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Murljashka [212]
3 years ago
14

One bond has a coupon rate of 5.4%, another a coupon rate of 8.2%. Both bonds pay interest annually, have 13-year maturities, an

d sell at a yield to maturity of 7.5%.a. If their yields to maturity next year are still 7.5%, what is the rate of return on each bond? (Do not round intermediate calculations. Enter your answers as a percent to 1 decimal place.) rate or returnbond 1 __________bond 2 ___________b. Does the higer-coupon bond give a higher rate of return? (yes or no)
Business
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

a. rate or return bond 1 <u>6.6%</u> bond 2 <u>7.71%</u>

b. Does the higher-coupon bond give a higher rate of return? <u>yes</u>

Explanation:

bond 1 has a coupon rate of 5.4%

bond 2 has a coupon rate of 8.2%

yield to maturity formula = {C + [(Face value - market value) / n]} / [(Face value + market value) / 2]

assume bond 1's face value = $1,000

coupon = 54

n = 13

YTM = 7.5%

0.075 = {54 + [(1,000 - M) / 13]} / [(1,000 + M) / 2]

0.075 x  [(1,000 + M) / 2] = 54 +  [(1,000 - M) / 13]

0.075 x (500 + 0.5M) = 54 + 76.92 - 0.0769M

37.50 + 0.0375M = 130.92 - 0.0769M

0.0375M + 0.0769M = 130.92 - 37.50

0.1144M = 93.42

M = 93.42 / 0.1142 = $818.04

rate of return = $54 / $818.04 = 0.066 = 6.6%

assume bond 2's face value = $1,000

coupon = 82

n = 13

YTM = 7.5%

0.075 = {82 + [(1,000 - M) / 13]} / [(1,000 + M) / 2]

0.075 x  [(1,000 + M) / 2] = 82 +  [(1,000 - M) / 13]

0.075 x (500 + 0.5M) = 82 + 76.92 - 0.0769M

37.50 + 0.0375M = 158.92 - 0.0769M

0.0375M + 0.0769M = 158.92 - 37.50

0.1144M = 121.42

M = 121.42 / 0.1142 = $1,063.22

rate of return = $82 / $1,063.22 = 0.07712 = 7.71%

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The question is incomplete. The complete question is :

A manufacturer believes that the cost function : $C(x) =\frac{5}{2}x^2+120 x+560$  approximates the dollar cost of producing x units of a product. The manu- facturer believes it cannot make a profit when the marginal cost goes beyond $210. What is the most units the manufacturer can produce and still make a profit? What is the total cost at this level of production?

Solution :

Given the cost function is :

$C(x) =\frac{5}{2}x^2+120 x+560$  

Now, Marginal cost = $\frac{d}{dx}C(x)$

So, if the marginal cost = $ 210, then the manufacturer also makes a profit and if it goes beyond $ 210 than the manufacturer cannot make a profit.

Therefore, we have to equate : $\frac{d}{dx}C(x)= \$ 210$

$\frac{d}{dx}C(x)= \frac{5}{2}(2x)+120 = 210$

$5x + 120 = 210$

$5x=210-120$

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Equipment was acquired on January 1, 2019 at a cost of $190,000. The equipment was originally estimated to have a salvage value
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Answer:

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Dec. 31, 2022:

Debit Depreciation Expense $18,600

Credit Accumulated Depreciation $18,600

To record depreciation expense for the year.

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2019: $16,800

2020: $16,800

2021: $16,800

Accumulated Depreciation to date = $50,400 ($16,800*3)

b) Book Value on January 1, 2022 = $139,600 ($190,000 - $50,400)

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d) There is adjusting journal entry.  Depreciation is an estimate based on judgement and past events.  Judgement can change to address current events.  So, there is no need adjusting the entries for the previous three years.

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