Chapter 13, Problem 19P

(a):

To determine

Calculate the usage maintenance cost.

Explanation of Solution

Construction cost of gravel road (CCG) is $1,025,000. Its maintenance cost (MCG) is $355,000. Time period (n) is 3 years. Construction cost of bituminous road (CCB) is $3,525,000. Time period (n1) is 10 years. Interest rate is 8%.

Maintenance cost of bituminous road (MCB) can be calculated as follows:

$\begin{array}{c}\text{CCB}\left(\frac{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n1}}}{{\left(\text{1}+\text{i}\right)}^{\text{n1}}-1}\right)+\text{MCB}=\text{CCG}\left(\frac{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}\right)+\text{MCG}\\ \text{3,525,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{10}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{10}}-1}\right)+\text{MCB}=\text{1,025,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{3}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{3}}-1}\right)+\text{355,000}\\ \text{3,525,000}\left(0.14903\right)+\text{MCB}=\text{1,025,000}\left(0.38803\right)+\text{355,000}\\ \text{MCB}=397,730.75+\text{355,000}-525,330.75\\ =227,400\end{array}$

Maintenance cost of bituminous road is $227,400.

(b):

To determine

Calculate the number of years.

Explanation of Solution

Maintenance cost of gravel road. Increases (MCGI) by 30%.

Time period for gravel road (n) can be calculated as follows:

$\begin{array}{c}\text{CCB}\left(\frac{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n1}}}{{\left(\text{1}+\text{i}\right)}^{\text{n1}}-1}\right)+\text{MCB}=\text{CCG}\left(\frac{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}\right)+\text{MCG}\times \left(1+\text{MCGI}\right)\\ \text{3,525,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{10}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{10}}-1}\right)+\text{227,400}=\text{1,025,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}\right)+\text{355,000}\left(1+0.3\right)\\ \text{3,525,000}\left(0.14903\right)+\text{227,400}=\text{1,025,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}\right)+461,500\\ \text{1,025,000}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}\right)=525,330.75+\text{227,400}-\text{461,5000}\\ \left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}\right)=\frac{291,230.75}{1,025,000}\\ \left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{n}}-1}\right)=0.28413\end{array}$

Substitute the time period as 4.3 years in the above calculation by trial and error method to verify the years.

$\begin{array}{l}\left(\frac{\text{0}\text{.08}{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{4}\text{.3}}}{{\left(\text{1}+\text{0}\text{.08}\right)}^{\text{4}\text{.3}}-1}\right)=0.28413\\ \left(\frac{\text{0}\text{.08}\left(1.39227\right)}{1.39227-1}\right)=0.28413\\ 0.28394\approx 0.28413\end{array}$

Since the calculated value is equal to the annual value factor to the present value, it is confirmed that the number of years is 4.3.