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Engineering Economics
Straight line method Declining balance method Sum-of-year digit method Sinking fund method Straight line method Declining balance method Sum-of-year digit method
Sinking fund method ANSWER DOWNLOAD EXAMIANS APP 1037 1043 1033 1053 1037 1043 1033 1053
ANSWER DOWNLOAD EXAMIANS APP Demand factor Sinking fund factor
Engineering Economics
A manufacturer produces certain items at a labor cost of P 115 each, material cost of P 76
each and variable cost of P 2.32 each. If the item has a unit price of P 600, how many units must be manufactured each month for the manufacturer to break even if the monthly overhead is P428,000 Engineering Economics
A mathematical expression also known as the present value of annuity of one is called ______.
Present worth factor
Load factor
Demand factor
Sinking fund factor
Present worth factor
Load factor
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Engineering Economics
What do you call a one-time credit against taxes?
Tax credit
Revenue credit
Due credit
Credible credit
Tax credit
Revenue credit
Due credit
Credible credit
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Engineering Economics
A farmer selling eggs at
50 pesos a dozen gains 20%. If he sells the eggs at the same price after the costs of the eggs rises by 12.5%, how much will be his new gain in percent?
0.0665
0.0658
0.0689
0.0612
0.0665
0.0658
0.0689
0.0612
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Engineering Economics
What refers to the interest rate at which the present work of the cash flow on a project is zero of the interest earned by an investment?
Return of investment
Rate of return
Yield
Economic return
Return of investment
Rate of return
Yield
Economic return
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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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