Why only ABG? Why not FC or something else? How do you determine that there are only three activities for crashing? Which activities are the most likely candidates for crashing $($i$.$e$.,$ which are the most cost $-$ effective to crash$)?$

$\backslash$begin$\{$tabular$\}\{|$c$|$c$|$c$|$c$|$c$|$c$|\}$

$\backslash$hline $\backslash$multirow$[$b$]\{2\}\{*\}\{$Activity$\}$ & $\backslash$multirow$[$b$]\{2\}\{*\}\{$Predecessors$\}$ & $\backslash$multicolumn$\{2\}\{|$c$|\}\{$Normal$\}$ & $\backslash$multicolumn$\{2\}\{|$c$|\}\{$Crashed$\}\backslash \backslash$

$\backslash$hline & & Duration & Cost & Duration & Cost $\backslash \backslash$

$\backslash$hline A & $-$ & $5$ days & $\backslash$$ $1,000$ & $3$ days & $\backslash$$ $1,500\backslash \backslash$

$\backslash$hline B & A & $7$ days & $700$ & $6$ days & $1,000\backslash \backslash$

$\backslash$hline C & A & $3$ days & $2,500$ & $2$ days & $4,000\backslash \backslash$

$\backslash$hline D & A & $5$ days & $1,500$ & $5$ days & $1,500\backslash \backslash$

$\backslash$hline E & $\backslash ($ C$,$ D $\backslash )$ & $9$ days & $3,750$ & $6$ days & $9,000\backslash \backslash$

$\backslash$hline F & B & $4$ days & $1,600$ & $3$ days & $2,500\backslash \backslash$

$\backslash$hline G & D & $6$ days & $2,400$ & $4$ days & $3,000\backslash \backslash$

$\backslash$hline H & E$,$ F$,$ G & $8$ days & $9,000$ & $5$ days & $15,000\backslash \backslash$

$\backslash$hline Total costs $\backslash (=\backslash )$ & & & $\backslash (\backslash$overline$\{\backslash$$ $22,450\}\backslash )$ & & $\backslash$$$37,500\backslash \backslash$

$\backslash$hline

$\backslash$end$\{$tabular$\}$ Rank the activities in ascending order of cost per day saved. Lower cost per day saved means the activity is more cost$-$effective to crash:

$1.$ Activity A: $\backslash$$$250/$day

$2.$ Activity B: $\backslash (\backslash$$ $300/\backslash )$ day

$3.$ Activity G: $\backslash (\backslash$$ $300/\backslash )$ day

$4.$ Activity F: $\backslash$$$900/$day

$5.$ Activity C: $\backslash$$$1500/$day

$6.$ Activity E: $\backslash$$$1750/$day

$7.$ Activity H: $\backslash$$$2000/$day

Explanation:

All in all, the steps taken to determine which activities should be crashed based on price are that crash cost per day saved is determined for every activity and prioritizing the lowest of them all, and since their crash costs are the least for $\backslash ($ A$,$ B $\backslash ),$ and $\backslash ($ G $\backslash ),$ it indicates that the former is the most economical means by which project duration can be cut. Through this, the project manager can be able to schedule the activities in an efficient way while holding costs, ensuring that the project will be completed in time without significantly exceeding the budget. Such an arrangement is seen to imply that time and cost management is one of the actions which applies in effective management.