Assume that you are the program manager of a project. You are provided with the table below that requires you to determine the crashing cost for each activity in the project of 4 months.

TABLE: PROJECT CRASHING

\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multirow{2}{}{Activity} & \multicolumn{2}{|c|}{NORMAL} & \multicolumn{2}{|c|}{CRASHED} & \multirow{2}{}{\begin{tabular}{c}
CRASHING\\
COST\\
Per day
\end{tabular}} \\
\cline{2-5}
& \begin{tabular}{c}
Duration \\
(Days)
\end{tabular} & \begin{tabular}{c}
Cost \\
(Million)
\end{tabular} & \begin{tabular}{c}
Duration \\
(Days)
\end{tabular} & \begin{tabular}{c}
Cost \\
(Million)
\end{tabular} \\
\hline
A & 5 & & (Days) & & \\
\hline
B & 17 & 100 & 3 & 1500 & \\
\hline
C & 1 & 700 & 6 & 1000 & \\
\hline
& 3 & 2500 & 2 & 4000 & \\
\hline
D & 5 & 1500 & 5 & 1500 & \\
\hline
E & 9 & & & & \\
\hline
F & & 3750 & 6 & 9000 & \\
\hline
& 4 & 1600 & 3 & 2500 & \\
\hline
G & 6 & 2400 & 4 & 3000 & \\
\hline
H & 8 & & & & \\
\hline
& & 9000 & 5 & 15000 & \\
\hline
\end{tabular}

Required:

Determine the crashing cost for each activity in the project, assuming that activity D cannot be shortened. (10 marks)



Answer :

Certainly! Let's determine the crashing cost per day for each activity in the project based on the provided data, under the assumption that Activity D cannot be shortened.

### Given Table Data
Here is the relevant data extracted from the table:

- Activity B
- Normal Duration: 17 days
- Normal Cost: 100 million
- Crashed Duration: 3 days
- Crashed Cost: 1500 million

- Activity C
- Normal Duration: 1 day
- Normal Cost: 700 million
- Crashed Duration: 6 days
- Crashed Cost: 1000 million

- Activity D
- Normal Duration: 5 days
- Normal Cost: 1500 million
- Crashed Duration: 5 days (cannot be shortened)
- Crashed Cost: 1500 million

- Activity G
- Normal Duration: 6 days
- Normal Cost: 2400 million
- Crashed Duration: 4 days
- Crashed Cost: 3000 million

### Step-by-Step Solution:

#### Activity B
1. Normal Duration: 17 days
2. Normal Cost: 100 million
3. Crashed Duration: 3 days (can reduce by 14 days)
4. Crashed Cost: 1500 million

- Crashing Cost per Day:
[tex]\[ \text{Crashing Cost per Day} = \frac{\text{Crashed Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crashed Duration}} \][/tex]
[tex]\[ \text{Crashing Cost per Day} = \frac{1500 - 100}{17 - 3} = \frac{1400}{14} = 100 \text{ million} \][/tex]

#### Activity C
In this case, the normal duration is less than the crashed duration (i.e., crashing implies an increase in the duration), hence calculations for crashing cost are not needed. We'll consider the crashing cost per day as 0.

#### Activity D
Since Activity D cannot be shortened, the crashing cost per day is 0.

#### Activity G
1. Normal Duration: 6 days
2. Normal Cost: 2400 million
3. Crashed Duration: 4 days (can reduce by 2 days)
4. Crashed Cost: 3000 million

- Crashing Cost per Day:
[tex]\[ \text{Crashing Cost per Day} = \frac{\text{Crashed Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crashed Duration}} \][/tex]
[tex]\[ \text{Crashing Cost per Day} = \frac{3000 - 2400}{6 - 4} = \frac{600}{2} = 300 \text{ million} \][/tex]

### Final Crashing Costs per Day:

- Activity B: 100 million per day
- Activity C: 0 million per day
- Activity D: 0 million per day (cannot be shortened)
- Activity G: 300 million per day

So, the crashing costs for each activity are:
[tex]\[ \{\text{B}: 100 \text{ million per day}, \text{C}: 0 \text{ million per day}, \text{D}: 0 \text{ million per day}, \text{G}: 300 \text{ million per day}\} \][/tex]

In conclusion,
[tex]\[ \boxed{\text{B} = 100 \text{ million/day}, \text{C} = 0 \text{ million/day}, \text{D} = 0 \text{ million/day}, \text{G} = 300 \text{ million/day}} \][/tex]