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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
Engineering Economics
The monthly demand for ice cans being manufactured by Mr. Camus is 3200 pieces. With a manual
operated guillotine, the unit cutting cost is P25.00. An electrically operated hydraulic guillotine was offered to Mr. Camus at a price of P275,000.00 and which cuts by 30% the unit cutting cost. Disregarding the cost of money, how many months will Mr. Camus be able to recover the cost of the machine if he decides to buy now? Engineering Economics
The ratio of current assets to current liabilities is known as
Liquidity ratio
Current ratio
Acid-Test (or Quick) ratio
Debts ratio
Liquidity ratio
Current ratio
ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics
Present worth Annuity (PWA) is generally known as
Future annuities
Income annuities
All of these
Premium annuities
Future annuities
Income annuities
All of these
Premium annuities
ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics
What is an accounting term that represents an inventory account adjustment?
Cost of goods sold
Cost accounting
Overhead cost
Standard cost
Cost of goods sold
Cost accounting
Overhead cost
Standard cost
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Engineering Economics
If there are many sellers and few buyers, the market situation is _________ .
Oligopoly
Oligopsony
Monopoly
Duopsony
Oligopoly
Oligopsony
Monopoly
Duopsony
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MORE MCQ ON Engineering Economics
Depreciation of Power Station Equipment
The 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 Depreciation
In 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,
X = Initial Value of Equipment
S = Scrap value after useful life
n = Useful life of equipement in years
r = Annual interest rate
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 Example
A 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 −
Initial cost of transformer, X = Rs.150000
Scrap value of transformer, S = Rs.10000
Useful life of transformer, n = 25 years
Annual rate of interest, r = 7$\%$
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
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