Part 2 TKM0844 11E IM Ch09

Topics: Bond, Bonds, Floating interest rate Pages: 22 (6473 words) Published: June 28, 2015
Chapter 9
Debt Valuation and Interest Rates
9-1.

Bensington Glass Company has a floating rate loan that calls for a rate of [LIBOR + 30 bp], where LIBOR is the rate for the previous week (so, for example, the rate for week 2 is based on LIBOR for week 1). However, the rate has a collar: It cannot be greater than 2.2% (the ceiling), and it cannot be less than 1.75% (the floor). Thus the loan rate is: 1.75%


loan rate t = LIBOR t −1 + 0.30%
2.2%


if [LIBOR t-1 +30 bp] < 1.75%
if 1.75% ≤ [LIBOR t −1 + 30 bp] ≤ 2.2%
if 2.2% < [LIBOR t −1 + 30 bp]

Now, given our LIBOR values, we can solve for the loan values, as shown in the spreadsheet below:

date
week 1
week 2
week 3
week 4
week 5
week 6
week 7
week 8
week 9

LIBOR (t)
1.98%
1.66%
1.52%
1.35%
1.60%
1.63%
1.67%
1.88%
1.93%

LIBOR (t-1) + spread
30
spread (bp)
2.28%
1.96%
1.82%
1.65%
1.90%
1.93%
1.97%
2.18%

2.20%
max
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%

1.75%
min
1.75%
1.75%
1.75%
1.75%
1.75%
1.75%
1.75%
1.75%
1.75%

rate
2.20%
1.96%
1.82%
1.75%
1.90%
1.93%
1.97%
2.18%

Note that in week 2, [LIBOR1 + 30bp] > 2.2%, so the loan rate is set at the ceiling of 2.2%; in week 5, [LIBOR4 + 30bp] < 1.75%, so the loan rate is set at the floor of 1.75%. In the other weeks, the loan rate is set by the index plus the margin. The graph below shows these values:

©2011 Pearson Education, Inc. Publishing as Prentice Hall

248

• Financial Management, Eleventh Edition

Titman/Keown/Martin

2.2%

2.1%

2.0%

1.9%

LIBOR

1.8%

LIBOR+spread
ceiling

1.7%

floor
loan rate

1.6%

1.5%

1.4%

1.3%

1

2

3

4

5

6

7

8

9

week

9-2.

Ace-Campbell Manufacturing has a floating rate loan that calls for a rate of [LIBOR + 40bp], where LIBOR is the rate for the previous week (so, for example, the rate for week 2 is based on LIBOR for week 1). However, the rate has a collar: It cannot be greater than 2.2% (the ceiling), and it cannot be less than 1.50% (the floor). Thus the loan rate is: 1.50%


loan rate t = LIBOR t −1 + 0.40%
2.2%


if [LIBOR t-1 + 40 bp] < 1.50%
if 1.50% ≤ [LIBOR t −1 + 40 bp] ≤ 2.2%
if 2.2% < [LIBOR t −1 + 40 bp]

Now, given our LIBOR values, we can solve for the loan values, as shown in the spreadsheet below:

date
week 1
week 2
week 3
week 4
week 5
week 6
week 7
week 8
week 9

LIBOR (t)
1.98%
1.66%
1.52%
1.35%
1.60%
1.63%
1.67%
1.88%
1.93%

LIBOR (t-1) + spread
40
spread (bp)
2.38%
2.06%
1.92%
1.75%
2.00%
2.03%
2.07%
2.28%

2.20%
max
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%
2.20%

1.50%
min
1.50%
1.50%
1.50%
1.50%
1.50%
1.50%
1.50%
1.50%
1.50%

©2011 Pearson Education, Inc. Publishing as Prentice Hall

rate
2.20%
2.06%
1.92%
1.75%
2.00%
2.03%
2.07%
2.20%

Solutions to End of Chapter Problems—Chapter 9

249

Note that in weeks 2 and 9, [LIBORt−1 + 40 bp] > 2.2%, so the loan rate is set at the ceiling of 2.2%. In the other weeks, the loan rate is set by the index plus the margin. The graph below shows these values:

2.2%

2.1%

2.0%

1.9%

LIBOR

1.8%

LIBOR+spread
ceiling

1.7%

floor
loan rate

1.6%

1.5%

1.4%

1.3%

1

2

3

4

5

6

7

8

9

week

9-3.

To find the price of a bond, we can use equation 9-2b:
 1 − 1 n  $1000
.
price = (coupon rate) ∗ ($1000) ∗  (1+ i )  +
n
 i  (1 + i )

Assume we have a bond with the following characteristics:
par = $1000
maturity (years) = 12
coupon rate = 8%
YTM = 12%

We will assume that this bond pays interest annually. Thus, we can substitute the values into equation 9-2 b:
 1 − 1 12
price = (8%) ∗ ($1000) ∗  (1.12)
 0.12

 $1000
+
12
 (1.12)

= $80 ∗ (6.1944) + $1000 ∗ (0.2567)
= $495.55 + $256.68 = $752.23.

©2011 Pearson Education, Inc. Publishing as Prentice Hall

250...
Continue Reading

Please join StudyMode to read the full document

You May Also Find These Documents Helpful

  • Part 2 TKM0844 11E IM Ch14 Essay
  • Part 2 TKM0844 11E IM Ch15 Essay
  • Part 2 TKM0844 11E IM Ch17 Essay
  • Part 2 TKM0844 11E IM Ch12 Essay
  • Part 2 TKM0844 11E IM Ch18 Essay
  • Essay on Part 2 TKM0844 11E IM Ch04
  • Part 2 TKM0844 11E IM Ch13 Essay
  • Part 2 TKM0844 11E IM Ch19 Essay

Become a StudyMode Member

Sign Up - It's Free