Chapter 6, Problem 43P

To determine

(a)

The methods to be used to produce the biofuels.

Answer to Problem 43P

The algae are good for the making of biofuels.

Explanation of Solution

Given:

Number of barrels of ethanol to be sold per year is $150000$.

The time period is $10$ years beginning in the end of year $4$.

The oil firm will pay NPB $\$10\text{M}$ annually beginning from $\text{EOY}0$ to $\text{EOY3}$ and then $\$110$ per barrel.

The NPB uses an interest rate of $15\%$.

Annual increase in O&M cost is $2\%$.

Annual increase in raw material cost is $3\%$.

Calculation:

Calculate the revenue earned in the next $10$ years.

$\begin{array}{c}\text{Revenue}\text{earned}=\$110\times 150000\\ =\$16500000\end{array}$

Calculate the annual amount.

$\begin{array}{c}A=\$10000000\left(\frac{A}{P},i\%,n\right)+\$16500000\left(\frac{A}{P},i\%,n\right)\\ =\$10000000\left(\frac{A}{P},15\%,3\right)+\$16500000\left(\frac{A}{P},15\%,10\right)\end{array}$ ...... (I)

Write the equation for the factor $\left(\frac{A}{P},i\%,n\right)$.

$\left(\frac{A}{P},i\%,n\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$ ...... (II)

Here, annual cost is A, present worth is P, rate of interest is i, and number of year is n.

Calculate the factor $\left(\frac{A}{P},15\%,3\right)$.

Substitute $15\%$ i and $3$ years for n in Equation (II).

$\begin{array}{c}\left(\frac{A}{P},15\%,3\right)=\left[\frac{0.15{\left(1+0.15\right)}^3}{{\left(1+0.15\right)}^3-1}\right]\\ =0.4380\end{array}$

Calculate the factor $\left(\frac{A}{P},15\%,10\right)$

Substitute $15\%$ i and $10$ years for n in Equation (II).

$\begin{array}{c}\left(\frac{A}{P},15\%,10\right)=\left[\frac{0.15{\left(1+0.15\right)}^{10}}{{\left(1+0.15\right)}^{10}-1}\right]\\ =0.1993\end{array}$

Substitute $0.4380$ for $\left(\frac{A}{P},15\%,3\right)$, and $0.1993$ for $\left(\frac{A}{P},15\%,10\right)$ in Equation (I).

$\begin{array}{c}A=\$10000000\times 0.4380+\$16500000\times 0.1993\\ =\$7668450\end{array}$

Therefore, the benefit from payment received for both alternatives is $\$7668450$.

Alternative Corn

Calculate the annual cost.

$\begin{array}{c}A=F\left(\frac{A}{F},i\%,n\right)\\ A=F\left[\frac{i}{{\left(1+i\right)}^n-1}\right]\end{array}$ ...... (III)

Here, future value is F.

Substitute $\$3000000$ for F, $15\%$ for i, and $13$ years for n in Equation (III).

$\begin{array}{c}A=\$3000000\left[\frac{0.15}{{\left(0.15+1\right)}^{13}-1}\right]\\ =\$3000000\times 0.0291\\ =\$87300\end{array}$

Therefore, salvage value will be $\$87300$ after $13$ years.

Calculate the equivalent uniform annual benefit (EUAB) which is the sum of total revenue earned during the first three years, next ten year plus the salvage value.

$\begin{array}{c}\text{EUAB}=\$7668450+87300\\ =\$7755750\end{array}$

It is required to calculate the annuity value of the corn alternative after $13$ years.

$A=P\left(\frac{A}{P},15\%,13\right)$ ...... (IV)

Calculate the factor $\left(\frac{A}{P},15\%,13\right)$.

Substitute $15\%$ i, and $13$ years for n in Equation (II).

$\begin{array}{c}\left(\frac{A}{P},15\%,13\right)=\left[\frac{0.15{\left(1+0.15\right)}^{13}}{{\left(1+0.15\right)}^{13}-1}\right]\\ =0.1791\end{array}$

Substitute $\$1900000$ for P, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (IV).

$\begin{array}{c}A=\$1900000\times 0.1791\\ =\$340290\end{array}$

Therefore, the value of land after $13$ years is $\$340290$.

Calculate the annuity value of facility construction cost.

Substitute $\$5300000$ for P, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (IV).

$\begin{array}{c}A=\$5300000\times 0.1791\\ =\$949230\end{array}$

Therefore, the construction cost after $13$ year will be $\$949230$.

Annual O&M expenses starting from the year $4$ are calculated by using the annul gradient uniform series as shown below.

$\begin{array}{c}A=P\left(\frac{A}{P},15\%,13\right)+G\left(\frac{A}{G},2\%,13\right)\\ A=P\left(\frac{A}{P},15\%,13\right)+G\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]\end{array}$ ...... (V)

Calculate the factor $\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]$

Substitute $2\%$ for i, and $13$ years for n.

$\begin{array}{c}\left(\frac{A}{G},2\%,13\right)=\left[\frac{1}{0.02}-\frac{13}{{\left(1+0.02\right)}^{13}-1}\right]\\ =5.5872\end{array}$

Substitute $\$2450000$ for P and G, $5.5872$ for $\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]$, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (V).

$\begin{array}{c}A=\left(\$2450000\times 0.1791\right)+\left(\$2450000\times 5.5872\right)\\ =\$14460390\end{array}$

Raw materials expenses starting from the year $4$ are calculated by using annul gradient uniform series as shown below.

Substitute $\$1500000$ for P and G, $5.5872$ for $\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]$, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (V).

$\begin{array}{c}A=\left(\$1500000\times 0.1791\right)+\left(\$1500000\times 5.5872\right)\\ =\$8649450\end{array}$

Calculate the equivalent uniform annual cost (EUAC) which is sum of land purchasing cost, amount spent in construction, annual O&M expenses, and cost of raw material.

$\begin{array}{c}\text{EUAC}=\$340290+\$949230+\$14460390+\$8649450\\ =\$24399360\end{array}$

Calculate the Equivalent uniform annual worth.

$\text{EUAW}=\text{EUAB}-\text{EUAC}$

Substitute $\$24399360$ for EUAC, and $\$7755750$ for EUAB.

$\begin{array}{c}\text{EUAW}=\$7755750-\$24399360\\ =-\$16643610\end{array}$

Alternative Algae

Calculate the annual cost.

Substitute $\$3600000$ for F, $15\%$ for i, and $13$ years for n in Equation (III).

$\begin{array}{c}A=\$3600000\left[\frac{0.15}{{\left(0.15+1\right)}^{13}-1}\right]\\ =\$3600000\times 0.0291\\ =\$104760\end{array}$

Calculate the equivalent uniform annual benefit (EUAB) which is sum of total revenue earned during first three year, next ten year plus salvage value.

$\begin{array}{c}\text{EUAB}=\$7668450+\$104760\\ =\$7773210\end{array}$

It is required to calculate the annuity value of land of algae alternative after $13$ years.

$A=P\left(\frac{A}{P},15\%,13\right)$

Substitute $\$3800000$ for P, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (IV).

$\begin{array}{c}A=\$3800000\times 0.1791\\ =\$680580\end{array}$

It is required to calculate the annuity value of construction facility of algae alternative after $13$ years.

$A=P\left(\frac{A}{P},15\%,13\right)$

Substitute $\$7100000$ for P, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (IV).

$\begin{array}{c}A=\$\$7100000\times 0.1791\\ =\$1271610\end{array}$

Annual O&M expenses starting from the year $4$ are calculated by using annul gradient uniform series as shown below.

Substitute $\$2800000$ for P and G, $5.5872$ for $\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]$, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (V).

$\begin{array}{c}A=\left(\$2800000\times 0.1791\right)+\left(\$2800000\times 5.5872\right)\\ =\$16024400\end{array}$

Raw material expenses starting from the year $4$ are calculated by using annul gradient uniform series as shown below.

Substitute $\$250000$ for P and G, $5.5872$ for $\left[\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right]$, and $0.1791$ for $\left(\frac{A}{P},15\%,13\right)$ in Equation (V).

$\begin{array}{c}A=\left(\$250000\times 0.1791\right)+\left(\$250000\times 5.5872\right)\\ =\$1441575\end{array}$

Calculate the equivalent uniform annual cost (EUAC) which is sum of land purchasing cost, amount spent in construction, annual O&M expenses, and cost of raw material.

$\begin{array}{c}\text{EUAC}=\$680580+\$1271610+\$16024400+\$1441575\\ =\$19418165\end{array}$

Calculate the Equivalent uniform annual worth.

$\text{EUAW}=\text{EUAB}-\text{EUAC}$

Substitute $\$19418165$ for EUAC, and $\$7773210$ for EUAB.

$\begin{array}{c}\text{EUAW}=\$7773210-\$19418165\\ =-\$11644955\end{array}$.

Conclusion:

The equivalent uniform annual worth for algae is more than that for corn.

Therefore, the algae are good for the making the biofuels.

To determine

(b)

The properties, types and use of the biofuels.

Explanation of Solution

Biofuels:

The fuel derived from the biomass product such as plants, fiber, algae, and animal waste is known as biofuels. Biofuels are renewable energy sources. They are the best alternatives for fossil fuels and petroleum products.

Ethanol and biodiesel are the final products of the biofuels and these products are blended with the gasoline which can be used as fuel in the automotive industries.

Pros of biofuels:

1. Biofuels are less expensive.
2. Large availability of source materials.
3. Renewability.
4. Economic stimulation.
5. Low carbon emission.

Cons of bifuels:

1. Biofuels have low energy outputs.
2. Refined biofuels are more expensive because of high energy outputs.
3. Biofuels are also responsible for increase in food prices.
4. Biofuels are also responsible for shortage of food.
5. Large amount of water is used in its production.

Non-auto industry uses of biofuels:

1. Used for energy generation.
2. Provide heat.
3. Charging electronics.
4. Used to clean oil spills and grease.
5. Used in cooking
6. As a lubricant.
7. Used to remove paints and adhesives.