Chapter 14, Problem 72P

To determine

(a)

The labour cost for the year $1$, $2$ and $3$.

Answer to Problem 72P

Labour cost for year $1$ is $\$89250$.

Labour cost for year $2$ is $\$93266$.

Labour cost for year $3$ is $\$97463$.

Explanation of Solution

Given:

The increase in the price after $12$ year is $55\%$.

Further increase in price for $8$ year is $25\%$.

Calculation:

Calculate the future cost of structural steel for year $1$.

$F=P{\left(1+i\right)}^n$ ...... (I)

Here, the future cost is F, the present value is P, the rate of interest is i and the time period is n.

Substitute $\$120000$ for $P$ and $0.043$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$120000{\left(1+0.043\right)}^1\\ =\$120000\times 1.043\\ =\$125160\end{array}$

Calculate the future cost of structural steel for year $2$.

Substitute $\$125160$ for $P$ and $0.032$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$125160{\left(1+0.032\right)}^1\\ =\$125160\times 1.032\\ =\$129165\end{array}$

Calculate the future cost of structural steel for year $3$.

Substitute $\$129165$ for $P$ and $0.066$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$129165{\left(1+0.066\right)}^1\\ =\$129165\times 1.066\\ =\$137690\end{array}$

Calculate the future cost of roofing material for year $1$.

Substitute $\$14000$ for $P$ and $0.02$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$14000{\left(1+0.02\right)}^1\\ =\$14000\times 1.02\\ =\$14280\end{array}$

Calculate the future cost of roofing material for year $2$.

Substitute $\$14280$ for $P$ and $0.025$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$14280{\left(1+0.025\right)}^1\\ =\$14280\times 1.025\\ =\$14637\end{array}$

Calculate the future cost of roofing material for year $3$.

Substitute $\$14637$ for $P$ and $0.03$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$14637{\left(1+0.03\right)}^1\\ =\$14637\times 1.03\\ =\$15076\end{array}$

Calculate the future cost of heating Equipment for year $1$.

Substitute $\$35000$ for $P$ and $0.016$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$35000{\left(1+0.016\right)}^1\\ =\$35000\times 1.016\\ =\$35560\end{array}$

Calculate the future cost of heating Equipment for year $2$.

Substitute $\$35560$ for $P$ and $0.021$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$35560{\left(1+0.021\right)}^1\\ =\$35560\times 1.021\\ =\$36307\end{array}$

Calculate the future cost of heating Equipment for year $3$.

Substitute $\$36307$ for $P$ and $0.036$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$36307{\left(1+0.036\right)}^1\\ =\$36307\times 1.036\\ =\$37614\end{array}$

Calculate the future cost of insulation material for year $1$.

Substitute $\$9000$ for $P$ and $0.058$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$9000{\left(1+0.058\right)}^1\\ =\$9000\times 1.058\\ =\$9522\end{array}$

Calculate the future cost of insulation material for year $2$.

Substitute $\$9522$ for $P$ and $0.06$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$9522{\left(1+0.06\right)}^1\\ =\$9522\times 1.06\\ =\$10093\end{array}$

Calculate the future cost of insulation material for year $3$.

Substitute $\$10093$ for $P$ and $0.075$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$10093{\left(1+0.075\right)}^1\\ =\$10093\times 1.075\\ =\$10850\end{array}$

Calculate the future cost of labour for year $1$.

Substitute $\$85000$ for $P$ and $0.05$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$85000{\left(1+0.05\right)}^1\\ =\$85000\times 1.05\\ =\$89250\end{array}$

Calculate the future cost of labour for year $2$.

Substitute $\$89250$ for $P$ and $0.045$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$89250{\left(1+0.045\right)}^1\\ =\$89250\times 1.045\\ =\$93266\end{array}$

Calculate the future cost of labour for year $3$.

Substitute $\$93266$ for $P$ and $0.045$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}F=\$93266{\left(1+0.45\right)}^1\\ =\$93266\times 1.045\\ =\$97463\end{array}$

Summarise the above calculated values in a tabular form.

Items	Year $1$	Year $2$	Year $3$
Structural concrete	$\$125160$	$\$129165$	$\$137690$
Roofing material	$\$14280$	$\$14637$	$\$15076$
Heating equipment	$\$35560$	$\$36307$	$\$37614$
Insulation material	$\$9522$	$\$10093$	$\$10850$
Labour	$\$89250$	$\$93266$	$\$97463$
Total	$\$273722$	$\$283467$	$\$298693$

Conclusion:

Labour cost for year $1$ is $\$89250$.

Labour cost for year $2$ is $\$93266$.

Labour cost for year $3$ is $\$97463$.

To determine

(b)

The average percentage increase in labour cost over the $3$ year period.

Answer to Problem 72P

The average percentage increase in labour cost over the $3$ year period is $9.8\%$.

Explanation of Solution

Calculation:

Calculate the average percentage increase of labour cost.

$\left(\text{Average}\text{Increase}\text{of}\text{labour}\text{cost}\right)=\frac{{\displaystyle \sum \left(\text{labour}\text{cost}\text{for}\text{year}\text{n}\right)}}{n}$ ...... (II)

Substitute $\$89250$ for labour cost for year $1$, $\$93266$ for Labour cost for year $2$, $\$97463$ for Labour cost for year $3$ and $3$ for n in Equation (II).

$\begin{array}{c}\text{Average}\text{Increase}\text{of}\text{labour}\text{cost}=\frac{\$89250+\$93266+\$97463}{3}\\ =\frac{\$279979}{3}\\ =\$93326\end{array}$

Calculate the average percentage increase of labour cost.

$\text{Average}\text{Increase}\text{of}\text{labour}\text{cost}=\frac{\left(\text{Increase}\text{cost}-\text{Initial}\text{cost}\right)}{\left(\text{Initial}\text{cost}\right)}\times 100\%$ .... (III)

Substitute $\$93326$ for $\text{Increase}\text{cost}$ and $\$85000$ for $\text{Initial}\text{cost}$ in Equation (III).

$\begin{array}{c}\text{Average}\text{Increase}\text{of}\text{labour}\text{cost}=\left(\frac{\$93326-\$85000}{\$85000}\right)\times 100\%\\ =\frac{\$8326}{\$85000}\times 100\%\\ =9.8\%\end{array}$.

Conclusion:

Thus, the average percentage increase in labour cost over the $3$ year period is $9.8\%$.

To determine

(c)

The present worth of the insulation cost.

Answer to Problem 72P

The present worth of the insulation cost is $\$19632$.

Explanation of Solution

Calculation:

Write the expression to calculate the present worth of the insulation cost.

$PW=\frac{F_1}{{\left(1+i\right)}^1}+\frac{F_2}{{\left(1+i\right)}^2}+\frac{F_3}{{\left(1+i\right)}^3}$ ...... (IV)

Here, the present worth is PW.

Substitute $\$9522$ for $F_1$, $\$10093$ for $F_2$, $\$10850$ for $F_3$ and $0.25$ for $i$ in Equation (IV).

$\begin{array}{c}PW=\frac{\$9522}{{\left(1+0.25\right)}^1}+\frac{\$10093}{{\left(1+0.25\right)}^2}+\frac{\$10850}{{\left(1+0.25\right)}^3}\\ =\$7617.6+\$6459.52+\$5555.2\\ =\$19632\end{array}$.

Conclusion:

Thus, the present worth of the insulation cost is $\$19632$.

To determine

(d)

The future worth of the labour and insulation material cost portion of the project.

Answer to Problem 72P

The future worth of the labour and insulation material cost portion of the project is $\$391843$.

Explanation of Solution

Calculation:

Write the expression to calculate the future worth of labour and insulation cost.

$FW=\left(P_{L1}+P_{I1}\right){\left(1+i\right)}^2+\left(P_{L2}+P_{I2}\right){\left(1+i\right)}^1+\left(P_{L3}+P_{I3}\right){\left(1+i\right)}^0$ ..... (V)

Here, the future worth is FW, the future cost of the labour is $P_L$ and the future cost of the insulation material is $P_I$.

Substitute $\$89250$ for $P_{L1}$, $\$9522$ for $P_{I1}$, $\$93266$ for $P_{L2}$, $\$10093$ for $P_{I2}$, $\$97463$ for $P_{L3}$, $\$100850$ for $P_{I3}$ and $0.25$ for $i$ in Equation (V).

$\begin{array}{c}FW=\left(\begin{array}{l}\left(\$89250+\$9522\right){\left(1+0.25\right)}^2+\left(\$93266+\$10093\right){\left(1+0.25\right)}^1\\ +\left(\$97463+\$100850\right){\left(1+0.25\right)}^0\end{array}\right)\\ =\$154331.25+\$129198.75+\$108313\\ =\$391843\end{array}$.

Conclusion:

Thus, the future worth of the labour and insulation material cost portion of the project is $\$391843$.

To determine

(e)

The present worth of the total construction project.

Answer to Problem 72P

The present worth of the total construction project is $\$553327.296$.

Explanation of Solution

Calculation:

Write the expression to calculate the present worth of the total construction project.

$PW=\frac{F_{T1}}{{\left(1+i\right)}^1}+\frac{F_{T2}}{{\left(1+i\right)}^2}+\frac{F_{T3}}{{\left(1+i\right)}^3}$ ...... (VI)

Here, the total construction cost is $F_T$.

Substitute $\$273722$ for $F_{T1}$, $\$283467$ for $F_{T2}$. $\$298693$ for $F_{T3}$ and $0.25$ for $i$ in Equation (VI).

$\begin{array}{c}PW=\frac{\$273722}{{\left(1+0.25\right)}^1}+\frac{\$283467}{{\left(1+0.25\right)}^2}+\frac{\$298693}{{\left(1+0.25\right)}^3}\\ =\$218977.6+\$181418.88+\$152930.816\\ =\$553327.296\end{array}$.

Conclusion:

Thus, the present worth of the total construction project is $\$553327.296$.