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Sav [38]
3 years ago
7

A company issues $16200000, 5.8%, 20-year bonds to yield 6% on January 1, 2020. Interest is paid on June 30 and December 31. The

proceeds from the bonds are $15825541. Using effective-interest amortization, how much interest expense will be recognized in 2020?
Business
1 answer:
bekas [8.4K]3 years ago
3 0

Answer:

The amount of interest expense which will be recognized in 2020 is $949,681.45.

Explanation:

The following are given in the question:

Bond value = $16200000

Bond interest rate = 5.8%

Proceed from bond = $15825541

Yield rate = 6%

The amount of interest expense which will be recognized in 2020 can now be calculated as follows:

Interest expense for January 1, 2020 to June 30, 2020 = Proceed from bond * Yield rate * (6 / 12) = $15825541 * 6% * (6 / 12) = $474,766.23

Discount amortized during first 6 months = Interest expense for January 1, 2020 to June 30, 2020 - (Bond value * Bond interest rate * (6 / 12)) = $474,766.23 - ($16200000 * 5.8% * (6 / 12)) = $474,766.23 - 469,800 = $4,966.23

Interest expense for July 1, 2020 to December 31, 2020 = (Proceed from bond + Discount amortized during first 6 months) * Yield rate * (6 / 12) = ($15825541 + $4,966.23) * 6% * (6 / 12) = $474,915.22

Interest expense to be recognized in 2020 = Interest expense for January 1, 2020 to June 30, 2020 + Interest expense for July 1, 2020 to December 31, 2020 = $474,766.23 + $474,915.22 = $949,681.45

Therefore, the amount of interest expense which will be recognized in 2020 is $949,681.45.

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UA Hamburger Hamlet (UAHH) places a daily order for its high-volume items (hamburger patties, buns, milk, and so on). UAHH count
Assoli18 [71]

Answer:

730 items

Explanation:

The objective of the given information is to determine the number of hamburgers UAHH should order for the following conditions:

Average daily demand 600

Standard deviation of demand 100

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Hamburger inventory 800

The formula for a given order quantity in a fixed period of time can be expressed as :

q = \overline d(L+T)+ z \sigma_{L+T}-I

where;

q =  order quantity = ???

\overline d = daily demand average = 600

L = lead time in days = 1

T = time taken = 1

z = no of standard deviation = ???

\sigma_{L+T} = standard deviation of usage in lead time and time taken = ???

I = present inventory level = 800

\sigma_{L+T} = \sqrt 2 × standard deviation of daily demand

\sigma_{L+T} = \sqrt{2} *100

\sigma_{L+T} = 1.4142 * 100

\sigma_{L+T} = 141.42 items

From the Desired service probability 99% = 0.99; we can deduce the no of standard deviation by using the excel function (=NORMSINV (0.99))

z = 2.33

From q = \overline d(L+T)+ z \sigma_{L+T}-I

q =600(1+1)+ 2.33*(141.42)-800

q =600(2)+ 2.33*(141.42)-800

q =1200+329.5086-800

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Therefore; the  number of hamburgers UAHH should order from the following given conditions = 730 items

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WARRIOR [948]

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Answer:

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