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We can use the below table to find the critical pathof the project:

The total time of each path is given below:

The longest timeof 25 days belongs to the path A-C-D-F-G and this path would be the critical pathof the project.

$$\begin{gathered} Totalcrashtime=Normaltime-crashtime \\ Totalcrash\cos t=Normal\cos t-crash\cos t \\ Crash\cos tperweek=\frac{Totalcrash\cos t}{Totalcrashtime} \end{gathered}$$

By using the above equation we can find the crash costsfor all the activities using excel:

role="math" localid="1650869487593" $$\begin{gathered} Totalnormal\cos t=\$\left(7000+5000+9000+3000+2000+4000+5000\right) \\ =\$35,000 \end{gathered}$$

Crash movement A is the best option because it has the lowest crash cost and is also on the critical path. It can only be crashed individually. For one day, crash it. The path duration after crashing is as follows:

$$\begin{gathered} Costofcrashing=Numberofweekscrashed×Crash\cos tperweek \\ =1×1000 \\ =1000 \\ Total\cos tcrashing=35,000+1000 \\ =\$36,000 \end{gathered}$$

Crash movement D is the best choice because it has the cheapest crash cost and is located on both critical paths. It can only be crashed individually. For one day, crash it. The path duration after crashing is as follows:

$$\begin{gathered} Costofcrashing=Numberofweekcrashed×Crash\cos tperweek \\ =1×1500 \\ =1500 \\ Total\cos tofcrashing=36000+1500 \\ =\$37,500 \end{gathered}$$

Crash movement F is the top pick because it has the cheapest crash cost and is also on the critical path. This can only be crashed individually. For one day, crash it. Ignoring the fact that it is much more pricey than E, it can minimize the duration of the two approaches. The path duration after crashing is as follows:

$$\begin{gathered} Costofcrashing=Numberofweekcrashed×Crash\cos tperweek \\ =1×1500 \\ =1500 \\ Total\cos tofcrashing=37500+1500 \\ =\$39,000 \end{gathered}$$

Crash movement C is the top pick and has the cheapest crash cost and is also on the critical track. Can only be crashed individually. Once per day, crash it. Because G has a larger cost than C, C should be crushed using the least crash cost model. The path duration after crashing is as follows:

$$\begin{gathered} Costofcrashing=Numberofweekscrashed×Crashed\cos tperweek \\ =1×1200 \\ =1200 \\ Total\cos tofcrashing=39000+1200 \\ =\$40,200 \end{gathered}$$