# Part 2 TKM0844 11E IM Ch09

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:

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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.

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250...

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