date is August 15, 2014, determine the

duration of the T-bond from the next table. From the duration formula, estimate

the price of the T-bond for a yield change of ±10 basis points. Note that the last

coupon date (LCD) was May 15, 2014 (92 days before the settlement date) and

the next coupon date is Nov. 15, 2014 (92 days after the settlement date).

C

y

Maturity

Date

Quoted

Price

D

10% 9.00% 11/15/2026 107.3043 xx.xx

Exercise 7.8: Reconsider Exercise 7.7. You may now estimate the prices at different

yield levels by using both duration and convexity. Calculate the convexity,

(using the formula for a round number of coupons remaining) compare the

estimated prices with the actual price and explain the deviations.

3

SOLUTIONS

Exercise 7.1: We have Face value = F = $1million, n = 182 days, discount yield = d

= 9.25%. The price is

P = F(1 ?

n d?

360

) = 1M(1 ?

182 0 0925

360

? .

) = $953,236.11.

Since n < 182 days, BEY =

100 95 3236

953236

? .

.

?

365

182

= 9.8386%.

If n = 275 days, and the discount yield is again d = 9.25%, then the price is

P = 1M(1 ?

n d?

360

) = 1M(1 ?

275 0 0925

360

? .

) = $929,340.28.

BEY =

? ? ?

?

? ? ? ?

?

2 275

365

275

365

2 2 275

365

100

92 934

2 275

365

2 1 1

1

( ) ( )( ). = 9.9258%.

Exercise 7.2:

(a) # days since last coupon = 5 = x ? z,

# days between coupon payments = 182 = x; coupon rate = 7.875% = c; y =

7.91%.

Now, as of the LCD there were 2 ? years remaining in the life of the bond; so, N = 5.

P = (1 +

y

2

)

x z

x

? C

y

C

y

y N

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

100

1

2

( )

= (1.03955

5

182) 5

7.8751007.875 0.0791

0.0791 (1.03955)

? ??? ?

?? ?

? ?

? ?? ?

= 100.02853.

(b) ai =

7 875

2

5

182

.

? = 0.108173.

(c) Cash price = quoted price + accrued interest, so

quoted price = Cash price – accrued interest = 100.02853 ? 0.108173

= 99.9204 (? 99 29

32

).

(d) For the duration, first we calculate the duration as of the last coupon date (LCD)

DLCD =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

=

)0791.0(2

0791.2 ?

)0791.0(2]1)

2

0791.01[()07875.0(2

)0791.007875.0(50791.2

5 ???

??

= 13.142 ? 10.824 = 2.318 years.

4

The exact duration is D = DLCD ? ?(

x z

x

?

) = 2.318 ? 0.5(

5

182

) = 2.30 years.

(e) Since the coupon rate is 7.875%, that means that C = 7.875. So, the

approximate convexity is

Cx =

?

?

2

2

P

y

1

P

,

where

?

?

2

2

P

y

=

N N

C

y

y N

( )( )

( )

? ?

? ?

1 100

4 1

2

2

?

CN

y

y N2 11

2

( )? ?

?

2

1

2

13

C

y

y N[( ) ]? ??

=

7)

2

0791.01(4

)

0791.0

875.7100)(6(5

?

?

?

62 )

2

0791.01()0791.0(

)5)(875.7(

?

? ]1)

2

0791.01[(

)0791.0(

)875.7(2 5

3 ??

?

= 2.530 – 4,986.494 – (?5,610.281) = 626.317

so, Cx = 626.317

100.02853

= 6.26.

(f) At the NCD we’ll receive a coupon of C/2 = 3.9375. Then, there will be 2 years

remaining in the life of the bond, so N = 4. Note that z = 182 – 5 = 177.

P = (1 +

y

2

)

z

x

? 100

2 (1 )

2

N

C

yC C

yy

? ??? ?

? ?? ?

? ??? ?? ?

= (1.03955

177

182)

?

4

7.8751007.875 0.07913.9375

0.0791 (1.03955)

? ??? ?

? ?? ?

? ?

? ?? ?

= 100.02853.

Exercise 7.3:

(a) Annual compounding: P =

100

11 10( . )

= 38.5543.

(b) Semi-annual compounding: P =

100

105 20( . )

= 37.6889.

(c) Monthly compounding: P =

100

1

010

12

120(

.

)?

= 36.9407.

(d) Daily compounding: P =

100

1

010

365

3650(

.

)?

= 36.7930.

Continuous compounding: P = 100e?(10?0.10) = 36.7879.

5

Note that the PV or price decreases. As the number of compounding periods

increases, the effective annual rate increases, so the PV decreases.

Exercise 7.4:

Bond 1: Since D = 4.8, we check years 2019, 2020, etc. By trial and error, using the

formula for round # of coupons, N remaining:

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

,

or the exact formula

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

?

1

2

(1 ?

z

x

),

where z = # of days to next coupon, and x = # of days between coupon payments.

Using 2019, the approximate formula gives (y = 9%, N = 11)

D =

2 0 09

2 0 09

? .

( . )

?

2 0 09 11 0 09 0 09

2 0 09 1

0 09

2

1 2 0 0911

? ? ?

? ? ?

. ( . . )

( . )[(

.

) ] ( . )

= 11.61 ? 7.15 = 4.45,

while the exact formula gives D = 4.46 ?

1

2

(1 ?

89

181

) = 4.20.

Using 2020, the formula gives (y = 9%, N = 13)

D =

2 0 09

2 0 09

? .

( . )

?

2 0 09 13 0 09 0 09

2 0 09 1

0 09

2

1 2 0 0913

? ? ?

? ? ?

. ( . . )

( . )[(

.

) ] ( . )

?

1

2

(1 ?

89

181

) = 4.8. (For 2021,

the approximate D = 5.6, the exact D = 5.4—definitely too large.)

Value of 0.01 = DV01(per 100% of par) = ?p?

1

100

, where

?p = ?

?

?

P

y

?

1

100

, and D = ?

?

?

P

y

?

( )1

2

?

y

P

. Combining these formulas gives DV01(per

100% of par) =

D P

y

?

?1

2

?

1

10 000,

;

Bond 1: DV01(per 100% of par) =

2

09.01

26.1028.4

?

? ?

1

10 000,

= 0.047.

Bond 2: For a zero coupon bond,

6

D = time to maturity = 6 years.

P =

100

1 0 045 12( . )?

= 58.97.

DV01(per 100% of par) =

6 59

1

0 09

2

?

?

. ?

1

10 000,

= 0.034.

Bond 3: DV01(per 100% of par) =

6 66 1065

1

0 09

2

. .

.

?

?

?

1

10 000,

= 0.0679.

To hedge the liability, we set

x1 + x2 = 1

and Dp = x1D1 + x2D2,

or 5.5 = x14.8 + x26.66,

which gives x1(4.8 – 6.66) = 5.5 – 6.66, so x1 =

8.466.6

5.566.6

?

? = 0.6237,

x2 = 0.3763. The present value of the liability is 25.5)045.01(

000,000,100

??

= $61,619,874.

So, we invest 61,619,874?0.6237 = 38,429,599 in Bond 1, and 23,190,275 in Bond 2.

The number of bonds required are n1 = 38,429,599/102,260 = 376, and n2 =

23,190,275/106,500 = 218. [I stored x1 and x2 into my calculator’s memory, so

that I avoided rounding off until the final step.]

Exercise 7.5:

Bond 1: (zero coupon bond) D = 4 years.

Bond 2: All information is given.

For the $100 million portfolio, we know that

Dp = x1D1 + x2D2,

where x1 = proportion of our wealth invested in Bond 1 (the zero coupon bond), and

x2 is the proportion invested in Bond 2. Here we require that

5 = x1(4) + x2(6), and 1 = x1 + x2.

Solving, we get x1(4 – 6) = 5 – 6, or x1 =

46

56

?

? = 0.50 = x2. So, we invest $50

million in the zero coupon bond and $50 million in the 9% bond to construct a

$100 million portfolio with a duration of five years. The number of bonds we

need to buy are n1 = 50M/700,000 = 71, and n2 = 50M/110,000 = 455.

7

Exercise 7.6: Since we’re given the bonds’ yields, we can calculate the cash prices

directly using the PLCD (the price on the last coupon date) and the formula

P = (1 +

y

2

)

x z

x

?

?PLCD = (1 +

y

2

)

x z

x

?

?[

C

y

+

100

1

2

?

?

C

y

y N( )

].

We first calculate the purchase price of the bond. This bond had a round number of

periods remaining; note that at date t = 0, y = 7.6%, x = 181, and x – z = 0:

P0 = [

7 5

0 076

.

.

+

100

7 5

0 076

1038 60

?

.

.

( . )

] = 98.8246.

Price at sale date (y = 7.5%, x = 184 days, z = 148 days, x ? z = 36 days, N = 59)

P1 = (1.0375 )

36

184 ?[

7 5

0 075

.

.

+

100

7 5

0 075

10375 59

?

.

.

( )

] = 100.7229.

Coupon interest = 7.5/2 = 3.75; Interest on coupon = 3.75?(5%)?36/365 = 0.0185.

So, for an initial investment of 98.8246, we receive in 217 days (i.e., 181 + 36 days),

100.7229 + 3.75 + 0.0185 = 104.4914, for an annualized return of

104 4914 988246

988246

365

217

. .

.

?

? = 9.645%.

Exercise 7.7: N = 25.

# days since last coupon = 92,

# days between coupon payments = 184. Coupon rate = 10%. ai =

10

2

92

184

? = 2.5.

Cash price = quoted price + accrued interest

= 107.3043 + 2.5 = 109.8043.

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

?

1

2

(1 ?

z

x

)

=

2 09

2 0 09

.

( . )

?

2 09 25 010 0 09

2 010 1

0 09

2

1 2 0 0925

. ( . . )

( . )[(

.

) ] ( . )

? ?

? ? ?

?

1

2

(1 ?

92

184

),

= 11.61 ? 4.03 ? 0.25 = 7.3342.

Now,??P ?

?

?

P

y

?y = ?D?

P

y( / )1 2? ??y.

Since ?y = 10 basis points = 0.001,

8

? ? ?P ? ?7.3342?

109 8043

1045

.

.

?y = ?0.7706.

So, for a ten basis point increase, (y = 9.1%) our estimated price is

109.8043 ? 0.7706 = 109.0337,

while, for a ten basis point decrease, (y = 8.9%) the estimated price is

109.8043 + 0.7706 = 110.5749.

(The actual prices would be 109.0375, and 110.5788.)

Exercise 7.8: The convexity (for round number of periods) is Cx =

?

?

2

2

P

y

1

P

,

?

?

2

2

P

y

=

N N

C

y

y N

( )( )

( )

? ?

? ?

1 100

4 1

2

2

?

CN

y

y N2 11

2

( )? ?

?

2

1

2

13

C

y

y N[( ) ]? ?? ,

where C is the cash flow per year, in this case; y = rate per year (with semi-

annual compounding) and N = #of coupon payments remaining. (C = 10,

N = 25, y = 9%)

So,

?

?

2

2

P

y

= 27)045.1(4

)

09.0

10100)(26(25 ?

? 262 )045.1()09.0(

)25)(10( ? ]1)045.1[(

)09.0(

)10(2 25

3 ?

?

= ?550.137 – 9,827.237 + 18,306.431 = 7,929.056.

(Cx = 72.21)

So, for ?y = +0.001,

?P ?

?

?

P

y

??y +

1

2

?

?

2

2

P

y

??y2 = ?0.7706 +

1

2

?7,929.056?(0.001)2

= ?0.7706 + 0.00396 = ?0.7666.

For ?y = ?0.001, ?P = +0.7706 + 0.00396 = 0.7746.

Note that the convexity effect is small.

So, the estimates for P are 109.0377 (at y = 9.1%) and 110.5789 (at y = 8.9%).

(Again, the actual prices would be 109.0375, and 110.5788—very close.)

學霸聯盟

CHAPTER 7: BOND MATHEMATICS

A word about BEY (i.e., Bond Equivalent Yield).

T-bills use semi-annual compounding. So, for n <

365

2

, we use simple interest.

Principle + interest is given by P(1 + y?

n

365

) = 100 at maturity.

Solving for y gives

y ? BEY =

100? P

P

?

365

n

.

For n =

365

2

, we have Principal + Interest = P(1 +

y

2

)

Thereafter, i.e., for n >

365

2

, we use simple interest again:

P(1 +

y

2

)(1 + y?

n ?

365

2

365

) = 100,

which is equivalent to the text’s equation on p. 132.

We can use the quadratic equation to solve for y:

BEY =

? ? ?

?

? ? ? ?

?

2

365 365

2 2

365

100

2

365

2 1 1

1

n n n

P

n

( ) ( )( )

.

Exercise 7.1: An investor buys a face amount $1 million of a six-month (182 days)

U.S. Treasury bill at a discount yield of 9.25%. What is the cost of purchasing

these bills? Calculate the bond equivalent yield. Indicate clearly the formula

you used and show all the steps in your calculations. Recalculate the bond

equivalent yield if the T-bill has a maturity of 275 days.

Exercise 7.2: On November 18, 2014, a 7 78 % U.S. T-bond maturing on May 15,

2017, was quoted for settlement on November 20, 2014. The last coupon was

paid on November 15, 2014. The number of days between coupon payments is

182. The bond’s YTM is 7.91%.

(a) What is the invoice price (cash price) of the T-bond?

(b) What is the accrued interest on the T-bond?

(c) What is the quoted price of the T-bond?

(d) What is the exact duration of the T-bond?

(e) What is the approximate convexity of the T-bond?

(f) Re-do part (a) using the next coupon date; show that the prices are the same.

Exercise 7.3: What is the price of a ten-year zero-coupon bond priced to yield 10%

under each of the following assumptions?

(a) Annual yield. (b) Semi-annual yield.

(c) Monthly yield. (d) Daily yield.

What is the continuous limit?

2

Exercise 7.4: In the table below, fill in the indicated blanks. Show all the steps in

your calculations. Duration is denoted by D, coupon by C, and yield by y

(compounded semi-annually). The settlement date is February 15, 2014. Now

suppose you have a liability of $100 million that will be paid as a lump sum in

5.5 years. Using the 9% coupon bond and the 10% bond how can you hedge

this liability? When calculating the number of bonds you need to purchase,

assume that the coupon bonds have a face value of $100,000.

C

y

Maturity

Date

Cash

Price

Exact

D

DV01

(per $100)

9% 9.00% 5/15/xx 102.26 4.8 xxxx

0% 9.00% 2/15/2020 xxx.xx x.x xxxx

10% 9.00% 2/15/2024 106.50 6.66 xxxx

Exercise 7.5: Supply the missing information in the following table. The settlement

date is February 15, 2014. Using this information, how would you use these

two bonds to hedge a liability with a present value of $100 million and a

Macaulay duration of 5.0? (Zero coupon bonds have a face value of

$1,000,000, and coupon bonds have a face value of $100,000.)

C

y

Maturity

Date

Cash

Price

D

0% 9.12% 2/15/2018 70.00 x.x

9% 7.33% 2/15/2022 110.00 6.0

Exercise 7.6: On November 15, 2014, you bought $10 million (face amount) of a

7.50% T-bond maturing on November 15, 2044, at a yield of 7.60%. All

coupons were reinvested at 5% yield (with simple interest for this yield only).

The bond was sold on June 20, 2015, (36 days after the last coupon payment,

and there are 184 days between coupon payments at the time that the bond is

sold) at a yield of 7.50%. Calculate the annualized rate of return from this

transaction using simple interest. (Note: there are 181 days from 15/11/2014 to

the first coupon payment on 15/05/2015.)

Exercise 7.7: Assuming that the settlement date is August 15, 2014, determine the

duration of the T-bond from the next table. From the duration formula, estimate

the price of the T-bond for a yield change of ±10 basis points. Note that the last

coupon date (LCD) was May 15, 2014 (92 days before the settlement date) and

the next coupon date is Nov. 15, 2014 (92 days after the settlement date).

C

y

Maturity

Date

Quoted

Price

D

10% 9.00% 11/15/2026 107.3043 xx.xx

Exercise 7.8: Reconsider Exercise 7.7. You may now estimate the prices at different

yield levels by using both duration and convexity. Calculate the convexity,

(using the formula for a round number of coupons remaining) compare the

estimated prices with the actual price and explain the deviations.

3

SOLUTIONS

Exercise 7.1: We have Face value = F = $1million, n = 182 days, discount yield = d

= 9.25%. The price is

P = F(1 ?

n d?

360

) = 1M(1 ?

182 0 0925

360

? .

) = $953,236.11.

Since n < 182 days, BEY =

100 95 3236

953236

? .

.

?

365

182

= 9.8386%.

If n = 275 days, and the discount yield is again d = 9.25%, then the price is

P = 1M(1 ?

n d?

360

) = 1M(1 ?

275 0 0925

360

? .

) = $929,340.28.

BEY =

? ? ?

?

? ? ? ?

?

2 275

365

275

365

2 2 275

365

100

92 934

2 275

365

2 1 1

1

( ) ( )( ). = 9.9258%.

Exercise 7.2:

(a) # days since last coupon = 5 = x ? z,

# days between coupon payments = 182 = x; coupon rate = 7.875% = c; y =

7.91%.

Now, as of the LCD there were 2 ? years remaining in the life of the bond; so, N = 5.

P = (1 +

y

2

)

x z

x

? C

y

C

y

y N

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

100

1

2

( )

= (1.03955

5

182) 5

7.8751007.875 0.0791

0.0791 (1.03955)

? ??? ?

?? ?

? ?

? ?? ?

= 100.02853.

(b) ai =

7 875

2

5

182

.

? = 0.108173.

(c) Cash price = quoted price + accrued interest, so

quoted price = Cash price – accrued interest = 100.02853 ? 0.108173

= 99.9204 (? 99 29

32

).

(d) For the duration, first we calculate the duration as of the last coupon date (LCD)

DLCD =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

=

)0791.0(2

0791.2 ?

)0791.0(2]1)

2

0791.01[()07875.0(2

)0791.007875.0(50791.2

5 ???

??

= 13.142 ? 10.824 = 2.318 years.

4

The exact duration is D = DLCD ? ?(

x z

x

?

) = 2.318 ? 0.5(

5

182

) = 2.30 years.

(e) Since the coupon rate is 7.875%, that means that C = 7.875. So, the

approximate convexity is

Cx =

?

?

2

2

P

y

1

P

,

where

?

?

2

2

P

y

=

N N

C

y

y N

( )( )

( )

? ?

? ?

1 100

4 1

2

2

?

CN

y

y N2 11

2

( )? ?

?

2

1

2

13

C

y

y N[( ) ]? ??

=

7)

2

0791.01(4

)

0791.0

875.7100)(6(5

?

?

?

62 )

2

0791.01()0791.0(

)5)(875.7(

?

? ]1)

2

0791.01[(

)0791.0(

)875.7(2 5

3 ??

?

= 2.530 – 4,986.494 – (?5,610.281) = 626.317

so, Cx = 626.317

100.02853

= 6.26.

(f) At the NCD we’ll receive a coupon of C/2 = 3.9375. Then, there will be 2 years

remaining in the life of the bond, so N = 4. Note that z = 182 – 5 = 177.

P = (1 +

y

2

)

z

x

? 100

2 (1 )

2

N

C

yC C

yy

? ??? ?

? ?? ?

? ??? ?? ?

= (1.03955

177

182)

?

4

7.8751007.875 0.07913.9375

0.0791 (1.03955)

? ??? ?

? ?? ?

? ?

? ?? ?

= 100.02853.

Exercise 7.3:

(a) Annual compounding: P =

100

11 10( . )

= 38.5543.

(b) Semi-annual compounding: P =

100

105 20( . )

= 37.6889.

(c) Monthly compounding: P =

100

1

010

12

120(

.

)?

= 36.9407.

(d) Daily compounding: P =

100

1

010

365

3650(

.

)?

= 36.7930.

Continuous compounding: P = 100e?(10?0.10) = 36.7879.

5

Note that the PV or price decreases. As the number of compounding periods

increases, the effective annual rate increases, so the PV decreases.

Exercise 7.4:

Bond 1: Since D = 4.8, we check years 2019, 2020, etc. By trial and error, using the

formula for round # of coupons, N remaining:

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

,

or the exact formula

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

?

1

2

(1 ?

z

x

),

where z = # of days to next coupon, and x = # of days between coupon payments.

Using 2019, the approximate formula gives (y = 9%, N = 11)

D =

2 0 09

2 0 09

? .

( . )

?

2 0 09 11 0 09 0 09

2 0 09 1

0 09

2

1 2 0 0911

? ? ?

? ? ?

. ( . . )

( . )[(

.

) ] ( . )

= 11.61 ? 7.15 = 4.45,

while the exact formula gives D = 4.46 ?

1

2

(1 ?

89

181

) = 4.20.

Using 2020, the formula gives (y = 9%, N = 13)

D =

2 0 09

2 0 09

? .

( . )

?

2 0 09 13 0 09 0 09

2 0 09 1

0 09

2

1 2 0 0913

? ? ?

? ? ?

. ( . . )

( . )[(

.

) ] ( . )

?

1

2

(1 ?

89

181

) = 4.8. (For 2021,

the approximate D = 5.6, the exact D = 5.4—definitely too large.)

Value of 0.01 = DV01(per 100% of par) = ?p?

1

100

, where

?p = ?

?

?

P

y

?

1

100

, and D = ?

?

?

P

y

?

( )1

2

?

y

P

. Combining these formulas gives DV01(per

100% of par) =

D P

y

?

?1

2

?

1

10 000,

;

Bond 1: DV01(per 100% of par) =

2

09.01

26.1028.4

?

? ?

1

10 000,

= 0.047.

Bond 2: For a zero coupon bond,

6

D = time to maturity = 6 years.

P =

100

1 0 045 12( . )?

= 58.97.

DV01(per 100% of par) =

6 59

1

0 09

2

?

?

. ?

1

10 000,

= 0.034.

Bond 3: DV01(per 100% of par) =

6 66 1065

1

0 09

2

. .

.

?

?

?

1

10 000,

= 0.0679.

To hedge the liability, we set

x1 + x2 = 1

and Dp = x1D1 + x2D2,

or 5.5 = x14.8 + x26.66,

which gives x1(4.8 – 6.66) = 5.5 – 6.66, so x1 =

8.466.6

5.566.6

?

? = 0.6237,

x2 = 0.3763. The present value of the liability is 25.5)045.01(

000,000,100

??

= $61,619,874.

So, we invest 61,619,874?0.6237 = 38,429,599 in Bond 1, and 23,190,275 in Bond 2.

The number of bonds required are n1 = 38,429,599/102,260 = 376, and n2 =

23,190,275/106,500 = 218. [I stored x1 and x2 into my calculator’s memory, so

that I avoided rounding off until the final step.]

Exercise 7.5:

Bond 1: (zero coupon bond) D = 4 years.

Bond 2: All information is given.

For the $100 million portfolio, we know that

Dp = x1D1 + x2D2,

where x1 = proportion of our wealth invested in Bond 1 (the zero coupon bond), and

x2 is the proportion invested in Bond 2. Here we require that

5 = x1(4) + x2(6), and 1 = x1 + x2.

Solving, we get x1(4 – 6) = 5 – 6, or x1 =

46

56

?

? = 0.50 = x2. So, we invest $50

million in the zero coupon bond and $50 million in the 9% bond to construct a

$100 million portfolio with a duration of five years. The number of bonds we

need to buy are n1 = 50M/700,000 = 71, and n2 = 50M/110,000 = 455.

7

Exercise 7.6: Since we’re given the bonds’ yields, we can calculate the cash prices

directly using the PLCD (the price on the last coupon date) and the formula

P = (1 +

y

2

)

x z

x

?

?PLCD = (1 +

y

2

)

x z

x

?

?[

C

y

+

100

1

2

?

?

C

y

y N( )

].

We first calculate the purchase price of the bond. This bond had a round number of

periods remaining; note that at date t = 0, y = 7.6%, x = 181, and x – z = 0:

P0 = [

7 5

0 076

.

.

+

100

7 5

0 076

1038 60

?

.

.

( . )

] = 98.8246.

Price at sale date (y = 7.5%, x = 184 days, z = 148 days, x ? z = 36 days, N = 59)

P1 = (1.0375 )

36

184 ?[

7 5

0 075

.

.

+

100

7 5

0 075

10375 59

?

.

.

( )

] = 100.7229.

Coupon interest = 7.5/2 = 3.75; Interest on coupon = 3.75?(5%)?36/365 = 0.0185.

So, for an initial investment of 98.8246, we receive in 217 days (i.e., 181 + 36 days),

100.7229 + 3.75 + 0.0185 = 104.4914, for an annualized return of

104 4914 988246

988246

365

217

. .

.

?

? = 9.645%.

Exercise 7.7: N = 25.

# days since last coupon = 92,

# days between coupon payments = 184. Coupon rate = 10%. ai =

10

2

92

184

? = 2.5.

Cash price = quoted price + accrued interest

= 107.3043 + 2.5 = 109.8043.

D =

2

2

? y

y

?

2

2 1

2

1 2

? ? ?

? ? ?

y N c y

c

y

yN

( )

[( ) ]

?

1

2

(1 ?

z

x

)

=

2 09

2 0 09

.

( . )

?

2 09 25 010 0 09

2 010 1

0 09

2

1 2 0 0925

. ( . . )

( . )[(

.

) ] ( . )

? ?

? ? ?

?

1

2

(1 ?

92

184

),

= 11.61 ? 4.03 ? 0.25 = 7.3342.

Now,??P ?

?

?

P

y

?y = ?D?

P

y( / )1 2? ??y.

Since ?y = 10 basis points = 0.001,

8

? ? ?P ? ?7.3342?

109 8043

1045

.

.

?y = ?0.7706.

So, for a ten basis point increase, (y = 9.1%) our estimated price is

109.8043 ? 0.7706 = 109.0337,

while, for a ten basis point decrease, (y = 8.9%) the estimated price is

109.8043 + 0.7706 = 110.5749.

(The actual prices would be 109.0375, and 110.5788.)

Exercise 7.8: The convexity (for round number of periods) is Cx =

?

?

2

2

P

y

1

P

,

?

?

2

2

P

y

=

N N

C

y

y N

( )( )

( )

? ?

? ?

1 100

4 1

2

2

?

CN

y

y N2 11

2

( )? ?

?

2

1

2

13

C

y

y N[( ) ]? ?? ,

where C is the cash flow per year, in this case; y = rate per year (with semi-

annual compounding) and N = #of coupon payments remaining. (C = 10,

N = 25, y = 9%)

So,

?

?

2

2

P

y

= 27)045.1(4

)

09.0

10100)(26(25 ?

? 262 )045.1()09.0(

)25)(10( ? ]1)045.1[(

)09.0(

)10(2 25

3 ?

?

= ?550.137 – 9,827.237 + 18,306.431 = 7,929.056.

(Cx = 72.21)

So, for ?y = +0.001,

?P ?

?

?

P

y

??y +

1

2

?

?

2

2

P

y

??y2 = ?0.7706 +

1

2

?7,929.056?(0.001)2

= ?0.7706 + 0.00396 = ?0.7666.

For ?y = ?0.001, ?P = +0.7706 + 0.00396 = 0.7746.

Note that the convexity effect is small.

So, the estimates for P are 109.0377 (at y = 9.1%) and 110.5789 (at y = 8.9%).

(Again, the actual prices would be 109.0375, and 110.5788—very close.)

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