Show
Engineering Economics Sinking fund method Declining balance method Straight line method Sum-of-year digit method Sinking fund method Declining balance method Straight line method
Sum-of-year digit method ANSWER DOWNLOAD EXAMIANS APP 13 months 11 months 10 months 12 months
13 months 11 months 10 months 12 months ANSWER
DOWNLOAD EXAMIANS APP Acid-Test (or Quick) ratio
Debts ratio Liquidity ratio Current ratio Acid-Test (or Quick) ratio Debts ratio Liquidity ratio Current ratio ANSWER DOWNLOAD EXAMIANS APP Future annuities Income annuities All of these Premium annuities Future annuities Income annuities All of these
Premium annuities ANSWER DOWNLOAD EXAMIANS APP Cost of goods sold Cost accounting Overhead cost Standard cost
Cost of goods sold Cost accounting Overhead cost Standard cost ANSWER
DOWNLOAD EXAMIANS APP Oligopoly
Oligopsony Monopoly Duopsony Oligopoly Oligopsony Monopoly Duopsony ANSWER DOWNLOAD EXAMIANS APP MORE MCQ ON Engineering Economics Depreciation of Power Station EquipmentThe reduction in the value of the equipment and other property of the power station every year is known as depreciation. Therefore, a suitable amount, called depreciation charge, must be set aside annually so that by the time the life span of the power plant is over, the collected amount equals to the cost of the replacement of the power plant. Sinking Fund Method of DepreciationIn the sinking fund method of depreciation, a fixed depreciation charge is made every year and the interest is compounded on it annually. The constant depreciation charge is such that the sum of annual investment and the interest accumulations is equal to the cost of replacement of equipment after its useful life. Explanation Let,
Therefore, the cost of replacement of the equipment is, $$\mathrm{\mathrm{Cost\: of\: replacement}\:=\:\mathit{X-S}}$$ Let an amount of p is set aside as depreciation charge every year and interest compounded on it so that an amount of (X-S), i.e. cost of replacement is available after n years. Therefore, the amount p at annual interest rate of r at the end of n years is given by, At the end of first year, $$\mathrm{\mathrm{Amount\:=\:}\mathit{p\:\mathrm{+}\:rp}\:=\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$ At the end of second year, $$\mathrm{\mathrm{Amount\:=\:}\mathrm{\left ( \mathit{p}+\mathit{rp} \right )}\:+\:\mathit{r}\mathrm{\left ( \mathit{p}+\mathit{rp} \right )}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{2}}}$$ Similarly, at the end of n years, $$\mathrm{\mathrm{amount\:=\:}\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}}$$ Now, the amount p deposited at the end of first year will earn compound interest for (n-1) years and it becomes, $$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:first\: year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n-\mathrm{1}}}}}$$ And the amount p deposited at the end of second year becomes, $$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:second\: year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n-\mathrm{2}}}}}$$ Similarly, amount p deposited at the end of (n-1) year becomes, $$\mathrm{\mathrm{Amount}\:\mathit{p}\:\mathrm{deposited\: at\: the\: end\: of \:}\mathrm{\left ( \mathit{n} -1\right )}\mathrm{year}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}\mathrm{\left ( \mathit{n}-1 \right )}}}\:=\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$ Therefore, the total fund after n years is given by, $$\mathrm{\mathrm{Total\: fund\: after\: n\: years}\:=\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-1}}\:+\:\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-2}}\:+\:...\:+\:\mathit{p}\mathrm{\left ( 1\:+\mathit{r} \right )}}$$ $$\mathrm{\Rightarrow \mathrm{Total\: fund\: after\: n\: years}\:=\:\mathit{p}\mathrm{\left [ \mathrm{\left ( 1\:+\mathit{r} \right )^{\mathit{n}-1}}\:+\:\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}-2}}\:+\:...\:+\mathrm{\left ( 1\:+\mathit{r} \right )} \right ]}}$$ As it is a geometric progression series and its sum is given by, $$\mathrm{\mathrm{Total \:fund\: after\: n \:years}\:=\:\frac{\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}-1}}{\mathit{r}}}$$ This total fund must be equal to the cost of replacement of the equipment, i.e., $$\mathrm{\mathit{X-S}\:=\:\frac{\mathit{p}\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}-1}}{\mathit{r}}}$$ Therefore, the amount of sinking fund is, $$\mathrm{\mathrm{Sinking \:fund,}\:\mathit{p}\:=\:\mathrm{\left ( \mathit{X-S} \right )}\mathrm{\left[ \frac{\mathit{r}}{\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}-1}\right ]}}$$ Where, $$\mathrm{\mathrm{Sinking \:fund\:factor}\:=\:\mathrm{\left[ \frac{\mathit{r}}{\mathrm{\left ( 1\:+\:\mathit{r} \right )^{\mathit{n}}}-1}\right ]}}$$ Numerical ExampleA transformer costs Rs 150000 and has a useful life of 25 years. If the scrap value of the transformer is 10000 and the rate of annual compound interest is 7$\%$. Then, calculate the amount to be saved annually for the replacement of the transformer after the end of 25 years. Solution Given data is −
Then, the annual amount for sinking fund is given by, $$\mathrm{\mathrm{Sinking\: fund,}\mathit{p}\:=\:\mathrm{\left ( \mathit{X-S} \right)}\mathrm{\left[\frac{\mathit{r}}{\mathrm{\left(1\:+\:\mathit{r}\right)^{\mathit{n}}-1}} \right ]}}$$ $$\mathrm{\mathit{p}\:=\:\mathrm{\left (150000-10000 \right )}\mathrm{\left[ \frac{0.07}{\mathrm{\left( 1\:+\:0.07 \right )^{25}}-1}\right ]}\:=\:\mathrm{Rs\:2212.18}}$$
Updated on 15-Feb-2022 09:23:42
In what method of computing depreciation where it assumes that a sinking fund is established?Sinking fund method is a method of calculating depreciation for an asset in which apart from calculating depreciation, it also keeps aside a fund for replacing the asset at the end of its useful life. This method is used when the assets that need to be replaced are of high cost.
What is the sinking fund method of depreciation?The sinking fund method is a technique for depreciating an asset while generating enough money to replace it at the end of its useful life. As depreciation charges are incurred to reflect the asset's falling value, a matching amount of cash is invested.
In what method of computing depreciation where it assumes that the loss in value is directly proportional to the age of the equipment or asset?The Straight Line Method : States that the loss in value is considered to be directly proportional to the age of the assets.
In what method of computing depreciation where it assumes that the annual cost?The method of computing depreciation where itassumes that annual cost of depreciation is a fixed. percentage of the book value at the beginning to year: 1 Sum-of year digit method.
|