Chapter 9, Problem 19P

(a):

To determine

Calculate the benefit–cost ratio.

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Item	Cash flow
First cost (F)	13,000,000
Benefit (B) per year	3,800,000
Future value of disbenefit (DB)	6,750,000
Maintenance cost (MC) per year	400,000
Time period (n)	20
Discount rate (i)	10%

Benefit–cost ratio (BC) can be calculated as follows:

$\begin{array}{c}\text{BC}=\frac{\text{B}-\text{DB}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)}{\text{F}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{MC}}\\ =\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{{\left(1+0.1\right)}^{\text{20}}-1}\right)}{\text{13,000,000}\left(\frac{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{20}}}{{\left(1+0.1\right)}^{\text{20}}-1}\right)+\text{400,000}}\\ =\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{6.7274999-1}\right)}{\text{13,000,000}\left(\frac{\text{0}\text{.1}\left(6.7274999\right)}{6.7274999-1}\right)+\text{400,000}}\\ =\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{5.7274999}\right)}{\text{13,000,000}\left(\frac{\text{0}.6727499}{5.7274999}\right)+\text{400,000}}\\ =\frac{\text{3,800,000}-\text{6,750,000}\left(0.0174596\right)}{\text{13,000,000}\left(0.1174596\right)+\text{400,000}}\\ =\frac{\text{3,800,000}-117,852.3}{1,526,974.8+\text{400,000}}\\ =\frac{3,682,147.7}{1,926,974.8}\\ =1.91\end{array}$

Benefit–cost ratio is 1.91. Since the benefit–cost ratio is greater than 1, the project is economically justified.

(b):

To determine

Minimum value of the first cost.

Explanation of Solution

The project is acceptable when at least the benefit–cost ration (BC) is equal to 1. If it is less than 1, then the project would not be accepted. The minimum value of the first cost (F) can be calculated as follows:

$\begin{array}{c}\text{BC}=\frac{\text{B}-\text{DB}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)}{\text{F}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{MC}}\\ 1=\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{{\left(1+0.1\right)}^{\text{20}}-1}\right)}{\text{F}\left(\frac{\text{0}\text{.1}{\left(1+0.1\right)}^{\text{20}}}{{\left(1+0.1\right)}^{\text{20}}-1}\right)+\text{400,000}}\\ 1=\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{6.7274999-1}\right)}{\text{F}\left(\frac{\text{0}\text{.1}\left(6.7274999\right)}{6.7274999-1}\right)+\text{400,000}}\\ 1=\frac{\text{3,800,000}-\text{6,750,000}\left(\frac{\text{0}\text{.1}}{5.7274999}\right)}{\text{F}\left(\frac{\text{0}.6727499}{5.7274999}\right)+\text{400,000}}\\ 1=\frac{\text{3,800,000}-\text{6,750,000}\left(0.0174596\right)}{\text{F}\left(0.1174596\right)+\text{400,000}}\\ 1=\frac{\text{3,800,000}-117,852.3}{\text{F}\left(0.1174596\right)+\text{400,000}}\\ 1\left(\text{F}\left(0.1174596\right)+\text{400,000}\right)=3,682,147.7\\ \text{F}\left(0.1174596\right)=3,682,147.7-\text{400,000}\\ \text{F}=\frac{3,282,147.7}{\left(0.1174596\right)}\\ =27,942,779.47\end{array}$

To accept this project, the first cost must be greater than $27,942,779.47. If the first cost is less than $27,942,779.47, then the project would not be accepted.