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A machine sells for \(\$ 10,000\) and has a salvage value of \(\$ 1000\) at the end of 10 years. The annual maintenance expense of the machine is \(\$ 500\). Assuming \(5 \%\) interest: a) Calculate the periodic charge of the asset. b) Calculate the capitalized cost of the asset.

Short Answer

Expert verified
The periodic charge of the asset is $900 per year and the capitalized cost of the asset is $6,000.

Step by step solution

01

Calculate the Periodic Charge:

To find the periodic charge, let's use the formula given above: \(P = \frac{C - S}{n}\) Plugging in the given values: \(P = \frac{10000 - 1000}{10}\) \(P = \frac{9000}{10}\) \(P = 900\) The periodic charge of the asset is $900 per year.
02

Calculate the Capitalized Cost:

Now we need to calculate the capitalized cost. The annual maintenance expense is given as $500. First, let's find the annual charge (which is the periodic charge multiplied by 10 years): \(A = P \times n\) \(A = 900 \times 10\) \(A = 9000\) Now, use the formula for the capitalized cost: \(CC = C + M - A\) Plugging in the given values: \(CC = 10000 + (500 \times 10) - 9000\) \(CC = 10000 + 5000 - 9000\) \(CC = 6000\) The capitalized cost of the asset is $6,000.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Depreciation
Depreciation is the gradual decrease in the value of an asset over time. This process acknowledges that most assets, such as machinery or vehicles, wear out and lose their value due to usage and aging.
To calculate depreciation, a common method is the straight-line depreciation formula, which is simply the initial cost of the asset minus its expected salvage value, divided by the useful life of the asset. The formula is: \[ P = \frac{C - S}{n} \]where:
  • \(P\) is the periodic charge or annual depreciation expense.
  • \(C\) is the original cost of the asset.
  • \(S\) is the salvage value—the estimated value of the asset at the end of its useful life.
  • \(n\) is the number of years the asset is expected to be used.
By consistently reflecting depreciation in financial statements, companies can more accurately report the value of their assets and their financial performance over time.
In the exercise, the machine's cost is \(\\(10,000\) and its salvage value at the end of 10 years is \(\\)1,000\). Using these values, the depreciation calculation provides an annual periodic charge of \(\$900\). This amount helps businesses allocate the cost of the asset over its useful life.
Capitalized Cost
Capitalized cost is a crucial concept when assessing the long-term value and expenses associated with an asset. It represents the total expense incurred to acquire an asset, including the purchase price, as well as costs necessary to extend the asset's useful life or enhance its value.To calculate the capitalized cost of an asset, the formula used is:\[ CC = C + M - A \]where:
  • \(CC\) is the capitalized cost.
  • \(C\) is the original cost of the asset.
  • \(M\) is the total maintenance costs over the asset's life.
  • \(A\) is the total accumulated depreciation or other adjustments.
In the provided exercise, the original cost of the machine is \(\\(10,000\). The annual maintenance expense is consistent at \(\\)500\), making the total maintenance expenses over 10 years \(\\(5,000\). Subtracting the total accumulated periodic charge of \(\\)9,000\) (\(10 \times \\(900\)) gives a capitalized cost of \(\\)6,000\).
Understanding the capitalized cost is vital as it allows for better budgeting and financial planning. By capitalizing an asset, businesses can spread out the expense over its useful life, improving financial accuracy and efficiency.
Salvage Value
Salvage value is an estimate of the residual value of an asset at the end of its useful life. It is often considered when calculating depreciation and projecting the potential return from an asset once its useful life is over.The salvage value of an asset can be an influencing factor in deciding whether to purchase an asset, as it affects the overall cost and depreciation. The formula to find annual depreciation takes the salvage value into account:\[ P = \frac{C - S}{n} \]where:
  • \(C\) is the initial cost of the asset.
  • \(S\) is the salvage value.
  • \(n\) is the useful life of the asset.
In our exercise, the machine has a salvage value of \(\\(1,000\) at the end of 10 years. This value is subtracted from the initial asset cost of \(\\)10,000\) to determine the total amount that will depreciate over a decade.
Salvage value provides important insight for businesses when deciding on asset replacements or sales, as it contributes to long-term financial planning and the accurate assessment of asset worth.

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Most popular questions from this chapter

Machine 1 sells of \(\$ 1000\) with a salvage value of \(\$ 50\) at the end of nine years. Machine 2 sells for \(\$ 1100\) with a salvage value of \(\$ 200\) at the end of nine years. At what rate of interest would a purchaser be indifferent between the two machines? Assume equal maintenance expenses for the two machines.

A piece of equipment that was purchased for \(\$ 15,000\) will have a salvage value of \(\$ 2000\) after 15 years. Its book value has been determined by depreciation in accordance with the compound interest method, using an interest rate of \(5 \%\) per annum. At the end of the 10 th year, the depreciation method is changed to the straight line method for the remaining 5 years. Determine the book value at the end of the 12 th year. Answer to the nearest dollar.

A loan of \(\$ 12,000\) is to be repaid in one year according to one of the following arrangements: \(A-\$ 1000\) payable at the end of each month in addition to a finance charge of \(\$ 1000\) payable when the loan is approved. \(\mathrm{B}-\) Repayment at the end of each month according to an amortization schedule with \(t^{(12)}=12 \%\) Find the difference in the amount of interest paid under options \(A\) and \(B\).

A construction firm buys \(\$ 1000\) worth of lumber. Find the maximum amount the firm would be willing to pay to treat the lumber to extend its life from 10 to 15 years if the salvage value in either case is \(\$ 50 .\) The effective rate of interest is 3 \(1 / 2 \%\)

A 15 -year mortgage has monthly payments of \(\$ 1000\) with interest convertible monthly. At the end of each month, the borrower makes a \(\$ 1000\) payment. In addition to the regular monthly payment, the borrower makes an additional payment equal to the amount of principal that would have been repaid in the next regular monthly payment. Under this arrangement the loan will be completely repaid after 90 payments. Show that the amount of interest saved over the life of the loan is equal to $$90,000-1000 \frac{\dot{a}_{\overline{180}}}{s_{\overline{2}\rceil}}$$ where the annuity symbols are computed at the monthly rate of interest o

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